96
Hajªasz spaces and their reflexivity
properties
Elefterios Soultanis
October 12, 2010
Graduate Thesis
Adviser: Eero Saksman
Inspectors: Eero SaksmanIlkka Holopainen
Faculty of Science
Department of Mathematics and Statistics
Contents
1 Introduction 3
2 Prerequisites 4
2.1 Some functional analysis . . . . . . . . . . . . . . . . . . . . . . . 42.2 Some theory of metric spaces . . . . . . . . . . . . . . . . . . . . 82.3 Some measure theory . . . . . . . . . . . . . . . . . . . . . . . . . 9
3 An alternative denition of the classical Sobolev space 14
4 The Hajªasz space 18
4.1 Basic properties and the density of Lipschitz functions . . . . . . 194.2 Embedding theorems . . . . . . . . . . . . . . . . . . . . . . . . . 24
5 The non-reexivity of the Hajªasz space of a Cantor type frac-
tal 30
6 Newtonian spaces and the Poincaré inequality 34
6.1 Rectiable curves . . . . . . . . . . . . . . . . . . . . . . . . . . . 346.2 The modulus of a family of curves . . . . . . . . . . . . . . . . . 376.3 Upper gradients . . . . . . . . . . . . . . . . . . . . . . . . . . . . 396.4 The Newtonian space . . . . . . . . . . . . . . . . . . . . . . . . . 40
7 Spaces supporting a Poincaré inequality 44
7.1 The Hajªasz space through Poincaré inequalities . . . . . . . . . 447.2 Lip and lip . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 467.3 Metric measure spaces supporting a Poincaré inequality . . . . . 527.4 Coincidence of N1,p(X) and M1,p(X) . . . . . . . . . . . . . . . 58
8 A dierentiable structure for spaces supporting a Poincaré in-
equality 59
8.1 Tangent spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . 618.2 Tangent functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 688.3 The dierential structure . . . . . . . . . . . . . . . . . . . . . . 758.4 Reexivity of N1,p(X) . . . . . . . . . . . . . . . . . . . . . . . . 838.5 Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88
2
1 Introduction
Analysis on metric spaces is a quite recent area in mathematics, and has beenmet with increasing interest since the last decades of the twentieth century. Bythen it had become apparent that many applications, especially in the theoryof quasiconformal mappings, required results from classical analysis in new,non-smooth settings. However, the class of general metric spaces is clearly toolarge for the purposes of a general theory of, say, dierentiability of Lipschitzfunctions. Thus it was that at an early stage attempts at generalizing someclassical tools in analysis focused on subdomains or submanifolds of Rn. Forinstance in [5] the authors got as far as dening Besov spaces for certain domainsof Rn (these spaces are beyond the reach of the general theory presented in thisthesis).
It soon became apparent that in order to avoid repeating similar arguments,a general theory was needed. Moreover, an intrinsic theory would reveal theproperties truly essential for the development of a theory of dierentiability. Ina geometric language, the existence of an abundance of rectiable curves betweenany two points in the given metric spaces turned out to play an essential rolein such a theory. The concept of a Loewner space (see [17], Chapter 8) directlyquanties this idea while also generalizing some classes of domains in Rn whereclassical results of Sobolev spaces hold. Other essential properties, discoveredearlier, were the doubling property of the space or of the measure in question.This was known to be related, roughly speaking, to the nite dimensionalityof the space. In particular, Assouad's embedding theorem ([17], Theorem 12.1,p. 98) and a result of Konyagin and Volberg (later extended in [30]) provide alink between complete doubling spaces and Euclidean spaces.
Groundbreaking articles on the subject include [16], where the notion of anupper gradient is introduced and the notion of a Poincaré inequality is related toLoewner spaces, and [12], [31] and [2] where dierent Sobolev type spaces overthe given metric space are dened. Shanmugalingam, in [31] in 2000, introducedwhat will in the sequel be referred to as Newtonian spaces, utilizing the notion ofan upper gradient. Four years earlier Hajªasz had dened The Hajªasz-Sobolevspace for arbitrary metric measure spaces, and proven very general embeddingtheorems in this context. For spaces satisfying the aforementioned essentialproperties of being complete and doubling, and supporting some Poincaré in-equality (which has largely replaced the notion of the Loewner space) Shanmu-galingam proved the coincidence of the Newtonian and Hajªasz-Sobolev spaces.An even more important advancement in this function-space approach to anal-ysis on metric spaces was obtained by Cheeger in [2] in 1999, when he provedthe reexivity of a (yet another) Sobolev type space over a complete doublingmetric space supporting a Poincaré inequality. This approach included the con-struction of a sort of dierentiable structure for these spaces, hence transferingmuch of the rst order calulus of Rn to the setting of these metric measurespaces. Shanmugalingam [31] proved that in the abovementioned context theSobolev space of Cheeger coincides with the other two.
In subsequent years there has been much interest in the properties of metricspaces satisfying a Poincaré inequality as well as in the development of vari-ational calculus in metric spaces. Also attempts at weakening the hypothesesunder which the metric space admits a dierential structure have been made,along with dierent approaches toward a denition of metric dierentiability.
3
This exposition serves as an introduction to the approach described above, withan emphasis on the Sobolev space perspective.
2 Prerequisites
2.1 Some functional analysis
Functional analysis lies at the very heart of many of the proofs presented inthis thesis and it is therefore advisable to have some backround knowledge ofit. Some concepts that will be needed, however, can be considered sucientlyadvanced (or rather specialized) so as to be beyond reasonable undergraduatelevel background. These concepts will be presented here, mostly without proof.Instead, references to material containing proofs are presented in connection toevery theorem or lemma.
The following two theorems are elementary in nature and can be found in mosttextbooks in functional analysis, one such being ([8], p. 160). They will proveuseful in section 4.
Lemma 2.1.1. Let X be a reexive Banach-space and Y a closed subspace ofX. Then Y as a Banach space with norm inherited from X is also reexive
Lemma 2.1.2. If X and Y are two Banach spaces which are isomorphic andX is reexive then Y is also reexive.
The third result is the so called Mazur's lemma which is a consequence of ageometric version of the Hahn-Banach theorem.
Theorem 2.1. Let (X, ‖ · ‖) be a Banach space and (xn) ⊂ X a sequence thatconverges weakly to x ∈ X. Then there is a sequence of convex combinations ofthe members of the sequence, that is, a sequence
yk :=mk∑i=1
aixki , ai ≥ 0, a1 + · · ·+ amk = 1
that converges to x in norm.
This convenient result enables one to construct strongly convergent seque-nces from weakly convergent ones. A proof can be found in ([32], p. 28).
Theorem 2.2. Let (Xk, ‖ · ‖k)k∈K (K ⊂ N) be a sequence of Banach spaces
and let X :=⊕`p(K)
Xk, 1 < p <∞, be the Banach space of all sequences
x = (xk)k∈K , xk ∈ Xk
for which the norm
‖x‖pX :=∑k∈K
‖xk‖pk
is nite. Then the dual X∗ of X is⊕`p′ (K)
X∗k . (Here p′ is the Hölder conjugate
exponent of p.) In particular if each Xk is reexive, then so is X.
4
Proof. Let ϕ : X → R be an element of X∗. Dene ϕk : Xk → R by ϕk(x) =ϕ(0, . . . , x, 0, . . .) where x is in the kth slot. This gives, for every k, an elementof X∗k for which ‖ϕk‖X∗k ≤ ‖ϕ‖X∗ . Now for every x = (xk)k∈K ∈ X ϕ(x) canbe written as
ϕ(x) =∑k∈K
ϕk(xk),
since (x1, . . . , xk, 0, . . .)→ x in X. Thus
|ϕ(x)| ≤∑k∈K
|ϕk(xk)| ≤∑k∈K
‖ϕ‖X∗k‖xk‖k ≤(∑k∈K
‖ϕk‖p′
X∗k
)1/p′ (∑k∈K
‖xk‖pk
)1/p
=
(∑k∈K
‖ϕk‖p′
X∗k
)1/p′
‖x‖X ,
yielding
‖ϕ‖X∗ = sup|ϕ(x)| : ‖x‖X ≤ 1 ≤
(∑k∈K
‖ϕk‖p′
X∗k
)1/p′
.
For the opposite inequality let ε > 0 be arbitrary and take for each k an elementxk ∈ Xk, ‖xk‖k ≤ 1 so that
ϕk(xk)p′+ ε/2k > ‖ϕk‖p
′
X∗k.
For this calculate∑k∈K
‖ϕk‖p′
X∗k<∑k∈K
ϕk(xk)p′+ ε =
∑k∈K
ϕk(xk)p′−1ϕk(xk) + ε
= ϕ
(∑k∈K
ϕk(xk)p′−1xk
)+ ε ≤ ‖ϕ‖X∗‖y‖X + ε (2.1.1)
where y =∑k∈K
ϕk(xk)p′−1xk. Its norm can be estimated by
‖y‖pX =∑k∈K
ϕk(xk)(p′−1)p‖xk‖pk ≤∑k∈K
ϕk(xk)p′≤∑k∈K
‖ϕk‖p′
X∗k.
Introducing the shorthand-notation∑k∈K
‖ϕk‖p′
X∗k= ‖ϕk‖p
′
`p′
equation (2.1.1) simplies to
‖ϕk‖p′
`p′≤ ‖ϕ‖X∗‖ϕk‖p
′/p
`p′+ ε
for arbitrary ε. This readily implies the desired inequality.
Next is a topic of high interest in itself. It is a condition the norm of aBanach space may or may not have, one that acts as a weak substitute for theparallellogram law of innerproduct norms. In particular this so called uniformconvexity condition salvages many properties inherent in Hilbert spaces. Thedenition is taken from ([24], p. 59)
5
Denition 2.3. Let (X, ‖ · ‖) be a Banach space. Dene the modulus of con-vexity δ(ε) for 0 < ε ≤ 2 by
δ(ε) := inf
1− ‖x+ y‖2
: ‖x− y‖ = ε and ‖x‖ = ‖y‖ = 1, x, y ∈ X.
The norm ‖·‖ is said to be uniformly convex if the following holds true: δ(ε) > 0for every 0 < ε ≤ 2.
Another statement of the notion of uniform convexity is that for every ε > 0there corresponds some δ so that whenever ‖x‖ = 1 = ‖y‖ and ‖x− y‖ = ε onehas ‖x + y‖ ≤ 2 − δ, or ‖x + y‖/2 ≤ 1 − δ/2. Geometrically this says that nomatter how close to each other two points on the boundary of the unit ball of Xare, the line segment stays well inside the unit ball, in the sense that there is auniform constant < 2 restricting the norm of the midpoint of the line segment.In particular the unit ball of a uniformly convex Banach space is strictly convex,i.e. its boundary does not contain line-segments. Examples of uniformly convexBanach spaces include Lp(µ),W k,p(Rn) for 1 < p < ∞ and any Hilbert space([10], p. 9). Here the notation Lp(µ) stands for the Banach space of p-integrableµ-measurable functions over a given measure space with given measure µ. Thespaces W k,p(Rn) will be revisited in the sequel.
Of the many properties uniformly convex spaces share with Hilbert spacesonly one will be mentioned. A proof for it may be found in ([24], p. 61).
Theorem 2.4. Let (X, ‖ · ‖) be a uniformly convex Banach space. Then it isreexive.
In particular the following result will be needed.
Theorem 2.5. The space Lp(µ) is uniformly convex for 1 < p <∞.
An original proof can be found in [3].
Another theme which will be needed in a small way is some elementary theoryof quasinormed spaces. The following denitions and theorems are taken from[1]
Denition 2.6. A pair (V, ‖ · ‖) is a quasinormed space if V is a vector spaceover a real or a complex scalar eld and ‖ · ‖ is a quasinorm, i.e.
1) ‖v‖ = 0 if and only if v = 0,
2) ‖av‖ = |a|‖v‖ for any v ∈ V and a from the scalar eld of V , and nally
3) there is a constant C ≥ 1 so that for every v, w ∈ V one has ‖v + w‖ ≤C(‖v‖+ ‖w‖).
The constant C is often referred to as the quasinorm constant of V (or ‖ · ‖)and V may also be called a C-quasinormed space.
Quasinorms give rise to a topology in the same manner as normed spaces,that is the balls B(x, r) = y ∈ V : ‖x − y‖ < r form a basis for a topologyused in V . This is the topology induced by the quasinorm. One could dene aconcept of completeness in the spirit of the theory of normed spaces. Another(equivalent) way to introduce a notion of completeness of V is through thefollowing result ([1], p. 59).
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Theorem 2.7. Let (V, ‖ · ‖) be a C-quasinormed space. Then there is an in-variant metric d on V satisfying
d(x, y) ≤ ‖x− y‖ρ ≤ Cd(x, y) for every x, y ∈ V (2.1.2)
where ρ is dened by (2C)ρ = 2. Furthermore d induces into V its originaltopology.
In accordance with theorem 2.7 (V, ‖ · ‖) is dened to be complete if themetric described in 2.7 is complete. Observe that any two metrics satisfying theconclusion of theorem 2.7 are equivalent. The next proposition states that, asin the case of normed spaces, completeness is characterized by the convergenceof absolutely summable series.
Theorem 2.8. A quasinormed space (V, ‖·‖) is complete if and only if whenever(xn) ⊂ V is a sequence with
∞∑n=1
‖xn‖ρ <∞
then the nite sums
sN :=N∑n=1
xn
converge (to some element which is then denoted
∞∑n=1
xn.)
Proof. Suppose rst that V is complete and let (xn) be a sequence satisfyingthe assumptions of the claim. Dene ‖ · ‖∗ := d(·, 0) given by the precedingtheorem (2.7). Note that the invariance of d implies ‖x − y‖∗ = d(x, y). Thesum
∞∑n=1
‖xn‖∗ ≤∞∑n=1
‖xn‖ρ <∞
and therefore the partial sums sN =N∑n=1
xn form a Cauchy sequence which then
converges to∞∑n=1
xn := s in the metric ‖ · ‖∗. Again using theorem 2.7 it can be
estimated that
||s−N∑n=1
xn‖ ≤ C1/ρ‖∞∑
n=N+1
xn‖1/ρ∗N→∞−→ 0
which proves the only if-part claim.For the opposite implication suppose that every sequence with the properties
stated in the claim converges. Let (xn) be a Cauchy sequence in the metric ‖·‖∗.Take a subsequence and relabel it (xk) so that ‖xk − xk−1‖∗ ≤ 2−k. Then, if(yk) is dened by y1 = x1 and yk = xk − xk−1 for k ≥ 2, this sequence satises
∞∑k=1
‖yk‖ρ ≤ C∞∑k=1
‖yk‖∗ <∞
7
so that sk =k∑j=1
yj converges to some s. But sk =k∑j=1
(xj − xj−1) = xk and
hence the subsequence (xk) converges to s. The Cauchy condition implies theconvergence of the whole sequence and hence completeness is guaranteed.
A particular special case of quasinormed spaces are the Lp(µ)-spaces for 0 <p < 1 and (X,µ) a given measure space. In these the quantity
‖f‖∗p =∫X
|f |pdµ = ‖f‖pp
denes a metric for which theorem 2.7 holds with equality in place of the rstinequality of (2.1.2). Here one can therefore take ρ = p.
2.2 Some theory of metric spaces
In this subsection some basic concepts of metric spaces are presented and certainconnections between them are pointed out. In the extensive understanding ofthe theory of analysis in metric spaces these concepts play an important role.Throughout the thesis B(x, r) = y ∈ X : d(x, y) < r will stand for the openball whereas the notation B(x, r) = y ∈ X : d(y, x) ≤ r is reserved for theclosed ball (unless otherwise stated). Note, in particular, that in this generalcontext the closure B(x, r) of the open ball is not necessarily the same as B(x, r).Rather one has the one sided inclusion B(x, r) ⊂ B(x, r).
Denition 2.9. Let (X, d) be a metric space. If, for every x ∈ X and r > 0,the closed ball B(x, r) is compact then (X, d) is said to be proper.
As an immediate consequence of the previous denition it can be seen that(X, d) is proper if and only the compact sets of X are precisely those which areboth closed and bounded. Moreover
Proposition 2.10. A proper metric space (X, d) is complete and locally com-pact.
Proof. Suppose (X, d) is proper. Then it is automatically locally compact. Tosee that it is complete take a Cauchy sequence (xn) ⊂ X. Since it is Cauchy itis also bounded, hence there is a closed ball B so that (xn) ⊂ B. It thereforehas a convergent subsequence (xnk) with limit x ∈ B. This is actually a limitfor the whole sequence (xn): if ε > 0 is given and k0 is such that d(x, xnk) < ε/2whenever k ≥ k0 and d(xn, xm) < ε/2 whenever n,m ≥ k0 then for all n ≥ k0
d(xn, x) ≤ d(x, xnk0 ) + d(xnk0 , xn) < ε.
Hence a proper space is locally compact and complete.
The converse implication fails to hold as can be seen by taking any inniteset equipped with the discrete metric.
Denition 2.11. A metric space (X, d) is called doubling with doubling con-stant CX > 0 if, for all r > 0, any ball of radius 2r can be covered by at mostCX balls of radii r.
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Naturally there is also a connection between the doubling property andproperness. Although the doubling property does not imply completeness ofthe space in question as can be seen by taking (X, d) = (Q, | · |) it does implythe properness (and thus local compactness) of a complete space.
Proposition 2.12. Let (X, d) be a complete metric space that has the doublingproperty dened in 2.11. Then it is proper (in particular also locally compact.)
Proof. Let B = B(x, r) be a closed ball. Since B is complete it suces, todemonstrate its compactness, to show B is precompact ([8], Theorem 3.5.6, p.109). To this end let ε > 0. By the doubling condition B can be covered byat most CX balls of radii r/2. Each of these can then be covered by at mostCX balls of radii r/4 which gives at most C2
X balls of radii r/4 covering B.Continuing this process k times, where k is the least integer for which r/2k < ε,implies that B can be covered by at most CkX balls of radii r/2k < ε. Thiscompletes the proof.
In the progress of this thesis several denitions involving the above conceptswill be introduced. Sometimes a certain denition will be presented in the con-text of complete locally compact spaces, sometimes in the context of proper met-ric spaces, other times in doubling proper or complete doubling spaces. Becauseof the interconnectedness of these concepts many of these contexts coincide.Therefore there should not be any confusion if for instance in some argumentconcerning a proper doubling space the completeness of that space is invoked.
Proposition 2.13. A complete doubling space is separable.
Proof. Let N be the doubling constant of the space and x a point x0 ∈ X.For each k ∈ N let Ek = xk1 , . . . , xkNk be the set of points, guaranteed by thedoubling condition, so that
B(x0, k) ⊂Nk⋃i=1
B(xki , k/2k).
Set E =∞⋃k=1
Ek. To prove the claim it suces to show that E is dense.
Let x ∈ X be xed. If k0 is the smallest integer so that x ∈ B(x0, k0) thenfor every k ≥ k0 there is some yk ∈ Ek so that d(yk, x) ≤ k/2k. Since k/2k → 0as k →∞ (yk)k≥k0 is a sequence of E converging to x.
2.3 Some measure theory
In this thesis outer measures will be used. Instead of writing the word 'outer',however, it is omitted. So it should be understood that when writing measure anouter measure is actually meant. The denitions appearing here can be foundin [25].
Let (X, d) be a metric space and µ a measure on X. µ is said to be
1. a Borel-measure if all Borel sets are µ-measurable, i.e. every Borel set Bsatises
µ(E) = µ(E \B) + µ(E ∩B)
for all E ⊂ X.
9
2. a Borel regular measure if, in addition to being a Borel-measure, for everyA ⊂ X there exists a Borel set B ⊃ A so that µ(A) = µ(B),
3. locally nite if 0 < µ(B) <∞ for every ball B.
4. doubling, if it is locally nite and if there is a constant C so that µ(2B) ≤Cµ(B) for every ball B ⊂ X. Here 2B denotes, as usual, the ball withsame centre and two times the radius of B.
5. s-regular, s > 0, if there is some b > 0 so that µ(B) ≥ brs for any ball Bwith radius r and, nally
6. completely s-regular (or Alhfors s-regular) if there is a constant c > 0 sothat rs/c ≤ µ(B) ≤ crs for any ball B of radius r.
A few remarks might be in order: by this terminology any locally nitemeasure is σ-nite, and a σ-nite Borel-measure µ has the property that Cc(X)∩Lp(µ) is dense in Lp(µ) for 1 ≤ p <∞, provided the measure space X is locallycompact ([29], p. 69). Also if µ is any measure on X, then the formula
µ(A) = infµ(B) : B ⊃ A, B Borel
denes a Borel-regular measure on X.The notions of s-regularity and complete s-regularity typically come across in
connection to Hausdor measures which form an important example of measureswith this property. Obviously complete s-regularity is a stronger condition thans-regularity. It is also easy to see that if µ is completely s-regular, then for everyball B with radius r
µ(2B) ≤ c(2r)s ≤ 2sc2µ(B),
so that µ is also doubling. A bit less trivial is the following result, taken from[14].
Proposition 2.14. If µ is doubling, then it is locally s-regular with s = log2 C,C being the doubling constant of µ, in the sense that if B is a ball with radiusR then for every x ∈ B and r ≤ R
µ(B(x, r))µ(B)
≥ 4−s( rR
)s.
Proof. Let B be a ball with radius R and let C be the doubling constant. Ifx ∈ B and r ≤ R, dene k to be the unique integer so that 2R ≤ 2kr < 4R.Then µ(B) ≤ µ(B(x, 2kr)) ≤ Ckµ(B(x, r)). Now
µ(B(x, r))µ(B)
≥ C−k = 2−ks ≥( r
4R
)s,
where s = log2 C.
The doubling constant, i.e. the constant that appears in the denition of adoubling measure, is not uniquely determined but one can take an inmum ofall admissible constants, thus obtaining the least possible constant. However,the above reasoning allows for any admissible constant.
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Denition 2.15. Any s > 0 satisfying the claim in proposition 2.14 is said tobe the homogeneity exponent of µ.
Given a general metric space X and a measure µ on X the support of µ isdened as the set
x ∈ X : µ(B(x, r)) > 0 for every r > 0
and denoted by sptµ. In separable metric spaces this coincides with the follow-ing denition, taken from [7]. If G is the collection of all closed sets F ⊂ X forwhich µ(X \ F ) = 0 then
sptµ =⋂G.
It is not dicult to see that sptµ is always a closed set. If a measure µ isdoubling then proposition 2.14 readily implies that sptµ = X.
Intuitively speaking, the doubling condition seems to indicate some sort of nitedimensional behaviour of the measure, and thus by the space which it actsupon. It is in fact true that if a metric space carries a doubling measure (i.e.some doubling measure can be dened on it) then the space itself is doubling.Although lacking in rigour this property portraits very strongly a sense of nitedimensionality of the space itself. Thus doubling measures lead naturally to aconcept of dimension, conveyed quantitatively by proposition 2.14. In the sequelthis will prove to be useful and much attention will be paid to metric measurespaces with a doubling measure.
Proposition 2.16. Let X be a metric space and µ a doubling measure on X.Then X is doubling. Further the doubling constant CX of the space can be takento be C5
µ, Cµ being the doubling constant of the measure.
Proof. LetB be any ball with radiusR. Denote by Cµ the doubling constant of µand s = log2 Cµ. Let x1 ∈ B be arbitrary and choose x2 ∈ B so that d(x1, x2) >R/2 and, in general, xk+1 ∈ B so that d(xk+1, xi) > R/2 for all i = 1, . . . k. Ápriori the acquired sequence may or may not be nite. However, if N ∈ N issuch that x1, . . . , xN as above can be chosen, the balls B(xi, R/4) are disjoint (ifx ∈ B(xj , R/4)∩B(xi, R/4) for i 6= j then d(xj , xi) ≤ d(xj , x)+d(xi, x) < R/2,a contradiction) and their union is contained in 5/4B ⊂ 2B. One then has theestimate
Cµµ(B) ≥ µ(2B) ≥N∑i=1
µ(B(xi, R/4)) ≥ N µ(B)(4R)s
(R
4
)sfrom which a uniform upper bound is found, of
N ≤ Cµ · 16s = C5µ
for the number of points it is possible to nd. Now if N is the largest integer
≤ C5µ then the B ⊂
N⋃i=1
B(xi, R/2). This is because by the maximality of N any
x ∈ B must have the property d(xj , x) < R/2 for some j, otherwise it would bethe (N + 1)th element in the constructed sequence. This proves the claim.
11
As a consequence of 2.16 and the remark following 2.14 the phrase dou-bling metric measure space which á priori is ambiguous becomes a bit less so(although strictly speaking it still is). The ambiguity arises from the fact thatthe phrase does not specify whether the doubling property refers to that of thespace or of the measure. As a consequence of 2.16 a doubling metric measurespace in the second sense (i.e. with a doubling measure) implies the doublingproperty of the space as well. However the opposite implication fails to hold by [30] it can be said that if the space is doubling then a doubling measure onit exists but it need not be the one appearing in the triplet. In the sequel thephrase might be used occasionally and it should be understood that a metricspace together with a doubling measure is meant. 1
Proposition 2.17. Let X be a doubling metric space with doubling constantCX and put s = log2 CX . Then dimX ≤ s where dimX denotes the Hausdordimension of X.
Proof. Let B a ball or radius R > 0 and let t > s. The goal is to proveHt(B) = 0 or, equivalently, limδ→0Htδ(B) = 0.
If δ > 0 take k to be the least integer for which 2−kR < δ. Then the ball Bcan be covered with at most CkX balls of radii 2−kR. Now
Htδ(B) ≤CkX∑i=1
d(Bi)t = CkX2−ktRt = RtCkX2−sk2−k(t−s) < RtCkXC−kX (δ/R)t−s.
ThusHtδ(B) ≤ Rsδt−s → 0
as δ → 0. The conclusion of this is that for a xed t > s, Ht(B) = 0 for
any ball B. Since X =∞⋃k=1
kB, where B is an arbitrary ball it follows that
Ht(X) ≤∑k
Ht(kB) = 0. Consequently dimX < t for any t > s, implying the
claim.
An immediate consequence of 2.16 and 2.17 is the following
Corollary 2.18. If a metric space supports a doubling measure then it has niteHausdor dimension.
Throughout the thesis the following notation will be used:
i) uA = −∫A
udµ =1
µ(A)
∫A
udµ stands for the average of a locally integrable
function u on a measure space X equipped with a measure µ.
ii) Mu(x) = supr>0|u|B(x,r) and MRu(x) = sup0<r<R |u|B(x,r) denote the
Hardy-Littlewood maximal function and its restricted version.
1In many cases the results stated hold regardless of the interpretation of the phrase.
12
One very important property of doubling measures is that for them the Lebesguedierentiation theorem is valid. The next theorem is a more general form ofthe Lebesgue dierentiation theorem. The proof of the standard form of theLebesgue dierentiation theorem for doubling measures is similar to the caseof the Lebesgue measure, see [14] and [17] where also more references can befound.
Theorem 2.19. Let (X, d, µ) be a complete doubling metric measure space (thatis, with doubling measure), let C > 0 be xed and let u ∈ L1
loc(X). For µ almostevery x ∈ X the following holds true. If xn is a sequence converging to x and(rn) ⊂ R is a sequence converging to zero such that rn > Cd(xn, x) for every nthen
u(x) = limn→∞
−∫B(xn,rn)
u(y)dµ(y).
Proof. For doubling measures the maximal function M has the property thatthere is a constant C so that if u ∈ L1(µ) then
µ(x :Mu(x) > t) ≤ C||u||L1
t(2.3.1)
for each t > 0. The proof of this can be found in [17]. Theorem 2.19 will followeasily from the more standard statement
limr→0−∫B(x,r)
|u(y)− u(x)|dµ(y) = 0 (2.3.2)
for almost every x ∈ X.
Let K ⊂ X be any compact set and dene an operator Λ : L1(K)→ R by
Λu(x) = lim supr→0
−∫B(x,r)
|u(y)− u(x)|dµ(y)
For continuous u it clearly holds that Λu(x) = 0 for every x ∈ K. Moreover Λobeys the following estimate
Λu(x) ≤M|u− v|(x) + Λv(x)
for any u, v ∈ L1(K). Given u ∈ L1loc(X) and any ε > 0 there exists a continuous
function g ∈ L1(K) so that ‖u− g‖L1(K) < ε. Now, an estimate on the amountof Λu(x) which exceeds some given t > 0 yields
µ(x ∈ K : Λu(x) > t) ≤ µ(x ∈ K :M|u− g|(x) > t/2)+µ(x ∈ K : Λg(x) > t/2).
The last term is zero since g is continuous and to estimate the rst term apply(2.3.1) to get
µ(x ∈ K :M|u− g|(x) > t/2) ≤2C‖u− g‖L1(K)
t≤ 2Cε
t
Choosing t = ε1/2 and letting ε→ 0 yields
Λu(x) = 0 for almost every x ∈ K.
13
Since a complete doubling space X can be written as a countable union ofcompact sets 2.3.2 follows.
To see how the claim of 2.19 follows from (2.3.2) note that B(xn, rn) ⊂ B(x, [1+C−1]rn) and µ(B(xn, rn)) ≥ c−1µ(B(x, [1 + C−1]rn)) by 2.14. Hence
lim supn→∞
−∫B(xn,rn)
|u(x)− u(y)|dµ(y) ≤ lim supn→∞
−∫B(x,[1+C−1]rn)
|u(x)− u(y)|dµ(y)
and the righthand side vanishes almost everywhere by (2.3.2).
