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Gerrit van DijkDistribution TheoryDe Gruyter Graduate Lectures

Gerrit van Dijk

Distribution Theory

Convolution, Fourier Transform, and Laplace Transform

Mathematics Subject Classification 201035D30, 42A85, 42B37, 46A04, 46F05, 46F10, 46F12

AuthorProf. Dr. Gerrit van DijkUniversiteit LeidenMathematisch InstituutNiels Bohrweg 12333 CA LeidenThe NetherlandsE-Mail: [email protected]

ISBN 978-3-11-029591-7e-ISBN 978-3-11-029851-2

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PrefaceThe mathematical concept of a distribution originates from physics. It was first usedby O. Heaviside, a British engineer, in his theory of symbolic calculus and then byP.A.M. Dirac around 1920 in his research on quantum mechanics, in which he in-troduced the delta-function (or delta-distribution). The foundations of the mathe-matical theory of distributions were laid by S. L. Sobolev in 1936, while in the 1950sL. Schwartz gave a systematic account of the theory. The theory of distributions hasnumerous applications and is extensively used inmathematics, physics and engineer-ing. In the early stages of the theory one used the term generalized function ratherthan distribution, as is still reflected in the term delta-function and the title of sometextbooks on the subject.

This book is intended as an introduction to distribution theory, as developed byLaurent Schwartz. It is aimed at an audience of advanced undergraduates or begin-ning graduate students. It is based on lectures I have given at Utrecht and LeidenUniversity. Student input has strongly influenced the writing, and I hope that thisbook will help students to share my enthusiasm for the beautiful topics discussed.

Starting with the elementary theory of distributions, I proceed to convolutionproducts of distributions, Fourier and Laplace transforms, tempered distributions,summable distributions and applications. The theory is illustrated by several exam-ples, mostly beginning with the case of the real line and then followed by examplesin higher dimensions. This is a justified and practical approach in our view, it helpsthe reader to become familiar with the subject. A moderate number of exercises areadded with hints to their solutions.

There is relatively little expository literature on distribution theory compared toother topics inmathematics, but there is a standard reference [10], and also [6]. I havemainly drawn on [9] and [5].

The main prerequisites for the book are elementary real, complex and functionalanalysis and Lebesgue integration. In the later chapters we shall assume familiaritywith some more advancedmeasure theory and functional analysis, in particular withthe Banach–Steinhaus theorem. The emphasis is however on applications, ratherthan on the theory.

For terminology and notations we generally follow N. Bourbaki. Sections witha star may be omitted at first reading. The index will be helpful to trace importantnotions defined in the text.

Thanks are due to my colleagues and students in The Netherlands for their re-marks and suggestions. Special thanks are due to Dr J. D. Stegeman (Utrecht) whosehelp in developing the final version of the manuscript has greatly improved the pre-sentation.

Leiden, November 2012 Gerrit van Dijk

ContentsPreface V

1 Introduction 1

2 Definition and First Properties of Distributions 32.1 Test Functions 32.2 Distributions 42.3 Support of a Distribution 6

3 Differentiating Distributions 93.1 Definition and Properties 93.2 Examples 103.3 The Distributions xλ−1+ (λ �= 0,−1,−2, . . .)* 123.4 Exercises 143.5 Green’s Formula and Harmonic Functions 143.6 Exercises 20

4 Multiplication and Convergence of Distributions 224.1 Multiplicationwith a C∞Function 224.2 Exercises 234.3 Convergence inD′ 234.4 Exercises 24

5 Distributions with Compact Support 265.1 Definition and Properties 265.2 Distributions Supported at the Origin 275.3 Taylor’s Formula forRn 275.4 Structure of a Distribution* 28

6 Convolution of Distributions 316.1 Tensor Product of Distributions 316.2 Convolution Product of Distributions 336.3 Associativity of the Convolution Product 396.4 Exercises 396.5 Newton Potentials and Harmonic Functions 406.6 Convolution Equations 426.7 Symbolic Calculus of Heaviside 456.8 Volterra Integral Equations of the Second Kind 47

VIII Contents

6.9 Exercises 496.10 Systems of Convolution Equations* 506.11 Exercises 51

7 The Fourier Transform 527.1 Fourier Transform of a Function on R 527.2 The Inversion Theorem 547.3 Plancherel’s Theorem 567.4 Differentiability Properties 577.5 The Schwartz Space S(R) 587.6 The Space of Tempered Distributions S′(R) 607.7 Structure of a Tempered Distribution* 617.8 Fourier Transform of a Tempered Distribution 637.9 Paley–Wiener Theorems on R* 657.10 Exercises 687.11 Fourier Transform in Rn 697.12 The Heat or Diffusion Equation in One Dimension 71

8 The Laplace Transform 748.1 Laplace Transform of a Function 748.2 Laplace Transform of a Distribution 758.3 Laplace Transform and Convolution 768.4 Inversion Formula for the Laplace Transform 79

9 Summable Distributions* 829.1 Definition and Main Properties 829.2 The Iterated Poisson Equation 839.3 Proof of the Main Theorem 849.4 Canonical Extension of a Summable Distribution 859.5 Rank of a Distribution 87

10 Appendix 9010.1 The Banach–Steinhaus Theorem 9010.2 The Beta and Gamma Function 97

11 Hints to the Exercises 102

References 107

Index 109

1 IntroductionDifferential equations appear in several forms. One has ordinary differential equa-tions and partial differential equations, equations with constant coefficients andwithvariable coefficients. Equations with constant coefficients are relatively well under-stood. If the coefficients are variable, much less is known. Let us consider the singulardifferential equation of the first order

xu′ = 0 . (∗)

Though the equation is defined everywhere on the real line, classically a solution isonly given for x > 0 and x < 0. In both cases u(x) = c with c a constant, differentfor x > 0 andx < 0 eventually. In order to find a global solution, we consider a weakform of the differential equation. Let ϕ be a C1 function on the real line, vanishingoutside some bounded interval. Then equation (∗) can be rephrased as

⟨xu′, ϕ

⟩ = ∞∫−∞xu′(x)ϕ(x)dx = 0 .

Applying partial integration, we get⟨xu′, ϕ

⟩ = − ⟨u, ϕ + xϕ′⟩ = 0 .

Take u(x) = 1 for x ≥ 0, u(x) = 0 for x < 0. Then we obtain

⟨xu′, ϕ

⟩ = − ∞∫0

[ϕ(x)+ xϕ′(x)

]dx = −

∞∫0

ϕ(x)dx +∞∫0

ϕ(x)dx = 0 .

Call this function H, known as the Heaviside function. We see that we obtain thefollowing (weak) global solutions of the equation:

u(x) = c1H(x)+ c2 ,

with c1 and c2 constants. Observe that we get a two-dimensional solution space. Onecan show that these are all weak solutions of the equation.

The functionsϕ are called test functions. Of course one can narrow the class oftest functions to Ck functions with k > 1, vanishing outside a bounded interval. Thisis certainly useful if the order of the differential equation is greater than one. It wouldbe nice if we could assume that the test functions areC∞ functions, vanishing outsidea bounded interval. But then there is really something to show: do such functionsexist? The answer is yes (see Chapter 2). Therefore we can set up a nice theory ofglobal solutions. This is important in several branches of mathematics and physics.Consider, for example, a point mass in R3 with force field having potential V = 1/r ,

2 Introduction

r being the distance function. It satisfies the partial differential equation ΔV = 0

outside 0. To include the origin in the equation, one writes it as

ΔV = −4πδ

with δ the functional given by 〈δ,ϕ〉 = ϕ(0). So it is the desire to go to globalequations and global solutions to develop a theory of (weak) solutions. This theoryis known as distribution theory. It has several applications also outside the theoryof differential equations. To mention one, in representation theory of groups, a well-known concept is the character of a representation. This is perfectly defined for finite-dimensional representations. If the dimension of the space is infinite, the concept ofdistribution character can take over that role.

2 Definition and First Properties of DistributionsSummaryIn this chapter we show the existence of test functions, define distributions, give someexamples and prove their elementary properties. Similar to the notion of support ofa function we define the support of a distribution. This is a rather technical part, butit is important because it has applications in several other branches of mathematics,such as differential geometry and the theory of Lie groups.

Learning Targets� Understanding the definition of a distribution.� Getting acquainted with the notion of support of a distribution.

2.1 Test Functions

We consider the Euclidean space Rn, n ≥ 1, with elements x = (x1, . . . , xn). Onedefines ‖x‖ =

√x2

1 + · · · + x2n , the length of x.

Letϕ be a complex-valued function on Rn. The closure of the set of points {x ∈Rn : ϕ(x) �= 0} is called the support ofϕ and is denoted by Suppϕ.

For any n-tuple k = (k1, . . . , kn) of nonnegative integers ki one defines the par-tial differential operator Dk as

Dk =(∂∂x1

)k1

· · ·(∂∂xn

)kn= ∂|k|

∂xk11 · · · ∂xknn

.

The symbol |k| = k1+· · ·+kn is called the order of the partial differential operator.Note that order 0 corresponds to the identity operator. Of course, in the special casen = 1 we simply have the differential operators dk/dxk (k ≥ 0).

A functionϕ : Rn → C is called a Cm function if all partial derivatives Dkϕ oforder |k| ≤m exist and are continuous. The space of all Cm functions on Rn will bedenoted by Em(Rn). In practice, once a value of n ≥ 1 is fixed, this space will besimply denoted by Em.

A functionϕ : Rn → C is called a C∞ function if all its partial derivatives Dkϕexist and are continuous. A C∞ function with compact support is called a test func-tion. The space of all C∞ functions onRn will be denoted by E(Rn), the space of testfunctions on Rn byD(Rn). In practice, once a value of n ≥ 1 is fixed, these spaceswill be simply denoted by E andD respectively.

It is not immediately clear that nontrivial test functions exist. The requirement ofbeing C∞ is easy (for example every polynomial function is), but the requirement ofalso having compact support is difficult. See the following example however.

4 Definition and First Properties of Distributions

Example of a Test FunctionFirst take n = 1, the real line. Letϕ be defined by

ϕ(x) =

⎧⎪⎨⎪⎩ exp(− 1

1− x2

)if |x| < 1

0 if |x| ≥ 1 .

Then ϕ ∈ D(R). To see this, it is sufficient to show that ϕ is infinitely many timesdifferentiable at the points x = ±1 and that all derivatives at x = ±1 vanish. Afterperforming a translation, this amounts to showing that the function f defined byf(x) = e−1/x (x > 0), f(x) = 0 (x ≤ 0) is C∞ at x = 0 and that f (m)(0) = 0 forallm = 0,1,2, . . .. This easily follows from the fact that limx↓0 e−1/x/xk = 0 for allk = 0,1,2, . . ..

For arbitraryn ≥ 1 we denote r = ‖x‖ and take

ϕ(x) =

⎧⎪⎨⎪⎩ exp(− 1

1− r 2

)if r < 1

0 if r ≥ 1 .

