Notes on Sobolev Spaces Peter Lindqvist Norwegian University of Science and Technology 1 L p -SPACES 1.1 Inequalities For any measurable function u : A → [−∞, ∞], A ∈ R n , we define ‖u‖ p = ‖u‖ p, A = A |u( x)| p dx 1 p and, if this quantity is finite, we say that u ∈ L p (A). In most cases of interest p ≥ 1. For p = ∞ we set ‖u‖ ∞ = ‖u‖ ∞, A = ess sup x∈A |u( x)|. The essential supremum is the smallest number M such that |u( x)|≤ M for a.e. x ∈ A. For example 1 , if u :[a, b] → R n , is continuous, then it is easy to see that b a |u( x)| p dx 1 p −→ p→∞ max a≤x≤b |u( x)|. The fundamental inequalities ‖uv‖ 1 ≤‖u‖ p ‖v‖ q , 1 p + 1 q = 1, H ¨ OLDER ‖u + v‖ p ≤‖u‖ p + ‖v‖ p , MINKOWSKI 1 An interesting application of this fact in connection with the heat equation u t = u xx is given in [BB, Ch.24, pp 145-150]. 1
Transcript
Notes on Sobolev Spaces
Peter Lindqvist
Norwegian University of Science and Technology
1 Lp-SPACES
1.1 Inequalities
For any measurable function u : A→ [−∞,∞], A ∈ Rn, we define
‖u‖p = ‖u‖p, A =
∫
A
|u(x)|p dx
1p
and, if this quantity is finite, we say that u ∈ Lp(A). In most cases of interest p ≥ 1.
For p = ∞ we set
‖u‖∞ = ‖u‖∞,A = ess supx∈A
|u(x)|.
The essential supremum is the smallest number M such that |u(x)| ≤ M for a.e.
x ∈ A.
For example1, if u : [a, b]→ Rn, is continuous, then it is easy to see that
b∫
a
|u(x)|p dx
1p
−→p→∞
maxa≤x≤b
|u(x)|.
The fundamental inequalities
‖uv‖1 ≤ ‖u‖p‖v‖q,1
p+
1
q= 1, HOLDER
‖u + v‖p ≤ ‖u‖p + ‖v‖p, MINKOWSKI
1An interesting application of this fact in connection with the heat equation ut = uxx is given
in [BB, Ch.24, pp 145-150].
1
where p ≥ 1, can be derived in many ways. For example, taking the “elementary
inequality”
ab ≤ap
p+
bq
q(a, b ≥ 0, p + q = pq) YOUNG
for granted, we obtain the Holder inequality (choose a = u(x)/‖u‖p, b = v(x)/‖u‖qand integrate the resulting inequality). The Holder inequality implies the Minkowski
inequality.
Remarks:
1) The special case
∫
A
|uv| dx ≤√√∫
A
u2 dx
√√∫
A
v2 dx CAUCHY
is called the Cauchy inequality.
2) If 1 > p > 0, then the Minkowski inequality is reversed for positive func-
tions! That is why one usually has p ≥ 1.
3) If mes (A) < ∞, then the Holder inequality shows that Lp1 (A) ⊂ Lp2 (A), if
p1 ≥ p2. This is not true in general, if mes (A) = ∞. (Find a simple ex.!)
However, if 1 ≤ p1 < p < p2 ≤ ∞, then the Holder inequality implies that
‖u‖p,A ≤ ‖u‖λp1,A‖u‖1−λp2,A
where1
p=λ
p1
+1 − λ
p2
.
That is, if u ∈ Lp1 ∩ Lp2 , then u ∈ Lp for all intermediate p.
4) Suppose that 0 < mes (A) < ∞. The function
φ(p) =
1
mes (A)
∫
A
|u(x)|p dx
1p
, 0 < |p| < ∞,
φ(0) = exp
1
mes (A)
∫
A
log |u(x)| dx
,
φ(−∞) = ess infA|u(x)|, φ(+∞) = ess sup
A
|u(x)|
2
is increasing2 as −∞ ≦ p ≦ +∞.
Let us finally mention that
Λ
1
mes (A)
∫
A
u(x) dx
≤
1
mes (A)
∫
A
Λ(u(x)) dx JENSEN
whenever Λ : R→ R is a convex function (0 < mes (A) < ∞, u : A→ [−∞,∞]
is measurable).
Example:
e
1∫
0
u(x) dx
≤1∫
0
eu(x) dx
1
mes (A)
∫
A
|u(x)| dx
p
≤ 1
mes (A)
∫
A
|u(x)|p dx
Remark: Discrete versions of the above inequalities are
∣∣∣∣∣∣∣
∑
k
akbk
∣∣∣∣∣∣∣
≤ p
√∑
k
|ak|p q
√∑
k
|bk|q;
n a1a2 . . . an ≤ |a1|n + |a2|n + . . . + |an|n
orn√
q1q2 . . . qn︸ ︷︷ ︸
GEOMETRIC MEAN
≤q1 + q2 + . . . + qn
n︸ ︷︷ ︸
ARITHMETIC MEAN
(qk ≥ 0);
p
√∑
k
|ak + bk|p ≤ p
√∑
k
|ak|p + p
√∑
k
|ak|p.
Especially, limp→∞
p√∑
k |ak|p = supk
|ak|.
Example: Draw the curves p√
|x|p + |y|p = 1 (0 < p ≦ ∞) in the xy-plane.
2J. Moser’s celebrated method to relate the maximum and the minimum of the solution to a
partial differential equations is based on this.
3
For p = 2 we have the parallelogram3 law
‖u + v‖22 + ‖u − v‖22 = 2‖u‖22 + 2‖v‖22.
Remark: So-called reverse Holder inequalities like
1
mes (Q)
∫
Q
|u|
1p
≤C
mes (Q)
∫
Q
|u|
valid for every cube with some fixed p > 1 (!) play a central role for solutions u to certain
partial differential equations. (Such an inequality cannot hold for arbitrary functions.)