A particular (and equivalent) consequence of this is that in a metric measuresetting with doubling measure for a measurable set A ⊂ X almost every pointof A is a density point of A.
3 An alternative denition of the classical Sobo-
lev space
In the classical case a measurable function f : Rn → R (or C) is said to belongto the Sobolev spaceW k,p(Rn), where k ∈ N and 1 ≤ p ≤ ∞, if f ∈ Lp(Rn) and,in addition, for every multi-index α with |α| ≤ k there is an Lp(Rn)-function gαsuch that the equation∫
Rnf(x)∂αφ(x)dx = (−1)|α|
∫Rngα(x)φ(x)dx
holds for every φ ∈ C∞0 (Rn). The functions gα are more customarily denotedby ∂αf , and are called the αth weak derivatives of f .
The norm (or quasinorm) ||u||W 1,p(Rn) = ||u||p + ||∇u||p makes W 1,p(Rn)into a Banach space (or a complete quasinormed space). In the case p > 1 thefollowing surprising characterization holds [14].
Theorem 3.0.1. Let u ∈ Lp(Rn). Then u ∈W 1,p(Rn) if and only if there existsan (almost) everywhere positive g ∈ Lp(Rn) and a set N ⊂ Rn of measure zerosuch that the inequality
|u(x)− u(y)| ≤ |x− y|(g(x) + g(y)) (3.0.3)
holds for every x, y ∈ Rn \ N .Furthermore ||∇u||p ≈ infg∈D(u) ||g||p where theinmum is taken over the set of the admissible functions g in the conditions ofthe theorem, denoted by D(u).
It is said that the above inequality holds almost everywhere
Proof. To prove one direction suppose that (3.0.3) holds for some g ∈ Lp(Rn)and N ⊂ Rn. Fix some 1 ≤ j ≤ n. First redene g on N to be innity. Then theinequality (3.0.3) holds for every x, y ∈ Rn. From Fubini's theorem it followsthat for almost every line ` parallel to the jth axis, the restriction of g to ` is inLp(R). Consequently the function uj , dened pointwise as
uj(x) = lim supr→0
1r
∫ r
0
u(x+ tej)− u(x)t
dt
14
belongs to Lp(Rn). This is since by (3.0.3)
1r
∫ r
0
|u(x+ tej)− u(x)|t
dt ≤ g(x) +1r
∫ r
0
g(x+ tej)dt.
The measurability of uj follows from the fact that the lim sup can be taken overthe set of rationals, since for almost every x the mapping
r 7→ 1r
∫ r
0
u(x+ tej)− u(x)t
dt
is continuous. Passing to lim sup r → 0 and using the one dimensional versionof Lebesgue's dierentiation theorem and the Fubini theorem one obtains
|uj(x)| ≤ 2g(x) (3.0.4)
for almost every x ∈ Rn. Also the following lemma will be of use.
Lemma 3.0.2. Suppose f(x) = lim supr→0 fr(x) is measurable along with eachfr, r > 0 and that the mapping r → fr(x) is continuous for almost every x.Then, for each x there is a sequence rk(x)→ 0 so that
f(x) = limk→∞
frk(x)(x) for almost every x
and x 7→ frk(x)(x) is measurable for all k.
Proof of lemma. For xed x ∈ Rn and k ∈ N, let
sk(x) = sup0 < r < 2−k : sups≤r
fs(x)− f(x) ≤ 2−k.
The lower semicontinuity of gr(x) := r 7→ sups≤r fs(x) (for almost every xedx) implies 0 ≤ sup
s≤sk(x)
fs(x)− f(x) ≤ 2−k. Now set
rk(x) = sup0 < s < sk(x) : supt≤sk(x)
ft(x)− fs(x) ≤ 2−k.
Let us prove that rk(x) satises the properties in the claim.
For almost every x the continuity of r 7→ fr(x) implies that
frk(x)(x) = supt≤sk(x)
ft(x)− 2−k (3.0.5)
and hence the estimate
|f(x)− frk(x)(x)| ≤ |f(x)− supt≤sk(x)
ft(x)|+ | supt≤sk(x)
ft(x)− frk(x)(x)| ≤ 2−k+1
shows the (almost everywhere) pointwise convergence. To prove the measurabil-ity result it suces to demonstrate that of x 7→ sk(x). This is since by 3.0.5 themeasurability of x 7→ frk(x)(x) follows from that of h(x) := x 7→ supt≤sk(x) ft(x).Since gr(x) is lower semicontinuous in the parametre r and measurable in x themeasurability of x 7→ sk(x) will yield the measurability of h. But the measura-bility of sk(·) (with xed k) is evident, since for each α ∈ R, α ≤ 2−k
x : sk(x) ≥ α = x : gα(x)− f(x) ≤ 2−k.
15
This lemma enables uj to be expressed as a limit of a sequence of func-tions (the dependence of the sequence on x will be omitted in the subsequentnotation),
uj(x) = limk→∞
1rk
∫ rk
0
u(x+ tej)− u(x)t
dt =: limk→∞
ukj (x)rk
.
Take any ϕ ∈ C∞0 (Rn) and let R be such that sptϕ ⊂ B(0, R) and rk ≤ R forall k. Compute∫
Rnϕ(x)uj(x)dx =
∫B(0,R)
ϕ(x) limk→∞
1rk
∫ rk
0
u(x+ tej)− u(x)t
dtdx
= limk→∞
∫B(0,R)
ϕ(x)1rk
∫ rk
0
u(x+ tej)− u(x)t
dtdx.
Here the dominated convergence theorem can be applied since
|ukj (x)| ≤ g(x) +1rk
∫ rk
0
g(x+ tej)dt ≤ g(x) + sup0<r<R
1r
∫ r
0
g(x+ sej)ds
for almost every x ∈ Rn and the rightmost function is inLp(B(0, t)) ⊂ L1(B(0, t)) for any t > 0.
To see this estimate the integral
∫B(0,t)
:=MjRg(x1,...,xn)p︷ ︸︸ ︷(
sup0<r<R
1r
∫ r
0
g(x+ sej)ds)p
dx
≤∫ t
−t· · ·∫ t
−t︸ ︷︷ ︸n−1
[∫ t
−tMj
Rg(x1, . . . , xn)pdxj
]dx1 · · · ˆdxj · · · dxn
≤ C∫ t
−t· · ·∫ t
−t︸ ︷︷ ︸n−1
[∫ t
−tg(x1, . . . , xn)pdxj
]dx1 · · · ˆdxj · · · dxn
≤ C∫
Rngpdx.
The middle inequality is a consequence of the boundedness ofMj
R : Lp([−t, t]) → Lp([−t, t]). MjRg is dened by xing all but the jth coor-
dinate in the argument of g and taking the restricted maximal function of theresulting function of one variable.
By changing the order of integration (both integrals are over a compact set)and making a suitable change of variables one has
limk→∞
∫B(0,R)
ϕ(x)1rk
∫ rk
0
u(x+ tej)− u(x)t
dtdx
= limk→∞
1rk
∫ rk
0
u(x)∫
Rn
ϕ(x− tej)− ϕ(x)t
dxdt
=∫
Rnu(x) lim
k→∞
1rk
∫ rk
0
ϕ(x− tej)− ϕ(x)t
dtdx = −∫
Rnu(x)∂jϕ(x)dx.
16
Again interchanging the order of integration and limit is justied by a similarargument as before. This completes the proof of one direction of (3.0.1) since itshows that uj is the jth weak derivative of u.
Note that from (3.0.4) the inequality ||∇u||p ≤ C infg∈D(u) ||g||p follows.The other direction of the proposition is explicitly given by the pointwise
inequality
|u(x)− u(y)| ≤ C|x− y|(M|x−y||∇u|(x) +M|x−y||∇u|(y))
for almost every x, y ∈ Rn (in the sense mentioned earlier). It is enough toprove this for C1 ∩W p-functions (for which the inequality holds everywhere)since their density will then imply the general case.
Let x ∈ Rn and R > 0 and denote by B the ball of radius R and centre x.For any y ∈ B one has B ⊂ B(y, 2R). Bearing this in mind compute
|u(y)− uB | ≤1|B|
∫B
|u(y)− u(z)|dz
=1|B|
∫B
∣∣∣∣∫ 1
0
(z − y) · ∇u(y + t(z − y))dt∣∣∣∣dz
≤ 2R|B|
∫B(x,R)
∫ 1
0
|∇u(y + t(z − y))|dtdz
≤ 2R|B|
∫ 1
0
∫B(y,2R)
|∇u(y + t(z − y))|dzdt =2R|B|
∫ 1
0
∫B(0,2R)
|∇u(y + tz)|dzdt
=2R|B|
∫ 1
0
t−n∫B(0,2tR)
|∇u(y + w)|dwdt ≤ 2n+1R
∫ 1
0
M2R|∇u|(y)dt
= 2n+1RM2R|∇u|(y).
Now if x, y ∈ Rn put z =x+ y
2and R =
|x− y|2
. Then x, y ∈ B = B(z,R) andone can compute
|u(x)−u(y)| ≤ |u(x)−uB |+|u(y)−uB | ≤ 2n|x−y|(M|x−y||∇u|(x)+M|x−y||∇u|(y)).
In particular 2nM|∇u| ∈ D(u) so
infg∈D(u)
||g||p ≤ 2n||M|∇u|||p ≤ C||∇u||p.
Thus the last part of the proposition is also proven.
This section is closed with the presentation of another, more neat proof of theonly if part of theorem 3.0.1. What the preceding proof loses in elegance,however, it gains in that it does not at any point use the reexivity of thespaces Lp(µ).
In fact this direction of the theorem follows quite easily from
Lemma 3.1. Let u ∈ Lp(Rn), 1 < p <∞. Then u ∈W 1,p(Rn) if and only if
sup|h|≤1
‖u(·+ h)− u(·)‖p|h|
<∞. (3.0.6)
17
Proof. The necessity is a rather straightforward consequence of the fact thatthe jth weak partial derivative of u, since it exists as an Lp-function is almosteverywhere given by the pointwise limit
∂ju(x) = limh→∞
u(x+ hej)− u(x)h
.
The more essential only if part can be seen by the following reasoning.Assuming (3.0.6) x some standard basis vector ej and denote, for h ∈ R, |h| ≤ 1
bjh(x) =u(x+ hej)− u(x)
h.
Then clearly for any sequence hk → 0, |hk| ≤ 1
supk‖bjhk‖p ≤ C <∞
and the reexivity of Lp(Rn) implies that there is a weakly convergent sub-sequence bjhkm with limit denoted by bj ∈ Lp(Rn). By the denition of weakconvergence for any ϕ ∈ C∞0 it holds that∫
Rnbjϕdx = lim
m→∞
∫Rnbjhkm
ϕdx = limm→∞
∫Rn
u(x+ hkmej)− u(x)hkm
ϕ(x)dx
= − limm→∞
∫Rnu(x)
ϕ(x+ hkmej)− ϕ(x)hkm
dx = −∫
Rnu∂jϕdx,
making bj the weak jth partial derivative of u.
Now assume (3.0.3). From this it follows that
‖u(·+ h)− u(·)‖p ≤ |h|‖g(·+ h) + g(·)‖p ≤ 2|h|‖g‖pwhich yields the sucient condition (3.0.6).
4 The Hajªasz space
Proposition 3.0.1 forms the basis for this section. It's most important virtueis that it gives a characterization of the Sobolev space W 1,p(Rn) without usinganything other than the metric and measure theoretic structure of Rn. Bearingthis in mind we will dene a more general Sobolev space. The denition is givenfor an arbitrary metric measure space by which we mean the triplet (X, d, µ),where X is a set, d a metric on X and µ a locally nite Borel measure on X.
Even though in the preceding section attention was paid only to the casep > 1 as indeed without this assumption (3.0.3) need not characterize theclassical Sobolev space the consideration in this section will include the wholerange of positive real numbers that is 0 < p ≤ ∞.
Denition 4.0.3. Let (X, d, µ) be a metric measure space and 0 < p ≤ ∞.The class of Hajªasz functions, M1,p(X), consists of those measurable functionsu : X → R that belong to Lp(X) and for which there exists a set N ⊂ X and afunction g ∈ Lp(X) so that µ(N) = 0, g ≥ 0 and
|u(x)− u(y)| ≤ d(x, y)(g(x) + g(y)) (4.0.7)
for every x, y ∈ X \N .
18
The set of admissible functions g in the denition will throughout this paperbe denoted by D(u). Note that by dening g = ∞ on N the inequality (4.0.7)can be assumed to hold everywhere.
4.1 Basic properties and the density of Lipschitz functions
A very straightforward generalization of (3.0.3), denition 4.0.3 does bring thesubject to a completely new, very abstract setting. Of course not very much canbe said about these Hajªasz spaces in this level of generality. It is evident fromthe denition that M1,p(X) ⊂ Lp(µ). In the spirit of the last statement of theproposition 3.0.1 a norm corresponding to the additional requirements in thedenition will now be introduced, one that will make M1,p(X) into a Banachspace or, when 0 < p < 1, a complete quasinormed space.
Theorem 4.1.1. The quantity
||u||M1,p(X) = ||u||p + infg∈D(u)
||g||p,
denes a norm on M1,p(X). The space M1,p(X) equipped with this norm is aBanach space when p ≥ 1 and a complete quasinormed space when 0 < p < 1.
Of course the elements of M1,p(X) equipped with this norm are actuallyequivalence classes of functions, as is usual in the Lp-theory. Hence when talkingabout an element u ∈M1,p(X) it is dened only up to a set of measure zero.
Proof. Suppose cp is the quasinorm-constant of the the quasinorm || · ||p. (Ob-viously || · ||p is a quasinorm also when p ≥ 1, the constant cp being one.) Letu, v ∈ W 1,p(X) and λ ∈ R. If gu, Nu and gv, Nv are as in denition 4.0.3 for uand v respectively then for all x, y ∈ X \ (Nu ∪Nv)
|(u+ v)(x)− (u+ v)(y)| ≤ |u(x)− u(y)|+ |v(x)− v(y)|≤ d(x, y)(gu(x) + gv(x) + gu(y) + gv(y))
so
||u+ v||W 1,p(X) ≤ ||u+ v||p + ||gu + gv||p ≤ cp(||u||p + ||v||p + ||gu||p + ||gv||p).
Taking inmum over admissible gu and gv
||u+ v||M1,p(X) ≤ cp(||u||M1,p(X) + ||v||M1,p(X))
is obtained. In a similar manner it is seen that
||λu||M1,p(X) = |λ| ||u||M1,p(X).
Finally ||u||p ≤ ||u||M1,p(X) so if the latter is zero, it implies that u is zeroalmost everywhere, hence zero as an element of M1,p(X). Thus || · ||M1,p(X)
is a quasinorm. As remarked when p ≥ 1, cp can be taken to be one, making|| · ||M1,p(X) a norm.
To prove completeness use theorem 2.8. Take a sequence (uk) ⊂ M1,p(X)so that
∞∑k=1
||uk||ρM1,p(X) <∞.
19
Here ρ = min1, p, see the remark following Theorem 2.8. Let ε > 0 and let gkbe such that ||gk||p ≤ infg∈D(u) ‖g‖p + 2−kε and gk satises (4.0.7) with respectto uk for some Nk ⊂ X. One has
∞∑k=1
||uk||ρp ≤∞∑k=1
||uk||ρM1,p(X) <∞ and
∞∑k=1
||gk||ρp ≤∞∑k=1
||uk||ρM1,p(X) + Cερ <∞,
thus by the completeness of Lp(µ) there are u, gε ∈ Lp(µ) such that
||u−n∑k=1
uk||pn→∞−→ 0 and ||gε −
n∑k=1
gk||pn→∞−→ 0.
By a standard argument passing to a subsequence and relabeling the partialsums this implies the existence of sets A,B ⊂ X of µ-measure zero so that
u(x) =∞∑k=1
uk(x) ∀x ∈ X \A, and (4.1.1)
gε(x) =∞∑k=1
gk(x) ∀x ∈ X \B. (4.1.2)
It remains to show that the function gn := gε −n∑k=1
gk satises
∣∣∣∣∣u(x)−n∑k=1
uk(x)− u(y) +n∑k=1
uk(y)
∣∣∣∣∣ ≤ d(x, y)(gn(x) + gn(y)), x, y ∈ X \N
for some N ⊂ X of µ-measure zero since then it follows that
||u−n∑k=1
uk||M1,p(X) ≤ ||u−n∑k=1
uk||p + ||gn||pn→∞−→ 0.
Set N = A∪B∪N1∪N2∪· · · . Clearly gn ≥ 0 and µ(N) = 0. If x, y ∈ X \Nthen the identities (4.1.1) and (4.1.2) are available so∣∣∣∣∣u(x)−
n∑k=1
uk(x)− u(y) +n∑k=1
uk(y)
∣∣∣∣∣ =
∣∣∣∣∣∞∑
k=n+1
uk(x)−∞∑
k=n+1
uk(y)
∣∣∣∣∣≤
∞∑k=n+1
|uk(x)− uk(y)| ≤∞∑
k=n+1
d(x, y)(gk(x) + gk(y))
= d(x, y)(gn(x) + gn(y)).
This completes the proof.
Examples According to Theorem 3.0.1 it holds that
M1,p(Rn) = W 1,p(Rn)
20
with equivalent norms when 1 < p < ∞. There are some cases, however,when M1,p(X) becomes trivial in the sense that M1,p(X) = Lp(µ). One suchexample is given by the following.
Take X = N with the discrete metric and µ =∑n∈N
δn. Then µ is Borel regular
and locally nite. Yet it is easy to see that if f ∈ Lp(X) for p > 1 then|f | ∈ D(f), hence Lp(X) = M1,p(X) with equivalent norms.
In another sense the space M1,p(X) is never trivial. Namely for any 0 < p <∞the inclusion
LIP0(X) ⊂M1,p(X)
holds. Here LIP0(X) denotes the set of Lipschitz-functions X → R with com-pact support.
To see this let u ∈ LIP0(X) and K = sptu. Then u is µ-measurable and∫X
|u|pdµ ≤ µ(K)‖u‖p∞ <∞,
hence u ∈ Lp(µ). Furthermore if g = LχK , where L is the Lipschitz constant ofu then g ∈ D(u).
The next proposition shows that for p ∈ (1,∞) there is a unique Hajªasz uppergradient satisfying (4.0.7) with which the inmum in the norm is obtained.
Proposition 4.1.2. Let u ∈ M1,p(X) with 1 < p < ∞. Then there exists aunique positive g ∈ Lp(µ) satisfying (4.0.7) so that
||u||M1,p(X) = ||u||Lp(µ) + ||g||Lp(µ).
Proof. Let u ∈ M1,p(X) and denote by D(u) the set of admissible functions,i.e. the positive Lp(µ)-functions that satisfy (4.0.7). If g1, . . . , gn ∈ D(u) anda1, . . . , an are positive real numbers summing up to one then
|u(x)− u(y)| =n∑k=1
ak|u(x)− u(y)| ≤n∑k=1
d(x, y)(akgk(x) + akgk(y))
= d(x, y)
(n∑k=1
akgk(x) +n∑k=1
akgk(y)
)a.e.,
that is, the convex combinationn∑k=1
akgk ∈ D(u). Hence D(u) is a convex set,
in particular it is closed if and only if it is weakly closed ([26], p. 216). ButD(u) is closed: if gn is a sequence of D(u) and gn → g (in norm) then thereis a subsequence that converges to g pointwise a.e. and so (4.0.7) follows for gas a pointwise limit. Now take a sequence (gn) ⊂ D(u) so that ||gn||Lp(µ) →infh∈D(u) ||h||Lp(µ). Since Lp(µ) is reexive the unit ball is compact in the weaktopology. Thus take a weakly convergent subsequence of this and call the weaklimit g. It follows that g ∈ D(u) and ||g||Lp(µ) ≤ limn→∞ ||gn||Lp(µ). Hence gminimizes the norm in the denition of || · ||M1,p(X).
For the uniqueness of g the uniform convexity of Lp(µ) is used. Suppose thatg′ ∈ D(u) is another function minimizing the norm, and assume that g 6= g′.
21
Then if ‖g − g′‖Lp(µ) := ε it follows that since ‖g‖Lp(µ) = ‖g′‖Lp(µ) := a thereis some δ(ε) for which
‖(g + g′)/2‖Lp(µ) ≤ a(1− δ(ε)) < a
which is a contradiction since (g + g′)/2 ∈ D(u) and its norm is strictly lessthan a. Therefore the minimizing element must be unique.
It is evident that in the case p = ∞ the space M1,p(X) essentially consistsof Lipschitz functions: if u ∈ W 1,∞(X) then there exists a positive g that isessentially bounded on sptµ such that (4.0.7) is satised. But then it followsthat
|u(x)− u(y)| ≤ d(x, y)2||g||∞ almost everywhere on sptµ
which is precicely the Lipschitz condition. It follows that u has an 2||g||∞-Lipschitz representative and so can be considered Lipschitz. To make this dis-cussion a little more rigorous one has
Lemma 4.1.3. Let E ⊂ X and u : E → R be L-Lipschitz. Then there existsa Lipschitz map u : X → R, called a Lipschitz extension of u, so that u isL-Lipschitz and u = u|E. Further, if µ(sptµ \ E) = 0 then any two extensionsof u agree on sptµ.
Proof. The extension is given by the formula
u(x) = infLd(x, y) + u(y) : y ∈ E, x ∈ X (4.1.3)
and it is straightforward to verify this. To see the second part of the claimsuppose its hypotheses and suppose u1 and u2 are two extensions of u andx ∈ sptµ \ E. Since x ∈ sptµ it follows that for every k ∈ N the ball B(x, 1/k)has positive measure. Since sptµ \ E has zero measure, there must exist xk ∈B(x, 1/k) \ (sptµ \ E) = (B(x, 1/k) ∩ E) \ sptµ ⊂ B(x, 1/k) ∩ E. Now usingthe extension properties of u1 and u2 and their continuity
u1(x) = limk→∞
u1(xk) = limk→∞
u(xk) = limk→∞
u2(xk) = u2(x)
is obtained. This proves the claim.
Corollary 4.1.4. If any two functions are identied whenever they agree outsidea set of measure zero one has the identity
M1,∞(X) = LIP(X) ∩ L∞(X).
Proof. Let u ∈M1,∞(X) and g ∈ D(u). If N ⊂ X is the set where g(x) > ‖g‖∞and the pointwise inequality (4.0.7) fails, then µ(N) = 0. Now u is Lipschitz onE := sptµ \N since for every x, y ∈ E
|u(x)− u(y)| ≤ d(x, y)(g(x) + g(y)) ≤ 2d(x, y)‖g‖∞.
If u1 and u2 are two Lipchitz extensions of u|E then u1(x) = u2(x) for allx ∈ sptµ, according to lemma 4.1.3. In particular u1 = u2 µ-almost everywhereon X. Let u be any Lipschitz extension of u|E . Now
esssupx∈X |u(x)| = esssupx∈sptµ |u(x)| = esssupx∈sptµ\N |u(x)| ≤ ‖u‖∞Hence u ∈ LIP(X) ∩ L∞(X). Moreover u = u almost everywhere since u = uon sptµ \N and µ(X \ (sptµ \N)) = 0.
22
This fact makes the space M1,∞ a bit more concrete, something M1,p(X) asof yet lacks for other values of p. To account for this a sort of approximationresult for M1,p(X) will now be proven, following [14]. The classes of innitelydierentiable or Schwartz functions are not available in metric spaces theeasiest thing is to use the class of Lipschitz functions.
Theorem 4.1.5. Let 0 < p ≤ ∞, u ∈M1,p(X) and ε > 0. Then there exists aLipschitz function uε ∈M1,p(X) so that
µ(x ∈ X : uε(x) 6= u(x)) < ε
||u− uε||M1,p(X) < ε.
Proof. Suppose p = ∞. By corollary 4.1.4 if u ∈ M1,p(X) then u| sptµ can beconsidered Lipschitz, in the sense that there is a Lipschitz function that agreeswith u| sptµ almost everywhere. Therefore uε = u, where u is any Lipschitzextension of u| sptµ, satises the claims of the theorem.
It can therefore be assumed that p < ∞. The symbol cp again denotes thequasinorm-constant of || · ||p. Let λ > 0. Let g and N be as in (4.0.7) and deneEλ = x ∈ X \N : g(x) ≤ λ, |u(x)| ≤ λ.
It follows that
λpµ(X \ Eλ) ≤∫X\Eλ
(gp + |u|p)dµ λ→0−→ 0
by the absolute continuity of (gp + |u|p)dµ and the fact that µ(X \Eλ)→ 0. Itis seen that u, restricted to Eλ, is 2λ-Lipschitz. Dene
uλ = u|Eλ and
uλ(x) =
λ if uλ(x) > λ,uλ(x) if |uλ(x)| ≤ λ,−λ if uλ(x) < −λ
It is easy to see that uλ is 2λ-Lipschitz on X, uλ|Eλ = u|Eλ and further|uλ| ≤ λ. Hence uλ extends u |Eλ , dened on Eλ, to a Lipschitz function on thewhole space X with the same Lipschitz constant.
Since Eλ ⊂ x : u(x) = uλ(x) it follows that
µ(x : u(x) 6= uλ(x)) ≤ µ(X \ Eλ) (4.1.4)
which can be made arbitrarily small. One also has the estimate∫X
|u− uλ|pdµ ≤∫X\Eλ
|u− uλ|pdµ ≤ 2p(∫
X\Eλ|u|pdµ+
∫X\Eλ
|uλ|pdµ
)
≤ 2p∫X\Eλ
|u|pdµ+ 2pλpµ(X \ Eλ)
≤ 2p∫X\Eλ
|u|pdµ+ 2p∫X\Eλ
(gp + |u|p)dµ λ→∞−→ 0. (4.1.5)
Now the proof is almost ready. To nish the argument dene
gλ(x) =
0 if x ∈ Eλg(x) + 3λ otherwise.
23
One can calculate that for almost every x, y ∈ X
|u(x)− uλ(x)− u(y) + uλ(y)| ≤ d(x, y)(gλ(x) + gλ(y)) :
If x, y ∈ Eλ then this is clear since u and uλ coincide on Eλ. If x ∈ Eλ andy /∈ Eλ then gλ(x) = 0 and gλ(y) = 3λ+ g(y). Consequently
|u(x)− uλ(x)− u(y) + uλ(y)| ≤ 2λd(x, y) + |u(x)− u(y)| ≤(2λ+ g(x) + g(y))d(x, y) ≤ (3λ+ g(y))d(x, y) = d(x, y)(gλ(y) + gλ(x)).
Finally if x, y /∈ Eλ then
|u(x)−uλ(x)−u(y)+uλ(y)| ≤ (2λ+g(x)+g(y))d(x, y) ≤ d(x, y)(gλ(y)+gλ(x)).
Again
||gλ||pp ≤ 2p∫X\Eλ
gpdµ+ 4pλpµ(X \ Eλ) ≤ C∫X\Eλ
gpdµ (4.1.6)
can be made arbitrarily small. (4.1.5) and (4.1.6) together imply that ||u −uλ||M1,p(X) can be made arbitrarily small and together with (4.1.4) the proof ofboth claims is complete.
4.2 Embedding theorems
The elaboration of the theory of the Hajªasz spaces, as said, cannot proceedwithout further assumptions of the underlying space (X, d, µ). Of particularinterest at this point is more intricate knowledge of the behaviour of the measureµ. To have additional structure, especially structure similar to the model caseX = Rn, one could expect the measure to be required to have similar propertieswith the Lebesgue measure in Rn. In connection with the embedding theorems(in Rn), a very signicant role is played by the notion of the dimension of theunderlying space. This signicance comes, albeit implicitly, from the behaviourof the measure with respect to the underlying metric. Consequently a notion ofdimension, or a concept relating to the quantitative behaviour of the measure,must be introduced in order to speak of embedding theorems. An obviouscandidate would be the complete s-regularity of the measure since this is actuallyused in the proof of the classical embedding theorems. However, as it turns out alocal version of the embedding theorem can be proven with the following notionof lower s-regularity.
Denition 4.2.1. A measure µ on a metric measure space (X, d, µ) is said tobe lower s-regular on A, for some s ≥ 0 and A ⊂ X, if there exists a constantb > 0 so that
µ(B(x, r)) ≥ brs
for all x ∈ A and 0 < r ≤ d(A)/2.
There is also a more technical property under which the embedding theoremcan be proven with slight elaborations of the arguments presented below. Formore on this see [14].
The quantity s will replace the concept of dimension. The proof of the embed-ding theorem is taken from [14] and [12].
24
Theorem 4.2.2. Fix some number σ > 1. Let (X,µ, d) be a metric measurespace whose measure µ is lower s-regular on some ball σB0 of centre x0 andradius r0. Let also 0 < p ≤ ∞. Then there exist positive constants C,C ′ andC ′′ depending only on σ, s, and p so that for any u ∈ M1,p(X) and g ∈ D(u)one of the three holds:
1. if 0 < p < s then, denoting p∗ =ps
s− p, u ∈ Lp∗(B0) and
infa∈R
(−∫B0
|u− a|p∗dµ)1/p∗
≤ C(µ(B0)brs0
)1/p
r0
(−∫σB0
gpdµ)1/p
. (4.2.1)
2. If p = s then
−∫B0
exp(C ′b1/s
|u− uB0 |||g||Lp(B0)
)dµ ≤ C ′′. (4.2.2)
3. If p > s then u ∈ L∞(B0) and
||u− uB0 ||L∞(B0) ≤ C(µ(B0)brs0
)1/p
r0
(−∫B0
gpdµ)1/p
. (4.2.3)
Furthermore in the last case u is Hölder-continous in B0 with exponent 1− s/p:for any x, y ∈ B0
|u(x)− u(y)| ≤ Cb−1/p||g||Lp(B0)d(x, y)1−s/p.