Thenϕ ∈ D(Rn).The spaceD is a complex linear space, even an algebra. And evenmore generally,

ifϕ ∈ D andψ a C∞ function, thenψϕ ∈ D. It is easily verified that Supp(ψϕ) ⊂Suppϕ ∩ Suppψ.

We define an important convergence principle inD

Definition 2.1. A sequence of functions ϕj ∈ D (j = 1,2, . . .) converges toϕ ∈ D ifthe following two conditions are satisfied:(i) The supports of allϕj are contained in a compact set, not depending on j,(ii) For any n-tuple k of nonnegative integers the functions Dkϕj converge uniformly

toDkϕ (j →∞).

2.2 Distributions

We can now define the notion of a distribution. A distribution on the Euclidean spaceRn (with n ≥ 1) is a continuous complex-valued linear function defined onD(Rn),the linear space of test functions on Rn. Explicitly, a function T : D(Rn) → C isa distribution if it has the following properties:a. T(ϕ1 +ϕ2) = T(ϕ1)+ T(ϕ2) for allϕ1,ϕ2 ∈ D,b. T(λϕ) = λT(ϕ) for allϕ ∈ D and λ ∈ C,c. Ifϕj tends toϕ inD then T(ϕj) tends T(ϕ).

Instead of the term linear function, the term linear form is often used in the literature.

Distributions 5

The set D′ of all distributions is itself a linear space: the sum T1 + T2 and thescalar product λT are defined by

〈T1 + T2, ϕ〉 = 〈T1, ϕ〉 + 〈T2, ϕ〉 ,〈λT , ϕ〉 = λ 〈T , ϕ〉

for allϕ ∈ D.One usually writes 〈T , ϕ〉 for T(ϕ), which is convenient because of the double

linearity.

Examples of Distributions1. A function on Rn is called locally integrable if it is integrable over every compact

subset of Rn. Clearly, continuous functions are locally integrable. Let f be a lo-cally integrable function on Rn. Then Tf , defined by⟨

Tf , ϕ⟩=∫Rn

f(x)ϕ(x) dx (ϕ ∈ D)

is a distribution on Rn, a so-called regular distribution. Observe that Tf is welldefined because of the compact support of the functions ϕ. One also writes inthis case 〈f , ϕ〉 for 〈Tf , ϕ〉. Because of this relationship between Tf and f , itis justified to call distributions generalized functions, which was customary at thestart of the theory of distributions. Clearly, if the function f is continuous, thenthe relationship is one-to-one: if Tf = 0 then f = 0. Indeed:

Lemma 2.2. Let f be a continuous function satisfying 〈f , ϕ〉 = 0 for allϕ ∈ D.Then f is identically zero.

Proof. Let f satisfy 〈f , ϕ〉 = 0 for all ϕ ∈ D and suppose that f(x0) �= 0 forsome x0. Then we may assume that Ref(x0) �= 0 (otherwise consider if ), evenRef(x0) > 0. Since f is continuous, there is a neighborhood V of x0 whereRef(x0) > 0. If ϕ ∈ D, ϕ ≥ 0, ϕ = 1 near x0 and Suppϕ ⊂ V , thenRe[

∫f(x)ϕ(x)dx] > 0, hence 〈f , ϕ〉 �= 0. This contradicts the assumption

on f . �

We shall see later on in Section 6.2 that this lemma is true for general locallyintegrable functions: if Tf = 0 then f = 0 almost everywhere.

2. The Dirac distribution δ: 〈δ, ϕ〉 = ϕ(0)(ϕ ∈ D). More generally for a ∈ Rn,〈δ(a), ϕ〉 = ϕ(a). One sometimes uses the terminology δ-function and writes

〈δ, ϕ〉 =∫Rn

ϕ(x)δ(x)dx , 〈δ(a), ϕ〉 =∫Rn

ϕ(x)δ(a)(x)dx

forϕ ∈ D. But clearly δ is not a regular function.


3. For any n-tuple k of nonnegative integers and any a ∈ Rn, 〈T , ϕ〉 = Dkϕ(a)(ϕ ∈ D) is a distribution on Rn.

We return to the definition of a distribution and in particular to the continuityproperty. An equivalent definition is the following:

Proposition 2.3. A distribution T is a linear function onD such that for each compactsubsetK of Rn there exists a constant CK and an integerm with

|〈T , ϕ〉| ≤ CK∑

|k|≤msup

∣∣∣Dkϕ∣∣∣for allϕ ∈ D with Suppϕ ⊂ K.

Proof. Clearly any linear function T that satisfies the above inequality is a distribu-tion. Let us show the converse by contradiction. If a distribution T does not satisfysuch an inequality, then for some compact set K and for all C andm, e. g. for C =m(m = 1,2, . . .), we can find a functionϕm ∈ D with 〈T , ϕm〉 = 1, Suppϕm ⊂ K,|Dkϕm| ≤ 1/m if |k| ≤m. Clearlyϕm tends to zero inD ifm tends to infinity, but〈T , ϕm〉 = 1 for allm. This contradicts the continuity condition of T . �

If m can be chosen independent of K, then T is said to be of finite order. Thesmallest possiblem is called the order of T . If, in addition, CK can be chosen inde-pendent of K, then T is said to be a summable distribution (see Chapter 9).

In the above examples, Tf , δ, δ(a) are of order zero, in Example 3 above the dis-tribution T is of order |k|, hence equal to the order of the partial differential operator.

2.3 Support of a Distribution

Let f be a nonvanishing continuous function on Rn. The support of f was definedin Section 2.1 as the closure of the set of points where f does not vanish. If now ϕis a test function with support in the complement of the support of f then 〈f , ϕ〉 =∫Rn f(x)ϕ(x)dx = 0. It is easy to see that this complement is the largest open setwith this property. Indeed, if O is an open set such that 〈f , ϕ〉 = 0 for all ϕ withSuppϕ ⊂ O, then, similar to the proof of Lemma 2.2, f = 0 on O, hence Suppf ∩O = ∅.

These observations permit us to define the support of a distribution in a similarway. We begin with two important lemmas, of which the proofs are rather terse, butthey are of great value.

Lemma 2.4. Let K ⊂ Rn be a compact subset of Rn. Then for any open setO contain-ing K, there exists a functionϕ ∈ D such that 0 ≤ ϕ ≤ 1, ϕ = 1 on a neighborhoodof K and Suppϕ ⊂ O.

Support of a Distribution 7

Proof. Let 0 < a < b and consider the function f on R given by

f(x) =

⎧⎪⎨⎪⎩ exp(

1x − b −

1x − a

)if a < x < b

0 elsewhere.

Then f is C∞ and the same holds for

F(x) =b∫x

f(t)dt/ b∫

a

f(t)dt .

Observe that F(x) ={

1 if x ≤ a0 if x ≥ b.

The functionψ on Rn given byψ(x1, . . . , xn) = F(x21 + · · ·x2

n) is C∞, is equalto 1 for r 2 ≤ a and zero for r 2 ≥ b.

Let B ⊂ B be two different concentric balls in Rn. By performing a linear trans-formation in Rn if necessary, we can now construct a function ψ inD that is equalto 1 on B and zero outside B.

Now consider K. There are finitely many balls B1, . . . , Bm needed to cover K andwe can arrange it such that

⋃mi=1 Bi ⊂ O. One can even manage in such a way that

the balls B1, . . . , Bm which are concentric with B1, . . . , Bm, but have half the radius,cover K as well. Letψi ∈ D be such thatψi = 1 on Bi and zero outside Bi. Set

ψ = 1− (1−ψ1)(1−ψ2) · · · (1−ψm) .Thenψ ∈ D, 0 ≤ ψ ≤ 1,ψ is equal to 1 on a neighborhood of K and Suppψ ⊂ O.

�

Lemma 2.5 (Partition of unity). Let O1, . . . ,Om be open subsets of Rn, K a compactsubset of Rn and assume K ⊂ ⋃m

i=1Oi. Then there are functions ϕi ∈ D withSuppϕi ⊂ Oi such that ϕi ≥ 0,

∑mi=1ϕi ≤ 1 and

∑mi=1ϕi = 1 on a neighborhood

of K.

Proof. Select compact subsets Ki ⊂ Oi such that K ⊂ ⋃mi−1Ki. By the previous lem-ma there are functions ψi ∈ D with Suppψi ⊂ Oi, ψi = 1 on a neighborhood ofKi, 0 ≤ ψi ≤ 1. Set

ϕ1 = ψ1,ϕi = ψi(1−ψ1) · · · (1−ψi−1) (i = 2, . . . ,m) .

Then∑mi=1ϕi = 1 − (1 −ψ1) · · · (1 −ψm), which is indeed equal to 1 on a neigh-

borhood of K. �

Let T be a distribution. Let O be an open subset of Rn such that 〈T , ϕ〉 = 0 forallϕ ∈ D with Suppϕ ⊂ O. Then we say that T vanishes onO. LetU be the unionof such open sets O. Then U is again open, and from Lemma 2.5 it follows that Tvanishes onU. ThusU is the largest open set on which T vanishes. Therefore we cannow give the following definition.


Definition 2.6. The support of the distribution T on Rn, denoted by SuppT , is thecomplement of the largest open set in Rn on which T vanishes.

Clearly, SuppT is a closed subset of Rn.

Examples1. Suppδ(a) = {a}.2. If f is a continuous function, then SuppTf = Suppf .

The following proposition is useful.

Proposition 2.7. Let T be a distribution on Rn of finite orderm. Then 〈T , ψ〉 = 0 forallψ ∈ D for which the derivativesDkψ with |k| ≤m vanish on SuppT .

Proof. Let ψ ∈ D be as in the proposition and assume that Suppψ is contained ina compact subset K of Rn. Since the order of T is finite, equal tom, one has

|〈T , ϕ〉| ≤ CK∑

|k|≤msup

∣∣∣Dkϕ∣∣∣for allϕ ∈ Dwith Suppϕ ⊂ K, CK being a positive constant.

Applying Lemma 2.4, choose for any ε > 0 a functionϕε ∈ D such thatϕε = 1

on a neighborhood of K ∩ SuppT and such that sup |Dk(ϕεψ)| < ε for all partialdifferential operatorsDk with |k| ≤m. Then

|〈T , ψ〉| = |〈T , ϕεψ〉| ≤ CK∑

|k|≤msup

∣∣∣Dk(ϕεψ)∣∣∣ ≤ CK (m + 1)nε .

Since this holds for all ε > 0, it follows that 〈T , ψ〉 = 0. �

Further Reading

Every book on distribution theory contains of course the definition and first proper-ties of distributions discussed in this chapter. See, e. g. [9], [10] and [5]. Schwartz’sbook [10] contains a large amount of additional material.

3 Differentiating DistributionsSummaryIn this chapter we define the derivative of a distribution and show that any distribu-tion can be differentiated an arbitrary number of times. We give several examples ofdistributions and their derivatives, both on the real line and in higher dimensions.The principal value and the finite part of a distribution are introduced. In higher di-mensions we derive Green’s formula and study harmonic functions only dependingon the radius r , both as classical functions outside the origin and globally as distribu-tions. These functions play a prominent role in physics, when studying the potentialof a field of a point mass or of an electron.