The method is due to F. W. GEHRING.
1.2 (Strong) convergence in Lp
Let A ∈ Rn and fix p ≥ 1. Then ‖ ‖p defines a semi-norm in the vector space Lp(A),
i.e.,
(i)’ 0 ≤ ‖u‖p < ∞, u ∈ Lp(A)
(ii) ‖λu‖p = |λ|‖u‖p, u ∈ Lp(A), −∞ < λ < ∞
(iii) ‖u + v‖p ≤ ‖u‖p + ‖v‖p, u, v ∈ Lp(A).
However, this is, strictly speaking, not a norm, the reason being that ‖u‖p,A =0 ⇐⇒ u(x) = 0 for a.e. x ∈ A. We agree to say that u = v in Lp(A), if u(x) = v(x)
for a.e. x ∈ A. With this convention ‖ ‖p is a norm, i.e. (i)’ can be replaced by
(i) 0 ≤ ‖u‖p < ∞, and ‖u‖p = 0⇐⇒ u = 0 in Lp.
(Strictly speaking, this Lp-space consists of equivalence classes of functions, but
here there is no point in maintaining this distinction.)
Theorem 1 (RIESZ-FISCHER) The Lp-spaces, p ≥ 1, are Banach spaces. That
is, if u1, u2, . . . is a Cauchy sequence in Lp(A), then there is a function u ∈ Lp(A)
such that ‖uk − u‖p,A → 0, as k →∞.3The counterpart for general exponents p is more involved. A pair of inequalities (CLARK-
SON’s inequalities) will replace the parallelogram law [A, p.37].
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Proof: Suppose that u1, u2, . . . is a Cauchy sequence in Lp(A), i.e., given ε > 0 there
is an index Nε such that ‖uk − u j‖p < ε, when k, j ≥ Nε. Let us consider the case p , ∞(the case p = ∞ is simpler, and its proof will be skipped). We can construct indices
1 ≤ n1 < n2 < . . . such that
‖unk− unk+1
‖p <1
2k(k = 1, 2, 3, . . .).
Define
gN(x) =
N∑
k=1
|unk− unk+1
|, g(x) = limN→∞
gN(x) =
∞∑
k=1
|unk− unk+1
|
Then
0 ≤ g1(x) ≤ g2(x) ≤ . . . ≤ g(x) ≦ ∞
and ∫
A
|g|p =∫
A
∣∣∣∣∣
limN→∞
gN
∣∣∣∣∣
p
=
∫
A
limN→∞|gN |p ≤ lim
N→∞
∫
A
|gN |p
by Fatou’s lemma. By the construction ‖gN‖p ≤ 12+
14+. . .+ 1
2N < 1 and so∫
A
|g|p ≤ 1. Thus
g(x) < +∞ for a.e. x ∈ A. This means that the series |un2(x)−un1
(x)|+ |un3(x)−un2
(x)|+ . . .converges for a.e. x ∈ A. So does a fortiori, the series
un2(x) +
∞∑
k=1
(unk+1− unk
).
The partial sums of this series are plainly un2(x), un3
(x), un4(x), . . . Hence
u(x) = limk→∞
unk(x)
exists and is finite for a.e. x ∈ A (If you insist on having an everywhere defined function,
set u(x)=0 in a subset of measure zero.)
By Fatou’s lemma (again!)
∫
A
|u(x) − u j(x)|p dx =
∫
A
(
limk→∞|unk
(x) − u j(x)|p)
dx ≤ limk→∞
∫
A
|unk(x) − u j(x)|p dx ≤ εp
whenever j > Nε. This shows that ‖u − u j‖p,A → 0 as j→ ∞.
The proof yields information about pointwise behaviour.
Corollary 2 Suppose that uk → u in Lp(A), i.e., ‖uk − u‖p,A → 0. Then there is a
subsequence that converges a.e. in A : u(x) = limk→∞
unk(x) for a.e. x ∈ A.
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1.3 The dual of Lp
If X is any Banach space, its dual X∗ is the collection of all continuous linear
functions (functionals) l : X → R . The norm of l is defined as
‖l‖X∗ = supx∈X
(
|l(x)|‖x‖X
)
= sup‖x‖X≤1
|l(x)|
and the continuity is equivalent to ‖l‖X∗ < ∞. Note that
|l(x)| ≤ ‖l‖X∗‖x‖.
Let 1 ≤ p < ∞ and fix a function g ∈ Lq(A). Then
l( f ) =
∫
A
f (x)g(x) dx ( f ∈ Lp(A))
is well-defined and linear. By Holder’s inequality |l( f )| ≤ ‖ f ‖p‖g‖q. Hence ‖l‖∗ ≤‖g‖q. Here equality is attained for the choice f = |q|q−2g. (The case q = ∞ is
slightly different.) Thus
‖l‖∗ = ‖g‖q.The essential fact here is that virtually all continuous functionals in Lp (p , ∞)
are of this form!
Theorem 3 (F. RIESZ’ representation thm) Let l : Lp(A) → R be a continuous
linear functional, 1 ≤ p < ∞. Then there is a unique g ∈ Lq(A) such that
l( f ) =
∫
A
f (x)g(x) dx
for all f ∈ Lp(A). Moreover, ‖l‖∗ = ‖g‖q,A.
Proof: The Radon-Nikodym theorem is used in most proofs of the represen-
tation theorem. A more direct proof in the one dimensional case is given in [R.
p.121].
Remark:
1) There are continuous linear functionals in L∞(A) that do not have this simple
form.
2) One says that Lq is the dual of Lp (1 ≤ p < ∞).
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1.4 Weak convergence in Lp
Weak convergence in Lp is important for many applications, for example, it leads
to the existence theorems for partial differential equations. The so-called direct
methods in the Calculus of Variations are based on this concept.
