Fourier and Sobolev JorgeAar˜ao June 2005 These notes are a companion to the lectures I will give in Las Cruces, preparatory for the CBMS conference lectures by Terry Tao. I will not cover all the material in these notes; in particular we will spend little time on sections 1–4. Please read these notes beforehand, concentrating on sections 5, 7, 8, 10, 11, 12, and 13. Most proofs are omitted from this text, since the objective here is to get you acquainted with this material from a user’s perspective. On the other hand, except for the proofs of the main theorems, most proofs are quite straightforward, and you may try your hand at them yourself. Or, consult any book on this subject; I recommend . Topics in functional analysis and applications, by S. Kesavan; . Partial differential equations, by L.C. Evans; . Analysis, by Lieb and Loss. 1
Transcript
Fourier and Sobolev
Jorge Aarao
June 2005
These notes are a companion to the lectures I will give in Las Cruces,preparatory for the CBMS conference lectures by Terry Tao. I will not cover allthe material in these notes; in particular we will spend little time on sections1–4. Please read these notes beforehand, concentrating on sections 5, 7, 8, 10,11, 12, and 13. Most proofs are omitted from this text, since the objective hereis to get you acquainted with this material from a user’s perspective. On theother hand, except for the proofs of the main theorems, most proofs are quitestraightforward, and you may try your hand at them yourself. Or, consult anybook on this subject; I recommend
. Topics in functional analysis and applications, by S. Kesavan;
. Partial differential equations, by L.C. Evans;
. Analysis, by Lieb and Loss.
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1 Introduction
Relatively speaking, Algebra is a new branch of mathematics (as compared toGeometry, Analysis, and Number Theory), and since the emergence of Algebramathematicians have tried to use its power to serve the other areas. It canbe argued that the material we will cover here is part of that program, tryingto view spaces of functions in a more algebraic way, in order to solve partialdifferential equations. Here is a parallel: we want to solve x2 − 1 = 0, and thesolutions are real numbers. But what of x2 + 1 = 0? We have two options.We can declare that the equation has no solutions, or we can declare that theequation has solutions, only these solutions are not real numbers. With thatwe create a new number system, define its (algebraic) rules, and show that theold number system (R) is somehow included in the new one (C). Likewise withdifferential equations. Sometimes, when we take the Fourier transform of adifferential equation, we may obtain that its solution u is such that, say, u = 1,an equation that is not satisfied by any function. We have two options. We candeclare that the differential equation has no solutions, or we can declare that ithas solutions, only those solutions are not functions anymore. It is then our jobto provide a framework for that new space of solutions, complete with algebraicstructure, and making explicit how to include the old structure into the newstructure. We start from the beginning.
2 Smooth functions, and Lp functions
We let Ω be a non-empty open set of Rn; K ⊂ Ω will always denote a compactset. We define the multi-index α = (α1, . . . , αn) to be an n-tuple of non-negative integers (n is the dimension of Rn!). The order of α is the integer|α| = α1 + · · ·+ αn. If x = (x1, . . . , xn) ∈ Rn, we write xα = xα1
1 . . . xαnn , and
Dα =∂|α|
∂x1α1 . . . ∂xn
αn.
A function u : Ω → C is said to be smooth if all of its partial derivatives,of any order, exist and are continuous. This space is denoted by C∞(Ω). Asmooth function u : Rn → C is said to be in the Schwartz space S(Rn) if forany multi-indices α and β there is a constant C = C(u, α, β) such that
supx∈Rn
∣∣xβDαu(x)∣∣ < C.
Exercise 1 Show that f(x) = exp(−|x|2) is in S(Rn).
A function f : Ω → C is said to be compactly supported if there isa compact K ⊂ Ω (where K may depend on f) such that f(x) = 0 for allx ∈ Ω\K. The support of f is the intersection of all compact sets K as above,and is denoted by supp f . The set of all such smooth, compactly supportedfunctions (defined over the same Ω) is denoted by C∞c (Ω).
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Exercise 2 Show that C∞c (Ω) contains more than one function.
From now on, we let p be a real number with 1 ≤ p < ∞; on occasion wewill allow p = ∞ as well.
For 1 ≤ p < ∞, the space Lp(Ω) consists of those functions f : Ω → C suchthat
(‖f‖p)p =∫
Ω
|f(x)|p dx < ∞.
The equality defines the symbol ‖f‖p, the norm of f in Lp(Ω). The integralwe used is the Lebesgue integral.
If two functions f and g in Lp(Ω) are such that ‖f − g‖p = 0, then we justidentify f and g. In that situation it can be shown that there is a set A ⊂ Ω,of measure zero, such that f(x) = g(x) for x ∈ Ω \ A; in this case we say thatf(x) = g(x) almost everywhere (a.e.).
When p = ∞ we define the space L∞(Ω) as consisting of those functions f :Ω → C which are equal almost everywhere to a bounded function g; equivalentlythere is a constant C = C(f) and a set A of measure zero such that
‖f‖∞ = supx∈Ω\A
|f(x)| < C.
Exercise 3 Let Ω be bounded, containing the origin; let 1 ≤ p < ∞. Show thatf(x) = |x|k is in Lp(Ω) if and only if k > −n/p.
Exercise 4 Show that if 1 ≤ p < ∞, then C∞c (Ω) is dense in Lp(Ω). Whathappens if p = ∞?
Exercise 5 Show that if 1 ≤ p ≤ ∞, then S(Rn) is contained in Lp(Rn).
Exercise 6 Let 1 ≤ p < q ≤ ∞. Show that, in general, neither of Lp(Ω) andLq(Ω) is contained in the other. On the other hand, show that if q < ∞, thenLp(Ω) ∩ Lq(Ω) is dense in both. What happens if q = ∞?
Exercise 7 Suppose Ω is bounded, and that 1 ≤ p < q < ∞. Show that Lq(Ω)is contained in Lp(Ω). Let i(f) = f be the inclusion map. Is it continuous?That is, is there a constant C (independent of f) such that ‖f‖p ≤ C‖f‖q?Prove, or give counter-examples. What happens if q = ∞?
Each Lp-space comes with a local version. Let 1 ≤ p < ∞. We say that fbelongs to Lp
loc(Ω) if for each compact K in Ω there is a constant C = C(K, f)such that
∫K|f(x)|p dx < C. Clearly Lp(Ω) is contained in Lp
loc(Ω).
Exercise 8 If 1 ≤ p < q < ∞, then Lqloc(Ω) ⊂ Lp
loc(Ω).
The above result shows that L1loc(Ω) is, in a sense, the most important of
the local integrability spaces. We will return to this space in the future.
Exercise 9 Is it true that⋃
p>1 Lploc(Ω) = L1
loc(Ω)?
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We have the following notions of convergence in Lploc(Ω): we say that fm
converges strongly to f in Lploc(Ω) if and only if for every compact K in Ω
we have that fm converges to f in Lp(K) (in the usual Lp sense).We say that fm converges weakly to f in Lp
loc(Ω) if and only if for everycompact K in Ω fm converges to f weakly in Lp(K). In practical terms, thismeans that if 1 < p, then
∫
K
fm(x) g(x) dx −→∫
K
f(x) g(x) dx
for every g in Lp′(K) (where 1p + 1
p′ = 1), and if p = 1, then the same integralconverges for every bounded g.