Learning Targets� Understanding that any distribution can be differentiated an arbitrary number of

times.� Learning about the principal value and the finite part of a distribution. Deriving

a formula of jumps.� What are harmonic functions and how do they behave as distributions?

3.1 Definition and Properties

If f is a continuously differentiable function on R and ϕ ∈ D(R), then it is easilyverified by partial integration, using thatϕ has compact support, that

⟨dfdx, ϕ

�=

∞∫−∞f ′(x)ϕ(x)dx = −

∞∫−∞f(x)ϕ′(x)dx = −

⟨f ,

dϕdx

�.

In a similar way one has for a continuously differentiable function f on Rn(n ≥ 1),⟨∂f∂xi

, ϕ�= −

⟨f ,∂ϕ∂xi

�for all i = 1,2, . . . , n and allϕ ∈ D(Rn). The following definition is then natural fora general distribution T on Rn: 〈∂T/∂xi, ϕ〉 = −〈T , ∂ϕ/∂xi〉. Notice that ∂T/∂xi(ϕ ∈ D(Rn)) is again a distribution. We resume:

Definition 3.1. Let T be a distribution on Rn. Then the ith partial derivative of T isdefined by ⟨ ∂T

∂xi, ϕ

�= −

⟨T ,

∂ϕ∂xi

� (ϕ ∈ D(Rn)) .

10 Differentiating Distributions

One clearly has

∂2T∂xi∂xj

= ∂2T∂xj∂xi

(i, j = 1, . . . , n)

because the same is true for functions inD(Rn).Let k be a n-tuple of nonnegative integers. Then one has⟨

DkT , ϕ⟩= (−1)|k|

⟨T , Dkϕ

⟩ (ϕ ∈ D(Rn)) .

We may conclude that a distribution can be differentiated an arbitrary number oftimes. In particular any locally integrable function is C∞ as a distribution. Also theδ-function can be differentiated an arbitrary number of times.

LetD = ∑|k|≤m ak Dk be a differential operator with constant coefficients ak ∈C. Then the adjoint or transpose tD of D is by definition tD = ∑|k|≤m(−1)|k| ak Dk.One clearly has ttD = D and

〈DT, ϕ〉 =⟨T , tDϕ

⟩ (ϕ ∈ D(Rn))

for any distribution T . If tD = D we say that D is self-adjoint. The Laplace operatorΔ =∑ni=1 ∂2/∂x2

i is self-adjoint.

3.2 Examples

As before, we begin with some examples on the real line.

1. Let Y(x) ={

0 if x < 0

1 if x ≥ 0.The function Y is called the Heaviside function, named after Oliver Heaviside(1850–1925), a self-taught British electrical engineer, mathematician, and physi-cist. One has Y ′ = δ. Indeed,

⟨Y ′, ϕ

⟩ = − ⟨Y , ϕ′⟩ = − ∞∫0

ϕ′(x)dx = ϕ(0) = 〈δ, ϕ〉 (ϕ ∈ D) .

2. For allm = 0,1,2, . . . one has 〈δ(m), ϕ〉 = (−1)mϕ(m)(0) (ϕ ∈ D). This iseasy.

3. Let us generalize Example 1 to more general functions and higher derivatives.Consider a function f : R → C that ism times continuously differentiable forx �= 0 and such that

limx↓0

f (k)(x) and limx↑0

f (k)(x)

exist for all k ≤ m. Let us denote the difference between these limits (the jump)by σk, thus

σk = limx↓0

f (k)(x)− limx↑0

f (k)(x) .

Examples 11

Call f ′, f ′′, . . . the distributional derivatives of f and {f ′}, {f ′′}, . . . the distribu-tions defined by the ordinary derivatives of f on R\{0}. For example, if f = Y ,then f ′ = δ, {f ′} = 0.One easily verifies by partial integration that⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

f ′ = {f ′}+ σ0 δ ,f ′′ = {f ′′}+σ0 δ′ +σ1 δ,...f (m) = {f (m)}+σ0 δ(m−1) + σ1 δ(m−2) + · · · + σm−1 δ .

So, for example, (Y(x) cosx

)′ = −Y(x) sinx + δ ,(Y(x) sinx

)′ = Y(x) cosx .

4. The distribution pv 1x .

We now come to an important notion in distribution theory, the principal value ofa function. We restrict ourselves to the function f(x) = 1/x and define pv(1/x)(the principal value of 1/x) by⟨

pv 1x , ϕ

⟩= lim

ε↓0

∫|x|≥ε

ϕ(x)x

dx (ϕ ∈ D) .

By a well-known result from calculus, we know that limx↓0 xε logx = 0 for anyε > 0. Therefore, log |x| is a locally integrable function and thus defines a regu-lar distribution. Keeping this in mind and using partial integration we obtain foranyϕ ∈ D

⟨pv 1

x , ϕ⟩= − lim

ε↓0

∫|x|≥ε

log |x|ϕ′(x)dx = −∞∫−∞

log |x|ϕ′(x)dx ,

and we see that pv(1/x) is a distribution since

pv 1x =

(log |x|)′

in the sense of distributions.5. Partie finie.

We now introduce another important notion, the so-called partie finie (or finitepart) of a function. Again we restrict to particular functions, on this occasion topowers of 1/x, which occur when we differentiate the function 1/x. The notionof partie finie occurs when we differentiate the principal value of 1/x. Let usdefine forϕ ∈ D

⟨Pf 1

x2 , ϕ⟩= lim

ε↓0

{ −ε∫−∞

ϕ(x)x2

dx +∞∫ε

ϕ(x)x2

dx − 2ϕ(0)ε

}.


Though this is a rather complicated expression, it occurs in a natural way whenwe compute [pv(1/x)]′ (do it!). It turns out that(

pv 1x

)′ = − Pf 1x2 .

Thus Pf(1/x2) is a distribution.Keeping this in mind, let us define

Pfx−n = − ddx

Pf

(x−(n−1)

n− 1

)(n = 2,3, . . .)

and we set Pf(1/x) = pv(1/x).For an alternative definition of Pfx−n, see [10], Section II, 2:26.

3.3 The Distributions xλ−1+ (λ �= 0,−1,−2, . . .)*Let us define for λ ∈ Cwith Reλ > 0,

⟨xλ−1+ , ϕ

⟩=

∞∫0

xλ−1ϕ(x)dx (ϕ ∈ D) . (3.1)

Since xλ−1 is locally integrable on (0,∞) for Reλ > 0, we can regard xλ−1 as a dis-tribution. Expanding xλ−λ0 = e(λ−λ0) logx (x > 0) into a power series, we obtainfor any (small) ε > 0 the inequality |xλ−1 − xλ0−1| ≤ x−ε |λ − λ0| |xλ0−1| | logx|(0 < x ≤ 1) for all λ with |λ− λ0| < ε. A similar inequality holds for x ≥ 1 with −εreplaced by ε in the power of x. One then easily sees, by using Lebesgue’s theoremon dominated convergence, that λ � 〈xλ−1+ , ϕ〉 is a complex analytic function forReλ > 0. Applying partial integration we obtain

ddx

xλ+ = λxλ−1+

if Reλ > 0. Let us now define the distribution xλ−1+ for all λ �= 0,−1,−2, . . . bychoosing a nonnegative integer k such that Reλ+ k > 0 and setting

xλ−1+ = 1

λ(λ + 1) · · · (λ+ k− 1)

(d

dx

)kxλ+k−1+ .

Explicitly one has

⟨xλ−1+ , ϕ

⟩= (−1)k

λ (λ+ 1) · · · (λ+ k− 1)

∞∫0

xλ+k−1 dkϕdxk

dx (ϕ ∈ D) . (3.2)

It is clear, using partial integration, that this definition agrees with equation (3.1)when Reλ > 0, and that it is independent of the choice of k provided Reλ + k > 0.It also gives an analytic function of λ on the set

Ω = {λ ∈ C : λ �= 0,−1,−2, . . .}.

The Distributions xλ−1+ (λ �= 0,−1,−2, . . .)* 13

Note that (d/dx)xλ+ = λxλ−1+ for all λ ∈ Ω. The excluded points λ = 0,−1,−2, . . .are simple poles ofxλ−1

+ . In order to compute the residues at these poles, we replace kby k+ 1 in equation (3.2), and obtain

Resλ=−k⟨xλ−1+ , ϕ

⟩= limλ→−k

(λ+ k)⟨xλ−1+ , ϕ

⟩

= (−1)k+1

(−k+ 1) · · · (−1)

∞∫0

dk+1ϕdxk+1

(x)dx = dkϕdxk

(0)/k! .

Hence Resλ=−k〈xλ−1+ , ϕ〉 = 〈(−1)k δ(k)/k!, ϕ〉(ϕ ∈ D).

The gamma function Γ (λ), defined byΓ (λ) = ∞∫

0

e−x xλ−1 dx ,

is also defined and analytic for Reλ > 0. This can be shown as above, with ϕ re-placed by the function e−x. We refer to Section 10.2 for a detailed account of the gam-ma function. Using partial integration, one obtains Γ (λ + 1) = λ Γ (λ) for Reλ > 0.This functional equation permits us, similar to the above procedure, to extend thegamma function to an analytic function on thewhole complex planeminus the pointsλ = 0,−1,−2, . . ., thus on Ω. Furthermore, the gamma function has simple poles atthese points with residue at the point λ = −k equal to (−1)k/k!. A well-known for-mula for the gamma function is the following:

Γ (λ) Γ (1− λ) = πsinπλ

.

We refer again to Section 10.2. From this formula wemay conclude that Γ (λ) vanishesnowhere on C\Z. Since Γ (k) = (k− 1)! for k = 1,2, . . ., we see that Γ (λ) vanishes inno point ofΩ. Define now Eλ ∈ D′ by

Eλ =xλ−1+Γ (λ) (λ ∈ Ω) ,

E−k = δ(k) (k = 0,1,2, . . .) .

Then λ� 〈Eλ, ϕ〉 (ϕ ∈ D) is an entire function on C. Note that

dEλdx

= Eλ−1

for all λ ∈ C.


3.4 Exercises

Exercise 3.2. Let Y be the Heaviside function, defined in Section 3.2, Example 1 and letλ ∈ C. Prove that in distributional sense the following equations hold:(

ddx

− λ)Y(x) eλx = δ ,

(d2

dx2+ω2

)Y(x) sinωx

ω= δ ,

dm

dxm

(Y(x)xm−1

(m − 1)!

)= δ form a positive integer.

Exercise 3.3. Determine, in distributional sense, all derivatives of |x|.

Exercise 3.4. Find a distribution of the form F(t) = Y(t) f(t), with f a two timescontinuously differentiable function, satisfying the distribution equation

ad2Fdt2

+ bdFdt

+ cF =mδ+nδ′

with a,b, c,m,n complex constants.Now consider the particular cases

(i) a = c = 1; b = 2; m = n = 1,(ii) a = 1; b = 0; c = 4; m = 1; n = 0,(iii) a = 1; b = 0; c = −4; m = 2; n = 1.