Suppose that u1, u2, u3, . . . are functions in Lp. We say that uk → u weakly in
Lp, if
limk→∞
l(uk) = l(u)
for every continuous linear functional l : Lp → R . Using Riesz’ representation
theorem we can state the definition in a more convenient form.
Definition 4 Let 1 ≤ p < ∞. Suppose that u1, u2, . . . and u are functions in Lp(A).
We say that uk → u weakly in Lp(A), if
limk→∞
∫
A
ukv dx =
∫
A
uv dx for each v ∈ Lq(A).
Remark:
1) Sometimes the notation uk u is used to indicate weak convergence.
2) Strong convergence implies weak convergence: if ‖uk − u‖p → 0, then
uk u weakly in Lp.
3) The weak limit is unique, that is, unique in Lp.
Example: The functions un(x) = sin(nx) converge weakly in L2([0, 2π]) to zero.
By the Riemann-Lebesgue lemma
limn→∞
2π∫
0
v(x) sin(nx) dx = 0
for every v in L2([0, 2π]). (This follows easily from Bessel’s inequality.)
Example: (Warning!) Suppose that f : [a, b] → R is a bounded measurable
function, for example, assume 0 ≤ f (x) ≤ 2. Then there is a sequence of functions
vn : [a, b]→ 0, 2 converging weakly in L2([a, b]) to f . Observe that vn(x) = 0 or
= 2 ( vn takes no other values!!)
This is not difficult to realise in the special case f (x) ≡ 1.
7
Example: Define ui :]0, 1[→ R by ui = iχ]0,i−3[ for i = 1, 2, 3, . . . Then ui → 0
strongly in Lp([0, 1]), if 1 ≤ p < 3. We have ‖ui‖3+ε = iε /3 for every ε > 0. As
we shall see, weakly convergent sequences are bounded in Lp-norm. Therefore
u1, u2, . . . does not converge weakly in Lp(]0, 1[), if p > 3. —Show that ui 0
weakly in L3(]0, 1[)! (This convergence is not strong.)
If ui u weakly in Lp(A), then there is a constant M such that ‖ui‖p,A ≤ M ≤∞ for all i = 1, 2, 3, . . . This follows from the following lemma.
Lemma 5 Suppose that u1, u2, . . . are functions in Lp(A), 1 < p < ∞. If the
sequence ‖ui‖p,A is unbounded, there is a w ∈ Lq(A) such that
limk→∞
∫
A
uik (x)w(x) dx = +∞
for some subsequence.
Proof: The function w is constructed and written down in [S, pp.25-28].
The most important fact about Lp-spaces seems to be the following weak com-
pactness property. (Not valid for p = 1).
Theorem 6 (Weak Compactness) Let u1, u2, . . . be functions in Lp(A), 1 < p <
∞. If there is a constant M such that ‖ui‖p,A ≤ M for each index i, then there exists
a function u ∈ Lp(A) such that uik u weakly in Lp(A) for some subsequence.
Moreover
‖ui‖p,A ≤ limk→∞‖uik‖p,A.
Proof: Advanced books on Functional Analysis usually contain a proof. For
example, the above mentioned book of Sobolev [S] gives a proof on pp. 29-30.
See also [A]. Usually, the lower semicontinuity comes as a by-product of the
proof, but the following simple argument also yields this property. Since |x|p is a
convex function of x, we have
|y|p ≥ |x|p + p|x|p−2x · (y − x), p ≥ 1.
Thus ∫
A
|uik |p dx ≥∫
A
|u|p dx + p
∫
A
|u|p−1u · (uik − u) dx.
Now |u|p−1u is in Lq(A), and so the last integral approaches zero as k→ ∞ (by the
weak convergence!). This gives us the desired lower semicontinuity.
8
Remarks:
1) The theorem is not true for p = 1.
2) The existence of solutions to partial differential equations is often a direct
consequence of the weak compactness.
3) Let ‖ui‖p ≤ M for i = 1, 2, 3, . . .According to the BANACH-SAKS theorem
there are indices i1 < i2 < i3 < . . .and a function u in Lp such that
∥∥∥∥∥
ui1 + ui2 + . . . + uiν
ν− u
∥∥∥∥∥
p
−→ 0
as ν → +∞, that is, the arithmetic means converge strongly. (The Banach-
Saks theorem is valid also for p = 1.)
1.5 Approximation in Lp and some other things
The essential fact is that the functions in Lp can be approximated in the Lp-normby smooth functions, if 1 ≤ p < ∞. These are constructed as convolutions. Let us
first state some auxiliary results.
Lemma 7 (“Continuity in the Lp-norm”) Let 1 ≤ p < ∞. If f ∈ Lp(A), then
limh→0
∫
A
| f (x + h) − f (x)|p dx = 0,
where f is regarded as 0 outside A.
The proof is not quite simple. Usually one uses the theorem of Lusin, valid for measurable
functions.
Theorem 8 (Lusin) Let f : A → [−∞,∞] be a measurable function that is finite a.e.
in A. Suppose that (A) < ∞. Given ε > 0, there is a compact set Kε ⊂ A such that the
restriction f |Kεis continuous and mes(A\Kε) < ε .
Proof: See, [EG, p.15].
If A is open, then Lusin’s theorem can be combined with the extension theorem of
Urysohn-Tietze [R, p.148].
9
Theorem 9 (Urysohn-Tietze) Suppose that A is open and mes (A) < ∞. Let K ⊂ A be a
compact set. If f : K → R is a continuous function, then there is a function ϕ ∈ C0(A)
such that ϕ(x) = f (x), when x ∈ K. Moreover, maxA|ϕ| = max
A| f |
Lemma 10 C0(Ω) is dense in Lp(Ω), 1 ≤ p < ∞. Here Ω is open and C0(Ω)
denotes all continuous functions with compact support on Ω. In other words, if
u ∈ Lp(Ω), then there are functions ϕk ∈ C0(Ω) such that
limk→∞‖u − ϕk‖p,Ω = 0.