Proof. None of the claims is aected by the addition of a constant to u. This
fact will be used later on. Likewise, by substituting g with g+(−∫σB0
gpdµ)1/p
it may be assumed that
g(x) ≥(−∫σB0
gpdµ)1/p
> 0 for all x ∈ σB0.
For each k ∈ Z dene Ek = x ∈ σB0 : g(x) ≤ 2k. Again it is evident thatu restricted to Ek is 2k+1-Lipschitz (or can be modied to be) so one can alsodene
ak = supEk∩B0
|u|.
Finally let rk = (2/b)1/sµ(σB0 \ Ek−1)1/s.The sets Ek satisfy Ek ⊂ Ek+1 and also µ(σB0 \Ek) ↓ 0 as k →∞ because
for almost every x ∈ σB0 g(x) ≤ 2k for some k (that is, g is nite almosteverywhere). Since 2k−1 < g(x) ≤ 2k for x ∈ Ek \ Ek−1 one has
∞∑k=−∞
2(k−1)pµ(Ek\Ek−1) ≤∞∑
k=−∞
∫Ek\Ek−1
gpdµ =∫σB0
≤∞∑
k=−∞
2kpµ(Ek\Ek−1),
in other words ∫σB0
gpdµ ≈∞∑
k=−∞
2kpµ(Ek \ Ek−1). (4.2.4)
25
Here and in the sequel the implied constants depend only on s, p and σ. Thedependence on b will be stated explicitly. For rk one has the estimate, comingfrom Chebychev's inequality
rk ≤ (2/b)1/s
(2−(k−1)p
∫σB0
gpdµ)1/s
= (2p+1/b)1/s2−kp/s||g||p/sp . (4.2.5)
Now let k0 ∈ Z be the least integer such that µ(Ek0−1) <µ(σB0)
2≤
µ(Ek0). This exists since µ(Ek) ↑ µ(σB0) and Ek = ∅ for k such that 2k ≤(−∫σB0
gpdµ)1/p
. Obviously Ek0 6= ∅.
On one hand there is the estimate(−∫σB0
gpdµ)1/p
≤ g(x) ≤ 2k0 for x ∈ Ek0
and on the other
µ(σB0)2
< µ(σB0 \ Ek0−1) ≤ 2p2−pk0 ||g||pp.
Together these lead to
2k0 ≈(−∫σB0
gpdµ)1/p
. (4.2.6)
Dene k1 to be the rst integer such that(µ(σB0)brs0
)1/p
max
(σ − 1)r02−kp/s,
(2p+1/b
)1/s1− 2−p/s
2−kp/s< (σ − 1)r0 (4.2.7)
whence k1 > 0. It also follows that
2k1 &
(1brs0
)1/p
. (4.2.8)
If k2 = k0 + k1 then combining (4.2.6) and (4.2.8) yields(1brs0
)1/p
‖g‖Lp(σB0) . 2k2 .
On the other hand the minimality of k1 together with (4.2.6) implies
2k2 .
(1brs0
)1/p
‖g‖Lp(σB0).
Together these last two yield
2k2 ≈(
1brs0
)1/p
‖g‖Lp(σB0). (4.2.9)
Next an estimate for ak from above is obtained. If k > k2 then (4.2.7) issatised for k, in particular
k∑l=k2
rk < (σ − 1)r0.
26
It can be assumed that B0 ∩ Ek 6= ∅ for if this is the case then ak = 0. Takexk ∈ B0 ∩Ek and consider the ball B(xk, rk) ⊂ σB0. The lower regularity of µyields
µ(B(xk, rk)) ≥ brsk = 2µ(σB0 \ Ek−1),
implying µ(B(xk, rk) ∩ Ek−1) = µ(B(xk, rk)) − µ(B(xk, rk) \ Ek−1) > 0. Inparticular an element xk−1 ∈ B(xk, rk) ∩ Ek−1 can be found. Repeating thisprocedure one gets a (nite) sequence xi ∈ B(xi+1, ri+1)∩Ei, i = k0, . . . , k− 1.Notice that B(xi, ri) ⊂ B(xk, rk0 + · · ·+ rk) ⊂ σB0 for each i = k0, . . . , k for alli = k0, . . . , k.
The purpose of all this was to allow for the following estimate:
|u(xk)| ≤ |u(xk0)|+k∑
l=k0+1
|u(xl)− u(xl−1)| ≤ |u(xk0)|+k∑
l=k0+1
2l+1d(xl, xl−1)
≤ |u(xk0)|+ 2(2p+1/b)1/s||g||p/sp
k∑l=k0+1
2l(1−p/s).
Now use the fact mentioned in the beginning of the proof that the claimsremain unaected if u is replaced by some u+ c, c a constant. This allows oneto assume that essinfEk2 |u| = 0. Thus there is a sequence (yi) ⊂ Ek2 for whichlimi→∞
u(yi) = 0. Then
|u(xk2)| = limi→∞
|u(xk2)− u(yi)| ≤ 2σr02k2+1.
Taking supremum over xk ∈ Ek ∩B0 a nal estimate
ak ≤ 2(
2p+1
b
)1/s
||g||p/sp
k∑l=k2+1
2l(1−p/s) + 4σr02k2 (4.2.10)
is obtained. This estimate holds for all k since if k ≤ k2 then ak ≤ ak2 ≤4σr02k2 .
Inequalities (4.2.9) and (4.2.10) are the main ingredient in the rest of the proof.To start with (4.2.1), suppose 0 < p < s. Then 1−p/s > 0 and one can estimatethe sum as
k∑l=k0+1
2l(1−p/s) ≤ 2k(1−p/s)
1− 2p/s−1.
Use (4.2.10) to estimate
∫B0
|u|p∗dµ ≤
∞∑k=−∞
ap∗
k µ(B0 ∩ (Ek \ Ek−1))
. b−p∗/s‖g‖pp
∗/sp
∞∑k=−∞
2k(1−p/s)p∗µ(B0 ∩ (Ek \ Ek−1))
+ rp∗
0 2k2p∗∞∑
k=−∞
µ(B0 ∩ (Ek \ Ek−1)).
27
The rst term equals by (4.2.4)
b−p∗/s||g||p
∗p/sp
∞∑k=−∞
2kpµ(B0 ∩ (Ek \ Ek−1))
≤ b−p∗p/s||g||p(p
∗/s+1)p = b−p
∗/s||g||p∗
whereas the second one is, using (4.2.9)
2k2p∗rp∗0 µ(B0) . b−p
∗/sµ(B0)brs0
||g||p∗
p .
Together with the lower s-regularity one has∫B0
|u|p∗dµ . bp
∗/s
(1 +
µ(B0)brs0
)||g||p
∗
p . b−p∗/sµ(B0)
brs0||g||p
∗
p
and (4.2.1) follows.
In the second case 1− p/s = 0 so (4.2.10) becomes
ak . b−1/s||g||p/sp (k − k2) + 4σr0(brs0)−1/p||g||p
if k ≥ k2 and ak ≤ 4σr0(brs0)−1/p||g||p if k ≤ k2. Put together,
ak . b−1/p||g||p maxk − k2, 1. (4.2.11)
Let at rst C ′ be any positive real. To estimate the left hand side of (4.2.2)use Jensen's inequality to obtain
−∫B0
exp(C ′b1/s
|u− uB0 |||g||Lp(σB0)
)dµ
≤ −∫B0
exp(C ′b1/s
||g||p−∫B0
|u(x)− u(y)|dµ(y))
dµ(x)
≤ −∫B0
−∫B0
exp(C ′b1/s
|u(x)− u(y)|||g||p
)dµ(y)dµ(x)
≤[−∫B0
exp(C ′b1/s|u|/||g||p
)dµ]2
.
This last integral is split into two parts, one over Ek2 ∩ B0 and the other overB0 \ Ek2 . First, however, make a choice of C ′ such that exp(CC ′) = 2p, whereC is the implied constant in (4.2.11). Now
−∫B0∩Ek2
exp(C ′b1/s|u|/||g||p
)dµ ≤ −
∫B0∩Ek2
exp(C ′b1/sak2/||g||p
)dµ
≤ −∫B0
exp(C ′Cb1/s−1/p||g||p/||g||p
)dµ = 2p.
28
The second integral can be estimated from above by (4.2.9) and (4.2.11):
1µ(B0)
∞∑k=k2+1
µ(B0 ∩ (Ek \ Ek−1)) exp(C ′b1/s
ak||g||p
)
≤ 1µ(B0)
∞∑k=−∞
µ(B0 ∩ (Ek \ Ek−1)) exp (C ′C(k − k2 + 1))
≤ 2p
µ(B0)2−pk2
∞∑k=−∞
µ(B0 ∩ (Ek \ Ek−1))2kp ≤ C ′′ brs0µ(B0)
||g||−pp ||g||pp ≤ C ′′.
The last inequality was again a result of the lower s-regularity of µ. Thus (4.2.2)is also proven.
Finally suppose ∞ > p > s. Here 1 − p/s < 0 and the sum in (4.2.10) can beestimated as
k∑l=k2+1
2l(1−p/s) . 2k2(1−p/s),
leading to a uniform estimate for ak, using again (4.2.11)
ak . b−1/s||g||p/sp 2k2(1−p/s) + r0(brs0)−1/p||g||p. b−1/s(brs0)−1/p(1−p/s)||g||1−p/s+p/sp + r0(brs0)−1/p||g||p= b−1/p||g||pr1−s/p
0 . (4.2.12)
This implies, in particular, that u ∈ L∞(B0). Using ||u − uB0 ||L∞(B0) ≤2||u||L∞(B0) = 2 supk∈Z ak and rearranging the exponents (4.2.3) is obtained.In case p =∞ the result is a direct consequence of the inculsions L∞ ⊂ Lq forany q and the fact that ||g||∞ = limq→∞ ||g||q.
The estimate (4.2.12) is also suitable for proving the stated Hölder continuityof u. Suppose x, y ∈ B0 and assume, at rst, that r := d(x, y) < (σ−1)r0/(2σ).Then B := B(x, 2r) ⊂ σB0 and also σB ⊂ σB0. Thus the embedding theoremis valid with B0 replaced by B and one can estimate
|u(x)− u(y)| ≤ 2||u− uB ||L∞(B) ≤ Cb−1/pr1−s/p||g||Lp(σB0).
If, on the other hand, d(x, y) ≥ (σ − 1)r0/(2σ) one has
|u(x)− u(y)| ≤ 2||u− uB0 ||L∞(B0)
≤ Cb−1/pr1−s/p0 ||g||p ≤ Cb−1/p||g||p
(4σσ − 1
)1−s/p
d(x, y)1−s/p.
All in all, the Hölder continuity of u with exponent 1−s/p follows and the proofof Theorem 4.2.2 is complete.
Having proved Theorem 4.2.2 with quite weak and technical assumptions,it is desirable to narrow down the class of admissible measures and gain morestructure by providing a narrower, less technical denition. From section 1 theconcept of a doubling measure ts this purpose, especially since as a directconsequense of proposition 2.14 a doubling measure µ is lower s-regular on anyball σB0, for a xed σ, with s = log2 C and b = µ(σB0)(4σr0)−s.
Therefore, if the measure of the metric measure space (X, d, µ) is assumedto be doubling Theorem 4.2.2 takes on a simpler form:
29
Corollary 4.2.3. Let 0 < p ≤ ∞ and σ > 1 be xed and denote s = log2 Cdwhere Cd is the doubling constant of µ. Then there exist positive constants C,C ′
and C ′′ depeding only on σ, p and Cd so that if B ⊂ X is any ball of radius rthen for any u ∈M1,p(X) and g ∈ D(u) one the three following holds:
1. if 0 < p < s then, denoting p∗ =ps
s− p, u ∈ Lp∗(B) and
infa∈R
(−∫B
|u− a|p∗dµ)1/p∗
≤ Cr(−∫σB
gpdµ)1/p
. (4.2.13)
2. If p = s then
−∫B
exp(C ′µ(σB)1/p
r
|u− uB |||g||Lp(σB)
)dµ ≤ C ′′. (4.2.14)
3. If p > s then u ∈ L∞(B) and
||u− uB ||L∞(B) ≤ Cr(−∫σB
gpdµ)1/p
. (4.2.15)
Furthermore u is Hölder-continous in B with exponent 1 − s/p: for anyx, y ∈ B
|u(x)− u(y)| ≤ Crs/p(−∫σB
gpdµ)1/p
d(x, y)1−s/p.
5 The non-reexivity of the Hajªasz space of a
Cantor type fractal
The result of this section provides an example of a metric space equipped with ameasure that is even completely regular (or Ahlfors-regular), the Hajªasz spacefor p > 1 of which still fails to be reexive. This result is due to Kari Astalaand Juha Rissanen, see [28] more on the non-reexitivity of Hajªasz spaces forself similar fractals. For this section p > 1 will be xed.
Let E denote the standard middle-thirds Cantor set and µ the log 2/ log 3-dimensional Hausdor measure on it. The measure µ is complelety regular onE. The nth step, En, in the construction of E consists of 2n disjoint intervalsEn,k, k = 1, . . . , 2n, each with length 3−n and measure µ(En,k) = 2−n. Denethe functions un for n ∈ N by
un =2n∑k=1
(−1)k+1χE∩En,k .
By construction for each natural number n the function un is 2 · 3n-Lipschitz,has absolute value 1 and integral average 0. Thus for example ||un||p = 1 and||un||M1,p(E) ≤ 1 + 3n.
For any a = (a1, a2, . . .) ∈ `∞ dene
ua =∞∑n=1
3−nanun.
The rst lemma of this section states that
30
Lemma 5.0.4. the map L = a 7→ ua is a linear bounded map from `∞ toM1,p(E).
Proof. The linearity of the map is of course obvious. The triangle inequality
yields the estimate ||ua||p ≤∞∑n=1
3−n|an| ≤ C||a||`∞ . To see that ua belongs to
the Hajªasz-classM1,p(E) take x, y ∈ E such that x 6= y. Let m be rst naturalnumber for which x and y are in dierent intervals Em,k and Em,k′ . Of coursenecessarily then k′ = k + 1 and, more importantly, 3−m ≤ |x− y| ≤ 3−m+1. Inparticular for all indices n less than m the values of un at x and y coincide that is un(x) = un(y) for all n < m. The dierence is therefore
|ua(x)− ua(y)| =
∣∣∣∣∣∞∑n=m
3−nan(un(x)− un(y))
∣∣∣∣∣ ≤ 2∞∑n=m
3−n|an|
≤ 2||a||`∞∞∑n=m
3−n ≤ C3−m||a||`∞ ≤ C|x− y|||a||`∞ .
This shows that ua is Lipschitz and thus infg∈D(ua)
||g||p ≤ C||a||`∞ . All in all one
has ||ua||M1,p(E) ≤ C||a||`∞ .
The next step is to show the injectivity of L. To this purpose it is convenientto derive an expression for recovering the coecients an of the bounded sequencea from the function ua.
Lemma 5.0.5. For a given ua the coecients an can be recovered by
3−nan =∫E
unuadµ. (5.0.16)
In particular, L is an injective map.
Proof. By Lebesque's dominated convergence∫E
unuadµ =∞∑m=1
3−mam∫E
unumdµ.
In order to prove (5.0.16) it therefore suces to show that∫E
unumdµ = δnm.
If n = m one has unun = |un|2 = 1 so this case is obvious. In case n 6= m itcan be assumed that n < m since the other option (n > m) follows from thisby interchanging the roles of n and m. But it is easy to see that∫
E∩En,kumdµ = 0
for any k = 1, . . . , 2n. Consequently∫E
unumdµ =2n∑k=1
(−1)k+1
∫E∩En,k
umdµ = 0.
31
Let F denote the image of `∞ under L. The aim of the ongoing constructionis ultimately to prove that F is isomorphic to `∞. This will imply in particularthat the Hajªasz space of E is not reexive. So far it has been established thatL is an injective map from `∞ to F . The remaining part is to show that theinverse L−1 : F → `∞ is also bounded. This is formulated explicitly by the nextlemma.
Lemma 5.0.6. The inverse of L : `∞ → F , given by
S := u 7→(
3n∫E
unudµ)∞n=1
, (5.0.17)
is bounded.
Proof. The expression (5.0.17) is a direct consequence of the previous lemma.Albeit valid it is not a very useful tool in understanding the boundedness of S.Two slightly dierent expressions will be derived instead, the combination ofwhich will yield the desired result.
Denote An = E ∩2n−1⋃k=1
En,2k−1 and Bn = E ∩⋃2n−1
k=0 En,2k. In a few words
An is the set E from which every even-indexed (with respect to k) interval En,kof the nth step is removed and Bn likewise with odd-indexed intervals removed.Therefore un = 1 on An. As in the proof of 5.0.5 for n < m one has∫
E∩En,kumdµ = 0
for all k = 1, . . . , 2n so∫An
umdµ =2n−1∑k=1
∫E∩En,2k−1
umdµ = 0. Unlike in 5.0.5,
the situation here is not symmetric with respect to n and m because the set thefunctions are integrated over depends on n. However, if n > m, each intervalEm,k is divided into 2n−m intervals En,k′ half of which belong to the set An.Then∫An∩Em,k
umdµ = (−1)k+1µ(An∩Em,k) = (−1)k+1·2n−m−1·2−n = (−1)k+12−m−1,
yielding ∫An
umdµ =2n∑k=1
2−m−1(−1)k+1 = 0.
Finally∫An
undµ = µ(An) = 1/2 so all in all
∫An
umdµ =12δnm. (5.0.18)
As before this identity enables S to be written in a slightly dierent formcompared to (5.0.17). As promised one other form will be found, based on theexpression ∫
An
um(x+ 2 · 3−n)dµ(x) = −12δnm. (5.0.19)
32
In fact the translation in the integrand in (5.0.19) is the reason for taking theintegral over An instead of the whole space E. The problem with E is demon-strated by the following: if x ∈ E ∩ En,2k−1 then x + 2 · 3−n ∈ E ∩ En,2k butif, instead, x ∈ E ∩ En,2k then x + 2 · 3−n /∈ E (thus the intervals En,2k wereexcluded in An).
To prove (5.0.19) notice that un(x+ 2 · 3−n) = −1 on An, giving∫An
un(x+ 2 · 3−n)dµ(x) = −1/2.
If m > n then∫E∩En,2k−1
um(x+ 2 · 3−n)dµ(x) =∫E∩En,2k
umdµ = 0
for all k = 1, . . . , 2n−1 while in the case n > m the partition of Em,k into 2n−m
separate intervals En,k′ can be used so that∫An∩Em,k
um(x+ 2 · 3−n)dµ(x) =∫Bn∩Em,k
umdµ = (−1)k+12−m−1.
Both these cases imply, in a similar manner than before the desired result(5.0.19).
The identities (5.0.18) and (5.0.19) easily imply, respectively, the expressions
− 12
3−nan =∫An
ua(x+ 2 · 3−n)dµ(x)
12
3−nan =∫An
uadµ
for recovering the coecients from the function. These two combined yield thefollowing expression for S: if u ∈ F and Sn(u) denotes the nth term of thesequance S(u) then
Sn(u) = 3n∫An
[u(x)− u(x+ 2 · 3−n)]dµ(x).
Now let u ∈ F and g ∈ D(u). Estimate
|Sn(u)| ≤ 3n∫An
|u(x)− u(x+ 2 · 3−n)|dµ(x)
≤ 2∫An
[g(x) + g(x+ 2 · 3−n)]dµ = 2∫E
gdµ ≤ 2||g||p.
Thus ||S(u)||`∞ ≤ 2||u||M1,p(E).
The last theorem completes the proof of the fact that `∞ ∼= F . Supposecontrary to what has been already said, that the Hajªasz space M1,p(E) isreexive. With the aid of lemmas 2.1.1 and 2.1.2 this implies that F , and thus`∞, is also reexive. This contradiction in turn shows the ultimate result: theHajªasz space for p > 1 over the Cantor set is not reexive.
33
6 Newtonian spaces and the Poincaré inequality
Another possible generalization of the classical Sobolev space of exponent p ≥ 1to the metric measure theoretic setting is the so called Newtonian space N1,p(X)which utilizes the notion of upper gradients and p-weak upper gradients. It wasrst dened in [31]. This approach is more geometric and in general the spaceis better behaved than the Hajªasz space. Upper gradients have perhaps moreto do with the geometry of the underlying space whereas Hajªasz gradientssomehow merge the measure theoretic structure into the metric one so as toremove some oddities arising from the pure geometric structure of the underlyingspace. For instance, given any space (X, d, µ) any set of µ-measure zero can beremoved fromX leaving the corresponding Hajªasz space unaected (this followsdirectly from denition 4.0.3). This is not the case for N1,p(X).
In the context of this section various assumptions about the metric measurespace will be employed. These will mostly be specied in the beginning of eachsubsection. This exposition largely follows [14], wherein additional informationand more references on the subject can be found.
6.1 Rectiable curves
In this subsection (X, d) will unless otherwise specied be a separable metricspace with no extra structure. µ, a measure on X, when present will be assumedto be Borel regular with respect to the metric d but no other conditions will beimposed.
Given two points x, y ∈ X, curve γ joining (or connecting) them is a continuousmapping f : [a, b] → X for which f(a) = x and f(b) = y. The mapping f iscalled a parametrization of γ and is by no means unique.
Denition 6.1.1. Two continuous mappings f : [a, b]→ X and g : [a′, b′]→ Xare said to parametrize the same curve if there exists a homeomorphism ϕ :[a, b]→ [a′, b′] so that g = f ϕ.
A rectiable curve, then, is a curve for which there exists a rectiable paramet-rization, i.e. a continuous f : [a, b] → γ, f(a) = x, f(b) = y such that thefollowing quantity is nite:
`(f) = supn∑i=1
d(f(ai), f(ai−1)) : a = a0 < · · · < an = b is a partition of [a, b].
This is called the length of f . It is in fact independent of parametrization inother words if f and g are two parametrizations of the same curve then theyhave the same length, `(f) = `(g). Consequently the notation `(γ) will be usedfor the length of a curve γ and no parametrization need be xed.
` can be thought of as the one dimensional Hausdor measure with oneexception. Since a parametrization can overlap itself in a set of positive H1
measure ` measures not the image of the parametrization as such (which iswhat H1 does) but rather the curve as the parametrization sees it. ` still hasa property of additivity.
34
Lemma 6.1.2. Lemma Let f : [a, b]→ γ be a curve in X and a ≤ t ≤ s ≤ u ≤ b.Then `(f |[t,u]) = `(f |[t,s]) + `(f |[s,u])
Proof. If t = a0 < · · · < an = s and s = b0 < · · · < bm = u are partitions of [t, s]and [s, u], respectively, then t = a0 < · · · an < b1 · · · bm = u forms a partition of[t, u], hence
`(f |[t,u]) ≥n∑i=1
d(f(f(ai), f(ai−1)) +m∑j=1
d(f(bi), f(bi−1))
which yields `(f |[t,u]) ≥ `(f |[t,s]) + `(f |[s,u]).The other inequality follows from the observation that any partition t =
a0 < · · · < an = u can be split into partitions t = a0 < · · · am ≤ s ands < am+1 < · · · < an = u. Again this leads to
n∑i=1
d(f(ai), f(ai−1)) ≤m∑i=1
d(f(ai), f(ai−1)) + d(f(s), f(am))
+ d(f(am+1), f(s)) +n∑
j=m+2
d(f(ai), f(ai−1)) ≤ `(f |[t,s]) + `(f |[s,u])
and `(f |[t,u]) ≤ `(f |[t,s]) + `(f |[s,u]).
Denition 6.1.3. If γ : [a, b]→ X is a curve in X the function sγ : [a, b]→ R,dened as
sγ(t) = `(γ|[a,t]),
is called the length function associated to γ.
Theorem 6.1.4. The length function associated to a given parametrization f ofa curve is continuous. Furthermore there is a unique 1-Lipschitz parametrizationf : [0, `(γ)] → γ so that f sf = f and, in addition `(f |[0,t]) = t for all0 ≤ t ≤ `(γ)
Proof. Let c ∈ (a, b] and take a sequence tk → c, tk < c for all k. Sincethe length function is obviously non-decreasing the limit limk→∞ sf (tk) =: αexists and is bounded from above by sf (c). For a given ε > 0 and partitiona = a0 < · · · < an = c for which
sf (c) < ε+n∑i=1
d(f(ai), f(ai−1))
let k0 be such that an−1 < tk < an = c whenever k ≥ k0. The string of estimates
sf (c) < ε+n−1∑i=1
d(f(ai), f(ai−1)) + d(f(tk), f(an−1)) + d(f(c), f(tk))
≤ ε+ sf (tk) + d(f(c), f(tk))
which holds for any k ≥ k0 yields,upon passing to the limit k →∞, sf (c) < ε+α.Since ε was arbitrary it follows that α = sf (c). Similarly it can be shown thatlimt→c− sf (t) = sf (c) for c ∈ [a, b) and the continuity of sf follows.
35
Next dene f as f(t) = f(s−1f (t)), t ∈ [0, `(γ)]. To prove that this is well
dened, suppose x, y ∈ s−1f (t), t ∈ [0, `(γ)] and assume x < y. By lemma 6.1.2
t = sf (y) = `(f |[a,y]) = `(f |[a,x]) + `(f |[x,y]) = t+ `(f |[x,y]),
hence d(f(x), f(y)) ≤ `(f |[x,y]) = 0. Clearly f satises f sf (t) = f(t) for everyt in the domain of f . For s < t ∈ [0, `(γ)] and y ∈ s−1
f (s), x ∈ s−1f (t) one clearly
has f |[s,t] = f |[y,x], in particular `(f |[0,t]) = `(f |[0,x]) = sf (x) = t. This, in turn,implies that
d(f(t), f(s)) ≤ d(f |[s,t]) ≤ `(f |[s,t]) = `(f |[y,x]) = `(f |[0,x])− `(f |[0,y])
= `(f |[0,t])− `(f |[0,s]) = t− s.
The last remaining task is to prove the uniqueness of f . Suppose on the contrarythat g and v are two dierent functions for which g sf = f = v sf . Then, ifx = sf (t),
d(g(t), v(t)) = d(g sf (x), v sf (x)) = d(f(x), f(x)) = 0.
This is possible for any t since sf is a surjection. Thus g and v have to be thesame.
Denition 6.1.5. Given a curve f , the unique parametrization f as assured bytheorem 6.1.4 is called the arc length parametrization of f .
The next theorem shows that even dierent parametrizations of a curvelead to the same arc length parametrization. Thus the notation γ where noreference is made to the particular parametrization is justied and will beused hereafter.
Proposition 6.1.6. Suppose f and g parametrize the same curve in the senseof 6.1.1. Then f = g.
Proof. Let t ∈ [0, `(γ)] (here `(γ) denotes the common value of `(f) and `(g))and suppose ϕ is a homeomorphism such that g = f ϕ. Then
s−1g (t) = ϕ−1s−1
f (t)
as sets and therefore from the denition of the arc-length parametrization
g(t) = g(s−1g (t)) = g(ϕ−1s−1
f (t)) = f(s−1f (t)) = f(t).
This completes the proof of the claim.
By proposition 6.1.6 it is quite obvious that the following denition theculmination of this subsection is, indeed, independent of any particular para-metrizations.
Denition 6.1.7. If γ : [a, b] → X is a given curve and ρ : γ[a, b] → [0,∞] aBorel function, the integral of ρ over (or along) γ is dened to be∫
γ
ρ =∫ `(γ)
0
ρ(γ(t)) dt.
This concept of line-integral will have a very important role in what follows.
36
6.2 The modulus of a family of curves
Having dened integration along rectiable curves it is now possible to consider,for instance, the question whether some Borel function has nite integral overall curves or perhaps no curves at all. However, in connection with merelymeasurable or Borel functions one should be able to speak about almost allcurves for example a measurable function belongs to Lp(R) if and only if it isp-integrable over almost all lines parallel to the coordinate axes.
To this purpose it is convenient to introduce a measure in the family of allrectiable curves in a given metric space, denoted M(X). In other words
M(X) = γ : [a, b]→ X continuous : `(γ) <∞.
Denition 6.2.1. Let 1 ≤ p <∞ and Γ ⊂M(X) a set of curves in X. Denoteby F (Γ) the set of Borel functions ρ : X → [0,∞] so that∫
γ
ρ ≥ 1 for all γ ∈ Γ.
The p-modulus of Γ is
Modp(Γ) = infρ∈F (Γ)
(∫X
ρpdµ)1/p
.
Note that if Γ = ∅ then F (Γ) is the set of all Borel functions so that thedenition above makes sense in this case as well. As promised, Modp turns outto be an outer measure.
Proposition 6.2.2. Modp is a(n outer) measure on M(X).
Proof. For the empty set F (∅) consists of all Borel functions and consequentlyModp(∅) = 0. If Γ ⊂ ∆ then F (∆) ⊂ F (Γ), hence Modp(Γ) ≤ Modp(∆).