Exercise 3.5. Define the following generalizations of the partie finie:

Pf

∞∫0

ϕ(x)x

dx = limε↓0

[ ∞∫ε

ϕ(x)x

dx +ϕ(0) log ε],

Pf

∞∫0

ϕ(x)x2

dx = limε↓0

[ ∞∫ε

ϕ(x)x2

dx − ϕ(0)ε

+ϕ′(0) log ε].

These expressions define distributions, denoted by pv(Y(x)/x) and Pf(Y(x)/x2).Show this and determine the derivative of the first expression.

3.5 Green’s Formula and Harmonic Functions

We continue with examples in Rn for n ≥ 1.

Green’s Formula and Harmonic Functions 15

1. Green’s formula.Let S be a hypersurface F(x1, . . . , xn) = 0 in Rn. The function F is assumedto be a C1 function such that gradF does not vanish at any point of S. Sucha hypersurface is called regular. Therefore, in each point of S a tangent surfaceexists and with a normal vector on it. Choose at each x ∈ S as normal vector

�n(x) = gradF(x)|gradF(x)| .

Clearly �n(x) depends continuously on x. Given x ∈ S and �e ∈ Rn, the linex + t�e with (�e, �n(x)) �= 0 does not belong to S if t is small, t �= 0. Verify it!Notice also that S has Lebesgue measure equal to zero.Let f be a C2 function onRn\S such that for eachx ∈ S and each partial deriva-tiveDkf on Rn\S (|k| ≤ 2) the limits

limy → xF(y) > 0

Dkf(y) and limy → xF(y) < 0

Dkf(y)

exist. Observe that F increases in the direction of �n and decreases in the directionof−�n.The difference will be denoted by σk:

σk(x) = limy → xF(y) > 0

Dkf(y) − limy → xF(y) < 0

Dkf(y) (x ∈ S) .

Assume that σk is a nice, e. g. continuous, function on S. Let us regard f asa distribution,Dkf as a distributional derivative, {Dkf} as the distribution givenby the functionDkf on Rn\S. Forϕ ∈ D we have⟨

∂f∂x1

, ϕ�= −

⟨f ,

∂ϕ∂x1

�

= −∫f(x)

∂ϕ∂x1

dx

= −∫· · ·

∫f(x1, . . . , xn)

∂ϕ∂x1

(x1, . . . , xn)dx1 · · ·dxn

= −∫

dx2 · · ·dxn∫f(x1, x2, . . . , xn)

∂ϕ∂x1

(x1, x2, . . . , xn)dx1 .

From Section 3.2, Example 3, follows, with σ 0 = σ(0,...,0) and �e1 = (1,0, . . . ,0):⟨∂f∂x1

, ϕ�=∫S

σ 0(x1, . . . , xn) sign(�e1, �n(x1, . . . , xn)

)ϕ(x1, . . . , xn) dx1 . . .dxn

+∫

dx2 . . .dxn

∞∫−∞

∂f∂x1

ϕdx1 .


Let ds be the surface element of S atx ∈ S. One has cosθ1 dx1∧ds = d�n(x)∧ds = dx1 ∧ dx2 . . . ∧ dxn, so dx2 . . .dxn = | cosθ1|ds, θ1 being the anglebetween �e1 and �n(x). We thus obtain∫

S

σ 0(x1, . . . , xn) sign(�e1, �n(x1, . . . , xn)

)ϕ(x1, . . . , xn) dx1 . . .dxn =

∫S

σ 0(s)ϕ(s) cosθ1(s)ds .

Notice that 〈T , ϕ〉 =∫S σ 0(s)ϕ(s) cosθ1(s)ds (ϕ ∈ D) defines a distribu-

tion T . Symbolically we shall write T = (σ 0 cosθ1)δS . Hence we have fori = 1, . . . , n, in obvious notation,

∂f∂xi

={∂f∂xi

}+ (σ 0 cosθi

)δS .

Differentiating this expression once more gives

∂2f∂x2

i={∂2f∂x2

i

}+ ∂∂xi

{(σ 0 cosθi)δS

}+ (σi cosθi)δS .

Here σi = σ(0,...,0,1,0,...,0), with 1 on the ith place.

Let Δ = n∑i=1

∂2

∂x2ibe the Laplace operator. Then we have

Δf = {Δf}+ n∑i=1

∂∂xi

{(σ 0 cosθi

)δS}+ n∑

i=1

(σi cosθi

)δS .

Observe that

a.n∑i=1

σi cosθi is the jump ofn∑i=1

cosθi∂f∂xi

= ∂f∂ν

,

the derivative in the direction of �n(s). Call it σν .

b.⟨ n∑i=1

∂∂xi

{(σ 0 cosθi

)δS}, ϕ

⟩=−

∫S

n∑i=1

cosθi∂ϕ∂xi

σ 0 ds=−∫S

∂ϕ∂ν

σ 0 ds .

Symbolically we will denote this distribution by (∂/∂ν)(σ 0 δS).

We now come to a particular case.Let S be the border of an open volume V and assume that f is zero outside theclosure V of V . Let ν = ν(s) be the direction of the inner normal at s ∈ S. Set

∂f∂ν(s) = lim

x → sx ∈ V

∂f∂ν(x)


and assume that this limit exists. Then one has⟨Δf , ϕ⟩ = ⟨f , Δϕ⟩ = ∫

V

f(x)Δϕ(x)dx

=⟨{Δf}, ϕ⟩+⟨∂f

∂νδS, ϕ

�+⟨ ∂∂ν

(fδS), ϕ�

=∫V

Δf(x)ϕ(x)dx +∫S

∂f∂ν(s)ϕ(s)ds −

∫S

f(s)∂ϕ∂ν(s)ds .

Hence we get Green’s formula in n-dimensional space∫V

(f Δϕ −Δf ϕ)dx =

∫S

(ϕ∂f∂ν

− f ∂ϕ∂ν

)ds (ϕ ∈ D) ,

where still ν is the direction of the inner normal vector. Observe that this formulais even correct for any C2 functionϕ.

2. Harmonic functions.Let f be a C2 function on Rn\{0}, only depending on the radius r = ‖x‖ =√x2

1 + · · · + x2n. Notice that r = r(x1, . . . , xn) is not differentiable at x = 0.

The function f is called a harmonic function if Δf = 0 on Rn\{0}. What is thegeneral form of f , when writing f(x) = F(r)? We have for r �= 0

∂f∂xi

= dFdr

∂r∂xi

= xir

dFdr

,

∂2f∂x2

i= ∂∂xi

(xir

dFdr

)= xir

∂∂xi

(dFdr

)+ dF

dr

(r − x2

i /rr 2

)

= x2ir 2

d2Fdr 2

+(

1r− x

2ir 3

)dFdr

,

hence Δf(x) = d2Fdr 2

+ n− 1r

dFdr

.

We have to solve d2F/dr 2 + (n− 1)/r (dF/dr) = 0. Set u = dF/dr . Thenu′ + [(n− 1)/r] ·u = 0, hence u(r) = c/rn−1 for some constant c and thus

F(r) =

⎧⎪⎨⎪⎩Arn−2

+ B if n �= 2 ,

A log r + B if n = 2 ,

A and B being constants.The constant functions are harmonic functions everywhere, while the functions1/rn−2 (n > 2) and log r (n = 2) are harmonic functions outside the origin x =0. Atx = 0 they have a singularity. Both functions are however locally integrable


(see below, change to spherical coordinates), hence they define distributions. Weshall compute the distributional derivative Δ(1/rn−2) for n �= 2. Therefore weneed some preparations.

3. Spherical coordinates and the area of the (n− 1)-dimensional sphere Sn−1.Spherical coordinates inRn(n > 1) are given by

x1 = r sinθn−1 · · · sinθ2 sinθ1

x2 = r sinθn−1 · · · sinθ2 cosθ1

x3 = r sinθn−1 · · · cosθ2

...xn−1 = r sinθn−1 cosθn−2

xn = r cosθn−1 .

Here 0 ≤ r <∞; 0 ≤ θ1 < 2π, 0 ≤ θj < π(j �= 1).Let rj ≥ 0, r 2

j = x21 + · · · + x2

j and assume that rj > 0 for all j �= 1. Then⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎩

cosθj =xj+1

rj+1,

sinθj =rjrj+1

(j �= 1), sinθ1 =x1

r2,

r = rn .

One has dx1 . . .dxn = rn−1 J(θ1, . . . , θn−1)dr dθ1 . . .dθn−1 with

J(θ1, . . . , θn−1

) = sinn−2 θn−1 sinn−3 θn−2 · · · sinθ2 .

These expressions can easily be verified by induction on n. Forn = 2 andn = 3

they are well known. Let Sn−1 denote the area of Sn−1, the unit sphere in Rn.Then

Sn−1 =2π∫0

π∫0

· · ·π∫0

J(θ1, . . . , θn−1

)dθ1 . . .dθn−1 .

To compute Sn−1, we can use the explicit form of J. Easier is the followingmethod. Recall that

∫∞−∞ e−x

2dx = √π . Then we obtain

πn2 =

∫Rn

e−(x2

1+···+x2n

)dx1 . . .dxn =

∞∫0

rn−1 e−r2

dr .Sn−1

= 12

∞∫0

e−t tn2 −1 dt .Sn−1 = 1

2Sn−1 Γ (n

2

).

Hence Sn−1 = (2πn/2)/Γ (n/2). If we set S0 = 2 then this formula for Sn−1 isvalid for n ≥ 1. For properties of the gamma function Γ , see Section 10.2.


4. Distributional derivative of a harmonic function.We return to the computation of the distributional derivative Δ(1/rn−2) forn �= 2. One has⟨Δ 1

rn−2, ϕ

�=⟨

1rn−2

, Δϕ�

=∫Rn

1rn−2

Δϕ(x)dx = limε→0

∫r≥ε

1rn−2

Δϕ(x)dx (ϕ ∈ D) .

Let us apply Green’s formula to the integral∫r≥ε(1/rn−2)Δϕ(x)dx. Take V =

Vε = {x : r > ε}, S = Sε = {x : r = ε}, f(x) = 1/rn−2 (r > ε). Then∂/∂ν = ∂/∂r . Furthermore Δf = 0 on Vε. We thus obtain∫

Vε

1rn−2

Δϕ(x)dx = −∫r=ε

1rn−2

∂ϕ∂r

ds +∫r=ε

ϕ(s)−(n− 2)εn−1

ds .

One has ∣∣∣∣∣∂ϕ∂r∣∣∣∣∣ =

∣∣∣∣∣n∑i=1

xir∂ϕ∂xi

∣∣∣∣∣ ≤ n supi.x

∣∣∣∣∣ ∂ϕ∂xi (x)∣∣∣∣∣ .