Exercise 10 Let fm(x) = sin(mx), f(x) ≡ 0. Show that fm converges weakly,but not strongly, to f in L2
loc(R).
3 Some tools of the trade
Let 1 < p < ∞, and let p′ be the conjugate exponent of p, that is, 1p + 1
p′ = 1.If f is in Lp(Ω) and g is in Lp′(Ω), then fg is in L1(Ω), and we have Holder’sinequality
∫
Ω
|f(x) g(x)| dx ≤(∫
Ω
|f(x)|p dx
) 1p
(∫
Ω
|g(x)|p′ dx
) 1p′
,
or simply ‖fg‖1 ≤ ‖f‖p ‖g‖p′ .If 1 ≤ p ≤ ∞, and both f and g are in Lp(Ω), Minkowski’s inequality
states that‖f + g‖p ≤ ‖f‖p + ‖g‖p.
Now let Ω be Rn, and let p and p′ be conjugate exponents. Let f be inLp(Rn), and g be in Lp′(Rn). Note that, if x ∈ Rn is fixed, then the functiony 7→ f(x− y) is such that ‖f‖p = ‖f(x− ·)‖p.
The convolution of f and g, given by
(f ∗ g)(x) =∫
Rn
f(x− y)g(y) dy,
is well-defined, and by Holder’s inequality we have ‖f ∗ g‖∞ ≤ ‖f‖p ‖g‖p′ .Observe that f ∗ g = g ∗ f .
Exercise 11 Show that supp (f ∗ g) ⊂ supp f + supp g, where the sum of twosets is defined in the standard way. In particular, if f and g are compactlysupported, then so is f ∗ g.
Exercise 12 Show that if f ∈ Lp(Rn) and g ∈ Lp′(Rn), then f∗g is continuous.Is it true that lim|x|→∞(f ∗ g)(x) = 0?
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The convolution of f and g is defined for every x ∈ Rn under the abovehypotheses. We may extend this definition in many ways, and pay only a smallprice.
Exercise 13 If f and g are in L1(Rn), then ‖f∗g‖1 ≤ ‖f‖1 ‖g‖1. In particular,(f ∗ g)(x) is finite a.e. Is it true that lim|x|→∞(f ∗ g)(x) = 0?
And here is Young’s inequality:
Theorem 1 If f ∈ Lp(Rn), g ∈ Lq(Rn) and 1p + 1
q = 1 + 1r , then
‖f ∗ g‖r ≤ ‖f‖p ‖g‖q.
In particular, if q = 1 then ‖f ∗ g‖p ≤ ‖f‖p ‖g‖1.One of the nice things about the convolution is that it inherits the good
differentiability properties of f and g.
Exercise 14 Let 1 ≤ p ≤ ∞. If f ∈ Lp(Rn) and g ∈ S(Rn), then f ∗ g ∈C∞(Rn), and Dα(f ∗ g) = f ∗ Dαg. In addition, if Dαf is in Lp(Rn), thenDα(f ∗ g) = (Dαf) ∗ g.
An approximation of the identity, or mollifier, is a one-parameter fam-ily of functions φt : Rn → R with the following properties:
1) For all t > 0 we have∫Rn φt(x) dx = 1;
2) For all δ > 0 φt converges uniformly to zero outside a ball of radius δcentered at the origin, as t goes to zero;
3) φt(x) ≥ 0 for all x.For example, if φ ≥ 0 is a radially symmetric function decaying to zero as x
approaches infinity, and with total area equal to 1, then φt(x) = 1tn φ(x/t) is a
mollifier. More specifically, we may take φ(x) = exp(−|x|2/2)/(2π)n/2, and inmany cases we want to take φ to be in C∞c (Ω).
Theorem 2 Let 1 ≤ p < ∞, let f belong to Lp(Rn), and let φt ∈ C∞(Rn) be amollifier. Then ft = f ∗ φt converges to f in Lp(Rn), as t goes to zero.
The result in the above theorem is very useful; for instance, you might wantto revisit some of the previous exercises in light of this theorem. On the otherhand, approximation in Lp in principle only tells us that for some sequencetm 0 we have ftm(x) → f(x) a.e. It can be shown that more is true, namelythat ft(x) → f(x) a.e. (if the mollifier is nice), but we won’t do it here.
Exercise 15 Show that if f is continuous and bounded, then ft converges to funiformly on compact sets, as t goes to zero. In particular ft(x) converges tof(x), for every x.
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4 The Fourier Transform
Let f be a function in L1(Rn). The Fourier Transform of f is a functionf : Rn → C, given by
f(ξ) =∫
Rn
f(x) e−2πix·ξ dx.
Here x ·ξ is the usual dot product in Rn. Notice that f(ξ) is defined for every ξ,and in fact ‖f ‖∞ ≤ ‖f‖1. So the Fourier Transform is a continuous operatorfrom L1(Rn) to L∞(Rn).
Exercise 16 (Riemann-Lebesgue Lemma) If f ∈ L1(Rn), then f is con-tinuous, and lim|ξ|→∞ f(ξ) = 0.
Exercise 17 Let 1[−1,1] be the function that is equal to 1 if −1 ≤ x ≤ 1, andzero otherwise. Compute its Fourier transform, and conclude that 1[−1,1] is notin L1(R).
Theorem 3 For t > 0 let gt(x) = exp(−πt|x|2), for x ∈ Rn. Then gt(ξ) =t−n/2 exp(−π|ξ|2/t).
Observe that gt in Theorem 3 is a mollifier.
Theorem 4 If f and g are in L1(Rn), then (f ∗ g)(ξ) = f(ξ) g(ξ).
Theorem 6 (Fourier Inversion Theorem) If f is in S(Rn), then for all x
f(x) =∫
Rn
f(ξ) e2πiξ·x dξ.
Proof : With the same gt as in Theorem 3, and using the result of Exercise 15,we have∫
Rn
f(ξ) e2πix·ξ dξ =∫
Rn
[∫
Rn
f(y) e−2πiξ·y dy
]e2πix·ξ dξ
= limt→0+
∫
Rn
[∫
Rn
f(y) e−2πiξ·y dy
]e2πix·ξ e−πt|ξ|2dξ
= limt→0+
∫
Rn
f(y)[∫
Rn
e−πt|ξ|2e−2πi(y−x)·ξ]
dy
= limt→0+
∫
Rn
f(y)tn/2e−π|x−y|2/t dy
= limt→0+
(f ∗ gt)(x) = f(x).
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¤
Exercise 18 (Fourier Inversion Theorem, 2) Show that if f is in L1(Rn),then for almost every x we have
f(x) = limt→0+
∫
Rn
f(ξ) e2πiξ·x e−πt|ξ|2 dξ.
Theorem 7 If f and g are in S(Rn), then∫
Rn
f(y) g(y) dy =∫
Rn
f(y) g(y) dy.
In particular ‖f‖2 = ‖f‖2.