Finally for a sequence Γi ⊂ M(X) and an arbitrary ε > 0 take, for each
i ∈ N ρi ∈ F (Γi) so that
(∫X
ρpi dµ)1/p
< Modp(Γi) + ε/2i. If ρ :=∑i
ρi then
ρ ∈ F (⋃i
Γi) which is seen as follows: for any γ ∈⋃i Γi
∫γ
ρ =∑i
∫γ
ρi ≥ 1
since γ belongs to some particular Γi. Then the estimate
Modp(⋃i
Γi) ≤(∫
X
ρpdµ)1/p
≤∑i
(∫X
ρpi dµ)1/p
≤∑i
Mod(Γi) + ε
for arbitrary ε demonstrates the countable subadditivity.
If a property holds for all curves except for some set with p-modulus zerothen this property is said to hold p-almost everywhere. For the null sets withrespect to the p-modulus there is a nice characterization.
37
Proposition 6.2.3. A set Γ ⊂ M(X) has p-modulus zero if and only if thereexists a non-negative Borel measurable function ρ ∈ Lp(µ) so that∫
γ
ρ =∞ for every γ ∈ Γ.
Proof. Suppose rst that Modp(Γ) = 0. Then there is a sequence ρn ∈ F (Γ) sothat ||ρn||p < 2−n. The function dened by ρ := ρ1 + ρ2 + · · · has the desiredproperty: it is non-negative, ||ρ||p ≤
∑i ||ρn||p = 1 and for any γ ∈ Γ∫
γ
ρ =∑n
∫γ
ρn ≥∑n
1 =∞.
For the other implication assume ρ is a positive p-integrable function for with∫γ
ρ =∞ for every γ ∈ Γ. Then the same holds for ρ/n and hence ρ/n ∈ F (Γ),
yielding
Modp(Γ) ≤ ||ρ||pn
for all n ∈ N.
This implies Modp(Γ) = 0.
The following corollary demonstrates the typical use of the notion of themodulus.
Corollary 6.2.4. If g is a Borel function and g ∈ Lp(µ) then∫γ|g| is nite for
p-almost every curve.
Proof. Let Γ be the set of curves for which∫γ|g| = ∞. Then, according to
proposition 6.2.3 Modp(Γ) = 0.
Proposition 6.2.5. If un is a sequence of p-integrable Borel functions converg-ing to a Borel function u in the p-norm, there exists a subsequence so that∫
γ
|unk − u|k→∞−→ 0
for p-almost all curves γ.
Proof. Take the subsequence unk so that∫X
|unk − u|pdµ ≤ 2−2pk and denote
by Γk set of curves γ for which∫γ
|unk − u| ≥ 2−k. Then 2k|unk − u| ∈ F (Γk)
yielding
Modp(Γk) ≤(∫
X
[2k|unk − u|]pdµ)1/p
≤ 2k−2k = 2−k
If Γ =⋂n
⋃i≥n
Γi then for any γ /∈ Γ there is some n so that γ /∈⋃i≥n
, i.e.
∫γ
|uni − u| ≤ 2−i,
38
for all i ≥ n. Hence for any γ /∈ Γ∫γ
|unk − u|k→∞−→ 0.
Since Modp(Γ) ≤ Modp(⋃i≥n
Γi) ≤ 21−n for all n the proof of the proposition is
completed.
6.3 Upper gradients
Earlier in this paper what are sometimes called the Hajªasz derivatives or Ha-jlasz upper gradients were introduced.2 The following denition has as a startingpoint a weak form of the fundamental theorem of calculus
|u(x)− u(y)| ≤∫ 1
0
|x− y||∇u(y + t(x− y))dt,
and, as it turns out, it yields a less severe condition than the pointwise inequality(4.0.7).
Denition 6.3.1. An upper gradient g of a locally integrable function u in ametric measure space (X, d, µ) is a non-negative Borel function such that
|u(γ(b))− u(γ(a))| ≤∫γ
g (6.3.1)
for every rectiable curve γ ∈M(X), γ : [a, b]→ X. Note that here the symbolγ is used both for the curve and its parametrization. If (6.3.1) holds for p-almostall curves γ g is said to be a p-weak upper gradient for u.
When speaking of Lp(µ)-functions two functions coinciding outside a setof measure zero dene the same Lp(µ)-element but, unfortunately, what is anupper gradient for one may fail to be so for the other. Similarly if an uppergradient of a given function is changed on a set of measure zero then it mayfail to remain an upper gradient of the same function. As it turns out no suchdiculties occur with p-weak upper gradient another demonstration of theusefulness of the concept of p-a.e..
Proposition 6.3.2. If g is an upper gradient for a given u and g′ agrees withg almost everywhere then g′ is a p-weak upper gradient for u.
Proof. To prove the rst claim it is sucient to demonstrate that∫γ
g =∫γ
g′
for p-almost every curve or, more generally that∫γ
|g−g′| = 0 for p-almost every
2The pointwise inequality in 4.0.3 constitutes the requirement for a Hajªasz upper gradient.In this terminology a HajªIsz upper gradient therefore need not be an element of Lp(µ).Functions satisfying the conditions of 4.0.3 will be referred to as an Lp(µ)- Hajªasz uppergradient.
39
curve. For this purpose consider the set of curves Γk for which∫γ
|g−g′| > 2−k.
Obviously 2k|g − g′| ∈ F (Γk), hence
Modp(Γk) ≤ 2k(∫
X
|g − g′|pdµ)1/p
= 0.
If Γ denotes the set of curves for which∫γ|g− g′| 6= 0 then Γ =
⋃k Γk and thus
Modp(Γ) = 0.
The situation is not so favourable if u is altered in a set of measure zerofor neither g nor g′ need then be even a p-weak upper gradient of the newfunction. This fact will make it necessary to talk about versions of functionsin the theory of Newtonian spaces. However proposition 6.2.3 yields a niceapproximation result relating upper and p-weak upper gradients.
Proposition 6.3.3. If g is an upper gradient of an almost everywhere nitefunction u then for every ε > 0 there exists an upper gradient gε ≥ g for which
||gε − g||p ≤ ε.
Proof. Suppose Γ is the set of curves for which |u(γ(b)) − u(γ(a))| >∫γg. By
assumption Modp(Γ) = 0 so, according to proposition 6.2.3 there exists a non-negative Lp(µ)-function ρ such that∫
γ
ρ =∞ for every γ ∈ Γ.
Set gε = g +ερ
||ρ||p. Then clearly gε ≥ g and ||gε − g||p ≤ ε. Also |u(γ(b)) −
u(γ(a))| ≤∫γ
gε for all γ ∈ M(X) with innity on the right-hand side for
γ ∈ Γ.
6.4 The Newtonian space
Let 1 ≤ p <∞. A measurable function u : X → R is said to belong to the classN1,p(X)1,p if there exists a non-negative Lp(µ)-function g which is a p-weakupper gradient for u. Introduce in N(X)1,p the norm-like expression
||u||1,p = ||u||p + infg∈G(u)
||g||p
where G(u) denotes the set of all p-weak upper gradients of u. The spaceN(X)1,p in itself exhibits some quite unwanted for properties as discussed justbefore. For instance the norm ||u||1,p does not behave well under the normalequivalence relation u v v i u = v a.e.. To construct the Newtonian space aslightly modied equivalence relation will be dened.
Denition 6.4.1. Dene an equivalence relation between elements of N(X)1,p:u v v if and only if ||u−v||1,p = 0. The Newtonian space N1,p(X) is the quotientN(X)1,p/ v. The space will be equipped with the norm ||u||N1,p(X) := ||u||1,p.
40
In particular when inquiring whether or not a given Lp(µ)-function belongsto N1,p(X) the question really is: does there exist a version of u i.e. somefunction v which agrees with u outside a set of measure zero which belongsto N1,p(X). While altering a function in a null set may result in the modiedversion no longer being an element of N1,p(X) it is still true that
Proposition 6.4.2. If u, v ∈ N1,p(X) and u = v almost everywhere then ||u−v||1,p = 0.
Proof. Let N denote the set where u and v dier. Then µ(N) = 0 and g =∞ · χN ∈ Lp(µ). By corollary 6.2.4
∫γg < ∞ for p-almost every curve. Since
u−v ∈ N1,p(X) there exists, by proposition 6.3.3, an upper gradient g′ ∈ Lp(µ)of w := u− v. Here also
∫γg′ <∞ for p-almost every curve. Along such curves
|w(γ(β)) − w(γ(α))| ≤∫ β
α
g′(γ(t))dt < ∞ for every α and β in the domain of
γ. Thus by the absolute continuity of the integral w γ is continuous (in factabsolutely continuous). If a curve satises both
∫γg < ∞ and
∫γg′ < ∞ then
the rst condition implies that |γ−1(N)| = 0 because g(γ(t)) = ∞ on γ−1(N).(Here |A| denotes the Lebesgue measure of the set A.) Thus also u − v = 0almost everywhere on γ with respect to the Lebesgue measure. By the secondcondition w is continuous along γ, implying that wγ = 0 on the whole of [a, b].
Since p-almost all curves indeed do satisfy both conditions above it followsthat g = 0 is a p-weak upper gradient for w = u− v, hence ||u− v||1,p = 0.
The following lemma relates the limit of upper gradients of a sequence offunctions to the upper gradient of the limiting function.
Lemma 6.4.3. Suppose that un is a sequence of functions converging to u inLp(µ). Let gn be p-weak upper gradients of un and let g be such that gn → g inLp(µ). Then there is a version of u so that g is a p-weak upper gradient of u.
Proof. Proposition 6.3.3 allows one to replace the p-weak upper gradients gk ofuk by upper gradients g′k. For each k let g′k be an upper gradient of uk suchthat ‖gk − g′k‖p ≤ 2−k. Then ‖g − g′k‖p → 0 as k →∞. Take a subsequence g′j
so that∫γ
|g′j − g|j−→ 0 for p-almost every curve and denote by Γ0 the set of
curves for which this convergence fails to hold. From this subsequence take yet
another, indexed by jk, k ≥ 1, for which ujkk−→ u pointwise almost everywhere.
The functions ukj and u can be assumed to be everywhere nite. Dene u =lim infk→∞ ujk . Then u = u almost everywhere. If E ⊂ X denotes the set wherelim infk→∞ |ujk | =∞ (whence µ(E) = 0) let Γ′ = γ ∈M(X) : L1(γ−1E) > 0and Γ′′ = γ ∈M(X) :
∫γg =∞. Since ρ =∞χE satises
∫γρ =∞ for every
γ ∈ Γ′ it follows from 6.2.3 that Modp(Γ′) = 0. Hence all these curve-familieshave p-Modulus zero and if Γ = Γ′ ∪ Γ′′ then Modp(Γ) = 0.
If γ /∈ Γ, γ : [a, b] → X, then |u γ(α)| < ∞ for L1- almost every α ∈ [a, b].To see that, in fact, |u γ(α)| < ∞ for every α ∈ [a, b] notice that for every kujk γ is absolutely continuous,
|ujk γ(β)− ujk γ(α)| ≤∫ β
α
g′jk γ(t)dt,
41
and moreover ∫ β
α
|g′jk γ(t)− g γ(t)|dt ≤∫γ
|g′jk − g|k→∞−→ 0
for every a ≤ α ≤ β ≤ b. Now if α ∈ [a, b] is arbitrary, estimate for every k
|ujk γ(α)| ≤ |ujk γ(β)−ujk γ(α)|+ |ujk γ(β)| ≤∫ β
α
g′jk γ(t)dt+ |ujk γ(β)|.
Since γ−1E ⊂ [a, b] has zero L1-measure there must exist some β ∈ [a, b]\γ−1E,that is, lim infk→∞ |ujk γ(β)| <∞, and therefore
|u γ(α)| ≤ lim infk→∞
|ujk γ(α)| ≤ lim infk→∞
∫ β
α
g′jk γ(t)dt+ lim infk→∞
|ujk γ(β)|
≤∫γ
g + lim infk→∞
|ujk γ(β)| <∞
Therefore, if γ /∈ Γ then u γ <∞ on [a, b] and
|u(γ(b))− u(γ(a))| = limk→∞
| infn≥k
ujn(γ(b))− infn≥k
ujn(γ(a))|
≤ lim supk→∞
|ujk(γ(b))− ujk(γ(a))| ≤ lim supk→∞
∫γ
g′jk =∫γ
g.
These preliminaries allow one to formulate a theorem stating that
Theorem 6.4.4. The space N1,p(X) is a Banach space for 1 ≤ p <∞.
Proof. That || · ||N1,p(X) is a norm is obvious. As with the Hajlasz spacescompleteness will be proven via the convergence of absolutely summable series.
Let then un be a sequence in N1,p(X) so that∑n
||un||N1,p(X) < ∞. For
each n and given ε > 0 let gn ∈ G(un) and ||gn||p < infg∈G(un) ||g||p + ε/2n.Since both ∑
n
||un||p <∞ and∑n
||gn||p < ε+∑n
infg∈G(un)
||g||p <∞
the following convergences hold in Lp(µ) (by completeness of Lp(µ)):
vk :=k∑n
unk→∞−→
∑n
un := v,
gkε :=k∑n
gnk→∞−→
∑n
gn := gε.
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Clearly gkε is a p-weak upper gradient of vk, hence by lemma 6.4.3 gε is a p-weakupper gradient of a version of v and thus v ∈ N1,p(X). It remains to show thatvk → v in N1,p(X). To this purpose note that gε − gkε is also a p-weak uppergradient of v − vk and therefore
||v−vk||N1,p(X) ≤ ||v−vk||p+||gε−gkε ||p ≤∞∑
n=k+1
||un||p+∞∑
n=k+1
infg∈G(un)
||g||p+ε.
The sums on the left-hand side converge to zero while ε is arbitrary andhence vk → v in N1,p(X).
In general N1,p(X) does not need to coincide with M1,p(X). If the un-derlying space X doesn't contain any rectiable curves then self-evidently anynon-negative Borel function in particular the zero function is an upper gra-dient for any Lp(µ)-function u. Thus N1,p(X) = Lp(µ) for such spaces (withequal norms). For instance the cantor set E encountered earlier illustrates onesuch example. But, as was seen, the Hajªasz space M1,p(X) is still not trivial.However the spaces N1,p(X) and M1,p(X) are not completely without relation.In this spirit it shall shortly be proven that the Hajªasz space continuously em-beds into the Newtonian one. Roughly speaking, the richer the structure of thecurves of the underlying space the closer the resemblance between N1,p(X) andM1,p(X).
Theorem 6.4.5. M1,p(X) embeds into N1,p(X) continuously. More specicallyM1,p(X) ⊂ N1,p(X) and
||u||N1,p(X) ≤ 2||u||M1,p(X) for all u ∈M1,p(X).
Proof. Let u ∈ M1,p(X) be a Lipschitz map with Lipschitz constant C. Thenfor any curve γ and its arc-legth parametrization γ : [0, L]→ γ the compositionu γ : [0, L] → R is Lipschitz and hence almost everywhere dierentiable. Inparticular
u(γ(a))− u(γ(b)) =∫ b
a
ddtu γ(t)dt, 0 ≤ a ≤ b ≤ L.
Let then g is a Hajªasz upper gradient of u. By redening g to be, say, C in aset of µ-measure zero it can be assumed that the inequality
|u(x)− u(y)| ≤ d(x, y)[g(x) + g(y)]
holds for every x, y ∈ X and that g is a Borel function. For any n ∈ N Luzin'stheorem implies the existence of an open set Un ⊂ [0, L] for which |Un| < 1/nand g γ is continuous in [0, L]\Un =: Cn. The sets Un can, of course, be chosenso as to form a decreasing sequence U1 ⊃ U2 ⊃ U3 . . .. Let B be the set of density
points of∞⋃n=1
Cn. Evidently |[0, L] \ B| = 0 since |[0, L] \ B| ≤ |Un| < 1/n for
every n. If t ∈ B then
|⋃∞n=1[t− 1/k, t+ 1/k] ∩ Cn|
2/kk→∞−→ 1,
43
in particular there is for each k (suciently large) some sk ∈ [−1/k, 1/k] \ 0so that t+ sk ∈
⋃∞n=1[t− 1/k, t+ 1/k]∩Cn. It then follows that t+ sk → t and
g γ(t+ sk)→ g γ(t).Letting A denote the intersection of B with poits of dierentiability of u γ
can be seen that for any t ∈ A∣∣∣∣ ddtu γ(t)
∣∣∣∣ = limk→∞
|u γ(t+ sk)− u γ(t)||sk|
≤ limk→∞
d(γ(t+ sk), γ(t))|sk|
[g(γ(t+ sk)) + g(γ(t))]
≤ limk→∞
[g(γ(t+ sk)) + g(γ(t))] = 2g(γ(t)).
Consequently
|u γ(b)− u γ(a)| ≤∫ b
a
∣∣∣∣ ddtu γ(t)
∣∣∣∣dt ≤ ∫ b
a
2g(γ(t))dt =∫γ
2g.
Having thus demonstrated that 2g is an upper gradient of u whenever u isLipschitz, take any u ∈ M1,p(X) and un ∈ M1,p(X) a sequence of Lipschitzfunctions converging to u in M1,p(X)-norm. If gn are Hajªasz upper gradientsof un, g a Hajªasz upper gradient of u and ||un − u||p → 0 and ||gn − g||p → 0then, according to lemma 6.4.3 there is a version of g (which can be assumed tobe Borel) which is a p-weak upper gradient of a version of u. Therefore
infg∈D(u)
||g||p ≤ 2 infg∈G(u)
||g||p
and the theorem is proven.
7 Spaces supporting a Poincaré inequality
Both the Newtonian and the Haªjasz spaces can be dened over any metricmeasure space (X, d, µ). However if X is too thin like the Cantor set then N1,p(X) fails to take into account any dierentiability properties of itselements. The Hajªasz space M1,p(X) on the other hand then may fail to bereexive, see section 3. This section is devoted to the study of a class of metricmeasure spaces with a rich enough structure of curves, so that one can hope toavoid the phenomena described above. Again [14] is used as a reference, unlessanother reference is specied.
7.1 The Hajªasz space through Poincaré inequalities
If (X, d, µ) is any metric measure space and p ≥ 1, u ∈ M1,p(X) and g ∈ D(u)then integrating inequality (4.0.7) in the denition of the Hajªasz space over aball B = B(x, r) yields
|u(x)− uB | ≤ r(g(x) + gB)
and, upon second integration using Jensen's inequality
−∫B
|u− uB |dµ ≤ 2r(−∫B
gpdµ)1/p
.
44
This is called the (strong) p-Poincaré inequality for the pair (u, g). Thus anypair (u, g) of a M1,p(X)-function and its Hajªasz upper gradient g satises thep-Poincaré inequality. In fact,
Theorem 7.1.1. If (X, d, µ) is a metric measure space with a doubling measureand p > s/(s + 1) ,s being the homogeneity exponent of µ (see 2.15), thenu ∈M1,p(X) if and only if there exists a non-negative g ∈ Lp(µ) for which
−∫B
|u− uB |dµ ≤ Cr(−∫σB
gqdµ)1/q
(7.1.1)
for every ball B, where r is the radius of B, for some s/(s+ 1) ≤ q < p and forsome constants C > 0 and σ ≥ 1 depending only on p and the doubling constantof µ.
Proof. Suppose that u ∈ M1,p(X), B is a ball of radius r and g ∈ Lp(µ) isa Hajªasz upper gradient for u. Then u ∈ W 1,s/(s+1)(B) and since s/(s +1) < s inequality (4.2.13) from the embedding theorem 4.2.3 is applicable, with(s/(s+ 1))∗ = 1. It yields
−∫B
|u− uB |dµ ≤ Cr(−∫σB
gs/(s+1)dµ)(s+1)/s
.
This proves the only if-part of the claim.
Conversely, suppose 0 ≤ g ∈ Lp(µ) satises (7.1.1) for xed constants C, σand some s/(s+ 1) ≤ q < p. Since µ is doubling the Lebesgue's dierentiationtheorem holds. If x and y are Lebesgue points of u denote ri = 2−id(x, y) andBi = B(x, ri). Then the equality
uBi − uB0(x) =i−1∑k=0
[uBk − uBk+1 ]
leads to
|u(x)− uB0(x)| ≤∞∑k=0
|uBk − uBk+1 |
upon passing to the limit i→∞. Here B0 is written as B0(x) to emphasize thefact that it has centre x. On the other hand
|uBk − uBk+1 | = −∫Bk+1
|uBk − uBk+1 |dµ ≤ −∫Bk+1
|u− uBk |dµ
+−∫Bk+1
|u− uBk+1 |dµ ≤ Crk+1
(−∫σBk+1
gqdµ
)1/q
+ Crk
(−∫σBk
gqdµ)1/q
≤ Crk (Mgq(x))1/q = C2−kd(x, y) (Mgq(x))1/q.
Combining these two inequalities leads to
|u(x)− uB0(x)| ≤ C (Mgq(x))1/q∞∑k=0
2−kd(x, y) = Cd(x, y) (Mgq(x))1/q.
45
A similar argument gives the same result with x replaced by y. Since |u(x) −u(y)| ≤ |u(x)−uB0(x)|+ |u(y)−uB0(y)|+ |uB0(y)−uB0(x)| the pointwise estimate
|u(x)− u(y)| ≤ Cd(x, y)[(Mgq(x))1/q + (Mgq(y))1/q]
will be established as soon as the last term in the right-hand side of the triangleinequality-estimate is handled. For it the triangle inequality yields
|uB0(y) − uB0(x)| ≤ |uB0(x) − u2B0(x)|+ |uB0(y) − u2B0(x)|
and, for each term separately
|uB0(y) − u2B0(x)| ≤ −∫B0(y)
|u− u2B0(x)|dµ ≤1
µ(B0(y))
∫2B0(x)
|u− u2B0(x)|dµ
≤C2µ
µ(2B0(x))
∫2B0(x)
|u− u2B0(x)|dµ ≤ CC2µd(x, y) (Mgq(x))1/q
.
(The term |uB0(x) − u2B0(x)| is estimated in the same way.) These inequalitiescomplete the proof of the claim, since
‖ (Mgq)1/q ‖pp ≤ C‖gq‖q/pp/q = C‖g‖pp
In fact as can easily be seen in the above proof the maximal functions Mcan be replaced byM2σd(x,y) because 2σd(x, y) is the radius of the largest ballanything is integrated over. Also, the proof of the if part of theorem 7.1.1goes through even with q = p. Hence a more careful study of this proof yields
Corollary 7.1.2. Suppose the pair (u, g) satises the p-Poincaré inequality,p ∈ (s/(s+ 1),∞)
−∫B
|u− uB |dµ ≤ Cr(−∫σB
gpdµ)1/p
for any ball B = B(x, r). Then
|u(x)− u(y)| ≤ Cd(x, y)[(M2σd(x,y)g
p(x))1/p +
(M2σd(x,y)g
p(y))1/p]
µ-almost everywhere.
7.2 Lip and lip
Recall that a function F : (X, d) → (Y, d′) between two metric spaces (X, d)and (Y, d′) is called Lipschitz if there is a constant 0 < L <∞ so that
d′(f(x), f(y)) ≤ Ld(x, y) for every x, y ∈ X.
The collection of all Lipschitz-functions X → R is denoted as LIP(X) andthe Lipschitz constant of u ∈ LIP(X) is LIP(u) (unless some other symbol isspecied). Given any u ∈ LIP(X) the Lip and lip of u are functions X → Rdened everywhere as follows.
46
Denition 7.2.1. Let u ∈ LIP(X). Then for every x ∈ X
1.
Lipu(x) = lim supr→0
supd(x,y)<r
|u(x)− u(y)|r
2.
lipu(x) = lim infr→0
supd(x,y)<r
|u(x)− u(y)|r
.
Proposition 7.2.2. For every r > 0 and Lipschitz-function u the function
Lru(x) = supd(x,y)<r
|u(x)− u(y)|r
is lower semicontinuous.
Proof. It suces to show that for every t ∈ R the set At = x ∈ X : u(x) > tis open. Of course t can be assumed to be non-negative. Let x ∈ At, that is,
supd(x,y)<r
|u(x)− u(y)|r
> t.
Thus there is some y′ ∈ B(x, r) so that |u(x)−u(y′)| > rt. Choose an ε > 0 forwhich |u(x) − u(y′)| > rt + ε. On the other hand by the continuity of u thereexists some δ ∈ (0, r − d(x, y′)] so that
|u(x)− u(y)| < ε for every y ∈ B(x, δ).
Now if y ∈ B(x, δ) then d(y, y′) ≤ d(y, x)+d(x, y′) < δ+d(x, y′) < r−d(x, y′)+d(x, y′) = r and hence
|u(y)− u(y′)| ≥ |u(y′)− u(x)| − |u(x)− u(y)| > rt+ ε− ε = rt.
Thus Lru(y) > t for each y ∈ B(x, δ) which implies that At is open.
Corollary 7.2.3. Given any u ∈ LIP(X) the functions Lipu and lipu are Borelfunctions.
Proof. This follows directly from 7.2.2 since Lip and lip can be expressed as
Lipu(x) = lim supr→0r∈Q
Lru(x)
andlipu(x) = lim inf
r→0r∈Q
Lru(x).
To see this x some x ∈ X. In the rst case note that, since ut(x) := t 7→
supr<t
supy∈B(x,r)
|u(x)− u(y)|r
is increasing, one has
Lipu(x) = inft>0
ut(x)
47
and similarly
lim supr→0r∈Q
Lru(x) = inft∈Q+
supr<tr∈Q
supy∈B(x,r)
|u(x)− u(y)|r
.
But in fact
supr<tr∈Q
supy∈B(x,r)
|u(x)− u(y)|r
= ut(x)
by an easy continuity argument Consequently
lim supr→0r∈Q
Lru(x) = inft∈Q+
ut(x)
and from this it is clear that
Lipu(x) ≤ lim supr→0r∈Q
Lru(x).
On the other hand, if ε > 0 is arbitrary and t ∈ Q+ is such that lim supr→0r∈Q
Lru(x)+
ε > ut(x) take some 0 < t′ < t. Utilizing the monotonicity of t 7→ ut(x) yields
Lipu(x) ≤ ut′(x) ≤ ut(x) < lim supr→0r∈Q
Lru(x) + ε
which proves the rst identity. The second one is proven along the same lines.
For Lipu there is another characterization, namely
Proposition 7.2.4. If u ∈ LIP(X) then for every x ∈ X Lipu is given by
Lipu(x) = lim supr→0
supd(x,y)<r
|u(x)− u(y)|d(x, y)
.
Proof. That Lipu(x) ≤ lim supr→0
supd(x,y)<r
|u(x)− u(y)|d(x, y)
is obvious since
supd(x,y)<r
|u(x)− u(y)|r
≤ supd(x,y)<r
|u(x)− u(y)|d(x, y)
for any r > 0. Let ε > 0 and suppose z is such that d(x, z) < r and
supd(x,y)<r
|u(x)− u(y)|d(x, y)
<|u(x)− u(z)|
d(x, z)+ ε ≤ sup
d(x,y)<r0
|u(x)− u(y)|r0
+ ε
where r0 := d(x, z) < r. The claim follows by the denition of lim sup.
The following lemma will be used in the sequel.
Lemma 7.2.5. For xed r > 0 the map x 7→ µ(B(x, r)) satises
lim supy→x
µ(B(y, r)) ≤ µ(B(x, r)).
In particular it is measurable.
48
Proof. Let δ > 0 be arbitrary. Since for any y ∈ B(x, δ) one has B(y, r) ⊂B(x, r + δ) it follows that
supy∈B(x,δ)
µ(B(y, r)) ≤ µ(B(x, r + δ))
from which the claim follows by taking limδ→0.
To connect Lip (and later lip) to the innitesimal behaviour of Lipschitzfunctions the following lemma will be used. As will be seen in the last section thetheorem below can be used used to prove an abstract relation which is satisedby spaces supporting a Póincare inequality. Consequently it is ultimately astatement about the underlying metric space. The proof follows [21].
Theorem 7.2.6. Let (X, d, µ) be a complete measure space with a doublingmeasure. Then there exists a constant 0 < C <∞ so that the inequality
1C
Lipu(x) ≤ lim supr→0
1r−∫B(x,r)
|u− uB(x,r)|dµ ≤ 2 Lipu(x)
holds for µ-almost every x ∈ X.
Proof. The righthand side inequality follows from the denition of the limitsupremum: x an arbitrary ε > 0. Then there exists δ > 0 so that
Lipu(x) ≤ supr<δ
1r
supd(y,x)<r
|u(x)− u(y)| < ε+ Lipu(x).
Thus1r|u(x)− u(y)| ≤ ε+ Lipu(x)
for all y ∈ B(x, r) and r < δ. Now if y, y′ ∈ B(x, r) this leads to
1r|u(y′)− u(y)| ≤ 1
r|u(x)− u(y)|+ 1
r|u(x)− u(y′)| ≤ 2(ε+ Lipu(x)),
from which one obtains
lim supr→0
−∫B(x,r)
|u− uB(x,r)|dµ
≤ lim supr→0
−∫B(x,r)
−∫B(x,r)
1r|u(y′)− u(y)|dµ(y)dµ(y′) ≤ 2(ε+ Lipu(x)).
This implies the rightmost inequality since ε > 0 is arbitrary.