Hence ∣∣∣∣∣∫r=ε

1rn−2

∂ϕ∂r

ds

∣∣∣∣∣ ≤ n supi,x

∣∣∣∣∣ ∂ϕ∂xi∣∣∣∣∣ .εSn−1 → 0

when ε → 0. Moreover∫r=ε

ϕ(s)−(n− 2)εn−1

ds =∫r=ε

ϕ(0)−(n− 2)εn−1

ds

+∫r=ε

[ϕ(s)−ϕ(0)] −(n− 2)

εn−1ds ,

and ∫r=ε

ϕ(0)−(n− 2)εn−1

ds = −(n− 2) Sn−1ϕ(0) .

For the second summand, one has

|ϕ(s)−ϕ(0)| ≤ ‖s‖√n sup‖x‖≤ε1≤i≤n

∣∣∣∣ ∂ϕ∂xi∣∣∣∣ ≤ ε√n supi,x

∣∣∣∣ ∂ϕ∂xi∣∣∣∣ ,

for ‖s‖ = ε. Hence∣∣∣∣∣∫r=ε

[ϕ(s)−ϕ(0)] −(n− 2)

εn−1ds

∣∣∣∣∣ ≤ const. ε .

Wemay conclude ⟨Δ 1rn−2

, ϕ�= −(n− 2) Sn−1ϕ(0) ,


or Δ 1rn−2

= −(n− 2)2πn/2Γ (n/2) δ (n �= 2) .

In a similar way one can show inR2

Δ (log r) = 2π δ (n = 2) .

Special Cases

n = 1: Δ |x| = d2

dx2|x| = 2δ.

n = 3: Δ 1r= −4π δ.

3.6 Exercises

Exercise 3.6. In the (x,y)-plane, consider the square ABCD with

A = (1,1) , B = (2,0) , C = (3,1) , D = (2,2) .

Let T be the distribution defined by the function that is equal to 1 on the square andzero elsewhere. Compute in the sense of distributions

∂2T∂y2

− ∂2T∂x2

.

Exercise 3.7. Show that, in the sense of distributions,(∂∂x

+ i ∂∂y

)(1

x + iy

)= 2πδ .

Exercise 3.8. Show that, in the sense of distributions,

Δ(log r) = 2πδ

in R2.

Exercise 3.9. In Rnone has, in the sense of distributions, forn ≥ 3,

Δ( 1rm

)={Δ 1rm

}if m< n− 2 .

Prove it.

Exercises 21

Exercise 3.10. In the (x, t)-plane, let

E(x, t) = Y(t)2√πt

e−x2/(4t) .

Show that, in the sense of distributions,

∂E∂t− ∂

2E∂x2

= δ(x)δ(t) .

Further Reading

Most of the material in this chapter can be found in [9], where many more exercisesare available. Harmonic functions only depending on the radius are treated in severalbooks on physics. Harmonic polynomials form the other side of the medal, they de-pend only on the angles (in other words, their domain of definition is the unit sphere).These so-called spherical harmonics arise frequently in quantum mechanics. For anaccount of these spherical harmonics, see [4], Section 7.3.

4 Multiplication and Convergence of DistributionsSummaryIn this short chapter we define two important notions: multiplication of a distributionwith a C∞ function and a convergence principle for distributions. Multiplication witha C∞ function is a crucial ingredient when considering differential equations withvariable coefficients.

Learning Targets� Understanding the definition of multiplication of a distribution with a C∞ func-

tion and its consequences.� Getting familiar with the notion of convergence of a sequence or series of distribu-

tions.

4.1 Multiplication with a C∞FunctionThere is no multiplication possible of two arbitrary distributions. Multiplication ofa distribution with a C∞ function can be defined.

Let α be a C∞ function on Rn. Then it it easily verified that the mapping ϕ �αϕ is a continuous linear mapping of D(Rn) into itself: if ϕj converges to ϕ inD(Rn), then αϕj converges to αϕ in D(Rn) when j tends to infinity. Therefore,given a distribution T on Rn, the mapping defined by

ϕ � 〈T , αϕ〉 (ϕ ∈ D(Rn))

is again a distribution onRn. We thus can define:

Definition 4.1. Let α ∈ E(Rn) and T ∈ D′(Rn). Then the distribution αT is definedby

〈αT, ϕ〉 = 〈T , αϕ〉 (ϕ ∈ D(Rn)) .

If f is a locally integrable function, then clearly αf = {αf} in the commonnotation.

Examples on the Real Line1. αδ = α(0)δ, in particular xδ = 0.2. (αδ)′ = α(0)δ′ +α′(0)δ.

The following result is important, it provides a kind of converse of the first example.

Theorem 4.2. Let T be a distribution onR. IfxT = 0 then T = c δ, c being a constant.

Exercises 23

Proof. Letϕ ∈ D and let χ ∈ D be such that χ = 1 near x = 0. Define

ψ(x) =

⎧⎪⎨⎪⎩ϕ(x)−ϕ(0)χ(x)

xif x �= 0

ϕ′(0) if x = 0 .

Then ψ ∈ D (use the Taylor expansion ofϕ at x = 0, cf. Section 5.3) and ϕ(x) =ϕ(0) · χ(x) + xψ(x). Hence 〈T , ϕ〉 = ϕ(0) 〈T , χ〉(ϕ ∈ D), therefore T = c δwith c = 〈T , χ〉. �

Theorem 4.3. Let T ∈ D′(Rn) and α ∈ E(Rn). Then one has for all 1 ≤ i ≤ n,

∂∂xi

(αT) = ∂α∂xi

T +α ∂T∂xi

.

The proof is left to the reader.

4.2 Exercises

Exercise 4.4.a. Show, in a similar way as in the proof of Theorem 4.2 , that T ′ = 0 implies that

T = const.b. Let T and T ′ be regular distributions, T = Tf , T ′ = Tg . Assume both f and g are

continuous. Show that f is continuously differentiable and f ′ = g.

Exercise 4.5. Determine all solutions of the distribution equation xT = 1.

4.3 Convergence inD′

Definition 4.6. A sequence of distributions {Tj} converges to a distributionT if 〈Tj,ϕ〉converges to 〈T , ϕ〉 for everyϕ ∈ D.

The convergence corresponds to the so-called weak topology onD′, known fromfunctional analysis. The spaceD′ is sequentially complete: if for a given sequence ofdistributions {Tj}, the sequence of scalars 〈Tj, ϕ〉 tends to 〈T , ϕ〉 for eachϕ ∈ D,when j tends to infinity, then T ∈ D′. This important result follows from Banach–Steinhaus theorem from functional analysis, see [1], Chapter III, §3, n◦ 6, Théorème 2.A detailed proof of this result is contained in Section 10.1.

A series∑∞j=o Tj of distributions Tj is said to be convergent if the sequence {SN}

with SN =∑Nj=0 Tj converges inD′.

Theorem 4.7. If the locally integrable functions fj converge to a locally integrable func-tion f for j → ∞, point-wise almost everywhere, and if there is a locally integrable func-

24 Multiplication and Convergence of Distributions

tion g ≥ 0 such that |fj(x)| ≤ g(x) almost everywhere for all j, then the sequence ofdistributions fj converges to the distribution f .

This follows immediately from Lebesgue’s theorem on dominated convergence.

Theorem 4.8. Differentiation is a continuous operation inD′: if {Tj} converges to T ,then {DkTj} converges toDkT for everyn-tuple k.

The proof is straightforward.

Examples1. Let

fε(x)

⎧⎨⎩ 0 if ‖x‖ > εn

εn Sn−1if ‖x‖ ≤ ε (x ∈ Rn) .

Then fε converges to the δ-function when ε ↓ 0.

2. The functions

fε(x) = 1

εn(√

2π)n e−x

2/(2ε2) (x ∈ R)

converge to δwhen ε ↓ 0.

4.4 Exercises

Exercise 4.9.a. Show that for everyϕ ∈ C∞(R) and every interval (a, b),

b∫a

sinλxϕ(x)dx andb∫a

cosλxϕ(x)dx

converge to zero when λ tends to infinity.b. Derive from this, by writing

ϕ(x) = ϕ(0)+ [ϕ(x)−ϕ(0)]that the distributions (sinλx)/x tend to π δ when λ tends to infinity.

c. Show that, for real λ, the mapping

ϕ � Pf

∞∫−∞

cosλxx

ϕ(x)dx = limε↓0

∫|x|≥ε

cosλxx

ϕ(x)dx (ϕ ∈ D)

defines a distribution. Show also that these distributions tend to zero when λ tendsto infinity.

Exercises 25

Exercise 4.10. Show that multiplication by a C∞ function defines a continuous linearmapping inD′. Determine the limits inD′, when a ↓ 0, of

ax2 + a2

and axx2 + a2

.

Further Reading

This chapter is self-contained and no reference to other literature is necessary.

5 Distributions with Compact SupportSummaryDistributions with compact support form an important subset of all distributions. Inthis chapter we reveal their structure and properties. In particular the structure ofdistributions supported at the origin is shown.

Learning Targets� Understanding the structure of distributions supported at the origin.

5.1 Definition and Properties

We recall the space E of C∞ functions on Rn. The following convergence principle isdefined in E:

Definition 5.1. A sequence ϕj ∈ E tends to ϕ ∈ E if Dkϕj tends to Dkϕ (j → ∞)for all n-tuples k of nonnegative integers, uniformly on compact subsets of Rn.

Thus, given a compact subset K, one has

supx∈K∣∣∣Dkϕj(x)−Dkϕ(x)

∣∣∣→ 0 (j → ∞)

for all n-tuples k of nonnegative integers. In a similar way one defines a convergenceprinciple in the space Em of Cm functions on Rn by restricting k to |k| ≤m. Noticethat D ⊂ E and the injection is linear and continuous: if ϕj, ϕ ∈ D and ϕj tendstoϕ inD, thenϕj tends toϕ inE.

Let {αj} be a sequence of functions inDwith the propertyαj(x) = 1 on the ball{x ∈ Rn : ‖x‖ ≤ j} and letϕ ∈ E. Then αjϕ ∈ D and αjϕ tends toϕ (j → ∞)in E. We may conclude thatD is a dense subspace of E.

Denote by E′ the space of continuous linear forms L on E. Continuity is definedas in the case of distributions: ifϕj tends toϕ in E, then L(ϕj) tends to L(ϕ).

Let T be a distribution with compact support and letα ∈ D be such thatα(x) =1 for x in a neighborhood of SuppT . We can then extend T to a linear form on E by〈L, ϕ〉 = 〈T , αϕ〉(ϕ ∈ E). This extension is obviously independent of the choiceof α. Moreover L is a continuous linear form on E and L = T on D. Because D isa dense subspace of E, this extension L of T to an element of E′ is unique. We have:

Theorem 5.2. Every distribution with compact support can be uniquely extended to acontinuous linear form on E.

Conversely, any L ∈ E′ is completely determined by its restriction to D. Thisrestriction is a distribution sinceD is continuously embedded into E.

Theorem 5.3. The restriction of L ∈ E′ toD is a distribution with compact support.