Theorem 8 Let fm be a sequence in S(Rn), let f be in L2(Rn), and supposethat ‖fm − f‖2 → 0. Then fm is a Cauchy sequence in L2(Rn), and thereforehas a limit in L2(Rn). We extend the definition of the Fourier Transform toL2(Rn) by defining f to be this limit. From the results of previous theorems andexercises the Fourier Transform is an invertible isometry from L2(Rn) to itself.Moreover
f(ξ) = limR→∞
∫
|x|≤R
f(x) e−2πix·ξ dx,
f(x) = limR→∞
∫
|ξ|≤R
f(ξ) e2πix·ξ dξ,
where both limits are in the L2 sense.
It can be shown (but it is much harder) that the equations above hold foralmost every x.
Exercise 19 Let 1 < p < 2. Show that any function f in Lp(Rn) may bewritten (non-uniquely) as f = f1 + f2, with fj in Lj(Rn), j = 1, 2. Definef = f1 + f2. Show that f is well defined (that is, its definition is independentof the particular way of splitting f into f1 and f2).
The previous exercise shows that it is possible (and easy) to define the FourierTransform over any Lp(Rn), 1 ≤ p ≤ 2. It is possible to obtain a better (andharder) result, namely that the Fourier Transform maps Lp(Rn) continuouslyinto Lp′(Rn), where 1 ≤ p ≤ 2 and p′ is the conjugate exponent of p.
What if p > 2? The following is a standard example.
Exercise 20 Let t = a+ ib, with a > 0, and define gt(x) = exp(−πt|x|2). Showthat if p > 2 and 1 < q < 2, then ‖gt‖q/‖gt‖p is not bounded (in t).
As a consequence the Fourier Transform cannot map Lp(Rn) boundedly intoany Lq(Rn), if 1 < q < 2 < p. As mentioned in the Introduction, we can give up
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now, or we can try to enlarge the domain of definition of the Fourier Transform(and its range, as well).
Let’s look at an application of the Fourier Transform. Suppose f is smoothand compactly supported, and we want to solve the equation ∆u = f on Rn,n ≥ 3. Assuming the solution u is a well-behaved function, we take the FourierTransform of both sides to obtain−|2πξ|2 u = f , and so u = −f/|2πξ|2. Becausef is in S(Rn), we know that f is also in S(Rn), and as a consequence the functionf/|2πξ|2 is in L1(Rn) (n ≥ 3!). Consequently, we may now define u to be
u(x) =∫
Rn
−f(ξ)|2πξ|2 e2πiξ·x dξ,
and go on to show that this u indeed solves the problem ∆u = f . But let mecall your attention to Theorem 4. Since u = −f/|2πξ|2, it is natural to thinkthat the solution u will be given by u = f ∗G, where G is a function such thatG = −1/|2πξ|2, except that G is not in any Lp(Rn). Again, we can either giveup, or . . .
5 Distributions
Here is what we want to accomplish: we want to enlarge the spaces Lp(Ω) toinclude objects that are not necessarily in Lp(Ω). We wrote ‘objects’ because inprinciple they may not even be functions – but we want our old Lp-spaces to benaturally included in this new space. The motivation we presented in the lastsection was that this new space would allow us to broaden the definition of theFourier Transform, but we will accomplish much more than that. Here is someextra motivation.
Given any f in L2(Ω) the mapping φ 7→ ∫Ω
φf dx is a bounded linear func-tional. The converse is true.
Theorem 9 (Riesz Representation) If T : L2(Ω) → C is a bounded linearfunctional, then there is a unique f ∈ L2(Ω) such that
T (φ) =∫
Ω
φ(x)f(x) dx
for every φ in L2(Ω).
So we can think of the bounded linear functionals on L2(Ω) as being exactlythe very functions in L2(Ω).
Exercise 21 Take Ω = [0, 1] ⊂ R. Show that if φ is in L3([0, 1]), then it is inL2([0, 1]), and ‖φ‖2 ≤ ‖φ‖3. Show that there is a function f not in L2([0, 1])such that T (φ) =
∫ 1
0φ(x)f(x) dx is a bounded linear functional on L3([0, 1]).
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What this exercise shows is that by restricting our function space (fromL2 to L3), we enlarged our dual space (that is, the space of bounded linearfunctionals). But in any case this enlarged space of functionals still contained,in a sense, the original space L2. So here is the idea: we will restrict thefunction space as much as possible, and correspondingly enlarge the dual space.The elements of this enlarged dual will be our distributions. We give yet anotherexample, then move to the theory itself.
Exercise 22 Let H = C∞c (Rn) ⊂ L2(Rn) be endowed with the following norm:if φ is in H, then ‖φ‖2H =
∑|α|≤1 ‖Dαφ‖22. Show that there is a bounded linear
functional T on H (that is, |Tφ| ≤ C‖φ‖H) that is not representable by anyfunction f in L2(Rn).
A function φ : Ω → C will be called a test function if it belongs to C∞c (Ω).We introduce a notion of convergence on this space: A sequence of test functionsφm will converge to a test function φ if and only if there is some fixed compactset K in Ω such that the supports of all φm and of φ are all contained in K,and moreover for every multi-index α we have that Dαφm converges uniformly(in K) to Dαφ. With this notion of convergence, the space of test functions isdenoted by D(Ω).
Notice that D(Ω) is contained in every space Lp(Ω), and moreover, if φm
converges to φ in D(Ω), then φm converges to φ in Lp(Ω) as well. (And all ofits derivatives converge in Lp(Ω) as well, to the corresponding derivative of φ.)As a consequence we expect that, in some sense, all spaces Lp(Ω) should becontained in the dual of D(Ω), and this inclusion map should be continuous.
The (topological) dual of D(Ω) is the space of distributions, and is denotedby D′(Ω). Its elements (the distributions themselves) are continuous linearfunctionals T : D(Ω) → C; that is:
1) T (φ1 + φ2) = T (φ1) + T (φ2),
2) T (λφ) = λT (φ),
3) T (φm) → T (φ) whenever φm → φ in D(Ω).
Exercise 23 (Dirac delta-function) If x0 ∈ Ω, show that δx0(φ) = φ(x0) isa distribution. Show that there is no function f in L1
loc(Ω) such that for all φin D(Ω)
δx0(φ) =∫
Ω
φ(x) f(x) dx.
(Since L1loc contains all other types of Lp, we are saying that this distribution
can’t be realized as an integrable function.) If x0 = 0, we simply write δ = δ0.
Exercise 24 If x0 ∈ Ω and α is any multi-index, show that Tx0,α(φ) = Dαφ(x0)is a distribution.
Exercise 25 Let f be in L1loc(Ω). Show that φ 7→ Tf (φ) =
∫Ω
fφ dx is a distri-bution.
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Theorem 10 If f and g are in L1loc(Ω) and they determine the same distribu-
tions, then f(x) = g(x) a.e.
Functions in L1loc are not the most general objects that can be viewed as
distributions.
Exercise 26 Let µ be a measure with bounded total variation. Then Tµ(φ) =∫φdµ defines a distribution.
We say that the distributions Tm converge to a distribution T if for eachφ in D(Ω) we have Tm(φ) → T (φ).