Remarks:
1) Of course, this implies that C(Ω) is dense in Lp(Ω).
2) Functions in L∞(Ω) cannot, in general, be uniformly approximated by con-
tinuous functions.
3) If ϕ : Rn → R is continuous, then the closure in Rn of the set where ϕ(x) , 0
is called the support of ϕ. Thus
supp(ϕ) = x ∈ Rn |ϕ(x) , 0.
If supp(ϕ) is compact (it is closed by definition) and if supp(ϕ) ⊂ Ω, then
we say that ϕ ∈ C0(Ω). In this case the distance between supp(ϕ) and the
boundary ∂Ω is positive.
4) A deep result for Lp-functions is related to the Lebesgue points. Suppose that
f ∈ Lploc(R
n), that is, f ∈ Lp(B) for every ball B in Rn . Then
limr→o+
1
mes (B(x, r))
∫
B(x,r)
| f (x) − f (y)|p dy = 0
for a.e. x.
Define
ρ(x) =
Ce− 1
1−|x|2 , |x| < 1
0, |x| ≥ 1
and choose the constant C > 0 such that∫
Rn
ρ(x) dx = 1. “Friedrichs’ mollifier” is
ρε(x) =1
εnρ
(x
ε
)
=
C
εne− ε2
ε2 −|x|2 , when |x| < ε,
0, when |x| ≥ ε .
10
The constant C depends only on the dimension n. Now we have
∫
Rn
ρε(x) dx = 1, ε > 0. (1)
Observe that ρε ∈ C∞0 (Rn). The support of ρε is the closed ball |x| ≤ ε .The convolution
uε(x) = (ρε ∗ u)(x) =
∫
Rn
ρε(x − y)u(y) dy
is well-defined for u ∈ L1loc
(Rn). If u is defined only in the domain Ω, then we
regard u as extended to zero outside Ω : u(x) = 0, when x ∈ Rn \Ω. Hence we can
calulate the convolution uε for any u in L1loc
(Ω).
Observe that uε is always a smooth function: uε ∈ C∞(Rn), if u ∈ L1loc
(Rn). For
differentiation we have the rule
Dαuε = (Dαρε) ∗ u. (2)
Here α = (α1, α2, . . . , αn) and
Dα=
∂α1+α2+...+αn
∂xα1
1∂x
α2
2. . . ∂x
αnn
.
If u was defined in the domain Ω, then the formula (2) holds for the original
(unextended) u at all points x with dist(x, ∂Ω) > ε . Analogously, if supp u ⊂ Ω,then uε ∈ C∞
0(Ω), when ε < dist(supp u, ∂Ω).
Lemma 11 If u ∈ Lp(Ω), 1 ≤ p < ∞, then uε ∈ Lp(Ω) and ‖ρε ∗ u‖p ≤ ‖u‖p.Moreover
limε→0+‖uε − u‖p = 0.
Lemma 12 If u ∈ C(Ω) and if K ⊂⊂ Ω is compact then
maxx∈K|uε(x) − u(x)| → 0 as ε→ 0+.
In other words, the convergence uε → u is uniform on compact subsets.
11
Proof: The locally uniform convergence for a continuous u follows form
u(x) − uε(x) = u(x) −∫
ρε(x − y)u(y) dy
=
∫
|x−y|<ε
(u(x) − u(y))ρε(x − y) dy
|u(x) − uε(x)| ≤ maxy
|x−y|≦ε
|u(x) − u(y)| · 1 (ε < dist(x, ∂Ω))
in accordance with Weierstrass theorem ( a function that is continuous on a com-
pact set us uniformly continuous).
If u ∈ Lp(Ω), then
|uε| ≤∫
ρε(x − y)|u(y)| dy ≤∫
ρε(x − y)|u(y)|p dy
1p∫
ρε(x − y) dy
1q
︸ ︷︷ ︸
=1
and so∫
|uε(x)|p dx ≤∫ (∫
ρε(x − y)|u(y)|p dy
)
dx
=
∫
|u(y)|p(∫
ρε(x − y) dx
)
︸ ︷︷ ︸
=1
dy =
∫
|u(y)|p dy.
This proves the contraction ‖uε‖p ≤ ‖u‖p.For the Lp-convergence, we first note that if ϕ ∈ C0(Ω), then it is easily seen
that∫
|ϕ(x) − ϕε(x)|p dx ≤ sup|x−y|≦ε
|ϕ(x) − ϕ(y)|p (ε < dist(supp ϕ, ∂Ω))
and so ‖ϕ − ϕε‖p → 0. (Doing some more work, we could replace C0 by C ∩ Lp.)
Now
‖u − uε‖p ≤ ‖u − ϕ‖p + ‖ϕ − ϕε‖p + ‖=(ϕ−u)ε︷ ︸︸ ︷
ϕε − uε ‖p︸ ︷︷ ︸
≤‖ϕ−u‖p
and so
limε→0+‖u − uε‖p ≤ 2‖u − ϕ‖p.
12
Since C∞0
(Ω) is dense in Lp(Ω), we can choose ϕ ∈ C0(Ω) so that ‖u − ϕ‖p is as
small as we please. This concludes the proof for the Lp-convergence ‖u−uε‖p → 0.
As an application we mention the Variational Lemma.
Lemma 13 (Variational Lemma) Let u ∈ L1loc
(Ω), Ω denoting an open set in Rn .
If∫
Ω
u(x)ϕ(x) dx = 0
whenever ϕ ∈ C∞0
(Ω), then u = 0 a.e. in Ω.
Proof: Take x ∈ Ω and choose ε so small that 0 < ε < dist(x, ∂Ω). Then
ρε(x − x) will do as a test function so that
uε(x) =
∫
u(x)ρε(x − x) dx = 0.