To prove the rst inequality it suces to replace the lim supr→0 withlim supr→0
r∈Q. This is because
lim supr→0r∈Q
1r−∫B(x,r)
|u− uB(x,r)|dµ ≤ lim supr→0
1r−∫B(x,r)
|u− uB(x,r)|dµ.
It can also be assumed that µ(X) <∞ by dividing it into countably many ballswith nite measure. Then, given ε > 0 Lusin's and Egoro's theorems easilyimply that there exists a set A ⊂ X which can be taken compact withµ(X \A) < ε and such that
49
(i) Lipu is uniformly continuous in A and
(ii) the convergence Lipu = limr→0r∈Q
supt<rt∈Q
supd(y,·)<t
|u(y)− u(·)|d(y, ·)
is uniform in A.
(iii) the convergengeµ(B(x, r) \A)µ(B(x, r))
r→0r∈Q−→ 0 is uniform in the set of density
points of A.
Now x ε > 0 and let δ1 > 0 be such that
|u(y)− u(z)|d(y, z)
< Lipu(y) + ε
whenever y ∈ A and z ∈ B(y, δ1) (utilizing condition (ii) above). Let δ2 > 0 besuch that
|Lipu(y)− Lipu(z)| < ε
whenever y, z ∈ A and d(y, z) < δ2 (utilizing condition (i) above). Finally letδ3 > 0 be such that
µ(B(y, r) \A)µ(B(y, r))
< ε
whenever y is a density point of A and 0 < r < δ3(utilizing condition (iii)above). Now suppose 0 < r < minδ1, δ2, δ3 := δ and x a density point x ofA. Then for any y ∈ B(x, r) ∩A and z ∈ B(y, r) ∩A
|u(y)− u(z)| < d(y, z)(Lipu(y) + ε) < d(y, z)(Lipu(x) + 2ε).
Integrating this yields
|u(y)− uB(y,r/4)| ≤ −∫B(y,r/4)
|u(y)− u(z)|dµ(z)
≤ 1µ(B(y, r/4))
[∫B(y,r/4)∩A
|u(y)− u(z)|dµ(z) +∫B(y,r/4)\A
|u(y)− u(z)|dµ(z)
]
≤ r/4(Lipu(x) + 2ε) + r/4Lµ(B(y, r/4) \A)µ(B(y, r/4))
≤ r/4(Lipu(x) + 2ε) + rLε, (7.2.1)
L being the Lipschitz constant of u.Now let s denote the homogeneity exponent of µ and let r > 0 be such that
r(1 + ε1/s) < δ. By denition one has
supr′<r
supy∈B(x,r)
|u(y)− u(x)|d(x, y)
> Lipu(x)− ε,
in particular there exists r′ < r, r′ ∈ Q and y′ ∈ B(x, r′) for which
|u(x)− u(y′)| > r′(Lipu(x)− ε).
50
On the other hand the ball B(y′, 8r′ε1/s) ⊂ B(x, r′(1 + 8ε1/s)) contains a pointof A:
µ(B(y′, 8r′ε1/s) ∩A)µ(B(x, r′(1 + 8ε1/s)))
=µ(B(y′, 8r′ε1/s))
µ(B(x, r′(1 + 8ε1/s)))− µ(B(y′, 8r′ε1/s) \A)µ(B(x, r′(1 + 8ε1/s)))
≥ 4−s(
8ε1/s
1 + 8ε1/s
)s− µ(B(x, r′(1 + 8ε1/s)) \A)
µ(B(x, r′(1 + 8ε1/s)))> ε
(2s
(1 + 8ε1/s)s− 1)> 0.
If y ∈ B(y′, 8r′ε1/s) ∩A is such a point then
|u(x)− u(y)| ≥ |u(x)− u(y′)| − |u(y)− u(y′)| ≥ r′(Lipu(x)− ε)− 8r′Lε1/s.
Since the set of density points of A is dense in B(y′, 8r′ε1/s) ∩ A it can beassumed that y is already a density point of A (contained in A).
These facts together with (7.2.1) yield
r′(Lipu(x)− ε− 8Lε1/s) < |u(y)− u(x)|≤ |u(y)− uB(y,r′/4)|+ |u(x)− uB(x,r′/4)|+ |uB(x,r′/4) − uB(y,r′/4)|≤ r′/2(Lipu(x) + 2ε) + |uB(x,r′/4) − uB(y,r′/4)|,
or
(Lipu(x)/2− 3ε− 8Lε1/s) ≤ 1r′|uB(x,r′/4) − uB(y′,r′/4)|. (7.2.2)
The right-hand side of (7.2.2) can be estimated by
|uB(x,r′/4) − uB(y′,r′/4)| ≤ |uB(x,2r′) − uB(y′,r′/4)|+ |uB(x,r′/4) − uB(x,2r′)|
and both of the right-hand side terms of this in turn by
|uB(x,2r′) − uB(y′,r′/4)| ≤ −∫B(y′,r′/4)
|u(z)− uB(x,2r′)|dµ(z)
≤ C−∫B(x,2r′)
|u(z)− uB(x,2r′)|dµ(z),
|uB(x,r′/4) − uB(x,2r′)| ≤ C−∫B(x,2r′)
|u(z)− uB(x,2r′)|dµ(z)
since B(y′, r′/4) ⊂ B(x, 2r′). Here the constant C depends only on the doublingconstant of µ.
Combining all the above results it can be concluded that for any r < δ thereexists r′ < r, r′ ∈ Q so that
(Lipu(x)/2− 3ε− 8Lε1/s) ≤ C
r′−∫B(x,2r′)
|u(z)− uB(x,2r′)|dµ(z).
The desired inequality follows by taking lim supr→0r∈Q
and letting ε→ 0. Since almost
every point of A is a density point and A can be taken to have measure arbitrarilyclose to full measure the proof is complete.
51
7.3 Metric measure spaces supporting a Poincaré inequal-
ity
The idea of the class of spaces about to be dened the spaces supporting ap-Poincaré inequality is simply to require an inequality similar to (7.1.1) tohold not only for Hajªasz upper gradients but arbitrary upper gradients. Thefollowing denition is from [14].
Denition 7.3.1. A doubling metric measure space (X, d, µ) is said to supporta (weak) p-Poincaré inequality if there are constants C > 0 and σ ≥ 1 suchthat for any locally integrable Borel function u : X → R and its arbitrary locallyintegrable upper gradient g the Poincaré inequality
−∫B
|u− uB |dµ ≤ Cr(−∫σB
gpdµ)1/p
(7.3.1)
is satised. In other words X supports the p-Poincaré inequality if there areconstants C > 0 and σ ≥ 1 so that for all locally integrable Borel functions thepair (u, g) satises the p-Poincaré inequality for any upper gradient g of u.
Note that while Theorem 7.1.1 exhibits a property of the space M1,p(X),denition 7.3.1 is a property concerning the underlying space X and, in par-ticular the space N1,p(X) does not appear anywhere in the denition.
This condition quanties the idea that if a function u has some pathwise regu-larity (i.e. has an upper gradient controlling it along curves) then that functionmust have some smoothness in a more traditional sense, i.e. the mean value ofit's oscillation is controlled by the same upper gradient that controls it alongpaths. The quantitative constant C is related to the doubling constant of theunderlying measure whereas p measures how well a function can be controlled inthe mean value sense in terms of the control one has over it through it's uppergradient. This connection between the behaviour of functions along curves andin the mean value sense turns out to be very fruitful for rst order calculus(larger values of p correspond to less control, or a weaker connection betweenthe two).
Example As an example it will be shown that if C ⊂ Rn is a closed convexset with positive Lebesgue measure then the metric measure space (C, | · |,Ln|C)supports a 1-Poincaré inequality. The measure will be abbreviated by µ. It isimportant to note that µ is Ahlfors n-regular, that is if BC(x, r) denotes a ballof C of radius r ≤ diam(C) and centre x ∈ C then
µ(B(x, r)) ≈ rn
where the implied constants depend only on C and n.
Let u : C → R be a locally integrable function and g it's (locally integrable)upper gradient. For any two points y, z ∈ C the line [y, z] = y + t(z − y) : t ∈[0, 1] connecting them lies in C and therefore
|u(z)− u(y)| ≤∫
[y,z]
g = |y − z|∫ 1
0
g(y + t(z − y))dt.
Now let x ∈ C, r > 0 and let B = B(x, r) ∩ C be a ball of C.
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The rst task is to estimate |u(y)− uB | for every y ∈ B:
|u(y)− uB | ≤1
µ(B)
∫B(y,2r)∩C
|u(y)− u(z)|dz
≤ 2rµ(B)
∫B(y,2r)∩C
∫ 1
0
g(y + t(z − y))dtdz. (7.3.2)
Denote by Lt the transformation Lt(z) = y + t(z − y) and note that
Lt(B(y, 2r) ∩ C) = B(y, 2tr) ∩ Lt(C) ⊂ B(y, 2tr) ∩ C.
Using this and the change of variables z 7→ Lt(z) yields∫B(y,2r)∩C
g(y + t(z − y))dt ≤ t−n∫B(y,2tr)∩C
g(z)dz.
Plugging this into (7.3.2) after interchanging the order of integration (justiedby the local integrability of g) gives
|u(y)− uB | ≤2rµ(B)
∫ 1
0
t−n∫B(y,2tr)∩C
g(z)dzdt.
Next integrate this with respect to y (over B) to get
−∫B
|u− uB |dµ ≤2r
µ(B)2
∫ 1
0
t−n∫B(x,r)∩C
∫B(y,2tr)∩C
g(z)dzdydt. (7.3.3)
To estimate this note that∫B(x,r)∩C
∫B(y,2tr)∩C
g(z)dzdy =∫B(x,r)∩C
∫B(x,3r)∩C
χB(y,2tr)∩C(z)g(z)dzdy
=∫B(x,3r)∩C
g(z)∫B(x,r)∩C
χB(y,2tr)∩C(z)dydz
=∫B(x,3r)∩C
g(z)Ln(B(z, 2tr) ∩B(x, r) ∩ C)dz . (tr)n∫B(x,3r)∩C
g(z)dz.
This combined with (7.3.3) then gives the estimate
−∫B
|u− uB |dµ .2r
µ(B)2
∫ 1
0
rn∫B(x,3r)∩C
g(z)dzdt ≈ r−∫
3B
gdµ.
Here 3B = B(x, 3r) ∩ C.The above reasoning is actually valid for any Ahlfors n-regular measure on
C. It does not, however, extend directly to non-convex domains, and in generalthe claim fails for arbitrary domains D in Rn when the measure is the restrictionof Ln to D. [17]
Continuing with general spaces X it is not dicult to see that if u ∈ LIP(X)then lip(u) is an upper gradient of u.
Proposition 7.3.2. Let (X, d, µ) be a metric measure space and u : X → R aLipschitz function. Then lipu is an upper gradient of u.
53
Proof. Let γ : [0, L] → X be the arc-length parametrization of a rectiablecurve (if no rectiable curves exist then any function is an upper gradient of u)and consider the function u γ : [0, L]→ R. This obeys
u(γ(L))− u(γ(0)) =∫ L
0
ddtu γ(s)ds,
in particular
|u(γ(L))− u(γ(0))| ≤∫ L
0
∣∣∣∣ ddtu γ(s)
∣∣∣∣ds.But at any point s ∈ [0, L] of dierentiability∣∣∣∣ d
dtu γ(s)
∣∣∣∣ = lim infh→0
|u(γ(s+ h))− u(γ(s))||h|
≤ lim infh→0+
supy∈B(γ(s),h)
|u(y)− u(γ(s))|h
= lipu(γ(s)).
This is since d(u(γ(s+h)), u(γ(s))) ≤ |h| so always u(γ(s+h)) ∈ B(u(γ(s)), |h|).Consequently
|u(γ(L))− u(γ(0))| ≤∫γ
lipu.
A natural question is whether or not lipu (or Lip) is in some sense an optimalupper gradient. Both indeed turn out to be optimal when the underlyingspace supports a Poincaré inequality, in the sense that there is a constant sothat any upper gradient times the constant is a pointwise upper bound of Lipup to a set of measure zero. These special upper gradients are also adequate forthe denition of a space supporting a Poincaré inequality: in [21] a denitionusing lip is presented and these two denitions coincide at least whenever themeasure µ is doubling and the space complete.
Theorem 7.3.3. A complete metric measure space X, whose measure µ isdoubling, supports the p-Poincaré inequality if and only if there are C > 0 andσ ≥ 1 so that
−∫B
|u− uB |dµ ≤ Cr(−∫σB
(lipu)pdµ)1/p
for any Lipschitz function u and any ball B of radius r.
The proof of this theorem can be found in for instance in [20].
Denition 7.3.4. A metric space X is said to be C-quasiconvex, C > 0 aconstant, if it is path-connected and for any two points x, y ∈ X there exists arectiable curve γ joining them so that
`(γ) ≤ Cd(x, y).
When the particular constant is unimportant it is said merely that X is quasi-convex.
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Theorem 7.3.5. A complete path connected space X supporting a Póincareinequality for all pairs (u, g) of a Lipschitz function and a continuous uppergradient g of u is quasiconvex.
Proof. Let γ : [0, 1]→ X be a path (not necessarily rectiable!) connecting anytwo points x and y. For each k ∈ N and any partition τ = 0 = t0 < . . . tn = 1of [0, 1] set
sτk(γ) =n∑i=1
min`(γ|[ti−1,ti]), kd(γ(ti), γ(ti−1)).
Further let `k(γ) = infτ sτk(γ). Notice that if γ happens to be rectiable thensτk ≤
∑ni=1 `(γ|[ti−1,ti]) = `(γ) so that `k ≤ `. Also `k(γ) ≤ kd(γ(0), γ(1)) =
kd(x, y). Finally when k ≥ 1, bymin`(γ|[ti−1,ti]), kd(γ(ti), γ(ti−1)) ≥ d(γ(ti), γ(ti−1)) and the triangle inequal-ity it follows that
sτk ≥n∑i=1
d(γ(ti), γ(ti−1)) ≥ d(x, y).
Fix an arbitrary point x0 ∈ X. Dene uk(x) = infγ `k(γ) where the inmumis taken over all paths joining x0 and x, for arbitrary x ∈ X. The observationsmade above yield d(x0, x) ≤ uk(x) ≤ kd(x0, x) for k ≥ 1. Let ε > 0 and x, y ∈ Xand assume, as can be done without loss of generality, that uk(x) > uk(y). Letγy be a path joining x0 and y such that `k(γy)− ε < uk(y). Let γx be any pathjoining x0 and x and γ a path from x to y. Then
uk(x)− uk(y) ≤ `k(γx)− `k(γy) + ε ≤ `k(γ) + ε.
Since ε is arbitrary and `k(γ) ≤ kd(x, y) the k-Lipschitz continuity of uk follows.From the same inequality it follows that g ≡ 1 is an upper gradient for uk, forany k. (Since
∫γ
1 = `(γ) ≥ `k(γ).) Employing corollary 7.1.2 it then followsthat
|uk(x)− uk(y)| ≤ C[Mg(x) +Mg(y)]d(x, y) = 2Cd(x, y)
almost everywhere and hence everywhere. Thus there is a uniform Lipschitzconstant for uk with respect to k. If, in particular y is chosen to be x0 theinequality
uk(x) ≤ Cd(x0, x) (7.3.4)
is obtained. Hence there also exists a constant C > 0 (perhaps dierent fromthat in equation (7.3.4)) depending only on the constants in the Poincaréinequality so that for every k there is a path γk for which
`k(γk) ≤ Cd(x0, x)
and, consequently a constant C > 0 so that for every k there corresponds apartition τk for which
sk := sτkk (γk) ≤ Cd(x0, x).
Consider the segments [tk,i−1, tk,i], i = 1, . . . , n(k) determined by the par-titions τk. Construct a new metric space as follows. For each index-pair k, i,if
`(γk|[tk,i−1,tk,i]) > kd(γk(tk,i−1), γk(tk,i)) =: kd(xk,i−1, xk,i) (7.3.5)
55
then add a line segment Ik,i connecting the points xk,i−1 and xk,i, equippedwith the metric l inherited from the real line multiplied by a factor ofd(xk,i−1, xk,i))tk,i − tk,i−1
. Thus the line segment has length d(xk,i−1, xk,i). Denote by X
the result of this process after going through every k and i. For two arbitrarypoints x, y ∈ X dene a distance by
d(x, y) =
d(x, y) if x, y ∈ Xl(x, y) if x and y are in the
same line-segmentminl(x, xk,i) + d(xk,i, y),l(x, xk,i−1) + d(xk,i−1, y) if x ∈ Ik,i and y ∈ X
minl(x, x∗) + d(x∗, y∗) + l(y∗, y) :x∗ = xk,i, xk,i−1; y∗ = xr,j , xr,j−1 if x ∈ Ik,i, y ∈ Ir,j ,
(k, i) 6= (r, j).
That this indeed denes a metric is an easy albeit somewhat lengthy calculation.(Of course, the case where y ∈ Ik,i and x ∈ X is dened likewise as in above.)In X replace the curves γk by γk where
γk|[tk,i−1,tk,i] = γk|[tk,i−1,tk,i]
if (7.3.5) doesn't hold. Otherwise γk|[tk,i−1,tk,i] consists of the line segmentIk,i from xk,i−1 to xk,i. Therefore
˜(γk) ≤ sk ≤ Cd(x0, x)
and consequently kl(Ik,i) ≤ ˜(γk) ≤ Cd(x0, x). Note that ˜ is used to denotethe lenght in the new metric d and l the usual euclidean length/metric.
The new space X is proper. To see this it is convenient to demonstraterst that it is complete. To this end let (xm) ⊂ X be a Cauchy sequence. Ifinnitely many of its element lie in either X or in nitely many of the segmentsIk,i (whence innitely many elements lie in one segment Ik,i) then it is clearthat the sequence is convergent (because X and the line segments are complete).Hence it may be assumed that there is a subsequence, relabeled (xm) so thatxm ∈ Ikm,im where the line-segments are pairwise disjoint. But then
dist(Ikm,im , Ikl,il) ≤ d(xm, xl).
Here
dist(Ir,j , Ik,i) :=mind(xr,j , xk,i), d(xr,j−1, xk,i), d(xr,j , xk,i−1), d(xr,j−1, xk,i−1).
This implies that, passing to a subsequence if necessary, either xkm,im or xkm,im−1
is a Cauchy sequence whence there is a limit x′ ∈ X.
lim supm→0
d(xm, x′) = lim supm→0
l(xm, xkm,∗) ≤ lim supm→∞
d(xm, xl)
for every l. This proves xm → x′ in d and hence the completeness of X.Now assume that X is not proper let B = B(x, r) be a closed ball of X
which is not compact. Then there is a sequence (xn) ⊂ B and a > 0 so that
56
d(xn, xm) ≥ a if n 6= m. Since X is proper all but a nite amount of the pointsin the sequence must lie in the line segments and further no line segment cancontain more than a nite number of these points. Hence there is a subsequence(xn) so that xn ∈ Ikn,in with k1 < k2 < . . .. From the expression of d it can beestimated that
0 < a ≤ d(xn, xm) ≤ l(Ikn,in) + l(Ikm,im) + d(xkn,in , xkm,im).
But since xkn,in ∈ X there is a convergent subsequence. Hence the rst twoterms in the inequality above must stay above a xed positive number. This,however, is not possible either since the number of the segments is innite andl(Ik,i) ≤ C/k. This nishes the argument that X is proper.
As a result of this trick the situation is now such that there is a sequenceof curves γk parametrized from 0 to 1 and obeying ˜(γk) ≤ Cd(x0, x). Thenext task is to substract a subsequence that converges, in a certain sense, toa curve γ having the property ˜(γ) ≤ Cd(x0, x). Begin by observing that ifthe curves are parametrized by arc-length and then these parametrizations arere-scaled back to [0, 1] then all the parametrizations are ˜(γk)-Lipschitz where˜(γk) ≤ Cd(x0, x) the uniform Lipschitz constant implying, in particular, theequicontinuity of the sequence. Next let A = q1, q2, . . . be a countable densesubset of [0, 1] and denote ak,n = γk(qn). For (ak,1) choose a subsequenceconverging to a1 ∈ X. From this subsequence take another, so that ak,2 → a2
and so on. Dene the function γ on the dense subset A by γ(qn) := an. Ifqn, qm ∈ A with n > m then using the uniform Lipschitz constant we get
d(γ(qn), γ(qm)) = limk→∞
d(γk(qn), γk(qm)) ≤ C|qn − qm|.
The limit is with respect to the subsequence for the nth step in the aboveconstruction. This estimate and the completeness of X allow for an arbitraryt ∈ [0, 1] the following denition: γ(t) := limn→∞ γ(qn) where (qn) is anysequence converging to t. This is of course well dened. The resulting functionγ is a Cd(x0, x)-Lipschitz curve. Thus for any partition τ = 0 = t0 < · · · <tn = 1
n∑i=1
d(γ(ti), γ(ti−1)) ≤ Cd(x0, x)n∑i=1
(ti − ti−1) = Cd(x0, x)
and hence ˜(γ) ≤ Cd(x0, x).It remains to show that γ actually lies in X (from which it follows that
˜(γ) = `(γ)). Suppose γ(t) ∈ Ik,i for some t ∈ [0, 1], k, i. By the continuity ofγ it must attain all the values in the segment. However each of the curves γrattains values only in segments Ir,i whose length converges to zero as r → ∞.Now if γ attains values in some Ik,i then for suciently large the curves γr mustalso attain values in Ik,i which was observed to be impossible. Thus γ cannotattain values in any segment of xed positive length and hence γ(t) ∈ X forevery t. The quasiconvexity of X follows.
The proof of this result is taken from [13]. Here the path connectedness ofX was part of the assumptions. In [14] the quasiconvexity is proven withoutthis assumption. In fact spaces supporting a Poincaré inequality are much more
57
than quasiconvex. In such spaces there are always ample curves joining twogiven points, excluding in particular any behaviour akin to the Cantor set insection three.
7.4 Coincidence of N1,p(X) and M1,p(X)
Throughout this subsection it will be assumed that, in addition to being com-plete and doubling, the metric measure spaces (X, d, µ) appearing below admita p-Poincaré inequality
−∫B
|u− uB |dµ ≤ CP r(−∫σB
gpdµ)1/p
(7.4.1)
for some p > 1. The aim of this subsection is to prove the isomorphy of thespaces N1,p(X) andM1,p(X). Note, however that the case p = 1 is not includedin the discussion as, indeed, even in the model case of X = Rn the coicidenceof these two spaces is not true. (In this case N1,1 coincides with the classicalSobolev space and M1,1 is a smaller space, namely the Hardy Sobolev spaceover Rn. For details see [23].)
It is easy to demonstrate a partial result akin to the aim of this subsection. Theessential work has been done in the two previous sections and what remains isto connect the dots. Recall Theorem 6.4.5 which states that M1,p(X) embedscontinuously into N1,p(X). If p is as in (7.3.1) then X satises the q-Poincaréinequality with the same constants for any q ≥ p (by Jensen's inequality). Thusby Theorem 6.4.5 one can deduce that M1,q(X) → N1,q(X) continuously forall such q. Next, for any q > p consider any function u ∈ Lq(X) and its uppergradient g ∈ Lq(X) (so that in fact u ∈ N1,q(X)). Recall, from Corollary 7.1.2that, since the pair (u, g) now satises the p-Poincaré inequality
−∫B
|u− uB |dµ ≤ Cr(−∫σB
gpdµ)1/p
for every ball, then almost everywhere the pointwise estimate
|u(x)− u(y)| ≤ Cd(x, y)[(M2σd(x,y)g
p(x))1/p +
(M2σd(x,y)g
p(y))1/p]
holds so (Mgp)1/p is a Hajªasz gradient for u. Since q > p > 1 it follows that
||(Mgp)1/p||q ≤ C||gp||q/p ≤ C||g||q
and hence
||u||M1,p(X) ≤ ||u||Lp(µ)+ infg∈G(u)
||Mg||p ≤ ||u||p+C infg∈G(u)
||g||p ≤ C||u||N1,p(X),
that is, N1,q(X) embeds continuously into M1,q(X). Note that this result onlyholds for p strictly greater than 1 as the maximal operator is not bounded fromL1 → L1.
Combining the above reasoning and 6.4.5 yields
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Theorem 7.4.1. With xed p > 1 in (7.4.1) the Hajªasz space M1,q(X) isisomorphic to N1,q(X) for every p < q <∞.
With the aid of the following notable result the end-point case of the abovetheorem can be addressed.
Theorem 7.4.2. If the complete doubling metric space (X, d, µ) admits a p-Poincaré inequality for some p > 1 then there exists some 0 < ε < p − 1 sothat (X, d, µ) admits a p−ε-Poincaré inequality (and consequently a q-Poincaréinequality for any q > p− ε).
The proof of this will be omitted; it can, however be found from [22]. Nowit easily follows that
Corollary 7.4.3. with xed p > 1 in (7.4.1) the Hajªasz space M1,p(X) isisomorphic to N1,p(X).
Proof. Apply 7.4.1 to p− ε.
8 A dierentiable structure for spaces supporting
a Poincaré inequality
In the beginning of the previous section it was stated that in spaces supportinga Poincaré inequality a notion of dierentiability akin to the Euclidean casearises. It is the purpose of this section to dene what exactly that notion isand then to prove that, indeed, in these spaces such a notion is possible and ispossessed of many nice properties.
The Poincaré inequality was already linked to the innitesimal behaviour ofLipschitz functions in particular to the operator lip through theorem 7.3.3.For the rest of this section the Poincaré property will be replaced by anotherone using Lip and lip. Throughout this section (X, d, µ) will always be assumedto be complete and doubling. The rest of this exposition is based on the paperof Stephen Keith, [21].
Theorem 8.0.4. Suppose (X, d, µ) supports a p-Poincaré inequality for somep ≥ 1. Then there exists a constant K > 0 depending only on the constant CPin the Poincaré inequality so that for each u ∈ LIP(X)
Lipu(x) ≤ K lipu(x) for almost every x ∈ X.
Proof. Fix u ∈ LIP(X). According to Proposition 7.2.6 there is a constantC > 0 such that
1C
Lipu(x) ≤ lim supr→0
1r−∫B(x,r)
|u− uB(x,r)|dµ
for almost every x ∈ X. Since lipu is an upper gradient of u (Proposition 7.3.2it follows that
1r−∫B(x,r)
|u− uB(x,r)|dµ ≤ CP
(−∫B(x,σr)
(lipu)pdµ
)1/p
59
for every x ∈ X and r > 0. Lebesgue's dierentiation theorem then asserts thatfor almost every x ∈ X
1C
Lipu(x) ≤ lim supr→0
1r−∫B(x,r)
|u− uB(x,r)|dµ
≤ CP lim supr→0
(−∫B(x,σr)
(lipu)pdµ
)1/p
= CP lipu(x).
Hence K = CPC.
This more elegant condition leads to a natural notion of generalized linearityof tangent functions on tangent spaces at a given point x ∈ X through whichthe dierential structure (to be shortly dened) will be developed.
Denition 8.0.5. Given a metric measure space (X, d, µ) and a vector spaceV ⊂ LIP (X), a denumerable collection (Xk, ϕk), k ∈ K of pairs consisting ofa measurable set Xk and a function
ϕk = (ϕ1k, . . . , ϕ
n(k)k ) : Xk → Rn(k)
where ϕik ∈ V , 1 ≤ i ≤ n(k) is said to be a (strong) measurable structure withrespect to V if
1. each Xk has positive measure and X =⋃k∈K
Xk.
2. There exists a natural number n (possibly zero) such that 0 ≤ n(k) ≤ nfor every k ∈ K.
3. For any function u ∈ V and every k ∈ K there exists a measurable functiondku : Xk → Rn(k) for which
limy→xy 6=x
|u(y)− u(x)− dku(x) · (ϕk(x)− ϕk(y))|d(y, x)
= 0 (8.0.2)
for µ-almost every x ∈ Xk.
Furthermore, the sets Xk can be taken so that the following holds: there existpositive numbers δk > 0 so that for each k ∈ K and almost every x ∈ Xk theinequality
Lip(λ · ϕk)(x) ≥ δk|λ|
holds for all λ ∈ Rn(k). Here it is understood that |λ| refers to the standardEuclidean norm of the vector λ.
As one might guess, not every metric measure space X and every V ⊂LIP(X) can be equipped with a dierential structure of this sort. One suchexample, when V = LIP(X), is given by the Cantor set appearing in section 3.3 The purpose of this subsection and a main purpose of the whole exposition isto demonstrate that metric measure spaces supporting a p-Poincaré inequality
3If the Cantor set C could be equipped with a dierential structure as dened above thenaccording to the main result of this section the Hajªasz space over C would be reexive. This,however was seen not to be the case.
60
do admit such a dierential structure with respect to the vector space of allLipschitz functions.
Before introducing some machinery used in sequel a few remarks on Deni-tion 8.0.5 are in order.
i) The dierential structure is called degenerate if n(k) = 0 for some k ∈ K.If it is not degenerate then it is called non-degenerate.
ii) If x is an isolated point then equation (8.0.2) does not pose any restrictions.
iii) The smallest integer n satisfying n(k) ≤ n for all k ∈ K is referred to asthe dimension of the dierentiable structure.
iv) The last condition is a technical one, added for later convenience. It doesnot appear in the denition in [21].