Exercise 27 Show that the map f 7→ Tf from L1loc(Ω) to D′(Ω) is continuous,
that is, if fm → f weakly, then Tfm → Tf .
Exercise 28 Let φt be a mollifier. Show that Tφt converges to δ.
Next we define what we mean by the derivative of a distribution. Thedefinition is such that the formula for integration by parts is still valid.
Let T be a distribution, and α be a multi-index. The derivative DαT isitself a distribution, given by
(DαT )(φ) = (−1)|α| T (Dαφ).
The key to understanding this definition is to see that the integration byparts formula still works in some sense, as we pointed out before. So, for in-stance, if f is a locally integrable function such that f ′ is also a locally integrablefunction, then for any φ in D(R) we have
−∫
Rf ′ φdx =
∫
Rf φ′ dx.
Now, if f ′ is not locally integrable, the left hand side makes no sense, but theright hand side still does. So we extend the meaning of the left hand side:instead of f ′ we write (Tf )′ (the derivative of the distribution determined by f),and the formula becomes
−(Tf )′(φ) =∫
Rf φ′ dx = Tf (φ′).
Exercise 29 Let f be a smooth function on Ω. Justify the equality
(−1)|α|∫
Ω
(Dαφ) f dx =∫
Ω
(Dαf) φdx,
and thus verify that DαTf = TDαf .
Exercise 30 Show that DαT is indeed a distribution; that is, show it is linearand continuous.
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Exercise 31 Working over R, let H(x) be 1 if x ≥ 0, and 0 if x < 0. (H iscalled the Heaviside function.) Let TH be the distribution determined by H.Show that the derivative of the distribution TH is the delta function δ0.
From now on we will abandon the cumbersome notation Tf to representthe distribution determined by the function f , and simply write f , unless itis imperative to distinguish between the two. We will also write distributionalderivative to indicate that the derivative is being viewed as a distribution. Theprevious exercise, rephrased, says that the distributional derivative of H is δ0.
Next we want to define the Fourier Transform of a distribution, but it turnsout that not every distribution will have a Fourier Transform. The reason liesin the following result.
Exercise 32 If f and g are in S(Rn), then∫
Rn
f(z) g(z) dz =∫
Rn
f(z) g(z) dz.
Interpreted in terms of distributions determined by functions, this exercise wantsto say that if f is a distribution, then f is a distribution given by
f(g) = f(g).
This would indeed be the case, except for the fact that we can’t have both g andg in D(Rn) (by Heisenberg’s principle). So instead we look at distributions thatare defined on S(Rn); those will be called tempered distributions. Here is howit is done.
Define the following notion of convergence in S(Rn): fm ∈ S(Rn) convergesto f ∈ S(Rn) if and only if for any compact set K and any multi-index αthe sequence Dαfm converges uniformly to Dαf in K. With this notion ofconvergence the inclusion D(Rn) → S(Rn) is continuous. We denote by S ′(Rn)the (topological) dual space of S(Rn), that is, continuous linear functionalsf : S(Rn) → C. The elements of S ′(Rn) are called tempered distributions;they are distributions since there is a continuous inclusion S ′(Rn) → D′(Rn).
Let T be a tempered distribution. We define the Fourier Transform of Tas being the (tempered) distribution T given by
T (φ) = T (φ ),
for all φ in S(Rn).
Exercise 33 Show that T is a tempered distribution.
Thanks to Exercise 32, we know that Tf (φ) = Tf (φ), showing that thedefinition of Tf is compatible with the definition of f .
Exercise 34 Show that the delta function δ0 is a tempered distribution, andthat δ0 = 1 (the distribution determined by the constant function 1). Also,1 = δ0.
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Exercise 35 Show that if f is locally integrable, and grows no faster than apolynomial at infinity, then Tf is a tempered distribution.
Exercise 36 Give an example of a distribution that is not a tempered distribu-tion.
Exercise 37 Show that if T is a tempered distribution, then DαT is a tempereddistribution.
6 A bit of fun
In this short section let’s pretend we are Euler. Say f : R→ C is a 2π-periodicfunction (it is your job to put extra hypotheses here). Then
f(ξ) =∫
Rf(x) e−2πixξ dx.
Of course, the integral is not convergent, but that never stopped Euler, and itshouldn’t stop us. We compute
f(ξ) =∞∑
j=−∞
∫ 2π(j+1)
2πj
f(x) e−2πixξ dx
=∞∑
j=−∞
∫ 2π
0
f(y) e−2πiyξ dy e−2πiξ(2πj)
=∫ 2π
0
f(y) e−2πiyξ dy ·∞∑
j=−∞e−2πiξ(2πj)
= g(ξ) ·∞∑
j=−∞e−2πiξ(2πj).
Well, g is a function, and we want to know what that sum is. Writing η = 4π2ξ,we get
∞∑
j=−∞e−iηj = lim
r→1−
∞∑
j=−∞r|j|e−iηj .
Since r approaches 1 from below, we can actually sum the last series (it is thesum of two convergent geometric series).
Exercise 38 Show that∞∑
j=−∞r|j|e−iηj =
1− r2
1− 2r cos η + r2= 2πPr(η),
where the last equality defines Pr, the Poisson kernel. Show that Pr has thefollowing properties:
12
1. Pr(η) ≥ 0;2. Pr is 2π-periodic in η;3.
∫ π
−πPr(η) dη = 1;
4. For all ε ∈ (0, π), Pr converges uniformly to zero on the set ε < |η| < π;5. Pr(0) →∞ as r → 1−.
In other words, as r goes to 1, Pr behaves like an approximation of theidentity on each interval of the form [2πj − π, 2πj + π]. Therefore
∞∑
j=−∞e−iηj = 2π
∞∑
j=−∞δ(η − 2πj).
Thus we obtain
f(ξ) = 2π g(ξ)∞∑
j=−∞δ(4π2ξ − 2πj).
Using the inversion formula, we have
f(x) =∫
Rf(ξ) e2πixξ dξ
=∞∑
j=−∞2π
∫
Rg(ξ)δ(4π2ξ − 2πj) e2πixξ dξ
=∞∑
j=−∞
12π
∫
Rg
(z + 2πj
4π2
)δ(z) eixje
ixz2π dz
=∞∑
j=−∞
12π
g(j/2π) eixj .
In other words, we obtain the Fourier series expansion of f , by way of theFourier transform of f , and distributions.
Exercise 39 Justify the steps above.
7 Sobolev Spaces
Some distributions are representable by functions, and when that happens thefunction is uniquely defined a.e., as we saw. We are now concerned with func-tions whose distributional derivatives are representable by functions. Here is anexample.
Exercise 40 On R, let f(x) = max0, x. Show that the distributional deriva-tive f ′ is representable by the Heaviside function.
13
When a distributional derivative DαT is representable by a function in Lp(Ω)we will simply write DαT ∈ Lp(Ω), identifying the distribution with its repre-sentation, and denoting both by the same symbol. In this situation the symbol‖DαT‖p will denote the Lp-norm of the function which represents DαT .