Let B be any closed ball in Ω, i.e. B ⊂⊂ Ω. Then
‖u‖1,B = ‖u − uε‖1,B → 0
as ε→ 0+. Hence u = 0 a.e. in B and hence a.e. also in Ω.
Remark: The lemma is fundamental in the Calculus of Variations. If u is con-
tinuous, the proof is more elementary and u ≡ 0 in this case. We shall need the
lemma to establish that the Sobolev derivatives of a function are unique up to sets
of measure zero.
Example: (Hermann WEYL) Let u ∈ L1loc
(Rn) and suppose that∫
u(x)∆ϕ(x) dx = 0
whenever ϕ ∈ C∞0 (Rn). Then u has a continuous representative which is a harmonic
function, i.e. the continuous u belongs to C2 and ∆u = ∂2u
∂x21
+ . . .+ ∂2u
∂x2n= 0. (In fact,
u ∈ C∞(Rn).)
Take ϕ(x) = ρε(x − z) for some fixed z. Then ∆uε(z) = 0. Indeed,
uε(z) = (ρε ∗ u)(z) =
∫
ρε(z − x) u(x) dx
∆uε(z) = (∆ρε ∗ u)(z) =
∫
∆zρε(z − x) u(x) dx
=
∫
[∆xρε(z − x)]u(x) dx = 0
13
by assumption. Note that
∂2u
∂x2k
ρε(z − x) =∂2u
∂z2k
ρε(z − x)
by direct calculation (ρε(z − x) is a function of |z − x|2). Since z was arbitrary,
∆uε ≡ 0. Thus the function uε is harmonic.
One does not seem to get any further without using some deeper property of
harmonic functions. By the mean value property
uε(x) =1
mes (B(x, r))
∫
B(x,r)
uε(y) dy (Actually uε = u!)
Since uε → u in L1(B(x, r)), we have
limε→0+
uε(x) =1
mes (B(x, r))
∫
B(x,r)
u(y) dy.
For a.e. x, u(x) = lim uε(x), at least when ε approaches zero through a subse-
quence ε1, ε2, ε3, . . . (Corollary 2). Redefining u in a set of measure zero, we get a
function that satifies the mean value property. Hence (the redefined) u is harmonic
in Rn.
14
2 SOBOLEV SPACES
The situation with the derivatives belonging to some Lp-space was studied by
Tonelli, B. Levi, Sobolev, Kondrachev et consortes. The corresponding spaces are
named after Sobolev.
2.1 Wm,p and Hm,p
Throughout this chapter Ω denotes a domain or an open subset of Rn . Suppose
u ∈ C1(Ω), where Ω. Then integration by parts yields
∫
Ω
u(x)∂ϕ
∂xk
dx = −∫
Ω
∂u
∂xk
ϕ(x) dx
when ϕ ∈ C∞0 (Ω). This formula is the starting point for the definition of weak
for all ϕ ∈ C∞0 (Ω), we write vα = Dαu. This is a weak derivative of order |α|.We say that u ∈ Wm,p(Ω), if u ∈ Lp(Ω) and if Dαu exists and belongs to Lp(Ω)
for each multi-index α with |α| ≤ k. Provided with the norm
‖u‖Wm,p(Ω) = ‖u‖m,p,Ω =∑
|α|≤m
∫
Ω
|Dαu|p dx
1p
,
the Sobolev space Wm,p(Ω) is a Banach space. The term with index α = (0, 0, . . . , 0)
is interpreted as
∫
Ω
|u|p dx
1p
. The integer m counts the order of the highest weak
derivative.
Theorem 15 Assume that u ∈ Wm,p(Ω) for some 1 ≤ p < ∞. Then there exists a
sequence of functions ϕk ∈ C∞(Ω) such that ϕk → u in Wm,p(Ω), i.e.,
‖u − ϕk‖Wm,p(Ω)
k→∞→ 0.
About the proof. The case Ω = Rn is relatively simple. The general case was is in
[A, pp.52-53]. Let us just mention that the proof4 uses the partition of unity.
Partition of unity
The partition of unity is frequently used in the theory of distributions. Suppose that Ω ⊂∞∪j=1
U j where each U j is open. Then there are functions ϕ ∈ C∞0
(Rn) such that:
1) 0 ≤ ϕ ≤ 1.
2) ϕ1(x) + ϕ2(x) + . . . = 1 at each point x in Ω .
3) If K ⊂ Ω is any compact set, then only finitely many of the functions ϕk are not
identically zero in K.
4This was proved by N. Meyers and J. Serrin in 1964. However, this was not, as it were, the
first proof.
17
4) Each ϕ j ∈ C∞0
(Uk) for some k = k( j).
We define Hm,p(Ω) as the completion of C∞(Ω) in the norm ‖ ‖m,p,Ω. Often Hm
means Hm,2, p = 2 being the most important special case. To be more precise,
u ∈ Lp(Ω) belongs to the space Hm,p(Ω), if there are functions ϕi ∈ C∞(Ω) such
that ϕi → u in Lp(Ω) and Dαϕi is a Cauchy sequence in Lp(Ω) for each multi-index
α, |α| ≤ m.
It is not difficult to see that Dαu exists, |α| ≤ m, if u ∈ Hm,p(Ω) . Moreover,
Dαϕi → Dαϕ in Lp(Ω) . The central result is [A, pp.52-53]:
Theorem 16 Hm,p(Ω) =Wm,p(Ω), 1 ≤ p < ∞, m = 1, 2, 3, . . .
There is also a characterization of the Sobolev space in terms of integrated
difference quotients. To this end, let ei = (0, . . . , 1, . . . , 0) denote the unit vector
in the ith direction. If u ∈ W1,p(Ω), then
∫
Ω′
∣∣∣∣∣
u(x + hei) − u(x)
h
∣∣∣∣∣
p
dx ≤∫
Ω
|Diu|p dx
for any subdomain Ω′ ⊂⊂ Ω, when 0 < h < dist(Ω′, ∂Ω).