Distributions Supported at the Origin 27

Proof. The proof is by contradiction. Assume that the restriction T of L does not havecompact support. Then we can find for each natural numberm a functionϕm ∈ Dwith Suppϕm∩{x : ‖x‖ <m} = ∅ and 〈T , ϕm〉 = 1. Clearlyϕm tends to 0 inE,so L(ϕm) tends to 0 (m →∞), but L(ϕm) = 〈T , ϕm〉 = 1 for allm. �

Similar to Proposition 2.3 one has:

Proposition 5.4. Let T be a distribution. Then T has compact support if and only ifthere exists a constant C > 0, a compact subsetK and a positive integerm such that

|〈T , ϕ〉| ≤ C∑

|k|≤msupx∈K

∣∣∣Dkϕ(x)∣∣∣for allϕ ∈ D.

Observe that if T satisfies the above inequality, then SuppT ⊂ K. We also mayconclude that a distribution with compact support has finite order and is summable.

5.2 Distributions Supported at the Origin

Theorem 5.5. Let T be a distribution on Rn supported at the origin. Then there existan integerm and complex scalars ck such that T =

∑|k|≤m ck Dkδ.

Proof. We know that T has finite order, saym. By Proposition 2.7 we have 〈T , ϕ〉 = 0

for all ϕ ∈ D with Dkϕ(0) = 0 for |k| ≤ m. The same then holds for the uniqueextension of T toE. Nowwrite forϕ ∈ D, applying Taylor’s formula (see Section 5.3)

ϕ(x) =∑

|k|≤m

xk

k!Dkϕ(0)+ψ(x) (x ∈ Rn)

defining xk = xk11 xk2

2 · · ·xknn , k! = k1!k2! · · ·kn! and taking ψ ∈ E such thatDkψ(0) = 0 for |k| ≤m. Then we obtain

〈T , ϕ〉 =∑

|k|≤m

⟨T ,(−1)|k| xk

k!

⟩ ⟨Dkδ, ϕ

⟩(ϕ ∈ D) ,

or T =∑|k|≤m ckDkδwith

ck =⟨T ,(−1)|k|xk

k!

⟩. �

5.3 Taylor’s Formula for Rn

We start with Taylor’s formula forR. Let f ∈ D(R). Thenwe canwrite for anym ∈ N

f(x) =m∑j=0

xj

j!f (j)(0)+ 1

m!

x∫0

(x − t)m f(m+1)(t)dt ,

28 Distributions with Compact Support

which can be proved by integration by parts. The change of variable t → xs in theintegral gives for the last term

xm+1

m!

1∫0

(1− s)m f(m+1)(xs)ds .

This term is of the form (xm+1/m!)g(x) in which, clearly, g is a C∞ function on R,hence an element of E(R). In particular

f(1) =m∑j=0

f (j)(0)j!

+ 1m!

1∫0

(1− s)m f(m+1)(s)ds . (5.1)

Now letϕ ∈ D(Rn) and let t ∈ R. Then we can easily prove by induction on j,

1j!

dj

dtjϕ(xt) =

∑|k|=j

xk

k!

(Dkϕ

)(xt) .

Taking f(t) = ϕ(xt) in formula (5.1), we obtain Taylor’s formula

ϕ(x) =∑

|k|≤m

xk

k!

(Dkϕ

)(0)+

∑|k|=m+1

xk

k!ψk(x)

with

ψk(x) = (m + 1)1∫0

(1− t)m (Dkϕ)(xt)dt .

Clearlyψk ∈ E(Rn).

5.4 Structure of a Distribution*

Theorem5.6. LetT be a distribution onRn andK an open bounded subset of Rn. Thenthere exists a continuous function f onRn and an-tuple k (both depending onK) suchthat T = Dkf on K.

Proof. Without loss of generality we may assume that K is contained in the cubeQ = {(x1, . . . , xn) : |xi| < 1 for all i}. Let us denote by D(Q) the space of allϕ ∈ D(Rn)with Suppϕ ⊂ Q. Then, by the mean value theorem, we have

|ϕ(x)| ≤ sup∣∣∣∣ ∂ϕ∂xi

∣∣∣∣ (x ∈ Rn)

for allϕ ∈ D(Q) and all i = 1, . . . , n. Let R = ∂/∂x1 · · · ∂/∂xn and let us denotefor y ∈Q

Qy ={(x1, . . . , xn) : −1 < xi ≤ yi for all i

}.

Structure of a Distribution* 29

We then have the following integral representation forϕ ∈ D(Q):

ϕ(y) =∫Qy

(Rϕ)(x)dx .

For any positive integerm set

‖ϕ‖m = sup|k|≤m,x∣∣∣Dkϕ(x)∣∣∣ .

Then we have for allϕ ∈ D(Q) the inequalities

‖ϕ‖m ≤ sup∣∣∣Rmϕ∣∣∣ ≤ ∫

Q

∣∣∣Rm+1ϕ(x)∣∣∣ dx .

Because T is a distribution, there exist a positive constant c and a positive integermsuch for allϕ ∈ D(Q)

|〈T , ϕ〉| ≤ c ‖ϕ‖m ≤ c∫Q

∣∣∣Rm+1ϕ(x)∣∣∣ dx .

Let us introduce the function T1 defined on the image of Rm+1 by

T1(Rm+1ϕ

) = 〈T , ϕ〉 .This is a well-defined linear function on Rm+1(D(Q)). It is also continuous in thefollowing sense:

|T1(ψ)| ≤ c∫Q

|ψ(x)| dx .

Therefore, using the Hahn–Banach theorem, we can extend T1 to a continuous linearform on L1(Q). So there exists an integrable function g onQ such that

〈T , ϕ〉 = T1(Rm+1ϕ

) = ∫Q

g(x)(Rm+1ϕ

)(x)dx

for allϕ ∈ D(Q). Setting g(x) = 0 outsideQ and defining

f(y) =y1∫−1

. . .

yn∫−1

g(x)dx1 . . .dxn ,

and using integration by parts, we obtain

〈T , ϕ〉 = (−1)n∫Rn

f(x)(Rm+2ϕ

)(x)dx

for allϕ ∈ D(Q). Hence T = (−1)n+m+2Rm+2f on K. �

For distributions with compact support we have a global result.

30 Distributions with Compact Support

Theorem 5.7. Let T be a distribution with compact support. Then there exist finitelymany continuous functions fk onRn, indexed byn-tuples k, such that

T =∑kDkfk .

Moreover, the support of each fk can be chosen in an arbitrary neighborhood of SuppT .

Proof. Let K be an open bounded set such that SuppT ⊂ K. Select χ ∈ D(Rn) suchthat χ(x) = 1 on a neighborhood of SuppT and Suppχ ⊂ K. Then by Theorem 5.6there is a continuous function f on Rn such that T = Dlf on K for some n-tuple l.Hence

〈T , ϕ〉 = 〈T , χ ·ϕ〉 = (−1)|l|⟨f , Dl(χϕ)

⟩=⟨ ∑|k|≤|l|

Dkfk, ϕ⟩

for suitable continuous functions fk with Suppfk ⊂ K and for allϕ ∈ D(Rn). �

Further Reading

The proof of the structure theorem in Section 5.4, of which the proof can be omitted atfirst reading, is due to M. Pevzner. Several other proofs are possible, see Section 7.7 forexample. See also [10]. Distributions with compact support occur later in this book atseveral places.

6 Convolution of DistributionsSummaryThis chapter is devoted to the convolution product of distributions, a very usefultool with numerous applications in the theory of differential and integral equations.Several examples, some related to physics, are discussed. The symbolic calculus ofHeaviside, one of the eldest applications of the theory of distributions, is discussedin detail.

Learning Targets� Getting familiar with the notion of convolution product of distributions.� Getting an impression of the impact on the theory of differential and integral equa-

tions.

6.1 Tensor Product of Distributions

Set X = Rm, Y = Rn, Z = X × Y(� Rn+m). If f is a function on X and g one on Y ,then we define the following function on Z:

(f ⊗ g)(x,y) = f(x)g(y) (x ∈ X,y ∈ Y) .

We call f ⊗ g the tensor product of f and g.If f is locally integrable on X and g locally integrable on Y , then f ⊗ g is locally

integrable on Z. Moreover one has foru ∈ D(X) and v ∈ D(Y),

〈f ⊗ g, u⊗ v〉 = 〈f , u〉〈g, v〉 .

If ϕ ∈ D(X × Y) is arbitrary, not necessarily of the form u ⊗ v, then, by Fubini’stheorem,

〈f ⊗ g, ϕ〉 = 〈f(x)⊗ g(x), ϕ(x,y)〉 (notation)

= 〈f(x), 〈g(y), ϕ(x,y)〉〉

= 〈g(y), 〈f(x), ϕ(x,y)〉〉 .The following proposition, of which we omit the technical proof (see [10], Chapter IV,§§ 2, 3, 4), is important for the theory we shall develop.

Proposition 6.1. Let S be a distribution onX and T one onY . There exists one and onlyone distributionW on X ×Y such that 〈W, u⊗v〉 = 〈S, u〉〈T , v〉 for all u ∈ D(X),v ∈ D(Y). W is called the tensor product of S and T and is denoted byW = S ⊗ T .It has the following properties. For fixedϕ ∈ D(X × Y) set θ(x) = 〈Ty, ϕ(x,y)〉.Then θ ∈ D(X) and one has

〈W, ϕ〉 = 〈S, θ〉 = 〈Sx, 〈Ty, ϕ(x,y)〉〉 ,

32 Convolution of Distributions

and also〈W, ϕ〉 = 〈Ty , 〈Sx, ϕ(x,y)〉〉 .

Proposition 6.2. SuppS ⊗ T = SuppS × SuppT .

Proof. SetA = SuppS, B = SuppT .(i) Letϕ ∈ D(X×Y) be such that Suppϕ∩(A×B) = ∅. There are neighborhoods

A′ ofA andB′ ofB such that Suppϕ∩(A′×B′) = ∅ too. Forx ∈ A′ the functiony � ϕ(x,y) vanishes on B′, so, in the above notation, θ(x) = 0 for x ∈ A′,hence 〈Sx, θ(x)〉 = 0 and thus 〈W, ϕ〉 = 0. Wemay conclude SuppW ⊂ A×B.

(ii) Let us prove the converse. Let (x,y) ∉ SuppW . Then we can find an open“rectangle” P ×Q with center (x,y), that does not intersect SuppW . Now as-sume that for some u ∈ D(X) with Suppu ⊂ P one has 〈S, u〉 = 1. Then〈T , v〉 = 0 for all v ∈ D(Y)with Suppv ⊂ Q. HenceQ∩B = ∅, and therefore(x,y) ∉ A× B. �

Examples and More Properties1. DkxDly (Sx ⊗ Ty) = DkxSx ⊗DlyTy .2. δx ⊗ δy = δ(x,y).3. A function on X × Y is independent of x if it is of the form 1x ⊗ g(y), with g

a function on Y . Similarly, a distribution is said to be independent of x if it is ofthe form 1x ⊗ Ty with T a distribution on Y . One then has

⟨1x ⊗ Ty, ϕ(x,y)

⟩=∫X

⟨Ty, ϕ(x,y)

⟩dx =

⟨Ty,

∫X

ϕ(x,y)dx⟩

forϕ ∈ D(X × Y).We remark that the tensor product is associative(

Sx ⊗ Ty)⊗Uξ = Sx ⊗ (Ty ⊗ Uξ) .