Let 1 ≤ p ≤ ∞. The Sobolev space W k,p(Ω) consists of those functions fin Lp(Ω) such that all the distributional derivatives of f of order at least k arealso in Lp(Ω), or
W k,p(Ω) = f ∈ Lp(Ω) ; Dαf ∈ Lp(Ω) for all |α| ≤ k .
For 1 ≤ p < ∞, we put the following norm on W k,p(Ω):
‖f‖W k,p(Ω) =
∑
|α|≤k
‖Dαf‖pp
1/p
.
We have that D(Ω) ⊂ W k,p(Ω) ⊂ Lp(Ω), and ‖f‖p ≤ ‖f‖W k,p(Ω). Moreover,the first inclusion (as well as the second) is continuous.
Exercise 41 (Alternative definition) Denote by H the subspace of Lp(Ω)consisting of smooth functions f such that for all multi-indices α we have Dαfalso in Lp(Ω). Endow H with the norm ‖f‖H = (
∑|α|≤k ‖Dαf‖p
p)1/p. Show
that the completion of H in this norm is W k,p(Ω).
We may also define local Sobolev spaces W k,ploc (Ω); for these spaces we
have notions of convergence (both strong and weak), but not a norm.For 1 ≤ p < ∞, the space W k,p(Ω) has an important subspace, denoted by
W k,p0 (Ω). This space is the completion of D(Ω) in W k,p(Ω) with respect to the
norm of W k,p(Ω). Hence f ∈ W k,p(Ω) is in W k,p0 (Ω) if and only if there is a
sequence of functions fm in D(Ω) such that ‖fm − f‖W k,p(Ω) → 0.
Exercise 42 If 1 ≤ p < ∞ and Ω 6= Rn, then W k,p0 (Ω) ( W k,p(Ω).
Theorem 11 If 1 ≤ p < ∞, then W k,p0 (Rn) = W k,p(Rn).
Proof : We want to show that for any f ∈ W k,p(Rn) there is a sequence fm inD(Rn) such that fm → f in W k,p(Rn).
It is easy to obtain a sequence of smooth functions in Lp with all the con-vergence properties we want, except that they are not compactly supported.Indeed, let φt be a smooth mollifier; then φt is in L1(Rn), and by Young’s in-equality ft = f ∗ φt is in Lp(Rn). Moreover, for all multi-indices α with |α| ≤ kwe have Dαft = (Dαf) ∗ φt ∈ Lp(Rn). Using standard properties of mollifiers,ft → f and Dαft → Dαf in Lp(Rn), so that ft converges to f in W k,p(Rn). Asmentioned before, these ft are not compactly supported.
14
Now pick a function η ∈ D(Rn) such that 0 ≤ η(x) ≤ 1 for all x, η(x) ≡ 1if |x| ≤ 1, and η(x) ≡ 0 for |x| ≥ 2. Define gt(x) = ft(x)η(tx). For any multi-index α we have Dαgt(x) = Dαft(x) if |x| ≤ 1/t, and Dαgt(x) ≡ 0 if |x| ≥ 2/t.Thus, with |α| ≤ k,
‖Dαft −Dαgt‖pp =
∫
|x|>1/t
|Dαft(x)−Dαgt(x)|p dx.
Now comes the technical part of the proof. We may write Dαgt in the form
Dαgt =∑
|β|≤|α|ϕβ,t(x)Dβft(x) = η(tx) Dαft +
∑
|β|<|α|ϕβ,t(x)Dβft(x),
where the functions ϕβ,t depend on η as well. For example, if α = (1, 0, 1, 1),β1 = (0, 1, 0, 0), and β2 = (1, 0, 1, 0), then ϕβ1,t(x) ≡ 0, and ϕβ2,t(x) = t ∂η
∂x4(tx).
In any case, it is clear that the functions ϕβ,t (and there are finitely many ofthem) are all bounded by the same constant, and in fact since t will be goingto zero, they will converge to zero as well. Thus when we write
Dαft(x)−Dαgt(x) = Dαft(x) (1− η(tx))−∑
|β|<|α|ϕβ,t(x) Dβft(x),
we obtain, for some constant C which depends only on p, that
‖Dαft −Dαgt‖pp ≤ C
∫
|x|>1/t
2p |Dαft(x)|p +∑
|β|<|α||ϕβ,t(x)|p|Dβft(x)|p dx.
As t goes to zero, this integral goes to zero. ¤
8 Hk(Ω), particularly Hk(Rn)
The case p = 2 is special, since W k,2(Ω) is a Hilbert space; for that special casewe use the notation W k,2(Ω) = Hk(Ω).
Now let Ω = Rn. Thanks to the fact that the Fourier Transform is anisometry on L2(Rn), we have that
‖Dαf‖2 = ‖Dαf‖2 = (2π)|α|‖ξαf‖2,
and therefore
‖f‖2Hk(Rn) =∑
|α|≤k
(2π)|α|∫
Rn
|ξα|2 |f(ξ)|2 dξ
=∫
Rn
∑
|α|≤k
(2π)|α||ξα|2 |f(ξ)|2 dξ.
15
Exercise 43 There are constants M1 and M2, depending only on n and k, suchthat
M1(1 + |ξ|2)k ≤∑
|α|≤k
(2π)|α||ξα|2 ≤ M2(1 + |ξ|2)k.
Theorem 12 (Equivalence of norms) The norm
(∫
Rn
(1 + |ξ|2)k |f(ξ)|2 dξ
)1/2
is equivalent to the norm of Hk(Rn).
Henceforth we will use both norms interchangeably, and all bounds will beup to a multiplicative constant when translating into the other norm.
Theorem 13 (Transform characterization of Hk) Suppose that f ∈ L2(Rn),and that ∫
Rn
(1 + |ξ|2)k |f(ξ)|2 dξ
is finite. Then f is in Hk(Rn).
Proof : For each multi-index α with |α| ≤ k we consider the distributionalderivative Dαf . From the hypotheses, ‖ξαf‖2 is finite, and so there is a functiong ∈ L2(Rn) such that g = (−2πi)|α|ξαf . We claim that Tg = Dαf :
(Dαf)(φ) = (−1)|α|Tf (Dαφ) = (−1)|α|∫
Rn
f(x) Dαφ(x) dx
= (−1)|α|∫
Rn
f(ξ)Dαφ(ξ) dξ = (−1)|α|∫
Rn
f(ξ)(2πi)|α|ξαφ(ξ) dξ
=∫
Rn
g(ξ) φ(ξ) dξ =∫
Rn
g(x)φ(x) dx. ¤
This theorem suggests the definition of Sobolev spaces for non-integer in-dices. We define, for s ≥ 0,
Notice that s < t implies that Hs ⊃ Ht. We will come back to these spaceslater.
16
9 A word from the wise
At this point I can’t do better than to give you the gist of what is on pages56–57 in Kesavan’s book Topics in Functional Analysis and Applications.
Approximation theorems. The idea is to prove results for smooth functions,and obtain these results for Sobolev functions via density, so one object of studyis how to approximate functions in W k,p(Ω) by smooth functions. We got a tasteof that in Theorem 11.