8.1 Tangent spaces
One way to get to the concepts in denition 8.0.5, pursued in this paper, isto analyze the existence and properties of tangent spaces (and functions) at agiven point. This seems natural enough given the similarity of these concepts tothe classical manifold structure. However, the tangent spaces can no longer begiven such a direct denition as in the classical case and therefore will have tobe aproached through the more general notion of convergence of metric spaces.This subsection then starts with a (rather long and boring) list of denitionsfollowed by a few results that will be helpful in the context at hand.
Denition 8.1.1. Let (Z, d) be a complete locally compact metric space andFn, F non-empty closed subsets of Z. Fn is said to converge to F , abbreviatedFn → F , if
limn→∞
supz∈Fn∩B(q,R)
d(z, F ) = 0 and
limn→∞
supz∈F∩B(q,R)
d(z, Fn) = 0
for every q ∈ Z and R > 0.
(If it so happens that some of the sets Fn ∩ B(q,R), F ∩ B(q,R) vanishthen the suprema are interpreted to equal zero.) Convergence by this denitionmeans a sort of local Hausdor convergence of the sets in question.
Denition 8.1.2. Let (X, d) and (Y, g) be complete locally compact metricspaces, Fn and F nonempty closed subsets of X, Gn and G nonempty closedsubsets of Y and fn : Fn → Gn, f : F → G functions. The sequence fn issaid to converge to f if Fn → F and Gn → G in the sense of denition 8.1.1and the following holds: whenever xn ∈ Fn is a sequence for which there existsx ∈ F so that xn → x the values of xn under fn converge to f(x), that isg(fn(xn), f(x))→ 0 as n→∞.
The next denition concerns two properties of a sequence of (not necessarilyconvergent) functions. Sequences satifying these are later seen to have niceproperties.
61
Denition 8.1.3. Let fn : Fn → Gn be a sequence of functions, Fn ⊂ (X, d),Gn ⊂ (Y, g) nonempty closed subsets of the complete locally compact metricspaces (X, d) and (Y, g). The sequence (fn) is said to be equicontinuous if forevery ε > 0 there exists some δ > 0 so that whenever x, y ∈ Fn, d(x, y) < δ theng(fn(x), fn(y)) < ε.
Furthermore (fn) is said to be uniformly bounded on bounded sets if
supn
supz,w∈Fn∩B(xn,R)
g(fn(z), fn(w)) <∞
for all xn ∈ Fn and R > 0.
For the next denition it is useful to recall that if (X, d) is any metric spaceand 0 < α ≤ 1, then the space (X, dα), where dα(x, y) = d(x, y)α, is again ametric space, called the snowaked (or α-snowaked) version of X.
Denition 8.1.4. A sequence (Xn, dn) of complete locally compact metric spa-ces is said to converge to a complete locally compact metric space (X, d) if thereexists some m ∈ N and some α ∈ (0, 1] so that there are C-bi-Lipschitz embed-dings ın : (Xn, d
αn)→ (Rm, | · |), ı : (X, dα)→ (Rm, | · |) for some C > 0 so that
ın(Xn)→ ı(X) as subsets of Rm. In addition the following property should hold.If xn, yn ∈ Xn and x, y ∈ X are such that |ın(xn)− ı(x)| → 0 and similarly fory then dn(xn, yn) → d(x, y). In other words dn → d in the sense of denition8.1.2 (with respect to the metric space (ı(X)× ı(X), d =
√d2
1 + d22)).
The motivation behind this denition is, in part, Assouad's embedding the-orem. This remarkable and beautiful result and its proof can be found forinstance in [17].
Theorem 8.1.5. (Assouad's embedding theorem.) Let (X, d) be a complete anddoubling space. Then for every 0 < α < 1 there corresponds a natural numberm(α) so that there is a bi-Lipschitz mapping from (X, dα) to Rm(α) (whichis equipped with the usual Euclidean metric), that is, (X, dα) is bi-Lipschitzembeddable into (Rm(α), | · |).
A pointed metric space is a triplet (X, d, x) where (X, d) is a metric space andx some designated point of X.
Denition 8.1.6. A sequence (Xn, dn, pn) of complete locally compact pointedmetric spaces is said to converge to a complete locally compact pointed metricspace (X, d, p) if (Xn, dn)→ (X, d) in the sense of 8.1.4 and the embeddings inquestion can be chosen so as to respect the basepoints, that is ın(pn) = q = ı(p).
Denition 8.1.7. Suppose (X, d, x) is a complete locally compact pointed metricspace and f a real valued function on X. The triplet (Z, ρ, z) equipped with afunction g : Z → R is a tangent space for (X, d, x) and g a tangent functionof f if there exists a sequence rn > 0 of real numbers converging to zero and acompact set K containing a neighbourhood of x so that (K, d/rn, x) convergesto (Z, ρ, z) in the sense of denition 8.1.6 and if, in addition,
fn ı−1n → g ı−1
in the sense of denition 8.1.2. Here fn(·) =f(·)− f(x)
rnand ın and ı are the
respective embeddings for (K, (d/rn)α, x) and (Z, ρα, z) provided by denition8.1.6.
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Instead of talking about pointed metric spaces (X, d, x) and a function f :X → R separately it is more convenient to talk about the quartet (X, d, x, f)which will from now on be referred to as a space-function in accordance with[21]. Also the tangent space (Z, ρ, z) is said to be subordinate to (rn). Theclass of all tangent functions for a given space-function (X, d, x, f) is denotedby T (X, d, x, f).
Lemma 8.1.8. Let (Xn, dn) be a sequence of complete locally compact metricspaces with limit (X, d). Let (Rm, |·|) be the mutual space, α ∈ (0, 1] the constantappearing in denition 8.1.6 and ın, ı the respective embeddings of (Xn, d
αn) and
(X, dα) into Rm. If (xn) is a sequence such that xn ∈ Xn and z ∈ Rm satises
|ınxn − z|n→∞→ 0 then there is some x ∈ X so that z = ı(x).
Proof. From the denition of convergence and the assumptions above one has
limn→∞
dist(ınxn, ıX) ≤ limn→∞
supy∈ınXn∩B(z,R)
dist(y, ıX) = 0
for any R > 0. Now since dist(·, ıX) is a continuous mapping it follows thatdist(z, ıX) = 0 which, by the closedness of ıX implies z ∈ ıX.
Lemma 8.1.9. Let Fn be a sequence of non-empty closed subsets of Rm andsuppose that there exists r > 0 so that
Fn ∩B(0, r) 6= ∅
for all suciently large n. Then there is a convergent subsequence of Fn.
For a proof see [4].
Proposition 8.1.10. Let Fn, F ,a and F′ be nonempty closed subsets of a proper
space (Z, d) such that Fn → F and Fn → F ′. Then F = F ′.
To prove this the following lemma will be used.
Lemma 8.1.11. Let Fn be a convergent sequence of nonempty closed subsetsof a proper space (Z, d). Then for every x in the limit of the sequence and everyR > 0 there is n0 so that Fn ∩B(x,R) 6= ∅ for every n ≥ n0.
4
Proof. Let F = limn→∞ Fn in the sense of denition 8.1.1 and let x ∈ F , R > 0be arbitrary. Since
limn→∞
supz∈F∩B(x,R/2)
d(z, Fn) = 0
and F ∩B(x,R/2) 6= ∅ there are sequences (zn) ⊂ F ∩B(x,R/2) and yn ⊂ Fnsuch that d(zn, yn)→ 0. On the other hand by the compactness of F∩B(x,R/2)there is a subsequence (zn) and zR ∈ F ∩ B(x,R/2) so that zn → zR. Butthen also yn → zR, in particular there exists n0 so that d(x, yn) ≤ d(x, zR) +d(zR, yn) ≤ R/2 + R/2 whenever n ≥ n0. Hence every n ≥ n0 yn ∈ Fn ∩B(x,R).
4In particular this implies the existence of a sequence xn ∈ Fn such that xn → x in Z.This will be used repeatedly in the sequel.
63
Proof of 8.1.10. Take any x ∈ F and R > 0. Since
limn→∞
supz∈Fn∩B(x,R)
d(z, F ′) = 0
and Fn ∩ B(x,R) 6= ∅ for all but a nitely many n there exist sequences (zn)and (yn) with zn ∈ Fn ∩B(x,R), yn ∈ F ′ such that
d(yn, zn) ≤ 2 supz∈Fn∩B(x,R)
d(z, F ′)→ 0.
Again by the compactness-argument there is xR ∈ B(x,R) and subsequencesso that zn → xR and, consequently, yn → xR. This implies xR ∈ F ′ since F ′is closed. By taking R = 1/k for each k ∈ N a sequence (xk) ⊂ F ′ is obtained,and since xk ∈ B(x, 1/k) the sequence converges to x which then belongs toF ′. Thus F ⊂ F ′ and by reversing the roles of F and F ′ in the argument theopposite inclusion is obtained.
Proposition 8.1.12. (Equivalence of limits of metric spaces.) Let (X, d) and(Y, g) both be limits of a sequence of proper metric spaces (Xn, dn). Then thereis a bijective isometric map φ : X → Y .
Proof. Let ı′n : (Xn, (dn)α) → Rm, ı′ : (Xn, (dn)α) → Rm be the C-Lipschitzembeddings for some α and m provided by the denition of the convergenceof metric spaces, so that ı′nXn → ı′X. Let ′n : (Xn, (dn)β) → Rk and′ : (Y, gβ) → Rk be similarly for possibly dierent k and β. Let τ1 be thebi-Lipschitz embedding of (Rm, | · |β) into some (Rl, | · |) provided by Assouad'sembedding Theorem and likewise, τ2 the bi-Lipschitz embedding of (Rk, | · |α)into some (Rs, | · |). It can be assumed that l = s by replacing both byd = maxs, l. The embedding ı′ can be considered to be between the spaces(X, dαβ)→ (Rm, | · |β) and ′ between (Y, gαβ)→ (Rk, | · |α). The obvious ana-logues hold also for ı′n and ′n. Combine these embeddings with those providedby Assouad's theorem to obtain
ın := τ1 ı′n : (Xn, (dn)αβ)→ Rd, ı := τ1 ı′ : (X, dαβ)→ Rd
n := τ1 ′n : (Xn, (dn)αβ)→ Rd, := τ1 ′ : (Y, gαβ)→ Rd.
(The reference to the metric used in Rd is omitted for the rest of the proof.)
To see that nXn → Y as subsets of Rd calculate for arbitrary q ∈ Rd, R > 0
supz∈Y ∩B(q,R)
dist(z, nXn) ≤ C supw∈′Y ∩B(q,R)
dist(w, ′nXn)α n→∞−→ 0,
supz∈nXn∩B(q,R)
dist(z, Y ) ≤ C supw∈′nXn∩B(q,R)
dist(w, ′Y )α n→∞−→ 0.
The constant C appearing in the estimates above is the mutual bi-Lipschitzconstant of all the embeddings. In a similar manner it can be seen that ınXn →ıX as subsets of Rd.
Let ϕn := n ı−1n : ınXn → nXn. This is a C2-bi-Lipschitz bijection for all n.
With the aid of the following Proposition, an easy consequence of (Proposition5.1.9, [21]), it will be shown that passing to a subsequence ϕn has a limit in thesense of Denition 8.1.2.
64
Proposition 8.1.13. Let fn : Fn → Gn be a sequence of functions, where Fn isa nonempty closed subset of a complete doubling space (X, d) for each n, and Gnsimilarly for another complete doubling space (Y, g). Suppose the sequence (fn)is equicontinuous and uniformly bounded on bounded sets. Then there exists asubsequence (fnj ) and a mapping f : F → G with ∅ 6= F ⊂ X and ∅ 6= G ⊂ Yclosed such that fnj → f in the sense of Denition 8.1.2.
To use Proposition 8.1.13 the equicontinuity and uniform boundedness (onbounded sets) of (ϕn) needs to be veried. Equicontinuity is clear since theϕn's are Lipschitz with a uniform constant. If xn ∈ ınXn and R > 0 then
supz,w∈ınXn∩B(xn,R)
|ϕn(x)− ϕn(y)| ≤ supz,w∈ınXn∩B(xn,R)
C|z − w| ≤ 2CR
so that (ϕn) is also uniformly bounded on bounded sets.
Having veried the assumptions of Proposition 8.1.13 for (ϕn) it can be con-cluded that there is a convergent subsequence with limit ϕ : F → G. Here F andG are nonempty closed subsets of Rd. Denition 8.1.2 requires that ınXn → Fand nXn → G. Since it is known that ınXn → ıX and nXn → Y Proposition8.1.10 implies F = ıX and G = Y . Now dene φ : X → Y by φ = −1 ϕ ı.Take x, y ∈ X and xn, yn ∈ Xn so that |ınxn− ıx| → 0 and similarly for the y's.From Denition 8.1.4 it follows that
dn(xn, yn)→ d(x, y).
The fact that ϕnk → ϕ for some subsequence (ϕnk) implies, in particular, that|ϕnk ınkxnk − ϕıx| → 0 as k → ∞. Notice that |ϕnk ınkxnk − ϕıx| = |nkxnk −φx|. By applying the analogous fact for the y's and Denition 8.1.2 it can beconcluded that
dnk(xnk , ynk)→ g(φx, φy).
These two facts together imply
g(φx, φy) = d(x, y)
for all x, y ∈ X.
It remains to show the bijectivity of φ. To this end it is sucient to demonstratethat ϕ is bijective. For this consider the function sequence fn := ϕ−1
n : nXn →ınXn. As in the case of (ϕn) it can be seen that (fn) satises the assumptionsof Proposition 8.1.13 and hence, after passing to a subsequence, has a limitf : Y → ıX. Now let nk be the subscript of the mutual subsequence, so thatboth f and ϕ arise as limits of fnk and ϕnk . For any x ∈ ıX and a sequencexnk ∈ Xnk for which x = limk→∞ xnk one has ϕ(x) = limk→∞ ϕnk(xnk). Thenby the convergence fnk → f
f ϕ(x) = f(ϕ(x)) = limk→∞
fnk(ϕnk(xnk)) = limk→∞
xnk = x.
In a similar manner it is seen that ϕ f = idY
65
It may be noted that the proofs of the claims above transfer directly to thecase of pointed metric spaces. This is due to the fact that in that notion ofconvergence all the embeddings are required to respect basepoints, a featurewhich persists in the above discussions.
Theorem 8.1.14. Let (X, d, p, u) be a space-function where (X, d) is a completeand doubling space and u is L-Lipschitz. For any sequence (rn) of positive realnumbers converging to zero there exists a tangent space function(X∞, d∞, x∞, u∞) ∈ T (X, d, p, u) subordinate to a subsequence of (rn) such that(X∞, d∞) is doubling with doubling constant depending only on that of (X, d)and u∞ is L-Lipschitz.
In the proof of this the embedding theorem of Assouad will be used.
Proof of 8.1.14. Let (X, d, p) and u satisfy the assumptions of 8.1.14. Fix somer > 0 and a sequence rn of positive reals converging to zero. To prove thetheorem one needs to consider the sequence (Xn, dn, pn, un) with
Xn = B(p, r), dn = d/rn, pn = p and un(·) =u(·)− u(p)
rn
and demonstrate that this has a convergent subsequence. To accomplish thisx some 0 < α < 1 and let ı : (X, dα) → Rm be a C-bi-Lipschitz embedding(assured by Assouad's embedding theorem). Normalize ı so that ı(p) = 0.Then set ın = r−1
n ı whence ın : (Xn, (dn)α) → Rm remains a C-bi-Lipschitzembedding with ın(pn) = 0. The sets Fn := ınXn satisfy the following property:since
|ıny| = |ınpn − ıny| ≤ Cdn(p, y)α ≤ Crα
for y ∈ B(p, rnr) ⊂ B(p, r) (for suciently large n) it follows that Fn ∩B(0, Crα) 6= ∅ for suciently large n. Hence according to lemma 8.1.9 there isa subsequence convergent to some closed F ⊂ Rm. Denote this subsequence bythe subscript k. In a similar fashion Fk × Fk → F × F as k → ∞. 5 In anycase 0 ∈ F but if, for instance, B(p, r) is a discrete set F might contain onlyone point.
Next consider the map fk(x, y) = dk(ı−1k x, ı−1
k y)α for (x, y) ∈ Fk × Fk. Itsatises
|fk(x, y)− fk(z, w)| ≤ C||x− y| − |z − w|| ≤ C ′|(x, y)− (z, w)| and (8.1.1)
sup(x,y)∈B∩Fk
|fk(x, y)| ≤ C diam(B ∩ Fk) ≤ C diam(B)
for each bounded set B ∈ Rm.
The aim is to pass somehow to the limit and consider limk→∞ fk which wouldthen dene a metric on F ×F one could hope to be a limit of dαk . To accomplishthis x a countable dense subset E of F×F . For each (xj , yj) choose a sequence(xjk, y
jk)k∈N ⊂ Fk × Fk converging to (xj , yj). (Note that both Fk and F are
subsets of Rm.) (fk(x1k, y
1k)) is a bounded sequence and thus has a convergent
subsetquence with limit b1. From this extract another subsequence (labeled
5It is a general fact that convergence is preserved under cartesian products.
66
for notational reasons by the same subscripts) so that (fk(x2k, y
2k)) converges to
some b2 and so on. From the innite array
f1(x11, y
11) f2(x1
2, y12) . . . → b1
f1(x21, y
21) f2(x2
2, y22) . . . → b2
f1(x31, y
31) f2(x3
2, y32) . . . → b3
.... . . · · ·
choose the diagonal subsequence. Then for each j the sequence fk(xjk, yjk) where
k runs over the subscripts determined by the diagonal converges to bj . Denef on F × F as follows. Given two points x, y ∈ F let (xj , yj) be a sequence inE converging to (x, y), and let xjkj ∈ Fk so that |xj − xjkj | < 1/j (similarly fory). Set
f(x, y) = limj→∞
fkj (xjkj, yjkj )
1/α.
This is well dened since if (zj , wj) is another sequence in E converging to (x, y)the estimate (8.1.1) above yields
|fkj (xjkj, yjkj )− fkj (z
jkj, wjkj )| ≤ C
′|(xjkj − zjkj, yjkj − w
jkj
)| j→∞−→ 0
Each f1/αk = dk(ı−1
k ·, ı−1k ·) determines a metric in Fk and the properties of a
metric persist under pointwise limit, hence f denes a metric on F . Moreoverthe uniform estimates
1/C|x− y| ≤ fk(x, y) = dk(ı−1k x, ı−1
k y)α ≤ C|x− y|
also persist in the limit so that
1/C|x− y| ≤ f(x, y)α ≤ C|x− y| for x, y ∈ F.
Now it is easy to show that dk → f in the sense of 8.1.2: suppose x, y ∈ F aregiven and xk, yk ∈ Xk are sequences so that |x− ıkxk| → 0 and |y − ıkyk| → 0as k →∞. For each k choose xjk ∈ E ∩ B(x, |x− ıkxk|) ∩ Fk and similarly fory. then
lim supk→∞
|dk(xk, yk)α − f(x, y)α| = lim supk→∞
|fk(ıkxk, ıkyk)− fk(xjk , yjk)|
≤ C ′ lim supk→∞
|(ıkxk − xjk , ıkyk − yjk)| = 0.
Hence the sequence (Xk, dk, p) → (F, f, 0) in the sense of 8.1.6. By a similarargument that was was used to construct f it can be shown that, passing toyet another subsequence, the sequence uk : (Xk, dk, p) → R converges to someu : (F, f, 0) → R. The essential estimate (8.1.1) follows for uk using the factthat it is Lipschitz. Also u remains Lipschitz with the same constant by thepersistence of inequalities under pointwise limits.
Theorem 8.1.14 is a very useful one because in the context of doubling spacesit asserts the existence of tangent function-spaces at any given point. Thebusiness of the next subsection is to examine the additional properties satisedby the tangent spaces under the hypothesis that the space supports a p-Póincareinequality for some p ≥ 1 and link these to the dierential structure 8.0.5 denedin the previous section.
67
8.2 Tangent functions
In this section (X, d, µ) will always assumed to be a complete doubling metricspace.
Denition 8.2.1. A Lipschitz function u : X → R is said to be K-quasilinearif for every x ∈ X and 0 < r ≤ diam(X)
LIP(u) ≤ Kvarx,r
u
where
varx,r
u = Lru(x) = supy∈B(x,r)
|u(x)− u(y)|r
.
It can immediately be seen that K, if it exists, must be at least one sincevarx,r
u ≤ Lip(u) holds for any u ∈ LIP(X). The motivation behind the name in
the denition is that if X is a normed space, for instance X = Rn, then anylinear map u : X → R is 1-quasilinear.
The following Proposition constitutes the rst of several upcoming nite - di-mensionality results.
Proposition 8.2.2. Suppose V ⊂ LIP(X) is a vector space consisting of K-quasilinear functions for a given K, and a point x0 ∈ X is xed. If V has theproperty that every u ∈ V satises u(x0) = 0, then V is nite dimensional withdimension bounded above by a constant depending only on K and the doublingconstant for µ.
Proof. By [30] a complete doubling space carries a doubling Borel regular mea-sure µ on it the doubling constant of which depends only on that of the space.Throughout the proof µ will stand for such a xed measure on X.
Suppose x0 ∈ X is as in the claim and 1 > t > 0. Construct a maximal sequencex1, . . . , xn ∈ B(x0, 1) := B having the property that
xn+1 ∈ B \n⋃i=1
B(xi, t) and B(xn+1, t/2) ∩B(xi, t/2) = ∅ for all i ≤ n
akin to the proof of Proposition 2.16. This is nite, and using Proposition 2.14n can be estimated from above by
n ≤ C4t−s,
C being the doubling constant of the measure and s = log2 C. The balls Bi =B(xi, t) cover B (for all this see proof of 2.16). Now for any i = 1, . . . , n supposeMi is the set of indices j for which the balls Bj and Bi intersect. For any j ∈Mi
it follows that d(xi, xj) ≤ 2t, hence B(xi, 3t) contains each ball Bj . Since theballs 1/2Bj , j ∈Mi are disjoint one has
1 ≥µ(⋃j∈Mi
1/2Bj)µ(3Bi)
=∑j∈Mi
µ(1/2Bj)µ(3Bi)
≥ |Mi| · 4−s(t/23t
)s.
68
Thus for each i the quantity |Mi| has an upper bound 24s which depends onlyon the doubling constant of the measure. Let n =: n(t) and M be the largestintegers smaller than C4t−s and 24s, respectively. Dene ft : V → Rn(t),
ft(u) = (µ(B1)uB1 , . . . , µ(Bn(t))uBn(t)).
Clearly this is a linear map V → Rn(t). To prove the claim it suces to showthat there exists some t (depending on K and the doubling constant) so thatft is injective, since from that it is clear that any V satisfying the conditions inthe claim can have dimension at most N = n(t) = n(K,C).
For any u ∈ V there is a point x ∈ B(x0, 1/2) so that
LIP(u) ≤ K |u(x)− u(x0)|1/2
= 2K|u(x)|. Consequently for any y ∈ B(x, 1/(3K))
the estimate
|u(y)| = |u(x)− (u(x)− u(y))| ≥ |u(x)| − |u(x)− u(y)|
≥ LIP(u)2K
− LIP(u)d(x, y) ≥ LIP(u)6K
holds. Therefore
−∫B(x,1/(3K))
|u|dµ ≥ LIP(u)6K
.
The inclusion B(x, 1/(3K)) ⊂ B(x0, 1/2+ 1/(3K)) ⊂ B and the doubling prop-erty of µ imply that there is a constant L = L(C,K) so that
LIP(u)6K
≤ −∫B(x,1/(3K))
|u|dµ ≤ L−∫B
|u|dµ. (8.2.1)
Now estimate∫B
|u|dµ ≤n∑i=1
∫Bi
|u|dµ ≤n∑i=1
∫Bi
|u− uBi |dµ+n∑i=1
µ(Bi)|uBi |, (8.2.2)
and for each term in the rst sum∫Bi
|u− uBi |dµ ≤∫Bi
−∫Bi
|u(x)− u(y)|dµ(y)dµ(x)
≤∫Bi
−∫Bi
LIP(u)d(x, y)dµ(y)dµ(x) ≤ 2tLIP(u)µ(Bi).
Hence
n∑i=1
∫Bi
|u− uBi |dµ ≤ 2tLIP(u)n∑i=1
µ(Bi) ≤ 2tLIP(u)Mµ(n⋃i=1
Bi)
≤ 2tM LIP(u)µ(2B) ≤ 2tMC LIP(u)µ(B) ≤ 12tMCKL
∫B
|u|dµ. (8.2.3)
The last inequality is an application of (8.2.1) and the second one is a conse-quence of the fact that each ball Bi intersects at most M of the balls in the
69
covering B1, . . . , Bn(t). Now choose t =1
24MCKL< 1 and insert (8.2.3) to
(8.2.2) which then becomes∫B
|u|dµ ≤ 1/2∫B
|u|dµ+ |ft(u)|1,
where | · |1 denotes the 1-norm in Rn(t). Note also that t and thus n dependsonly on K and the doubling constant of the measure (which in turn depends onthe doubling constant of the space).
The last inequality eectively proves the injectivity of ft: if ft(u) = 0 then∫B
|u|dµ ≤ 0
implying that u = 0 on B but then the K-quasiconvexity of u implies thatLIP(u) ≤ K var
x0,1/2u = 0 and hence u = 0.
Proposition 8.2.3. Let (X, d, µ) be a locally compact doubling metric measurespace and let u ∈ LIP(X). Suppose (rn) is a sequence of positive real numbersconverging to zero. Then for almost every x ∈ X every tangent spacefunction(X∞, d∞, x∞, u∞) ∈ T (X, d, x, u) which is subordinate to a subsequence of (rn)satises
u∞(x∞) = 0 and
lipu(x) ≤ vary,s
u∞ ≤ LIP(u∞) ≤ Lipu(x)
for every y ∈ X∞ and s > 0.
Proof. Throughout the proof the notation of Denition 8.1.7 will be used.Let rn be any sequence converging to zero. X can be assumed to have nite
measure since it can be split into countably many sets of nite measure. Thenfor every n there exists, by Egoro's and Luzin's theorems, a measurable setAn so that µ(X \An) < 2−n and the sequences of functions
Ln := supr<rn
supd(·,y)<r
|u(·)− u(y)|d(·, y)
and ln := infr<rn
supd(·,y)<r
|u(·)− u(y)|d(·, y)
converge to Lipu and lipu uniformly on An, respectively, and both Lipu and
lipu are continuous on An. Let A =∞⋃n=1
An whence µ(X \A) = 0.
The rst task is to prove that for almost every x ∈ An, for any point y from(X∞, d∞, x∞, u∞) ∈ T (X, d, x, u) which is subordinate to a subsequence of (rn)(relabeled (rk)) there is a sequence (yk) ⊂ K ∩ An so that |ıkyk − ıy| → 0.Here K ⊂ X is a compact set containing a neighbourhood of x for which(K, d/rk, x, uk) =: (Xk, dk, xk, uk) → (X∞, d∞, x∞, u∞) and again uk(·) =(u(·)− u(x))/rk.
This is in fact true of every density point y of K ∩ An: by lemma 8.1.11there is a sequence yk ∈ K for which |ıkyk − ıy| → 0. In particular d(yk, x)α ≤Crk|ıkyk − ıy| ≤ Crk. Suppose there doesn't exist a sequence zk ∈ An ∩ K
70
with the desired property. Then there is some t > 0 and a sequence km so thatB(ykm , trkm) ∩ An ∩K = ∅ for all m.6 Indeed, were this not the case then forevery t > 0 the inequality d(An ∩K, yk)α ≤ trk would hold for suciently largek, leading to
lim supk→∞
dist(ık(An ∩K), ıkyk) ≤ C lim supk→∞
dk(An ∩K, yk)α < Ct
for every t > 0. Hence there would exist a sequence ak ∈ An ∩ K satisfyingρ(ıkak, ıy)→ 0.
However employing the inclusion B(ykm , trkm) ⊂ B(x, (t+C)rkm) and Pro-position 2.14 the following contradictory estimate can be deduced
µ(B(x, (t+ C)rkm) \ (An ∩K))µ(B(x, (t+ C)rkm))
≥ µ(B(ykm , trkm))µ(B(x, (t+ C)rkm))
≥ 4−s(
t
t+ C
)s> 0.7
For the rest of the proof x one density point of A and An and a tangent spaceof it, subordinate to a henceforth xed subsequence rk of rn. To prove the lastinequality it suces to show that if y, z ∈ X∞, y 6= z are arbitrary then
|u∞(y)− u∞(z)|d∞(y, z)
≤ Lipu(x).
Let yk, zk ∈ An ∩ K be sequences satisfying |ıkyk − ıy| → 0 and likewise forzk and z. This is possible by lemma 8.1.11 and the above discussion. An easycomputation yields
|u(yk)− u(zk)|d(yk, zk)
=|uk(yk)− uk(zk)|
dk(yk, zk)→ |u∞(y)− u∞(z)|
d∞(y, z).
The convergence here is assured by the assumption that the functions dk anduk converge to d∞ and u∞, respectively. Hence, having xed y and z, it sucesto prove that
lim supk→∞
|u(yk)− u(zk)|d(yk, zk)
≤ Lipu(x).