(For smooth functions ϕ this follows from the identity
ϕ(x + hei) − ϕ(x)
h=
1
h
h∫
0
Diϕ(x1, . . . , xi + t, . . . , xn) dt
and the general case follows by approximation.)
Remark: For smooth functions in convex domains the formula
ϕ(y) − ϕ(x) =
1∫
0
[
d
dtϕ(x + t(y − x))
]
dt =
1∫
0
(y − x) · ∇ϕ(x + t(y − x)) dt
is the source of many useful inequalities.
Theorem 17 Let u ∈ Lp(Ω), 1 < p < ∞. Suppose that for any subdomain
Ω′ ⊂⊂ Ω we have
∫
Ω′
∣∣∣∣∣
u(x + hei) − u(x)
h
∣∣∣∣∣
p
dx ≤ Kp < ∞
whenever 0 < h < dist(Ω′, ∂Ω). Then the weak derivative Diu exists and ‖Diu‖p,Ω ≤K.
18
Proof: See [GT, p.169].
In conclusion, there are three5 equivalent definitions for the Sobolev space:
I The definition based on the integration-by-parts formula.
II The definition based on approximation by smooth functions with respect to
the Sobolev norm.
III The characterization in terms of the integrability of difference quotients.
Often, II is used to prove auxiliary inequalities and imbedding theorems, III is
sometimes used to prove the existence of weak derivatives. But I is the Main
Definition.
When we say that u ∈ W1,p(Ω) is, for example, continuous, we mean that there
exists a continuous function ϕ such that u(x) = ϕ(x) for a.e. x ∈ Ω . Thus u can be
made continuous after a redefinition in a set of measure zero. (Remember that, in
general, functions in Lp are defined only almost everywhere.)
2.2 The Space W1,p
0(Ω)
We wish to introduce functions with boundary values zero in Sobolev’s sense.
Remember that functions in W1,p(Ω) can be approximated by functions in C∞(Ω)
with respect to the norm
‖u‖1,p,Ω = ‖u‖p,Ω +n∑
k=1
∥∥∥∥∥
∂u
∂xk
∥∥∥∥∥.
If the approximation can be done using merely functions with compact supportΩ,
then the function itself is in a closed subspace denoted by W1,p
0(Ω).
Definition 18 Suppose that u ∈ W1,p(Ω). We say that u ∈ W1,p
0(Ω), if, given ε > 0,
there is a function ϕε ∈ C∞0
(Ω) such that ‖u − ϕε‖1,p,Ω < ε .
Hence W1,p
0(Ω) is the closure of C∞
0(Ω) with respect to the corresponding
Sobolev norm. Clearly,
Cp
0(Ω) ⊂ W
1,p
0(Ω) ⊂ W1,p(Ω).
5IfΩ is the whole space Rn, a further definition is possible. It is based on the Fourier transform.
There is also a more advanced theory based on Bessel potentials.
19
There are functions in ⊂ W1,p
0(Ω), that do not have compact support in Ω . (Ex-
ample: n = 1, p = 2, Ω =]0, π[, u(x) = sin x, supp u = [0, π] ⊃ Ω .) If u ∈C(Ω) ∩W1,p(Ω) and if
limx→ξx∈Ω
u(x) = 0
at each boundary point ξ ∈ ∂Ω, then u ∈ W1,p
0(Ω). However, there are continu-
ous functions in the Sobolev space W1,p
0(Ω) that do not have “the right boundary
values” zero in the classical sense:
Example: Ω = x ∈ R3 | 0 < |x| < 1. Here |x| =√
x21+ x2
2+ x2
3. Then
u(x) = ln1
|x|, x ∈ Ω,
is in W1,2
0(Ω). The origin is an isolated boundary point. We have
limx→0
ln1
|x|= +∞ (, 0).
(At all the other boundary points the function has the right boundary values in the
classical sense.)
Remark: Let Ω ⊂ Rn and suppose that p > n. Then every function in W1,p
0(Ω) is
continuous and takes the boundary values zero in the classical sense. (In applica-
tions, one usually has p = 2 < 3 ≦ n, unfortunately.)
An extremely important property of the space W1,p
0(Ω) is that it is closed even
under weak convergence. That is, if u1, u2, . . . belong to W1,p
0(Ω) and if ui u and
∇ui ∇u weakly in Lp(Ω) (by definition u ∈ W1,p(Ω)), then u itself is in W1,p
0(Ω).
Let f ∈ W1,p(Ω) and suppose that u ∈ W1,p(Ω).We say that u has the boundary
values f in Sobolev’s sense, if u − f ∈ W1,p
0(Ω). (Sometimes this is written as
u ∈ f + W1,p
0(Ω).) —There is a theory of so-called Trace Spaces like Wk,α(∂Ω) for
general boundary value problems. In particular,the normal derivative ∂u∂n
makes sense.
Lemma 19 If u ∈ W1,p
0(Ω), then the function uχ
Ω+ 0χ
Rn \Ω is in W1,p
0(Rn).
If Ω ⊂ Rn, then the cases 1 ≤ p < n, p = n (the borderline case), and p > n
are very different in some essential estimates.
20
Theorem 20 (the Sobolev inequality) Suppose that u ∈ W1,p
0(Ω), where Ω is a
domain in Rn . If 1 ≤ p < n, then there is a constant C depending only on n and p
such that
∫
Ω
|u|np
n−p dx
n−p
np
≤ C
∫
Ω
|∇u|p dx
1p
. SOBOLEV
Thus u ∈ Lp∗(Ω); p∗ =np
n−p= the Sobolev conjugate.
Proof: Extending u as zero outside Ω we may assume that u ∈ W1,p
0(Rn).