Extension theorems. Some results are easier to prove over the whole of Rn;to prove them for Ω extend the setting from Ω to Rn, use the result there, thenrestrict it back again to Ω. For this to work we need to know if we can performthe extension, and if the extension, as a map from W k,p(Ω) to W k,p(Rn), iscontinuous.
Imbedding theorems. It turns out that many Sobolev spaces are naturallycontained in other well-known spaces, the reason being that the more integra-bility we require for the derivatives of a function f , the smoother f shouldbe.
Compactness theorems. Even better, some of the inclusions of Sobolev spacesinto other spaces are compact (in the functional analysis sense).
Trace theory. Sobolev spaces are often the natural spaces where to find so-lutions to partial differential equations, but these equations often come withboundary data. It becomes important to understand what is meant by restrict-ing a Sobolev function to the boundary of the domain. This restriction is calledthe trace of the function, and the goal of trace theory is to give meaning tothis restriction.
10 A warm-up
From now to the end our guiding light is the following principle: If a functionf has many distributional derivatives in Lp (that is to say, if f is in W k,p forlarge k), then f itself has to have a decent degree of smoothness (in the classicalsense). As a warm-up we have a famous result.
Theorem 14 Let I ⊂ R be an open interval, and suppose f ∈ W 1,1(I). Thenf is absolutely continuous.
Proof : We should have written that there is an absolutely continuous functiong such that g(x) = f(x) a.e., but we will gloss over such technicalities, unlessthey are really necessary. So fix x0 ∈ I and define
v(x) =∫ x
x0
f ′(x) dx,
which is absolutely continuous by definition. The Fundamental Theorem of Cal-culus asserts that v′(x) = f ′(x) a.e., and therefore in the sense of distributions
17
(f − v)′ = 0. After you do the next exercise, we will be able to conclude thatf(x) = v(x) + c a.e., for some constant c. ¤
Exercise 44 If T is a distribution on D(I), and T ′ = 0, then T = Tc for someconstant c.
Exercise 45 Let I ⊂ R be an open interval, and suppose f ∈ W k,1(I). Thenf ∈ Ck−1(I), and f (k−1) is absolutely continuous.
A consequence of Theorem 14 is that W 1,1(I) ⊂ C(I), the continuous func-tions on I. What sort of inclusion is that? Continuity of this inclusion wouldmean that ‖f‖∞ ≤ C ‖f‖W 1,1(I), but for this to make sense we’d better take Ito be a bounded interval. In that case W 1,p(I) ⊂ W 1,1(I) for 1 ≤ p < ∞, solet’s look at the inclusion j : W 1,p(I) → C(I). Fix y ∈ I. For any f ∈ W 1,p(I)we have
f(x) = f(y) +∫ x
y
f ′(t) dt,
and so|f(x)| ≤ |f(y)|+
∫
I
|f ′(t)| dt.
Integrating in y over I, and with |I| denoting the length of I, we obtain
|I| |f(x)| ≤∫
I
|f(y)| dy + |I|∫
I
|f ′(t)| dt.
For p = 1 we immediately obtain, for a constant C depending only on I,that ‖f‖∞ ≤ C‖f‖W 1,1(I). For p > 1 we first apply Holder’s inequality to theright-hand-side integrals with p and p′ to obtain ‖f‖∞ ≤ C‖f‖W 1,p(I), with aconstant C that depends on I and p. In any case the inclusion j is continuous.But we can say more about this inclusion. Look at the unit ball in W 1,p(I),
B = f ∈ W 1,p(I) ; ‖f‖W 1,p(I) ≤ 1 .
From the continuity of j we conclude that j(B) is a uniformly bounded set inC(I). Now, if p > 1 then |f(x) − f(y)| ≤ ‖f‖W 1,p(I) |x − y|1/p′ ≤ C|x − y|1/p′ .We conclude that j(B) is equicontinuous in C(I). Applying the Arzela-AscoliTheorem, we have just proved the following:
Theorem 15 If 1 < p < ∞, then the inclusion j : W 1,p(I) → C(I) is compact.
Thus any sequence fm bounded in W 1,p(I) has a subsequence fmlwhich
converges uniformly to a function f ∈ C(I). Pretty neat.If p = 1 the result is not true, as the example fm(x) = xm on I = (0, 1)
shows. Here equicontinuity fails to hold. Also, notice that for p > 1 thissequence fm is not bounded in W 1,p(I).
Notice another by-product of Theorem 14: If I = (a, b) and f ∈ W 1,1(I),then it makes sense to talk about f(a) and f(b). This is an example of a result
18
in trace theory. A consequence of this result is that f may be extended fromI to the whole line; if we extend it carefully, the extension procedure will giveus a continuous map from W 1,p(I) to W 1,p(R). Here is how this can be done.Given f ∈ W 1,p(I), I = (a, b), we define an extension f ∈ W 1,p(R) as follows:
f(x) =
0, x ≤ a− 1;
u(a) (x− a + 1), a− 1 ≤ x ≤ a;
u(x), a ≤ x ≤ b;
u(b) (b + 1− x), b ≤ x ≤ b + 1;
0, x ≥ b + 1.
Exercise 46 Show that f 7→ f defines a continuous map W 1,p(I) → W 1,p(R).
Exercise 47 Show that extension operators as defined above are not unique.
Finally, here is another way of looking at Theorem 14. It says that if westart with a function f that has a certain level of integrability p > 1, and forwhich f ′ also has that level of integrability, then f indeed has a higher level ofintegrability q > p. (But the theorem says nothing about improved integrabilityfor f ′.) This point of view will be a feature of the theory, that is, a functionwith a certain integrability p, and whose derivatives up to order k also havethat integrability, actually has a level of integrability q > p. This q depends onmany factors, including p, k, and the dimension n. So there is a trade-off: Ifwe are willing to give up some differentiability of f , we may view f as being ina higher integrability space, in the sense that if f ∈ W k,p, then f ∈ W 0,q forsome q > p.
11 Imbedding theorems
In contrast with other sections, here we state the main results first. In whatfollows, if 1 ≤ p < n, we define the exponent p∗ by
1p∗
=1p− 1
nor p∗ =
np
n− p,
and observe that p∗ > p.
Theorem 16 (Sobolev’s inequality) Let 1 ≤ p < n. Then the inclusionW 1,p(Rn) → Lp∗(Rn) is continuous. In other words, there is a positive constantC = C(p, n) such that for any f ∈ W 1,p(Rn) we have
‖f‖p∗ ≤ C ‖f‖W 1,p(Rn).
Theorem 17 If p = n and q ≥ n, then W 1,n(Rn) is contained in Lq(Rn).
19
Theorem 18 If n > p then we have the continuous inclusion W 1,p(Rn) →L∞(Rn). Moreover, there is a positive constant C = C(p, n) such that
|f(x)− f(y)| ≤ C ‖f‖W 1,p(Rn) |x− y|1−(n/p),
for all f in W 1,p(Rn). (So f is not only bounded, but also Holder continuous.)
Theorems 16, 17, and 18 offer a panorama of the imbedding situation whenΩ = Rn and k = 1. When Ω is not Rn we resort to extension operators(when feasible): start with f in W 1,p(Ω), extend to f in W 1,p(Rn), apply theappropriate theorem for that case, then restrict back to the original setting.The case when k > 1 is obtained more or less by iterating the case when k = 1.