Observe that⟨(Sx ⊗ Ty

)⊗ Uξ, u(x)v(y)t(ξ)⟩ = 〈S, u〉〈T , v〉〈U, t〉 .4. Define

Y(x1, . . . , xn) ={

1 if xi ≥ 0 (1 ≤ i ≤ n)0 elsewhere.

Then Y(x1, . . . , xn) = Y(x1)⊗ · · · ⊗ Y(xn). Moreover∂nY

∂x1 · · · ∂xn= δ(x1 ,...,xn) .

Convolution Product of Distributions 33

6.2 Convolution Product of Distributions

Let S and T be distributions on Rn. We shall say that the convolution product S ∗ Tof S and T exists if there is a “canonical” extension of S ⊗ T to functions of the form(ξ, η)� ϕ(ξ + η)(ϕ ∈ D), such that

〈S ∗ T , ϕ〉 =⟨Sξ ⊗ Tη, ϕ(ξ + η)

⟩(ϕ ∈ D)

defines a distribution.

Comments1. Sξ ⊗ Tη is a distribution onRn ×Rn.2. The function (ξ, η) � ϕ(ξ + η) (ϕ ∈ D) has no compact support in general.

The support is contained in a band of the form

��

��

��

��

��

��

��

��

��

��

��

η

ξ

3. The distribution S ∗ T exists for example when the intersection of Supp(S ⊗T) = SuppS×SuppT with such bands is compact (or bounded). The continuityof S∗T is then a straightforward consequence of the relations (∂/∂ξi)ϕ(ξ+η) =(∂ϕ/∂ξi)(ξ + η) etc. In that case T ∗ S exists also and one has S ∗ T = T ∗ S .

4. Case by case one has to decide what “canonical” means.

We may conclude:

Theorem 6.3. A sufficient condition for the existence of the convolution product S ∗ Tis that for every compact subsetK ⊂ Rn one has

ξ ∈ SuppS , η ∈ SuppT , ξ + η ∈ K ⇒ ξ and η remain bounded.

The convolution product S ∗ T has then a canonical definition and is commutative:S ∗ T = T ∗ S .

Special Cases (Examples)1. At least one of the distributions S and T has compact support. For example:

S ∗ δ = δ∗ S = S for all S ∈ D′(Rn).


2. (n = 1) We shall say that a subset A ⊂ R is bounded from the left if A ⊂ (a, ∞)for some a ∈ R. Similarly, A is bounded from the right if A ⊂ (−∞, b) for someb ∈ R.If SuppS and SuppT are both bounded from the left (right), then S ∗ T exists.Example: Y ∗ Y exists, with Y , as usual, the Heaviside function.

3. (n = 4) Let SuppS be contained in the positive light cone

t ≥ 0 , t2 − x2 −y2 − z2 ≥ 0 ,

and let SuppT be contained in t ≥ 0. Then S ∗ T exists. Verify that Theorem 6.3applies.

Theorem 6.4. Let S and T be regular distributions given by the locally integrable func-tions f and g. Let their supports satisfy the conditions of Theorem 6.3. Then one has(i)

∫Rn f(x−t)g(t)dt and

∫Rn f(t) g(x−t)dt exist for almost allx, and are equal

almost everywhere to, say, h(x);(ii) h is locally integrable;(iii) S ∗ T is a function, namely h.

The proof can obviously be reduced to the well-known case of L1 functions fand g with compact support.

Sometimes the conditions of Theorem 6.3 are not fulfilled, while the convolutionproduct nevertheless exists. Let us give two examples.1. If f and g are arbitrary functions in L1(Rn), then f ∗ g, defined by

f ∗ g(x) =∫Rn

f(t)g(x − t)dt

exists and is an element of L1(Rn) again. One has ‖f ∗ g‖1 ≤ ‖f‖1 ‖g‖1.2. If f ∈ L1(Rn), g ∈ L∞(Rn), then f ∗ g defined by

f ∗ g(x) =∫Rn

f(t)g(x − t)dt

exists and is an element of L∞(Rn) again. It is even a continuous function. Fur-thermore ‖f ∗ g‖∞ ≤ ‖f‖1 ‖g‖∞.Let now f and g be locally integrable on R, Suppf ⊂ (0, ∞) Suppg ⊂ (0, ∞).Then f ∗ g exists (see special case 2) and is a function (Theorem 6.4). One hasSupp(f ∗ g) ⊂ (0, ∞) and

f ∗ g(x) =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩0 if x ≤ 0

x∫0

f(x − t)g(t)dt if x > 0 .


Examples1. Let

Yαλ (x) = Y(x)xα−1

Γ (α) eλx (α > 0, λ ∈ C) .

One then has Yαλ ∗ Yβλ = Y

α+βλ , applying the following formula for the beta func-

tion:

B(α, β) =1∫0

tα−1 (1− t)β−1 dt = Γ (α) Γ (β)Γ (α+ β)(see Section 10.2).

2. LetGσ(x) =

1

σ√

2πe−x

2/(2σ2) (σ > 0) .

Then one hasGσ ∗Gτ = G√σ2+τ2 .Gσ is the Gauss distribution with expectation 0 and variance σ . If Gσ is the dis-tribution function of the stochastic variable x and Gτ the one of y, and if thestochastic variables are independent, then Gσ ∗ Gτ is the distribution functionof x +y . Its variance is apparently

√σ 2 + τ2.

3. LetPa(x) = 1

π2

ax2 + a2

(a > 0) ,

the so-called Cauchy distribution.One has Pa ∗ Pb = Pa+b. We shall prove this later on.

Theorem 6.5. Let α ∈ E and let T be a distribution. Assume that the distributionT∗{α} exists (e. g. if the conditions of Theorem 6.3 are satisfied). Then this distributionis regular and given by a function in E, namely x � 〈Tt, α(x − t)〉.

We shall write T ∗α for this function. It is called the regularization of T by α.

Proof. We shall only consider the case where T and {α} satisfy the conditions ofTheorem 6.3. Then the function x � 〈Tt, α(x − t)〉 exists and is C∞ since x �β(x) 〈Tt, α(x−t)〉 = 〈Tt, β(x)α(x−t)〉 is C∞ for every β ∈ D, by Proposition 6.1.


We now show that T ∗ {α} is a function: T ∗ {α} = {T ∗α}. One has forϕ ∈ D

〈T ∗ {α}, ϕ〉 =⟨Tξ ⊗α(η), ϕ(ξ + η)

⟩=⟨Tξ, 〈α(η), ϕ(ξ + η)〉

⟩=⟨Tξ,

∫Rn

α(η)ϕ(ξ + η)dη⟩

=⟨Tξ,

∫Rn

α(x − ξ)ϕ(x)dx⟩

=⟨Tξ, 〈ϕ(x), α(x − ξ)〉

⟩=⟨Tξ ⊗ϕ(x), α(x − ξ)

⟩.

Applying Fubini’s theorem, we obtain⟨Tξ ⊗ϕ(x), α(x−ξ)

⟩=⟨ϕ(x), 〈Tξ, α(x−ξ)〉

⟩=∫Rn

ϕ(x)⟨Tξ, α(x − ξ)

⟩dx

=∫Rn

(T ∗α)(x)ϕ(x)dx .

Hence the result. �

Examples1. If T has compact support or T = f with f ∈ L1. Then T ∗ 1 = 〈T , 1〉.2. Ifα is a polynomial of degree≤m, then T∗α is. Let us taken = 1 for simplicity.

Then(T ∗α)(x) =

⟨Tt, α(x − t)

⟩=∑k≤m

xk

k!

⟨Tt, α(k)(−t)

⟩by using Taylor’s formula for the functionx � α(x−t). Here T is supposed to bea distribution with compact support or T = f with xkf ∈ L1 for k = 0,1, . . . ,m.

Definition 6.6. Let T be a distribution on Rn and a ∈ Rn. The translation τaT of Tover a is defined by

〈τaT , ϕ〉 =⟨Tξ, ϕ(ξ + a)

⟩(ϕ ∈ D) .

Proposition 6.7. One has the following formulae:(i) δ∗ T = T for any T ∈ D′(Rn);(ii) δ(a) ∗ T = τ(a) T ; δ(a) ∗ δ(b) = δ(a+b);(iii) (n = 1) δ′ ∗ T = T ′ for any T ∈ D′.


The proof is left to the reader. In a similarway one hasδ(m)∗T = T(m) onR, and,if D is a differential operator on Rn with constant coefficients, then Dδ ∗ T = DT ,for all T ∈ D′(Rn). In particular, if

Δ = Δn = n∑i=1

∂2

∂x2i

(Laplace operator in Rn)

and= 1c2

∂2

∂t2−Δ3 (differential operator of d’Alembert inR4) ,

then Δδ∗ T = ΔT and δ∗ T = T .

Proposition 6.8. Let Sj tend to S (j → ∞) in D′(Rn) and let T ∈ D′(Rn). ThenSj ∗ T tends to S ∗ T (j → ∞) if either T has compact support or the SuppSj areuniformly bounded in Rn.

The proof is easy and left to the reader. As application consider a function ϕ ∈D(Rn) satisfying ϕ ≥ 0,

∫Rn ϕ(x)dx = 1, and set ϕk(x) = kn ϕ(kx) for k =

1,2, . . .. Then δ = limk→∞ ϕk and Suppϕk remains uniformly bounded in Rn.Hence for all T ∈ D′(Rn) one has

T = T ∗ δ = limk→∞

T ∗ϕk .

Notice that T ∗ ϕk is a C∞ function, so any distribution is the limit, in the sense ofdistributions, of a sequence of C∞ functions of the form T ∗ϕ withϕ ∈ D.

Now take n = 1. Then δ is also the limit of a sequence of polynomials.This follows from the Weierstrass theorem: any continuous function on a closed

interval can be uniformly approximated by polynomials. Select now the sequenceof polynomials as follows. Write (as above) δ = limk→∞ ϕk with ϕk ∈ D. Foreach k choose a polynomial pk such that supx∈[−k,k] |ϕk(x)−pk(x)| < 1/k by theWeierstrass theorem. Then δ = limk→∞ pk.

More generally, if T ∈ D′(R) has compact support, then, by previous results,T is the limit of a sequence of polynomials (because T ∗ α is a polynomial for anypolynomial α).

We just mention, without proof, that this result can be extended to n > 1.Let now f ∈ L1(Rn) and let again {ϕk} be the sequence of functions considered

above. Then f = limk→∞ f ∗ϕk in the sense of distributions. One can however showa much stronger result in this case with important implications.

Lemma 6.9. Let f ∈ L1(Rn). Then one can find for every ε > 0 a neighborhood V ofx = 0 inRn such that

∫Rn |f(x −y)− f(x)|dx < ε for all y ∈ V .