Now d(zk, yk) = rkdk(zk, yk) ≤ Crk. Fix some ε > 0. Then there is k1 so that
supr<rk
supd(w,w′)<r
|u(w)− u(w′)|d(w,w′)
< ε+ Lipu(w) for every w ∈ An
whenever k ≥ k1. There also exists k2 so that d(yk, zk) < rk1 whenever k > k2.For all such k it follows that
|u(yk)− u(zk)|d(yk, zk)
≤ supr<rk1
supd(yk,w′)<r
|u(yk)− u(w′)|d(yk, w′)
< ε+ Lipu(yk)
yielding
lim supk→∞
|u(yk)− u(zk)|d(yk, zk)
≤ ε+ limk→∞
Lipu(yk) = ε+ Lipu(x).
Since this inequality holds for arbitrary ε the latter inequality is obtained.
For the rst inequality in the claim suppose it was known that given y ∈ X∞and s > 0 the following holds true.
6B(ykm , trkm ) ⊂ X.7s denotes the homogeneity exponent of µ, see Denition 2.15 and Proposition 2.14.
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Lemma 8.2.4. There is a sequence (yk) with yk ∈ K ∩An so that
• |ıkyk − ıy|k→∞−→ 0 and
• lipu(x) ≤ lim infk→∞
supd(z,yk)<srk
|u(z)− u(yk)|srk
.
Then for arbitrary ε > 0 and every n ∈ N let zn ∈ B(yn, srn) so that
|u(zn)− u(yn)|srn
+ ε > supd(z,yn)<srn
|u(z)− u(yn)|srn
.
Since dn(yn, zn) < s it follows that |ınzn − ınyn| ≤ Csα and consequently thatınzn ∈ B(ıy, R) ⊂ Rm for some R. By the local compactness of Rm there is asubsequence and z′ ∈ B(ıy, R) so that |ınzn − ız′| → 0. According to Lemma8.1.8 there exists some z ∈ X∞ such that z′ = ız. Furtherd∞(z, y) = limn→∞ dn(zn, yn) ≤ s. Now
|u(zn)− u(yn)|srn
=|un(zn)− un(yn)|
s
n→∞→ |u∞(z)− u∞(y)|s
≤ vary,s
u∞.
This inequality, Lemma 8.2.4 and the choice of zn together yield
lipu(x) ≤ lim infn→∞
supd(z,yn)<srn
|u(z)− u(yn)|srn
≤
ε+ limn→∞
|u(zn)− u(yn)|srn
≤ ε+ vary,s
u∞
which implies the desired inequality since ε is arbitrary. Thus the proof of thisProposition is reduced to that of Lemma 8.2.4
Proof of 8.2.4. Let y ∈ X∞, s > 0 be given. Let (yk) be a sequence with theproperties yk ∈ K ∩ An and |ıkyk − ıy| → 0. Then d(x, yk)α = rαk dk(x, yk)α ≤Crαk |ıkx− ıkyk| ≤ Crαk for large enough k.
Fix some ε > 0. Then there exists some k1 so that
infr<rk
supd(w,z)<r
|u(w)− u(z)|r
+ ε > lipu(w) for all w ∈ An
whenever k ≥ k1. Choose k2 so that srk < rk1 whenever k > k2. For such k
lipu(yk) < ε+ infr<rk
supd(yk,z)<r
|u(yk)− u(z)|r
≤ ε+ infk>k2
supd(yk,z)<srk
|u(yk)− u(z)|srk
whence
lipu(x) = limk→∞
lipu(yk) ≤ ε+ lim infk→∞
supd(yk,z)<srk
|u(yk)− u(z)|srk
for arbitrary ε > 0. This proves the claim.
The second nite dimensionality result already presents a common tangent spacefor all tangent functions of certain type.
72
Proposition 8.2.5. Let (X, d, µ) be a complete and doubling metric space andu = (u1, . . . , un) : X → Rn so that ui ∈ LIP(X) for every 1 ≤ i ≤ n. Furtherlet (rn) be any sequence of positive reals converging to zero. Suppose there is aconstant K > 0 so that if λ ∈ Rn then
Lip(λ · u)(x) ≤ K lip(λ · u)(x) for almost every x ∈ X.
Then for almost every x ∈ X there is a tangent space (X∞, d∞, x∞) of(X, d, x) subordinate to a subsequence of (rn) with the following properties: foreach λ ∈ Rn there is a tangent function uλ,x : (X∞, d∞, x∞) → R of λ · u forwhich
i) uλ,x(x∞) = 0
ii) uλ,x is K-quasilinear.
For x ∈ X for which this space exists the dimension of the vector space Vx :=uλ,x : λ ∈ Rn has an upper bound. That is, there exists a constant N ∈ Ndepending only on K and the doubling constant of µ so that dimVx ≤ N .
Proof. Let u, K and (rk) be as in the assumptions. Denote by eini=1 thestandard basis of Rn. For almost every x ∈ X select a subsequence (r0
x,k)kof (rk) so that the claim of proposition 8.2.3 is satised for e1 · u. Then, bytheorem 8.1.14 select a subsequence (r1
x,k)k of (r0x,k)k so that there exists a
doubling tangent space (Xx, dx, x∞) of (X, d, x) and a tangent function u1,x ofe1 · u subordinate to (r0
x,k)k. Note that proposition 8.2.3 and the assumptionsof this proposition guarantee for almost every x ∈ X
vary,s
u1,x ≥ lip(λ · u)(x) ≥ 1/K Lip(λ · u)(x) ≥ 1/K LIP(u1,x)
for every y ∈ Xx and s > 0. Thus u1,x is K-quasilinear. From (r1x,k)k choose
another subsequence (r1,0x,k)k so that the claims of proposition 8.2.3 hold for
the function e2 · u and from this yet another (labeled (r2x,k)k), assured by
theorem 8.1.14 so that there is a tangent space function (X ′x, d′x, x′∞, u2,x) ∈
T (X, d, x, e2 · u) subordinate to (r2x,k)k. Again u2,x is K-quasilinear. Now
the pointed metric spaces (Xx, dx, x∞) and (X ′x, d′x, x′∞) are both subordinate
to (r2x,k)k. Hence, as limits of the respective sequences they are isometrically
equivalent and the domain of u2,x can be taken to be (Xx, dx, x∞). u2,x then re-mains K-quasilinear. After n repeated applications of this procedure the resultis a subsequence (rnx,k)k, a tangent space (Xx, dx, x∞) of (X, d, x) and tangentfunctions ui,x : Xx → R of ei ·u (1 ≤ i ≤ n) all subordinate to (rnx,k)k. Furtherthe functions ui,x are all K-quasilinear.
Suppose some x ∈ X which admits the above discussion is xed. For any λ ∈
Rn the function λ ·u can be written as λ ·u =n∑i=1
λiei ·u. The sumn∑i=1
λiui,x =:
uλ,x, dened on (Xx, dx, x∞) is a tangent function for λ · u (subordinate to(rnx,k)k). This is seen easily from the denition of tangent space functions: Iff∞ and g∞ are tangent functions of f and g respectively, subordinate to thesame sequence and (X∞, d∞, x∞) a tangent space of (X, d, x) their commondomain of denition then by considering the limits fn + gn on (Xn, dn, xn) it is
73
seen that f∞ + g∞ = limn→∞(fn + gn) = limn→∞ fn + limn→∞ gn is a tangentfunction for f + g.
Since ui,x(x∞) = 0 for all i ∈ 1, . . . , n then by the denition of the tangentfunction the same holds for uλ,x, λ ∈ Rn. Also, since uλ,x is a tangent functionof λ · u and the condition Lip(λ · u)(x) ≤ K lip(λ · u)(x) is satised (for almostevery x) it follows together with Proposition 8.2.3 that
LIP(uλ,x) ≤ Lip(λ · u)(x) ≤ K lip(λ · u)(x) ≤ Kvar(y,s)
uλ,x,
for all y ∈ (Xx, dx, x∞), s > 0 that is, uλ,x is K-quasilinear.Finally that the dimension of the space V := uλ,x : λ ∈ Rn has an upper
bound follows directly from proposition 8.2.2 which, indeed, provides an upperbound for vector spaces such as the above depending only onK and the doublingconstant of X∞ which, in turn, depends only on the doubling constant of X (asasserted by Theorem 8.1.14).
The obtained upper bound does not á priori constrict the degree of freedomn as such, since the vector space Vx could have dimension strictly less thann. (It is however at most n.) The following proposition states that under onemore hypothesis the dimension of the vector space Vx is preserved. This will becrucial in the follow-up.
Proposition 8.2.6. If, in the situation of the previous theorem there is a mea-surable set A ⊂ X with positive measure and a constant δ > 0 so that for eachλ ∈ Rn
Lip(λ · u)(a) ≥ δ|λ| for almost every a ∈ A, (8.2.4)
then the dimension of Va =: uλ,a : λ ∈ Rn equals dimVa = n for almost everya ∈ A.
Proof. If λ ∈ Rn is xed then there is a set Aλ ⊂ A so that µ(A \Aλ) = 0 and(8.2.4) holds for every a ∈ Aλ. However if some a ∈ A is xed then it is not truethat for every λ ∈ Rn equation (8.2.4) should hold. It would be true is a weretaken from the intersection of all Aλ, λ ∈ Rn but the problem here is that theintersection is uncountable. To circumvent this problem the intersection will betaken over Λ, a countable dense subset of Rn which from now forth is xed.
Denote AΛ =⋂λ∈Λ
Aλ. Then if a ∈ AΛ is xed equation (8.2.4) holds for every
λ ∈ Λ. Let B be the intersection of AΛ with the set of points where the claimof proposition 8.2.3 is valid for every λ ∈ Λ and with the set of points where Vacan be formed. Since all these requirements are valid almost everywhere the setB still satises µ(A \ B) = 0. Now x a ∈ B. For an arbitrary vector λ ∈ Rnlet λk be a sequence in Λ converging to λ. Then
|Lip(λk·u)(a)−Lip(λ·u)(a)| ≤ Lip((λk−λ)·u)(a) ≤n∑i=1
|λik−λi|Lipui(a) k→∞−→ 0
which implies
Lip(λ · u)(a) = limk→∞
Lip(λk · u)(a) ≥ lim supk→∞
δ|λk| = δ|λ|.
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in other words for a ∈ B equation (8.2.4) holds for every λ ∈ Rn.Now dene L := λ 7→ uλ,a : Rn → Va. It will be shown that this is a bijective
linear map; indeed linearity and surjectivity is quite clear so it suces to showthe injectivity of L. For this suppose 0 = L(λ) = uλ,a. Choose a sequence(λk) ⊂ Λ converging to λ. From Proposition 8.2.3 one has
lim supk→∞
LIP(uλk,a) ≥ lim supk→∞
lip(λk · u)(a) ≥ lim supk→∞
Lip(λk · u)(a)/K
≥ lim supk→∞
δ|λk|/K ≥ δ|λ|/K. (8.2.5)
On the other hand, from the assumptions and by Proposition 8.2.3
lim supk→∞
LIP(uλk,a) ≤ lim supk→∞
Lip(λk · u)(a) ≤ K lim supk→∞
lip(λk · u)(a)
≤ K lim supk→∞
vary,s
uλk,a (8.2.6)
for every y ∈ X∞ and s > 0. Now from the proof of proposition 8.2.5 it can beseen that uλ,a = λ · (u1,a, . . . , un,a) where ui,a is the tangent function of ei · u.Therefore uλk,a → uλ,a uniformly on compact subsets of X∞. In particular itis is easy to check that
lim supk→∞
vary,s
uλk,a ≤ vary,s
uλ,a
since B∞(y, s) ⊂ X∞ is compact. Now uλ,a = 0 implies vary,s
uλ,a = 0 and
consequently |λ| = 0 by (8.2.5) and (8.2.6) whence the injectivity of L follows.Hence in particular the dimension of Va equals n.
Propositions 8.2.5 and 8.2.6 immediately imply the following nite-dimensionalityresult.
Corollary 8.2.7. Let (X, d, µ) be a complete and doubling metric space andu = (u1, . . . , un) : X → Rn so that ui ∈ LIP(X) for every 1 ≤ i ≤ n. Supposethere exists a constant K > 0 so that if λ ∈ Rn then
Lip(λ · u)(x) ≤ K lip(λ · u)(x) for almost every x ∈ X,
and there is a measurable set A ⊂ X with positive measure and a constant δ > 0so that for each λ ∈ Rn
Lip(λ · u)(a) ≥ δ|λ| for almost every a ∈ A.
Then n cannot exceed a constant N depending only on K and the doublingconstant of X.
8.3 The dierential structure
The main result of this subsection will be the following theorem
Theorem 8.3.1. Let (X, d, µ) be a locally compact metric space and µ a dou-bling measure. If there exists some K so that
Lipu(x) ≤ K lipu(x) µ-almost everywhere
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for all u ∈ LIP(X), then there is a natural number N (depending only on Kand the doubling constant of the measure) such that for any µ-measurable Awith µ(A) > 0 there exists a µ-measurable V ⊂ A with µ(V ) > 0, a naturalnumber 0 ≤ n ≤ N , a positive number δ > 0 and a function
f = (f1, . . . , fn) : X → Rn
such that fi ∈ LIP(X) for 1 ≤ i ≤ n and further f has the following property:for any u ∈ LIP(X) there is a unique (up to a set of measure zero) measurablefunction du : V → Rn which satises
limy→x
|u(y)− u(x)− du(x) · (f(x)− f(y))|d(y, x)
= 0 for every x ∈ V.
Further for almost all x ∈ V the inequality
Lip(λ · f)(x) ≥ δ|λ|
holds for every λ ∈ Rn.
Before starting with the proof of Theorem 8.3.1, however, it will be shownhow Theorem 8.0.5 follows from 8.3.1. To this purpose the following Lemmawill be employed
Lemma 8.3.2. Suppose (X,µ) be a σ-nite measure space and P is some prop-erty dened for the measurable sets of X (i.e. a measurable set A ⊂ X eitherdoes or does not have the property P ) which obeys the following: Every mea-surable set A ⊂ X with positive measure contains a subset V ⊂ A of positivemeasure such that V has the property P . Then there is a countable disjointdecomposition
X = Z ∪⋃i
Vi
of X so that µ(Z) = 0, µ(Vi) > 0 for all i ∈ N and each Vi has the property P .
Proof. By the σ-niteness of X there is a countable collection U1, U2, . . . ofmeasurable subsets of X such that X =
⋃k
Uk and µ(Uk) <∞ for every k ∈ N.
It can also be assumed that µ(Uk) > 0 for every k ∈ N. Further the sets can betaken mutually disjoint. This sequence can also be nite. Fix a k ∈ N.
Let Ω be the set of all sequences (Vi)i<α, α < ω18 of measurable subsets of
Uk, indexed by countable ordinals, that satisfy µ(Vi) > 0, Vi has the property
P for all i and Vj+1 ⊂ Uk \j⋃i=1
Vi. Introduce an ordering ≤ in Ω by dening
(Vi)i<α ≤ (Wi)i<β
if α ≤ β and Vi = Wi for all i = 1, . . . , α. Here α and β are allowed to rangeover the countable ordinals. For any chain of sequences A of Ω the sequencewhich contains all the members of all the sequences of A is an upper boundfor A: if it weren't, then we would obtain an uncountable sequence of mutually
8ω1 denotes the rst uncountable ordinal. For a study of ordinals see [6]
76
disjoint measurable subsets with positive measure. This is impossible. Thusan application of Zorn's lemma yields a maximal sequence (Vi)i<α in Ω. Thenµ(Uk \
⋃i
Vi) = 0 because otherwise there would exist a set V ′ ⊂ Uk \⋃i<α
Vi with
positive measure having the property P . This would contradict the maximalityof (Vi)i<α (since α+1 > α for any ordinal α). Thus a mutually disjoint sequenceof positive measured sets with the property P is obtained, such that they coverUk apart from a set of measure zero.
By doing this to every k two sequences (Vi)∞i=1 and (Zk) are obtained, therst all having positive measure, being mutually disjoint having the property Pand the second satisfying µ(Zk) = 0 for all k, such that
X = Z ∪⋃i
Vi
where Z =⋃k
Zk is a set of measure zero. This completes the proof.
Theorem 8.3.1 then states that if the property P is taken to be the existenceof n,N ∈ N, f and du as in 8.3.1 then for any metric space satisfying certainconditions the property P obeys the assumption made in 8.3.2.
It can therefore be deduced that if theorem 8.3.1 holds true and (X, d, µ)is a locally compact doubling metric measure space then there exists a naturalnumberN and a countable collection (Vi) of sets of positive measure and positivenumbers δi so that for each i there is a natural number ni, functions f i : X →Rni that is a countable collection (Vi, f i) of pairs where f i = (f1
i , . . . , fnii )
and f ji ∈ LIP(X) so that for each u ∈ LIP(X) there is a function diu : Vi → Rni(for each i ∈ N), for which
limy→xy 6=x
|u(y)− u(x)− diu(x) · (f i(x)− f i(y))|d(y, x)
= 0 and (8.3.1)
Lip(λ · f i)(x) ≥ δi|λ| for every λ ∈ Rni
almost everywhere. This sequence of sets almost covers X in other wordsthere is a set Z of measure zero so that
X = Z ∪⋃i
Vi.
Since (8.3.1) is required to hold almost everywhere the null set Z can be absorbedinto one of the sets Vi. Hence to prove the existence of the dierential structure8.0.5 it suces to prove 8.3.1. A few more denitions used in the proof will nowbe presented.
Denition 8.3.3. For any point x in a metric space (X, d) the seminorm | · |xin the set LIP(X) is dened by
|u|x = Lipu(x).
Of course one has to prove that | · |x is, indeed, a seminorm.
Lemma 8.3.4. For any x ∈ X, | · |x is a seminorm.
77
Proof. Let u, v ∈ LIP(X) and t ∈ R. Then
Lip(tu)(x) = lim supr→0
supd(x,y)<r
|tu(x)− tu(y)|r
= |t| lim supr→0
supd(x,y)<r
|u(x)− u(y)|r
and similarly
Lip(u+ v)(x) ≤ lim supr→0
supd(x,y)<r
(|u(x)− u(y)|
r+|v(x)− v(y)|
r
)which leads to Lip(u+ v)(x) ≤ Lipu(x) + Lip v(x). These two are the requiredproperties of a seminorm.
Denition 8.3.5. Given any nite ordered set u ⊂ LIP(X), thought of as amapping u: X → Rn where n = #u and a positive number δ dene the sets
1) S(u, δ) = x ∈ X : |λ · u|x ≥ δ|λ| for all λ ∈ Rn and
2) S(u) = x ∈ X : |λ · u|x 6= 0 for all λ ∈ Rn \ 0.
Further for any measurable set A (of positive measure) put
3) SA(u, δ) = S(u, δ) ∩A and
4) SA(u) = S(u) ∩A
To see that these sets are measurable it is helpful to notice that
Lemma 8.3.6. for a xed u and x ∈ X the mapping a(x, ·) = λ 7→ |λ · u|x iscontinuous.
Proof. If λ and λ′ are elements in Rn where n = #u, x ∈ X is xed and u isdenoted as u = (u1, . . . , un) then∣∣∣∣|λ · u|x − |λ′ · u|x∣∣∣∣ ≤ |(λ− λ′) · u|x ≤ n∑
i=1
|λi − λ′i||ei · ui|x ≤ C|λ− λ′|1
Lemma 8.3.7. Both S(u, δ) and S(u) are measurable and further
S(u) =∞⋃i=1
S(u, 1/i).
Proof. Start by showing the measurability of S(u, δ) for δ > 0. By Corollary7.2.3 the map aλ := a(·, λ) = x 7→ |λ · u|x − δ|λ| is measurable X → R andS(u, δ) can be written as
S(u, δ) =⋂λ∈Rn
a−1λ ([0,∞]).
By the continuity of the map λ 7→ |λ · u|x the union can be taken over Λ whichis a countable dense subset of Rn: obviously⋂
λ∈Rna−1λ ([0,∞]) ⊂
⋂λ∈Λ
a−1λ ([0,∞])
78
and to see the opposite inclusion suppose x ∈ a−1λ ([0,∞]) for every λ ∈ Λ and
let λ ∈ Rn be arbitrary. Then there is a sequence (λk) ⊂ Λ converging to λ,hence having xed x ∈ X lemma 8.3.6 implies the persistence in the limit
a(x, λ) = limk→∞
a(x, λk) ∈ [0,∞].
Therefore S(u, δ) is measurable.Now it is evident that
S(u) ⊃∞⋃i=1
S(u, 1/i).
Again for the opposite inclusion let x ∈ S(u), i.e. |λ · u|x > 0 for all λ ∈Rn \ 0. The set |λ| = 1 is compact, therefore the continuity of a(x, ·)implies δ := min|λ|=1 |λ · u|x > 0. Then for any λ ∈ Rn
|λ · u|x ≥ δ|λ|
and hence x ∈ S(u, δ) ⊂ S(u, 1/i) for suciently large i. This completes theproof of the lemma.
Proof of 8.3.1. Let (X, d, µ) be complete and doubling and let K be such that
Lipu(x) ≤ K lipu(x) for almost every x ∈ X
for any u ∈ LIP(X). The condition
limy→x
|u(y)− u(x)− λ(x) · (f(x)− f(y))|d(y, x)
= 0
is equivalent to
lim supr→0
supd(x,y)<r
|u(y)− u(x)− λ(x) · (f(x)− f(y))|d(y, x)
= 0
since this last limit is just the denition lim supy→x. It can therefore beexpressed in terms of the seminorm | · |x:
|u(·)− λ(x) · f(·)|x = 0. (8.3.2)
Throughout this proof the measurable set A with positive measure will be xed.For any V ⊂ A measurable with positive measure and any ordered set f for
which #f > 0 and µ(SV (f)) > 0 (supposing such sets exist) utilize lemma 8.3.7to nd some δ > 0 for which µ(SV (f , δ)) > 0. The function f satises
Lip(λ · f)(x) ≤ K lip(λ · f)(x) for almost every x ∈ X
for any λ ∈ Rn because for each λ ∈ Rn, λ · f ∈ LIP(X). Also the set W :=x ∈ V : |λ · f |x ≥ δ|λ| for all λ ∈ Rn = SV (f , δ) has positive measure, namely
µ(W ) = µ(SV (f , δ)) > 0.
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Corollary 8.2.7 then ensures an upper bound for n.Now dene
n(A) = max#f : µ(SA(f)) > 0.
The maximum is taken over all nite ordered sequences f ⊂ LIP(X) (thoughtof as functions f : X → R#f ) for which µ(SA(f)) > 0.
There are two possibilities. One is that n(A) = 0 and the other that n(A) >0. Consider the rst one.
Set f : X → 0 to be the constant function in which case the condition(8.3.2) reduces to |u|x = 0. In other words every u ∈ LIP(X) should satisfyLipu(x) = 0 for almost every x ∈ V , V being any positive measured measurablesubset of A. Choose V = A. Then indeed Lipu(x) = 0 for almost every x ∈ Asince if Lipu(x) > 0 in a set B of positive measure then g := u ⊂ LIP(X)would constitute an ordered set satisfying #g > 0 and µ(SA(g)) = µ(B) > 0thus contradicting the assumption that n(V ) = 0 for all V ⊂ A. It has beenestablished, then, that if n = 0 then the constant function f : X → R0 andV = A satisfy the claims of proposition 8.3.1
It can therefore be supposed throughout the rest of this proof that n(A) > 0.By the discussion above n(A) is bounded above by a constant depending only onK and the doubling constant of the function. This upper bound for n ensuredby Corollary 8.2.7 is in fact the required upper bound in the claim of 8.3.1.
Now x some ordered set f ⊂ LIP(X) so that #f = n(A) =: n and let δ > 0and W be as above. It remains to prove that with this choice every u ∈ LIP(X)admits a dierential du : W → Rn, i.e. a measurable function that satises
|u(·)− du(x) · f(·)|x = 0 for almost every x ∈W.
To accomplish this consider the set
E = x ∈W : |u− λ · f |x 6= 0 for all λ ∈ Rn.
The measurability of this can be seen as in the proof of Lemma 8.3.7. Denef ′ = (f1, . . . , fn, u) : X → Rn+1. If x ∈ E and λ′ = (λ, λn+1) ∈ Rn+1 \ 0 then
|λ′ · f ′|x = |λn+1u+ λ · f |x =δ|λ| , λn+1 = 0|λn+1||λ/λn+1 · f − u|x , otherwise,
in particular x ∈ SW (f ′) ⊂ SA(f ′). But #f ′ = n + 1 > n(A) and thereforeµ(SV (f ′)) = 0.
It follows that the set E has measure zero and therefore that for almostevery x ∈W there is some λ ∈ Rn for which |λ · f − u|x = 0. Suppose λ and λ′
are two such vectors. Then
0 = |λ · f − u|x + |u− λ′ · f |x ≥ |(λ− λ′) · f |x ≥ δ|λ− λ′|
because the last inequality holds for all x ∈ W . Consequently for almost everyx ∈W there exists a unique λ = λ(x) ∈ Rn so that
|λ · f − u|x = 0. (8.3.3)
This denes a unique mapping du : W → Rn i.e. du(x) = λ(x) up to a setof measure zero which has the desired property.
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The proof will be complete as soon as du : W → Rn is shown to be measurable.The values of du in E do not have an inuence on the measurability of du. Letus dene du = 0 on E.
Let U ⊂ Rn be compact and consider the set
F = x ∈W : there exists some λ ∈ U so that |λ · f − u|x = 0.
This is almost the pre-image of U in du, in the sense that
du−1U =F if 0 /∈ UF ∪ E if 0 ∈ U.
Therefore it suces to show the measurability of F . Firstly
F =∞⋂n=1
Fn, Fn = x ∈W : |λ · f − u|x < 1/n for some λ ∈ U.
The inclusion ⊂ is clear. If, on the other hand, x ∈∞⋂n=1
Fn then there is
a sequence (λn) ⊂ U which, by the compactness of U can be assumed to beconvergent, so that |λn · f − u|x < 1/n for all n. Denoting by λ the limit of λn,the continuity of the mapping λ 7→ |λ · f − u|x implies |λ · f − u|x = 0.
Furthermore if Λ is some countable dense subset of U and if some x andλ′ ∈ U satisfy the relation |λ′ · f − u|x < 1/n then there exists some λ ∈ Λ forwhich |λ · f − u|x < 1/n holds. Hence Fn can be written as
Fn =⋃λ∈Λ
Fλn
where
Fλn = x ∈W : |λ · f − u|x < 1/n = [Lip(λ · f − u)]−1[0, 1/n).
These sets are cleary measurable (even Borel, since Lip v of a Lipschitz map vis Borel). Consequently
F =∞⋂n=1
⋃λ∈Λ
Fλn
is measurable (Borel).
In the preceding discussion it was established that each measurable setA ⊂ X with positive measure contains some measurable set W with positivemeasure for which the conclusions of 8.3.1 hold. In the proof the required setwent by the name W instead of V . Nevertheless after assuring the truthfulnessof 8.3.1 lemma 8.3.2 provides the actual dierential structure with respect toLIP(X). Concerning the non-degeneracy of the dierential structure we havethe following result.
Proposition 8.3.8. Let (X, d, µ) be k-quasiconvex for some some k > 0 andlet (Xλ, ϕλ)λ∈Λ be a dierential structure in X. Then the dierential structureis non-degenerate.
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Proof. Given a point y0 ∈ X dene
u(x) = infγ`(γ), x ∈ X.
The inmum is taken over all rectiable curves joining x to y0. This function isk-Lipschitz. To see this take x, y ∈ X and suppose u(x) ≥ u(y). For any ε > 0take γy to be a curve joining y0 and y so that u(y) + ε ≥ `(γy) and let γyx be acurve joining y and x so that `(γyx) ≤ kd(x, y). Then the composition γyγyx isa rectiable curve from y0 to x. Calculate
0 ≤ u(x)−u(y) ≤ `(γyγyx)−`(γy)+ε ≤ `(γy)+`(γyx)−`(γy)+ε ≤ kd(x, y)+ε.
To see that lipu(x) ≥ 1 consider the length metric on X dened by l(x, y) =infγ `(γ) where the inmum is taken over all rectiable curves joining x and y.Let x ∈ X and 0 < r < minl(y0, x), 1 be arbitrary. Let γ : [0, L] → X be arectiable curve joining y0 and x such that
`(γ) ≥ u(x) ≥ `(γ)− r2.
There exists yr = γ(t) so that `(γ|[0,t]) = `(γ)− r, whence
u(x) ≥ `(γ)− r2 = `(γ|[0,t]) + r − r2 ≥ u(yr) + r(1− r).
To estimate the distance of yr from x observe that r ≥ l(x, yr) ≥ d(x, yr). Thus
lim infr→0
supz∈B(x,r)
|u(x)− u(z)|r
≥ lim infr→0
|u(x)− u(yr)|r
≥ 1
These two inequalities together imply
1 ≤ lipu(x) ≤ Lipu(x) ≤ k
for all x ∈ X.
Next suppose that n(λ) = 0 for some coordinate patch Xλ. For such an indexλ ∈ Λ choose yλ ∈ Xλ and set uλ as above with y0 = yλ. Since the coordinatefunction ϕλ has range 0 condition (8.0.2) takes the form
|u|x = 0 for almost every x ∈ Xλ
for each u ∈ LIP(X) (similarly as in the proof of 8.3.1). This, however is clearlyuntrue in the case of uλ whose seminorm at any point x ∈ Xλ stays safely above1. Therefore the existence of a function such as the one constructed aboveguarentees that the dierential structure cannot be degenerate.
In connection with Theorem 7.3.3 Proposition 8.3.8 implies specically thatif (X, d, µ) admits a p-Poincaré inequality for some p ≥ 1 then this dierentialstructure is non-degenerate.