Being able to extend functions in W k,p(Ω) depends very much on the geom-etry of the boundary of Ω; here, smooth, bounded boundaries are best. Nev-ertheless, when Ω is uncooperative, we can still get good results for the classW k,p
0 (Ω), in a trade-off between good boundaries and functions which are zeroat the boundary. As such, these functions admit easy extensions to the wholespace. With this in mind, we state a more general imbedding theorem forΩ = Rn, pointing out at the end some instances when the theorem is valid ifΩ 6= Rn (because in those instances there are extension theorems).
Theorem 19 Let k ≥ 1 be an integer, and let 1 ≤ p < ∞.
i. If 1p − k
n > 0, then W k,p(Rn) ⊂ Lq(Rn), for 1q = 1
p − kn ;
ii. If 1p − k
n = 0, then W k,p(Rn) ⊂ Lq(Rn), for q ∈ [p,∞);
iii. If 1p − k
n < 0, then W k,p(Rn) ⊂ L∞(Rn).
In this last case k > (n/p). Let m be the integer part of k − (n/p), and θ =k − (n/p)−m. For each multi-index α with |α| ≤ m we have
‖Dαf‖∞ ≤ C ‖f‖W k,p(Rn).
If |α| = m, then
|Dαf(x)−Dαf(y)| ≤ C ‖f‖W k,p(Rn) |x− y|θ.
If |α| < m, then
|Dαf(x)−Dαf(y)| ≤ C ‖f‖W k,p(Rn) |x− y|.
Thus, for k > (n/p) we have the continuous inclusion W k,p(Rn) → Cm(Rn).All of the above results are valid for general Ω if we consider the spaces
W k,p0 (Ω) instead of W k,p(Ω).If Ω is the half-space Rn
+, or if the boundary of Ω is of class C1 and bounded,then the above results are valid for W k,p(Ω).
20
12 Compactness theorems
Exercise 48 Fix 1 ≤ p < ∞. Give an example of a sequence of functionsfm ∈ W 1,p(R) such that for all m we have ‖fm‖W 1,p(R) = 1, and such that forany 1 ≤ q ≤ ∞, and any j 6= k, we have ‖fj − fk‖q = c(q) > 0, where c(q) isa constant depending on q and the sequence fm. Conclude that no imbedding ofW 1,p(R) into Lq(R) can be compact. Generalize this result to Ω = Rn.
In view of this exercise we restrict our attention to bounded domains, so fromnow on Ω will denote a bounded domain in Rn. We wish to determine whichimbeddings in Theorem 19 are compact. The target spaces in that theoremare either Lq(Ω) or C(Ω), so it is important to understand compactness inthese spaces. The Arzela-Ascoli theorem, which we have used before, providescompactness information in C(Ω). We turn our attention to compactness inLq(Ω).
We say that Ω′ is relatively compact in Ω if Ω′ ⊂ Ω, and we write Ω′ ⊂⊂ Ω.In this case there is a δ > 0, depending only on Ω and Ω′, such that for all h ∈ Rn
with |h| < δ and all x ∈ Ω′, we have x − h ∈ Ω. (So, moving the whole of Ω′
by a quantity δ won’t leave Ω.) We denote by τ−h the shift by −h, and defineτ−hf(x) = f(x − h). We will also write τδ(Ω′) ⊂ Ω to indicate the fact thatshifts by δ keep Ω′ inside Ω.
If f ∈ Lq(Ω), we will still denote by f its restriction to Ω′. As such, ‖f‖Lq(Ω′)will make perfect sense. Moreover, if we take h ∈ Rn with |h| < δ, then therestriction to Ω′ of τ−hf also makes perfect sense, and we may compute thequantity
‖τ−hf − f‖Lq(Ω′).
Theorem 20 Let Ω ⊂ Rn be a bounded open set, and let 1 ≤ q < ∞. Let F bea bounded set in Lq(Ω). Assume both conditions below:
(i) For every ε > 0 and every Ω′ ⊂⊂ Ω, there is a δ > 0 such that τδ(Ω′) ⊂ Ω,and for all f ∈ F ,
|h| < δ =⇒ ‖τ−hf − f‖Lq(Ω′) < ε.
(ii) For every ε > 0 there is Ω′ ⊂⊂ Ω such that
f ∈ F =⇒ ‖f‖Lq(Ω\Ω′) < ε.
Then F is relatively compact in Lq(Ω); that is, every sequence fm ⊂ F has asubsequence that converges in Lq(Ω).
Using this criterion, we can prove the following.
Theorem 21 (Rellich-Kondrasov) Let Ω ⊂ Rn be a bounded open set ofclass C1. Then the following inclusions are compact:
i. If p < n, W 1,p(Ω) → Lq(Ω), 1 ≤ q < p∗;
ii. If p = n, W 1,n(Ω) → Lq(Ω), 1 ≤ q < ∞;
iii. If p > n, W 1,p(Ω) → C(Ω).
21
13 Once more, with feeling
In the introduction we promised that distributions would enlarge our functionalspaces, allowing us to find solutions to differential equations. How well did thispromise stack up? Consider again, for n ≥ 3, Poisson’s equation
∆u = f,
where f : Rn → C is a given function. As we saw at the end of Section 4, aslong as we can take the Fourier Transform on both sides of this equation, weobtain
u(ξ) = − 1|2πξ|2 f(ξ),
and it is natural to ask ourselves whether there is a tempered distribution Gsuch that G = −1/|2πξ|2. Notice that −1/|2πξ|2 is in L1
loc(Rn), and thereforecan be viewed as a tempered distribution. Therefore there is some tempereddistribution G such that G = −1/|2πξ|2.
Exercise 49 Let M be a square matrix and φ a function, both defined on Rn.We define the symbol (Mφ)(x) = φ(Mx). Notice that if φ is in S(Rn), thenMφ is also in S(Rn). A tempered distribution T is radial if for every orthogonalmatrix M we have T (Mφ) = T (φ). Show that T is radial if and only if T isradial.
Exercise 50 For every λ > 0 we define (φ λ)(x) = φ(λx). Again, noticethat if φ is in S(Rn), then φ λ is in S(Rn). A tempered distribution T ishomogeneous of degree k if for every λ > 0 we have T (φ λ) = λkT (φ). Showthat T is homogeneous of degree k if and only if T is homogeneous of degree−(n + k).
Since our G is radial and homogeneous of degree −2, then G is radial andhomogeneous of degree 2− n.
Exercise 51 Define G(x) = cn|x|2−n for some constant cn. Determine the cn
for which G = − 1|2πξ|2 .
Looking at the above, there’s nothing to prevent f from being a distribution.In particular if we take f = δ, then u = G is a solution to the equation. Ingeneral, if L is a linear partial differential operator, a solution G to LG = δis called a fundamental solution to L, and if we are looking for solutions toLu = f , then u = f ∗G is one such solution. Thus finding fundamental solutionsis a useful thing, although unfortunately not always easy.