By approximation it is sufficient to prove the inequality for ϕ ∈ C∞0
(Rn). Let
us, for instructive purposes, write down the proof in the two dimensional case
n = 2, x = (x1, x2). Multiply
ϕ(x1, x2) =
x1∫
−∞
∂ϕ(t, x2)
∂x1
dt, ϕ(x1, x2) =
x2∫
−∞
∂ϕ(x1, t)
∂x2
dt,
to get
|ϕ(x1, x2)|2 ≤∞∫
−∞
∣∣∣∣∣
∂ϕ(t1, x2)
∂x1
∣∣∣∣∣
dt1 ·∞∫
−∞
∣∣∣∣∣
∂ϕ(x1, t2)
∂x2
∣∣∣∣∣
dt2.
Integrate with respect to x1 :
∞∫
−∞
|ϕ(x1, x2)|2 dx1 ≤∞∫
−∞
∣∣∣∣∣
∂ϕ(t1, x2)
∂x1
∣∣∣∣∣
dt1
∞∫
−∞
∞∫
−∞
|D2ϕ| dx1 dx2.
Integrate with respect to x2 :
"
|ϕ|2 dx1 dx2 ≤"
|D2ϕ| dx1 dx2 ·"
|D2ϕ| dx1 dx2 ≤("
|∇ϕ| dx1 dx2
)2
.
Hence ‖ϕ‖2 ≤ ‖∇ϕ‖1. This proves the theorem in the case n = 2, p = 1.
For the general p, 1 < p < 2, set ψ = |ϕ|γ with γ > 1 as selected below. Then
the above inequality applied on ψ yields
"
|ϕ|2γ ≤ γ2
("
|ϕ|γ−1|∇ϕ|)2
HOLDER
≤ γ2
("
|ϕ|2γ)2
γ−12γ
("
|∇ϕ|2γγ+1
) γ+1γ
21
for γ > 1. Take 2γ =2p
2−p. Then, after some arithmetic,
("
|ϕ|2p
2−p
) 2−p
p
≤ γ2
("
|∇ϕ|p) 2
p
.
This is the desired inequality for 1 < p < n = 2.
See [EG, pp.138-140] or [GT, pp.155-156] for general n. A proof can also be
based on the formula
u(x) =1
ωn
∫
Ω
(x − y) · ∇u(y)
|x − y|ndy
valid for u ∈ W1,10
(Ω).
Remark: The case n < p < ∞. Let u ∈ W1,p(Ω), p > n. For each cube Q ⊂ Ωwe have
|u(x) − u(y)| ≤2pn
p − n|x − y|1−
np ‖∇u‖p,Q
for a.e. x, y ∈ Q. Hence u is locally Holder continuous in Ω with Holder exponent
α = 1 − np. If Ω is bounded, then u ∈ C(Ω) and ‖u‖∞,Ω ≤ C|Ω |1/n−1/p‖∇u‖p,Ω.
The case p = n (borderline case6) is very special.
For functions in W1,p (but not in W1,p
0) the corresponding inequalities are more
involved than the Sobolev inequality and they do require some additional regular-
ity of the domain in question (balls, cubes, domains, smooth domains, Lipschitz
domains etc).
1) 1 < p < n, u ∈ W1,p(Q), Q = a cube in Rn . Then
‖u‖p∗,Q ≤ p∗n − 1
n‖∇u‖p,Q + |Q|−1/n‖u‖p,Q
(The gain is that p∗ =np
n−p> p.)
2) Poincare’s inequality for u ∈ W1,p(Ω), 1 ≤ p < ∞.∫
Ω
|u(x) − uΩ|p dx ≤ C(n, p)|Ω |p/n∫
Ω
|∇u(x)|p dx.
6The Trudinger-Moser inequality
∫
Ω
exp
(
|u|c‖∇u‖n
)n/(n−1)
dx ≤ C mes(Ω)
holds.
22
Here Ω is a convex domain and uΩ =1|Ω |
∫
Ω
u(x) dx is the average of u over
Ω .
3) Poincare-Sobolev: 1 ≤ p < n, u ∈ W1,p(Ω), Ω = convex.
∫
Ω
|u(x) − uΩ|np
n−p dx
n−pnp
≤ Cn,p
∫
Ω
|∇u(x)|p dx
1/p
These inequalities constitute the main bulk in the Sobolev Imbedding Theorem.
For example, Lq(Ω) ⊃ W1,p
0(Ω), if q = p∗, and, for a domain of finite Lebesgue
measure, this is valid if 1 ≤ q ≤ p∗. The Rellich-Kondrachev theorem is even
stronger.
Theorem 21 (Rellich7-Kondrachev) Let Ω be an arbitrary bounded domain in Rn
and consider the space W1,p
0(Ω).
1 ≤ p < n. Then W1,p
0(Ω) is compactly imbedded in Lq(Ω), where 1 ≤ q <
p∗ =np
n−p(q need not be conjugate to p). In practical terms, if ui ∈ W
1,p
0(Ω), and
‖ui‖1,p,Ω ≤ M, i = 1, 2, . . . ,
then there exists a function u ∈ W1,p
0(Ω) and a subsequence such that
‖u − ui‖q,Ω → 0 as i→ ∞
for each fixed q, 1 ≤ q < p∗.
p > n. Then W1,p
0(Ω) is compactly imbedded in Cα(Ω), α = 1 − n
p. Now
|u(x) − u(y)| ≤ K|x − y|α.
p = n. This is the borderline case.
The same is true for W1,p(Ω), if the boundary of Ω is sufficiently regular.
Remarks:
• The functions, not their derivatives, converge strongly in Lq(Ω).
• The convergence uiν u need not be strong in Lp∗(Ω), but uiν u weakly
also in Lp∗ (Ω).
7The case p = 2 is credited to Rellich.
23
2.3 About W1,p
The first order Sobolev space has some special properties not valid for higher derivatives.