Corollary 8.3.9. Suppose (X, d, µ) is path connected and admits a p-Poincaréinequality for some p ≥ 1 and let (Xλ, ϕλ)λ∈Λ be a dierential structure in X.Then the dierential structure is non-degenerate.
As stated, this together with 8.0.4 and 8.3.1 immediately yields
Theorem 8.3.10. If (X, d, µ) is path connected and supports a p-Póincare in-equality for some p ≥ 1 then it admits a non-degenerate dierentiable structureas dened in 8.0.5.
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8.4 Reexivity of N1,p(X)
In this subsection the space of p-integrable vector elds over X, denoted Lp(X),will be used. In the classical case (where X = Rn) the space Lp(X) would cor-respond to Lp(X; Rn) the Rn-valued p-integrable (and measurable) mappingsu : X → Rn. With X a metric measure space supporting some Poincaré in-equality the situation is more complicated; Here Lp(X) consists of p-integrablevector elds over X. The precise denitions will be given shortly.
The aim of this subsection is to demonstrate that the Newtonian spaces (andconsequently the Hajªasz spaces) with exponent p > 1, dened over a metricmeasure space supporting a p-Poincaré condition are reexive. The strategyfor proving this is to construct a linear isomorphism between N1,p(X) and aclosed subspace of Lp(µ)×Lp(X) or, rather, a linear map between N1,p(X) andLp(µ) × Lp(X) that is bounded both above and below (in a similar fashion aswas done in section 3). The reexivity (or even uniform convexity) of N1,p(X)will then follow from that of Lp(µ)× Lp(X).
To dene vector elds over X one needs to talk about tangent spaces and -bundles. Let (Xk, ϕk)k∈K be a disjoint measurable structure as given by Lemma8.3.2 (the lemma states that the decomposition can be taken disjoint). Denethe disjoint union
TX :=⋃k∈K
(Xk ×Rn(k))
andTxX := π−1(x),
where π : TX → X is the natural projection. These are the tangent bundle andtangent space at the point x of X with respect to the dierentiable structure(Xk, ϕk)k∈K , given in analogue with the case where X would be some manifold.Since the union of Xk's covers only almost all of X the tangent space TxXexists for only almost every x ∈ X. Suppose k ∈ K and consider the functionϕk(x) = (ϕ1
k(x), . . . , ϕn(k)k (x)). Since
ϕik(x)− ϕik(y) = ei · (ϕk(x)− ϕk(y))
for every y ∈ Xk it follows that dkϕik(x) = ei for almost every x ∈ Xk, for each
1 ≤ i ≤ n(k). For such x then take
dkϕik(x)n(k)
i=1as the basis for TxX (which,
recall, is isomorphic with Rn(k)) and dene a norm in TxX by
||λ||x =
∣∣∣∣∣∣n(k)∑i=1
λiϕik
∣∣∣∣∣∣x
= |λ · ϕk|x
That this is a norm follows from the fact that |λ · ϕk|x > 0 for λ 6= 0 almosteverywhere in Xk.
With these notations a vector eld ω on X is a map ω : X → TX such thatω(x) ∈ TxX almost everywhere.
In particular at almost every point x ∈ Xk then the dierential dku of aLipschitz function u has a value in TxX;
dku(x) =n(k)∑i=1
diku(x)dkϕik(x) ∈ TxX
83
and thus dk is a mapping from Xk to TX.A measurable vector eld ω is a vector eld whose components in each patch
Xk are measurable in the usual sense. That is, if ω is restricted to any Xk
and written as ω(x) =n(k)∑i=1
ωi(x)dkϕik(x) then each ωi is a measurable mapping
X → R.Dene L(X) to be the set of measurable vector elds ω on X. For these
the map x 7→ ||ω(x)||x is measurable. Lp(X) then consists of those elementsω ∈ L(X) for which the mapping x 7→ ||ω(x)||x is p-integrable. For this integralthe following shorthand notation will be used(∫
X
‖ω(x)‖pxdµ(x))1/p
=: ‖ω‖Lp .
This denes, in the usual way, a norm in Lp(X) which makes it a Banachspace.
Lemma 8.4.1. Let p ∈ (1,∞). Then the space Lp(X) is a Banach space.Furthermore it is reexive.
Proof. Start by writing the norm of an element ω ∈ Lp(X) as
‖ω‖pLp(X) =∑k∈K
∫Xk
‖ω(x)‖pxdµ(x) =∑k∈K
‖ω|Xk‖Lp(Xk). (8.4.1)
Conversely every sequence (ωk)k∈K for which ωk ∈ Lp(Xk) and the right-handside of (8.4.1) is nite determines an element ω ∈ Lp(X) in the obvious way,such that (8.4.1) holds. Hence
Lp(X) =⊕`p(K)
Lp(Xk)
with equal norms. Now both claims will follow from Theorem 2.2 if the spacesYk := Lp(Xk) can be shown to be Banach and reexive.
To this purpose let
Lk = maxLIP(ϕik) : 1 ≤ i ≤ n(k)
and let δk be as in Denition 8.0.5, for k ∈ K xed. Regarding ω = (ω1, . . . , ωn(k))as a vector in Rn(k) the estimate
δk|ω(x)| ≤ ‖ω(x)‖x ≤n(k)∑i=1
|ωi(x)|‖ϕik(x)‖x ≤ n(k)Lk|ω(x)|
holds for almost every x ∈ Xk. Here | · | denotes the usual Euclidean norm inRn(k). As a consequence of this the norm ‖ · ‖Lp(Xk) is equivalent to‖ · ‖Lp(Xk;Rn(k)). In particular
Lp(Xk) ≈ Lp(Xk; Rn(k))
as Banach spaces. The reexivity of each Yk follows.
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The next proposition gives some more information of the dierential, thenon-trivial part of the isomorphism that is the purpose of this discussion. Therst natural domain of denition for this is the space LIP(X). The next lemmaglues together the pieces given above to establish this.
Proposition 8.4.2. Let (Xk, ϕk)k∈K be a dierential structure over (X, d, µ)as above. Then there is an operator d : LIP(X)→ L(X) satisfying the followingconditions.
a) For each u ∈ LIP(X) and k ∈ K one has |u− du(x) · ϕk|x = 0 for almostevery x ∈ Xk.
b) d(u + v) = du + dv and d(au) = adu almost everywhere for any u, v ∈LIP(X) and a ∈ R.
c) For each u ∈ LIP(X) the identity ‖d(x)‖x = Lipu(x) holds almost every-where.
d) If u ∈ LIP(X) and f : R→ R is a dierentiable mapping, then d(f u) =f ′(u)du almost everywhere.
Proof. Let u ∈ LIP(X). Dene du : X → TX almost everywhere by
du(x) := dku(x)
where k ∈ K is the unique index for which x ∈ Xk. This indeed gives a(n almosteverywhere dened) function du : X → TX. Hence d can be thought of as anoperator from LIP(X) to the set of (measurable) vector elds over X.
Condition a) is automatic because for any u ∈ LIP(X) almost every x ∈ Xbelongs to Xk for a unique k ∈ K du(x) = dku(x) and thus (8.0.2) holds almosteverywhere in Xk.
To prove b) take u, v ∈ LIP(X) and a ∈ R and suppose x is such thatcondition a) holds for both u and v. Let k be such that x ∈ Xk. In the proofof 8.3.1 it was seen that once the function ϕk is xed this condition determinesdku(x) uniquely. Bearing this in mind the computation
|u+ v − (du(x) + dv(x)) · ϕk|x = |u+ v − (dku(x) + dkv(x)) · ϕk|x≤ |u− dku(x) · ϕk|x + |v − dkv(x) · ϕk|x = 0
implies that d(u+ v)(x) = du(x) + dv(x). Similarly
|au− adku(x) · ϕk|x = |a||u− dku(x) · ϕk|x = 0,
hence d(au) = adu.For almost every x ∈ X one has
‖du(x)‖x =
∣∣∣∣∣∣n(k)∑i=1
diu(x)ϕik
∣∣∣∣∣∣x
= |du(x) · ϕk|x
On the other hand condition a) implies |u|x = |du(x) · ϕk|x almost everywheresince on every point where a) holds it leads to
||u|x − |du(x) · ϕk|x| ≤ |u− du(x) · ϕk|x = 0
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For d) let u ∈ LIP(X) and let x be a point where a) holds for u. Then f u(y) =f u(x) + f ′ u(x)[u(y)− u(x)] + o(d(x, y)), yielding
f u(y)− f ′ u(x)du(x) · ϕk(y) =f u(x)− u(x)f ′ u(x) + o(d(x, y)) + f ′ u(x)[u(y)− du(x) · ϕk(y)].
From this expression it is clearly seen that
|f u− f ′ u(x)du(x) · ϕk|x = f ′ u(x)|u− du(x) · ϕk|x = 0
Hence one can consider an exterior derivative d on LIP(X). The ultimate goalin the ongoing discussion is to extend d to an operator between Banach spaces.For this reason it is convenient to present a new equivalent norm on N1,p(X).This is rst done in a subset of N1,p(X). The whole space is then obtained bycompleting this subset. Consequently another characterization of N1,p(X) isattained and this will be used to construct the isomorphism discussed earlier.
Lemma 8.4.3. Consider the set
N1,pL (X) = u ∈ LIP(X) ∩ Lp(µ) : Lipu ∈ Lp(µ) =u ∈ LIP(X) ∩ Lp(µ) : du ∈ Lp(X).
On it dene a norm
‖u‖1,p := ‖u‖p + ‖Lipu‖p = ‖u‖Lp + ‖du‖Lp .
Then
1) ‖·‖1,p and ‖·‖N1,p(X) are equivalent on N1,pL (X), that is for some constant
C > 01/C‖u‖1,p ≤ ‖u‖N1,p(X) ≤ C‖u‖1,p
holds for every u ∈ N1,pL (X), and further
2) N1,p(X) is obtained as the completion with respect to the norm ‖ · ‖1,p.
In particular the norms ‖·‖N1,p(X) and ‖·‖1,p are equivalent and u ∈ N1,p(X) ifand only if there is a sequence un ∈ N1,p
L (X) and ω ∈ Lp(X) so that ||u−un||p →0 and ||dun − ω||Lp → 0 as n→∞.
Proof. By Proposition 7.2.6 for any v ∈ N1,pL (X) and its upper gradient g the
inequality
1/C Lip v(x) ≤ lim supr→0
1r−∫B(x,r)
|v − vB(x,r)| ≤ g(x)
holds almost everywhere. Hence in particular ‖v‖N1,p(X) ≥ ‖v‖p+C−1‖Lip v‖p.On the other hand the reverse inequality ‖v‖N1,p(X) ≤ ‖v‖p+‖Lip v‖p must holdsince Lip v itself is an upper gradient of v. Consequently the norms ‖ · ‖N1,p(X)
and ‖ · ‖1,p are equivalent on N1,pL (X). (In particular N1,p
L (X) is a subset ofN1,p(X).)
86
Denote by H the completion of N1,pL (X) with respect to ‖ · ‖1,p. For any
u ∈ H there exists a sequence un ∈ N1,pL (X) so that ‖u−un‖1,p → 0 as n→∞.
The sequence un is a Cauchy sequence in ‖ · ‖1,p and by the previous discussionalso for ‖ · ‖N1,p(X). Therefore un has a limit u ∈ N1,p(X). By passing toa subsequence it can be assumed that un → u and un → u pointwise almosteverywhere. Therefore u = u almost everywhere. This shows that H = N1,p(X)as sets. Further
‖u‖1,p = limn→∞
‖un‖1,p ≤ C limn→∞
‖un‖N1,p(X) = C‖u‖N1,p(X)
for a sequence (un) ⊂ N1,pL (X) converging to u. Note that convergence in
‖ · ‖N1,p(X) implies convergence in ‖ · ‖1,p and vice versa. A similar argumentcan be applied to the other inequality and consequently the norms are seen tobe equivalent.
The rest of the claim is a straightforward consequence of 1) and 2) and thefact that Lp(X) is complete.
By denition of the space N1,pL (X) the dierential d can be considered a
linear map from N1,pL (X) to Lp(X). Self-evidently it is also bounded so that it
can be extended to a bounded linear operator d : N1,p(X)→ Lp(X).
The representation of the dierential d as above is the main step in a construc-tion of a bi-Lipschitz linear map between N1,p(X) and Lp(µ) × Lp(X). Notethat the space Lp(µ) × Lp(X) can naturally be thought of as a normed vectorspace endowed with the norm
‖(u, ω)‖ := ‖(u, ω)‖Lp(µ)×Lp(X) := ‖u‖p + ‖ω‖Lp .
As a nite cartesian product of two Banach spaces Lp(µ)×Lp(X) is also Banach.The reexivity of Lp(µ)× Lp(X) follows from that of Lp(µ) and Lp(X).
Denition 8.4.4. Dene the linear map L : N1,p(X) → Lp(µ) × Lp(X) byL(u) = (u,du).
Denition 8.4.4 now implies the desired result by a very elementary argument
Corollary 8.4.5. The operator L is an isometry when N1,p(X) is endowed with‖ · ‖1,p. Consequently the space N1,p(X) is reexive.
Proof.‖Lu‖Lp(µ)×Lp(X) = ‖u‖p + ‖du‖p = ‖u‖1,p.
Hence N1,p(X) is isomorphic with a closed subspace of Lp(µ) × Lp(X) whichimplies the reexivity of N1,p(X).
Bearing in mind that the spaces N1,p(X) and M1,p(X) are isomorphic whenp > 1 the following main result is obtained.
Theorem 8.4.6. If (X, d, µ) is a complete doubling space and supports a p-Poincaré inequality for some p > 1 then the spaces N1,p(X) and M1,p(X) de-ned over X are isomorphic (to each other) and reexive.
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8.5 Remarks
The construction of the dierentiable structure could, with not much additionalwork, have been carried out for arbitrary families of Lipschitz functions underthe same hypotheses. Also the construction of the manifold structure of X atalmost every point does not require the disjointness of the charts Xk but this israther a convenient assumption which allows for a less messy treatment of thesubject (and is sucient for the purposes of proving the ultimate result). Thework done here was aimed at proving the reexivity of the spaces in question.In fact the space Lp(X) (and consequently N1,p(X)) can be equipped with anequivalent norm that is uniformly convex. To accomplish this some more workis needed.
Theorem 8.5.1. The space Lp(X) is uniformly convex.
To prove this the following lemma, known as John's ellipsoid theorem, willbe used. The claim is of elementary nature but it takes some doing to prove it.A proof can be found for example in [18], see also [11] and [9].
Lemma 8.5.2. Let K ⊂ Rm be a symmetric, compact convex set with nonemp-ty interior (that is, a symmetric body). Then there exists an ellipsoid D ⊂ K,symmetric about the origin so that K ⊂
√mD.
Proof of lemma 8.5.1. The objective is to construct, for almost every x ∈ Xequivalent norms given by an inner product in TxX so that the equivalenceconstants depend only on the dimension of TxX. These will lead to an equivalentnorm on Lp(X) and a version of the Hölder inequality that will imply thereexivity.
Fix some x ∈ X for which the normed space (TxX, ‖ · ‖x) can be constructed.Letm be the dimension of TxX that is, m = n(k) where k is such that x ∈ Xk.Every ellipsoid in Rm that is symmetric about the origin gives rise to an innerproduct in Rm. In particular if K is the unit ball with respect to the norm‖ · ‖x then K is a symmetric body and the ellipsoid D given by John's theoremdetermines an inner product, denoted 〈·, ·〉x. The inclusions D ⊂ K ⊂
√mD
transform to[a]x ≤ ‖a‖x ≤
√m[a]x for all a ∈ TxX
in the context of norms. Here [·]x is the norm determined by the inner prod-uct. Since m = n(k) has a uniform (in k) upperbound there exists a con-stant depending only on the dimension of the dierential structure such that1/C‖a‖x ≤ [a]x ≤ C‖a‖x for almost every x ∈ X and every a ∈ TxX. Inparticular ‖ · ‖Lp is equivalent with
‖ · ‖ := ω 7→(∫
X
[ω(x)]pxdµ(x))1/p
,
provided that the mapping x 7→ [ω]x is measurable for xed ω. For the momentthis will merely be assumed and this question will be visited in the sequel.
The following lemma will be used to prove that ‖ · ‖ is uniformly convex.
Lemma 8.5.3. For any two measurable vector elds ω and σ and for almostevery x ∈ X the following inequalities hold.
[ω(x) + σ(x)]px + [ω(x)− σ(x)]px ≤ 2p−1([ω(x)]px + [σ(x)]px) (8.5.1)
88
if p ≥ 2, and
([ω(x) + σ(x)]qx + [ω(x)− σ(x)]qx)p−1 ≤ 2p−1([ω(x)]px + [σ(x)]px) (8.5.2)
if 1 < p ≤ 2. Here q =p
p− 1is the Hölder conjugate exponent of p.
Proof. Note that in the case p = 2 equality holds in (8.5.1) by the usual paral-lellogram law of inner products. Let p ≥ 2 and estimate
([ω(x) + σ(x)]2x)p/2 + ([ω(x)− σ(x)]2x)p/2
≤ ([ω(x) + σ(x)]2x + [ω(x)− σ(x)]2x)p/2 = (2[ω(x)]2x + 2[σ(x)]2x)p/2
= 2p(
[ω(x)]2x + [σ(x)]2x2
)p/2.
Using the convexity of the mapping t 7→ tp/2 yields the nal result:
[ω(x)+σ(x)]px+[ω(x)−σ(x)]px ≤ 2p(
[ω(x)]2x + [σ(x)]2x2
)p/2≤ 2p
[ω(x)]px + [σ(x)]px2
.
For the case 1 < p ≤ 2 note that q ≥ 2. It is good to keep in mind the identities
p = q(p− 1)q = p(q − 1)(p− 1)(q − 1) = 1
which will be used without further mention. Start with the inequality
(s+ t)q + (s− t)q ≤ 2(sp + tp)q−1,
valid for 0 ≤ s ≤ t and 1 < p ≤ 2. The proof for this can be found in ([3],Th. 2). (Notice that this is a special case of (8.5.2).) To prove (8.5.2) choosean orthonormal basis for Rn with respect to the inner-product norm [·]x andwrite the norm (squared) of any vector eld as a sum of the squares of itscomponents. Raising both sides of (8.5.2) to the power q − 1 the claim thentakes the equivalent form(
n∑k=1
|ωi + σi|2)q/2
+
(n∑k=1
|ωi − σi|2)q/2
≤ 2
( n∑k=1
|ωi|2)p/2
+
(n∑k=1
|σi|2)p/2q−1
.
Write the left side of this as(n∑k=1
(|ωi + σi|q)2/q
)q/2+
(n∑k=1
(|ωi − σi|q)2/q
)q/2
89
and use the reverse Minkowski inequality ([15], pp. 146, Thm 198) for 2/q ≤ 1to estimate this by(
n∑k=1
(|ωi + σi|q + |ωi − σi|q)2/q
)q/2≤
(n∑k=1
22/q(|ωi|p + |σi|p)(q−1)2/q
)q/2= 2
(n∑k=1
(|ωi|p + |σi|p)2/p
)(q−1)p/2
.
Since 2/p ≥ 1, application of the normal Minkowski inequality yields an upperestimate of
2
( n∑k=1
(|ωi|p)2/p
)p/2+
(n∑k=1
(|σi|p)2/p
)p/2(q−1)
= 2
( n∑k=1
|ωi|2)p/2
+
(n∑k=1
|σi|2)p/2q−1
.
This nishes the proof.
With the aid of the inequality the above Lemma the uniform convexity of ‖ · ‖can be proven. For the case p ≥ 2 it immediately implies
‖ω + σ‖p + ‖ω − σ‖p ≤ 2p−1(‖ω‖p + ‖σ‖p)
for ω, σ ∈ Lp(X). Now if ‖ω‖ = ‖σ‖ = 1 and ‖ω − σ‖ = ε for a given 0 < ε ≤ 2then
‖ω + σ‖p + ‖ω − σ‖p ≤ 2p
which then implies‖ω + σ‖
2≤(
1−(ε
2
)p)1/p
.
Consequently
δ(ε) = inf
1− ‖ω + σ‖2
: ‖ω − σ‖ = ε, ‖ω‖ = ‖σ‖ = 1≥ 1−
(1−
(ε2
)p)1/p
.
To address the case 1 < p ≤ 2 one more Corollary is needed.
Corollary 8.5.4. For 1 < p ≤ 2 one has the inequality
‖ω + σ‖q + ‖ω − σ‖q ≤ 2(‖ω‖p + ‖σ‖p)q−1
for every ω, σ ∈ Lp(X). Here q =p
p− 1is the Hölder conjugate exponent of p.
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Proof. Note that q/p = q − 1 and p/q = p − 1 ≤ 1. Use a reverse Minkowskiinequality to obtain
‖ω + σ‖q + ‖ω − σ‖q =(∫
X
([ω + σ]q)p/qdµ))q/p
+(∫
X
([ω − σ]q)p/qdµ)q/p
≤(∫
X
([ω + σ]q + [ω − σ]q)p/qdµ)q/p
.
Now apply (8.5.2) to the integrand to get the desired result
≤(∫
X
2p−1([ω]p + [σ]p)dµ)q−1
= 2(‖ω‖p + ‖σ‖p)q−1.
The above Corollary implies in a similar manner to the case p ≥ 2, that
δ(ε) ≥ 1−(
1−(ε
2
)q)1/q
.
It still remains to show that if ω is measurable for some xed then the map-ping x 7→ [ω(x)]x obtained from the new norm is measurable. Loosely speakingthis means that the choice of the ellipsoid, or the innerproduct norm can bemade in a somehow measurable manner. The trouble is specifying what exactlyis meant by somehow measurable manner. To address the measurability ques-tion another version of John's lemma 8.5.2 will be of use. The proof and morerelated discussion of this can be found in [27] and [19].
Lemma 8.5.5. Let ‖ · ‖ be a norm in Rm, K ⊂ Rm its unit ball and [·] the dualnorm of ‖ · ‖∗, a norm in (Rm)∗ given by
‖u‖∗ =(−∫K
u(x)2dx)1/2
.
(Both ‖ · ‖∗ and [·] are given by an inner product.) Then there is a constantc = c(m) depending only on the dimension m so that
1c‖a‖ ≤ [a] ≤ c‖a‖ for every a ∈ Rm.
Returning to the measurability question consider the above norm [·]x forevery x ∈ Xk, (k ∈ K xed) for which (Rn(k), ‖ · ‖x) can be formed, and letm = n(k). Given ω ∈ L(X) the purpose is to show that x 7→ [ω(x)]x : Xk → Ris measurable, if [·]x is chosen as in Lemma 8.5.5. Let (aj)∞j=1 be an enumerationof Qm. For each n ∈ N dene the mappings
p1n : Xk → Rn, p1
n(x) = (‖a1‖x, . . . , ‖an‖x) and
p2n : Rn → P(Rm), p2
n(y1, . . . , yn) = convai : yi < 1, i = 1, . . . , n
for almost every x ∈ Xk. Clearly p1n can be modied to be Borel (dene it to be,
say, zero on a null set). The function p2n chooses those points aj ∈ Qm for which
91
the corresponding jth coordinate is less than one and forms the convex hull ofthose points that are chosen. Now for suciently large n the set p2
n p1n(x) is
convex with nonempty interior: the convexity is of course obvious since p2n(y) is
convex for every y ∈ Rn. Let L ∈ Q be larger than the maximum of the Lipschitzconstants of the functions ei ·ϕk, i = 1, . . . ,m whence ‖ei‖x < L. This constantdoes not depend on x. Since ‖ei/L‖x < 1 and furthermore ei/L,−ei/L ∈ Qm
for every i = 1, . . . ,m there exists some j0 so that p2n p1
n(x) is the convex hullof a set of points containing the ei/L's and the −ei/L's whenever n ≥ j0. Hencep2n p1
n(x) is a convex set with nonempty interior for all n ≥ j0.Denote by B(m) the set of all convex subsets of Rm with nonempty interior.
If B ∈ B(m) then
〈u, v〉B := −∫B
u(x)v(x)dx
denes an inner produt in (Rm)∗. The dual inner product of this can be iden-tied with a positive denite symmetric matrix p3(B). This construction alsodenes the mapping p3 : B(m) → sym+(m) where sym+(m) denotes the set ofpositive denite symmetric m by m -matrices. In particular p3 p2
n p1n(x) is
well dened for suciently large n (n ≥ j0). For all such n consider the imagep2n(Rn). This consists of sets convai1 , . . . , aik where i1, . . . , ik ⊂ 1, . . . , nand there are only nitely many of these. Hence p2
n(Rn) is a nite set and conse-quently p3p2
n(Rn) is a nite set. For anyM ⊂ sym+(m) the set (p3p2n)−1(M)
is then Borel: obviously (p3)−1(M) ∩ p2n(Rn) is nite, say B1, . . . , Br. If
Bi = convai1 , . . . , aik ∈ (p3)−1(M) ∩ p2n(Rn) then (p2
n)−1(B) is a cartesianproduct of intervals that are either (−∞, 1), [1,∞) or (−∞,∞) (if j 6= il forsome l then the interval corresponding to the jth coordinate is [1,∞) whereasfor j = il for some l the interval corresponding to the jth axis is either (−∞, 1)or,if it so happens that ail lies in the line-segment connecting two aj 's with jappearing in the denition of Bi, then it doesn't matter whether or not this el-ement is chosen in the denition of p2
n and therefore the interval correspondingto that is (−∞,∞)). The preimage can now be expressed as
(p3 p2n)−1(M) =
r⋃i=1
(p2n)−1(Br)
which is clearly Borel.
Hence p3 p2n and consequently pn := p3 p2
n p1n is a Borel function. To
accomplish the purpose of all this which was to prove the measurability ofx 7→ [ω]x it suces to prove that pn(x) → p(x) almost everywhere wherep : Xk → sym+(m) attaches to almost every x ∈ Xk the inner product norm [·]xin Rm described in 8.5.5. The suciency can be seen as follows: as a pointwiselimit of Borel functions p is measurable. Therefore its coordinate projectionspij : Xk → R, i, j = 1, . . . ,m are measurable. In particular, if 〈·, ·〉x denotes theinner product inducing [·]x, the coordinate projections satisfy
pij(x) = 〈ei(x), ej(x)〉xwhere ei(x) = dϕik(x), i = 1, . . . ,m is the standard basis of Rm. Therefore[ω(x)]x can be expressed as
[ω(x)]x =√∑
i,j
pij(x)ωi(x)ωj(x)
92
which shows the desired measurability since ωi is measurable for each i =1, . . . ,m.
Now to prove the pointwise convergence consider, for almost every x ∈ Xk
the open unit ball K ⊂ Rm of ‖ · ‖x and let Kn = p2n p1
n(x). Since Kn isby denition the smallest convex set containing the points ai, i = 1, . . . ,mfor which ‖ai‖x < 1 it follows that Kn ⊂ K. In addition the sequence Kn isincreasing in the sense that Kn ⊂ Kn+1 for all n. Now the density of (aj)j∈Nin Rm implies that
K =∞⋃n=1
Kn.
To see this take y ∈ K which then lies in the line segment between some ai, aj ∈Qm, which in turn lie in the haline ty : t ≥ 0, hence y ∈ Kn for n ≥ maxi, j.Then for all u, v ∈ (Rm)∗
−∫Kn
u(y)v(y)dy n→∞−→ −∫K
u(y)v(y)dy.
From the denition of the dual norm it is seen that likewise pn(x)(y, z) →p(x)(y, z) for all y, z ∈ Rm, that is pn(x)→ p(x) as n→∞.
This completes the proof of Theorem 8.5.1 by addressing the measurability ofx 7→ [ω(x)]x. In fact in the proof of 8.5.1 [·]x should mean the inner productnorm given by lemma 8.5.5. The quantitative results of course remain the same.Notice, however, that the ei's and ωi's appearing in the above discussion aredierent from those appearing in the proof 8.5.1.
It is now an immediate consequence of ([3], Th. 1) that if N1,p(X) =M1,p(X) =: H1,p(X) is equipped with the norm
‖u‖pH1,p =∫X
|u|pdµ+∫X
[du]pdµ
then H1,p(X) is uniformly convex.
In short, complete doubling metric spaces supporting some Poincaré inequalityenjoy a surprisingly large amount of the rst order smoothness usually associatedto Euclidean spaces (and their smooth subsets). In this thesis this is moststrongly displayed by the nice behaviour of the associated various Sobolev typespaces.
However for some results, supporting a Poincaré inequality is a needlesslystrong assumption. The construction of the dierential structure, as can readilybe noted, only requires that the metric space in question satises the conclusionof Theorem 8.0.4, for a uniform constant K. (Although this assumption doesnot guarantee the quasiconvexity of the space, and hence the non-degeneracy ofthe dierential structure.) The reexivity and the uniform boundedness, on theother hand, only requires that the space admits a dierentiable structure (an apriori weaker condition than even the inequality of Theorem 8.0.4). 9
9Actually the doubling property of the measure can also be weakened, see [21].
93
This notwithstanding, spaces supporting a Poincaré inequality already con-siderably generalize the class of admissible spaces for doing rst order calculus.For further study of some weaker conditions, a discussion of some open ques-tions, as well as a more extensive study of the dierent Sobolev type spaces, see[21] and [14] and the references therein.
94
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