Theorem 22 (Malgrange-Ehrenpreis) If L is a constant-coefficient linearpartial differential operator, then L has a fundamental solution.
22
This fundamental solution is a distribution, and doesn’t necessarily corre-spond to a function in L1
loc.We now turn to the other motivation we offered for the definition of distri-
butions, that of distributions as continuous linear functionals defined over allspaces Lp. In general terms, if V1 ⊂ V2, then the reverse relation is true for theirduals, and V ′
1 ⊃ V ′2 . Since D(Ω) is contained in all Lp(Ω), it follows that D′(Ω)
contains all Lp′(Ω). In particular, it contains the duals of all W k,p. This is allvery fine, but also very abstract, and it would be good to have more concreterepresentations for these dual spaces. For example, in Exercise 22 we asked foran explicit element in the dual of H1(Rn). Let’s denote by H−s(Rn) the dualof Hs(Rn). We have the following characterization of H−1(Rn).
Theorem 23 If T is in H−1(Rn), then there are functions f0, f1, . . . , fn inL2(Rn) such that for any φ in H1(Rn) we have
T (φ) =∫
Rn
(f0φ + f1
∂φ
∂x1+ . . . + fn
∂φ
∂xn
)dx.
The converse is also true.
Proof. The space H1(Rn) is a Hilbert space with inner product
〈φ, ψ 〉 =∫
φψ +∇φ∇ψ dx.
According to the Riesz Representation Theorem, there is a unique element u inH1(Rn) such that T (φ) = 〈φ, u 〉. In particular
T (φ) =∫
Rn
(uφ + ux1
∂φ
∂x1+ . . . + uxn
∂φ
∂xn
)dx,
which is what we wanted to prove, with f0 = u, fj = uxj . The converse is leftas an exercise. ¤
The theorem is true (with the same proof) if we replace H1(Rn) by H10 (Ω).
Exercise 52 Let n = 1. Find two functions f0 and f1 in L2(R) such that ifφ ∈ H1(R), then φ(0) =
∫R f0(x)φ(x) + f1(x)φ′(x) dx. Conclude that the delta
function is an element of H−1(R).
So, δ is in the dual of H1(R). We could have concluded the same by othermeans. Consider Theorem 19 with k = 1, n = 1, p = 2. That is case iii, and weconclude that H1(R) is contained in L∞(R) ∩C(R). Moreover, the inclusion iscontinuous. As a consequence
|δ(φ)| = |φ(0)| ≤ ‖φ‖∞ ≤ C ‖φ‖H1(R).
Again, we conclude that δ is in the dual of H1(R).The following theorem justifies our notation for the dual of Hs(Rn).
23
Theorem 24 Let s > 0. The elements of H−s(Rn) are those tempered distribu-tions T such that the tempered distribution U = (1 + |ξ|2)−s/2T is representableby a function in L2(Rn).
Here, U(φ) = T ((1 + |ξ|2)−s/2φ). We present a proof for s = 1.Proof (for s = 1). First, assume T ∈ H−1(Rn). Then
T (φ) =∫
f0φ +∑
fj∂φ
∂xjdx,
with fj ∈ L2(Rn). Taking φ in S ′(Rn), we see that T is a tempered distribution.Also,
T (ψ) = T (ψ) =∫
f0ψ +∑
fj∂ψ
∂ξjdξ
=∫
f0ψ +∑
fj(−2πi)ξjψ dξ
=∫
f0ψ −∑
(2πiξj fj)ψ dξ.
As a resultU(φ) =
∫(1 + |ξ|2)−1/2(f0 −
∑2πiξj fj)φ dξ,
showing that U is representable by a function in L2(Rn).
Now suppose T is a tempered distribution, and U = (1 + |ξ|2)−1/2T isrepresented by some f in L2(Rn). Let φ be in D(Rn), and ψ ∈ S(Rn) be suchthat ψ = φ. (Notice that ψ(ξ) = φ(−ξ).) Then
T (φ) = T (ψ) = T (ψ) = U((1 + |ξ|2)1/2ψ)
=∫
Rn
f(ξ) (1 + |ξ|2)1/2ψ(ξ) dξ
=∫
Rn
f(ξ) (1 + |ξ|2)1/2φ(−ξ) dξ
=∫
Rn
f(−ξ) (1 + |ξ|2)1/2φ(ξ) dξ.
Since φ is in H1(Rn), we use Holder’s inequality to obtain
This shows that T is in H−1(Rn). ¤Notice that since δ = 1, we have shown in yet another way that δ is in
H−1(R).The final technical tool we want to mention is the famous Poincare inequality,
or perhaps we should say inequalities, because there are several of them. Inessence, it says that on bounded domains the derivative of a function controlsthe function. The most common statement of the inequality is the following.
24
Theorem 25 (Poincare’s inequality) Let Ω be a bounded open set in Rn.Then there is a positive constant C = C(Ω, p) such that for every f ∈ W 1,p
0 (Ω)
‖f‖p ≤ C∑
|α|=1
‖Dαf‖p.
A few things to note about this theorem. First, the right hand side containsexactly first order derivatives, and the left hand side contains zero order deriva-tives. Second, a consequence of this theorem is that the right hand side may beused to define a norm in W 1,p
0 (Ω). Third, the result is not true for f in W 1,p(Ω),since in that case we can always add a constant to f to falsify the inequality. Ina sense, requiring f to be zero at the boundary ‘anchors’ the function. Thereare other ways to do this; see Exercise 53.
Proof. First we do the case Ω = (−a, a)n. Take f in D(Ω). Then, withx = (x′, xn), we have
f(x) =∫ xn
−a
∂f
∂xn(x′, t) dt.
Using Holder’s inequality, raising to the p-th power, and then increasing theinterval of integration produces
|f(x)|p ≤ |xn + a|p/p′∫ a
−a
∣∣∣∣∂f
∂xn(x′, t)
∣∣∣∣p
dt.
Now integrate in x to get∫
Ω
|f(x)|p dx ≤ (2a)1+p/p′∫
Ω
∣∣∣∣∂f
∂xn(x)
∣∣∣∣p
dx.
In particular ‖f‖p ≤ C ‖∂f/∂xn‖p ≤ C∑|α|=1 ‖Dαf‖p, with a constant de-
pending only on Ω and p. The result follows because D(Ω) is dense in W 1,p0 (Ω).
For a general bounded Ω, enclose it in a box of the form (−a, a)n, and extendf to be zero on (−a, a)n \ Ω. This extension preserves all norms of f (since fand its derivatives are identically zero on the boundary of Ω), and the theoremfollows. ¤
The attentive reader will notice that it is enough to have Ω bounded in justone direction for the above theorem to hold.
The final exercise is not just a simple extension (so to speak) of Poincare’sinequality.
Exercise 53 (Poincare-Wirtinger inequality) Let Ω be bounded, connected,and assume the boundary of Ω is C1. Denote by f the average of f over Ω; inother words f = (measure(Ω))−1
∫Ω
f(x) dx. Let 1 ≤ p ≤ ∞. Then there is aconstant C > 0, depending on Ω, n, and p, such that for all f ∈ W 1,p(Ω) wehave