If u ∈ W1,ploc (Ω), so does u+ and u−. As usually, u = u+ − u−, |u| = u+ + u−. Hence,
|u| ∈ W1,ploc
(Ω). We have
∇u+ =
∇u, when u > 0,
0, when u ≦ 0,
with similar rules for u− and |u|. For example, for |u| we use∫
(u2ν + ε
2)1/2∇ϕ dx = −∫
ϕuν∇uν
(u2ν + ε
2)1/2dx (ε , 0).
The passage to the limit under the integral sign can be justified [R, Ch.4,Thm.6]. Now
|ϕuν∇uν|(u1ν + ε
2)−1/2 ≤ |ϕ∇uν|.
Finally, letting ε→ 0, we have by the Dominated Convergence Theorem∫
|u|∇ϕ dx = −∫
ϕ[
1χu>0 − 1χu<0 + 0χu=0
]
∇u dx
i.e., ∇|u| = ∇u, when u > 0, = −∇u, when u < 0, and = 0, when u = 0.
Lemma 22 Let u ∈ W1,p(Ω). Then ∇u = 0 a.e. on any set where u is constant.
Lemma 23 u, v ∈ W1,p(Ω) =⇒ maxu, v, minu, v ∈ W1,p(Ω).
The last lemma is not true with W1,p replaced by Wk,p, if k ≥ 2. Example!?
3 The equation ∆u = 4αu3+ f (x), α ≥ 0
Let Ω be a bounded domain in Rn . Suppose that f ∈ L∞(Ω) and that α ≥ 0 is , for
simplicity, a constant. Fix a function ϕ ∈ W1,2(Ω). Usually, ϕ is continuous even
in Ω. It represents the boundary values.
Problem: Minimize the variational integral
I(u) =
∫
Ω
[
1
2|∇u|2 + αu4
+ f (x)u
]
dx
among all u ∈ W1,2(Ω) with boundary values ϕ (in Sobolev’s sense, i.e., u − ϕ ∈W1,2
0(Ω)).
The problem is interesting even for the case ϕ ≡ 0. Before proving the exis-
tence of a unique solution, let us observe that
24
[I]
The function u ∈ W1,2(Ω), u − ϕ ∈ W1,2
0(Ω), is minimizing if and only if
∫[
∇u · ∇η + 4αu3η + f (x)η]
dx = 0 EULER-LAGRANGE EQN in weak form
for all test-functions η ∈ W1,2
0(Ω). (Hence a partial integration and the Variational
lemma leads to the Eqn
∆u = 4αu3+ f (x),
provided that u has a second derivatives.)
To see this, note that for any real number ε
I(u + ε η) = I(u) + ε
∫
[∇u · ∇η + 4αu3η + f (x)η] dx +1
2ε2
∫
|∇η|2 dx
+ ε2
∫
α(6u2η2+ 4uη3 ε+η4 ε2) dx
(⇒) Suppose that u is minimizing. The u(x) + ε η(x) is admissible, and
I(u + ε η) ≥ I(u)
by assumption. We must have∫
(∇u · ∇η + 4αu3η + f (x)η) = 0. (If I(u) +
ε J + ε2(. . .) attains its minimum for ε = 0, then J = 0!!!)
(⇐) If Euler’s equation in weak form holds, then
I(u + 1η) = I(u) + 0 +1
2
∫
|∇η|2 dx + α
∫
η2[u2 · 6 + 4uη + η2] dx ≥ I(u),
since |∇η|2 ≥ 0 and 6u2+4uη+η2 ≥ 2u2 ≥ 0. This means that I(u+η) ≥ I(u),
in other words u is minimizing. (Any admissible function v can be written
as v = u + (v − u), v − u = η ∈ W1,2
0(Ω).)
[II]
The minimizing function u is unique (if it exists). Suppose that there are two
solutions, say u1 and u2. Then u1 − u2 and u2 − u1 are in W1,2
0(Ω). Choose the test
25
function η = u2 − u1 in the Euler equation for u2 and the test function u1 − u2 in
the Euler equation for u1. Adding the two equations we get (DO IT)
∫
Ω
|∇u1 − ∇u2|2 dx + 4
∫
Ω
α
≥(u1−u2)2 12
(u21+u2
2)
︷ ︸︸ ︷
(u31 − u3
2)(u1 − u2)︸ ︷︷ ︸
≥0
dx = 0.
Hence the first integral is zero and so ∇u1 = ∇u2 a.e. inΩ . Thus u1−u2 is constant
a.e. in Ω (you may wish to prove this in Sobolev’s space). The boundary values
will force this constant to be zero. This shows that u1 = u2.
[III]
There exists a minimizing function u. We need some estimates to begin with.
Now
∣∣∣∣∣
∫
f (x)v dx
∣∣∣∣∣=
∣∣∣∣∣
∫
f (x)(v − ϕ) dx +
∫
f (x)ϕ dx
∣∣∣∣∣
HOLDER
≤ ‖ f ‖2‖v − ϕ‖2 + ‖ fϕ‖1SOBOLEV
≤ ‖ f ‖2CΩ‖∇(v − ϕ)‖2 + ‖ fϕ‖1≦ ‖ f ‖2CΩ‖∇v‖2 + ‖ f ‖2CΩ‖∇ϕ‖2 + ‖ fϕ‖1= A‖∇v‖2 + B
for all v. This means that
I(v) ≥1
2‖∇v‖2
[
‖∇v‖2 −A
2
]
− B
and so I(v) ≥ −A2
32− B. Hence we have
−∞ < Infv
I(v) ≤ I(ϕ) < +∞.
Observe also that, for example,
‖∇v‖2 ≤ max2 +A
2, B + I(v) ≈ I(v).
By the definition of the infimum, there are admissible functions u1, u2, u3, . . .
such that
limi→∞
I(ui) = I0 = inf I(v).
26
This is called a minimizing sequence. We may assume that