Proof. Approximate f in L1-norm with a continuous function ψ with compact sup-port. Then the lemma follows from the uniform continuity ofψ. �


Proposition 6.10. Let f ∈ L1(Rn). Then f = limk→∞ f ∗ϕk in L1-norm.

Proof. One has

‖f − f ∗ϕk‖1 =∫Rn

∣∣∣∣ ∫Rn

[f(x)− f(x −y)]ϕk(y)dy

∣∣∣∣ dx

≤∫Rn

∫Rn

∣∣f(x) − f(y − x)∣∣ ϕk(y)dx dy .

Given ε > 0, this expression is less than ε for k large enough by Lemma 6.9. Hencethe result. �

Corollary 6.11. Let f ∈ L1(Rn). If f = 0 as a distribution, then f = 0 almost every-where.

Proof. If f = 0 as a distribution, then f ∗ϕk = 0 as a distribution for all k, because〈f ∗ϕk, ϕ〉 = 〈f , ϕk ∗ϕ〉 for all ϕ ∈ D(Rn), defining ϕk(x) = ϕk(−x)(x ∈Rn). Since f ∗ϕk is a continuous function, we conclude that f ∗ϕk = 0. Then byProposition 6.10, f = 0 almost everywhere. �

Corollary 6.12. Let f be a locally integrable function. If f = 0 as a distribution, thenf = 0 almost everywhere.

Proof. Let χ ∈ D(Rn) be arbitrary. Then χ f ∈ L1(Rn) and χ f = 0 as a distribu-tion. Hence χ f = 0 almost everywhere by the previous corollary. Since this holds forall χ, we obtain f = 0 almost everywhere. �

We continue with some results on the support of the convolution product of twodistributions.

Proposition 6.13. Let S and T be two distributions such that S∗T exists, SuppS ⊂ A,SuppT ⊂ B. Then SuppS ∗ T is contained in the closure of the setA+ B.

The proof is again left to the reader.

Examples1. Let SuppS and SuppT be compact. Then S ∗ T has compact support and the

support is contained in SuppS + SuppT .2. (n = 1) If SuppS ⊂ (a, ∞), SuppT ⊂ (b, ∞), then SuppS ∗ T ⊂ (a+ b, ∞).3. (n = 4) If both S and T have support contained in the positive light cone t ≥ 0,

t2 − x2 −y2 − z2 ≥ 0, then S ∗ T has.

Associativity of the Convolution Product 39

6.3 Associativity of the Convolution Product

Let R, S, T be distributions on Rn with supports A, B,C, respectively. We define thedistribution R ∗ S ∗ T by

〈R ∗ S ∗ T , ϕ〉 =⟨Rξ ⊗ Sη ⊗ Tζ, ϕ(ξ + η+ ζ)

⟩(ϕ ∈ D) .

This distribution makes sense if, for example,

ξ ∈ A, η ∈ B, ζ ∈ C, ξ + η+ ζ bounded ,

implies ξ, η and ζ are bounded. The associativity of the tensor product then givesthe associativity of the convolution product

R ∗ S ∗ T = (R ∗ S)∗ T = R ∗ (S ∗ T) .The convolution product may not be associative if the above condition is not satisfied.The distributions (R∗ S)∗T andR∗ (S ∗T)might exist without being equal. So is(1∗ δ′)∗ Y = 0 while 1∗ (δ′ ∗ Y) = 1∗ δ = 1.

Theorem 6.14. The convolution product of several distributions makes sense, is com-mutative and associative, in the following cases:(i) All, but at most one distribution has compact support.(ii) (n = 1) The supports of all distributions are bounded from the left (right).(iii) (n = 4) All distributions have support contained in t ≥ 0 and, up to at most one,

also in t ≥ 0, t2 − x2 −y2 − z2 ≥ 0.

The proof is left to the reader.Assume that S and T satisfy: ξ ∈ SuppS, η ∈ SuppT , ξ + η bounded, implies

that ξ and η are bounded. Then one has:

Theorem 6.15. Translation and differentiation of the convolution product S ∗ T goesby translation and differentiation of one of the factors.

This follows easily fromTheorem 6.14 by invoking the delta distribution and usingτa T = δ(a) ∗ T ,DT = Dδ∗ T if T ∈ D′(Rn), D being a differential operator withconstant coefficients on Rn and a ∈ Rn.

Example (n = 2)

∂2

∂x∂y(S ∗ T) = ∂2S

∂x∂y∗ T = S ∗ ∂2T

∂x∂y= ∂S∂x

∗ ∂T∂y

= ∂S∂y

∗ ∂T∂x

.

6.4 Exercises

Exercise 6.16. Show that the following functions are in L1(R):

e−|x| , e−ax2(a > 0) , xe−ax

2(a > 0) .


Compute(i) e−|x| ∗ e−|x|,(ii) e−ax

2 ∗ xe−ax2 ,

(iii) xe−ax2 ∗ xe−ax

2 .

Exercise 6.17. Determine all powers Fm = f ∗ f ∗ · · · ∗ f (m times), where

f(x) ={

1 if −1 < x < 1

0 elsewhere.

Determine also SuppFm and F ′m.

Exercise 6.18. In the (x,y)-plane we consider the set C : 0 < |y| ≤ x. If f and g arefunctions with support contained in C , write then f ∗ g in a “suitable” form. Let χ bethe characteristic function of C . Compute χ ∗ χ.

6.5 Newton Potentials and Harmonic Functions

Let μ be a continuous mass distribution in R3 and

Uμ(x1, x2, x3) =∫ ∫ ∫

μ(y1, y2, y3)dy1 dy2 dy3√(x1 −y1)2 + (x2 −y2)2 + (x3 −y3)2

the potential belonging to μ. Generalized to Rn we get

Uμ =∫Rn

μ(y)dy‖x −y‖n−2

or Uμ = μ ∗ 1rn−2

(n > 2) .

Definition 6.19. The Newton potential of a distribution T on Rn(n > 2) is the distri-bution

UT = T ∗ 1rn−2

.

In order to guarantee the existence of the convolution product, let us assume, forthe time being, that T has compact support.

Theorem 6.20. UT satisfies the Poisson equation

ΔUT = −NT with N = 2πn2

Γ (n2 ) .The proof is easy, apply in particular Section 3.5.An elementary or fundamental solution of ΔU = V is, by definition, a solution ofΔU = δ. Hence E = −1/(Nrn−2) is an elementary solution of the Poisson equation.Call a distribution U harmonic ifΔU = 0.

Newton Potentials and Harmonic Functions 41

Lemma 6.21 (Averaging theorem from potential theory).Let f be a harmonic function on Rn (i. e. f is a C2 function onRn with Δf = 0). Thenthe average of f over any sphere around x with radius R is equal to f(x):

f(x) =∫

‖y‖=Rf(x −y)dωR(y) =

∫Sn−1

f(x − Rω)dω

with

dω = |J(θ1, . . . , θn−1)|dθ1 · · ·dθn−1

Sn−1(in spherical coordinates) ,

dωR(y) = surface element of sphere around x = 0 with radius R ,such that total mass is 1 .

Proof. Let n > 2. In order to prove this lemma, we start with a C2 function f on Rn,a functionϕ ∈ D(Rn) and a volume V with boundary S, and apply Green’s formula(Section 3.5): ∫

V

(f Δϕ −ϕΔf)dx +

∫S

(f∂ϕ∂ν

−ϕ ∂f∂νi

)dS = 0 , (i)

with ν the inner normal on S.Suppose now f harmonic, soΔf = 0, and takeV = VR(x), being a ball aroundx

with radiusR and boundary S. Takeϕ ∈ D(Rn) such thatϕ = 1 on a neighborhoodof V . Then (i) gives ∫

S

∂f∂ν

dS = 0 . (ii)

Choose then ϕ ∈ D(Rn) such that ϕ(y) = 1/rn−2 for 0 < R0 ≤ r ≤ R and setψ(y) =ϕ(x −y).

Then (i) and (ii) imply∫‖x−y‖=R

f∂ψ∂ν

dS =∫

‖x−y‖=R0

f∂ψ∂ν

dS .

Now∂ψ∂ν(y) = ∂ϕ

∂ν(x −y) = −n− 2

rn−1(x −y) = − n− 2

‖x −y‖n−1.

Hence1

Rn−1

∫‖x−y‖=R

f(y)dS = 1

Rn−10

∫‖x−y‖=R0

f(y)dS ,

or ∫Sn−1

f(x − Rω)dω =∫

Sn−1

f(x − R0ω)dω .

Let finally R0 tend to zero. Then the left-hand side tends to f(x). �

The proof in the cases n = 1 and n = 2 of this lemma is left to the reader.


Theorem 6.22. A harmonic distribution is a C∞ function.

Proof. Choose α ∈ D(Rn) rotation invariant and such that∫α(x)dx = 1. Let f be

a C2 function satisfying Δf = 0. According to Lemma 6.21,

f ∗α(x) =∫Rn

f(x − t)α(t)dt = Sn−1

∞∫0

∫Sn−1

f(x − r ω)rn−1α(r)dr dω

= Sn−1

∞∫0

α(r) rn−1 dr · f(x) = f(x) .

Set ϕ(x) = ϕ(−x) for any ϕ ∈ D. Then also ϕ ∈ D. Let now T be a harmonicdistribution. Consider the regularization of T by ϕ: f = T ∗ ϕ. Then we have

〈T∗α, ϕ〉=(T∗α)∗ϕ(0)=(T∗ϕ)∗α(0)=f∗α(0)=f(0)=T ∗ϕ(0)=〈T , ϕ〉

for anyϕ ∈ D, since f = T ∗ ϕ is a harmonic function. Hence T ∗α = T and thusT ∈ E(Rn). Notice that the convolution product is associative (and commutative) inthis case, since both α and ϕ have compact support. �

We remark that, by applying more advanced techniques from the theory of differ-ential equations, one can show that a harmonic function on Rn is even real analytic.

The following corollary is left as an exercise.

Corollary 6.23. A sequence of harmonic functions, converging uniformly on compactsubsets ofRn, converges to a harmonic function.

6.6 Convolution Equations

We begin with a few definitions.An algebra is a vector space provided with a bilinear and associative product.A convolution algebra A′ is a linear subspace ofD′, closed under taking convolu-

tion products. The product should be associative (and commutative) and, moreover,δ should belong toA′. A convolution algebra is thus in particular an algebra, evenwith unit element. Let us give an illustration of this notion of convolution algebra.

Examples of Convolution Algebras(i) E′: the distributions with compact support on Rn.(ii) D′

+: the distributions with support in [0, ∞) on R; similarlyD′−.

(iii) In R4, the distributions with support contained in the positive light cone t ≥ 0,t2 − x2 −y2 − z2 ≥ 0.

Convolution Equations 43

LetA′ be a convolution algebra. A convolution equation inA′ is an equation ofthe form

A∗X = B .Here A, B ∈ A′ are given and X ∈ A′ has to be determined.

Theorem 6.24. The equationA∗X = B is solvable for any B inA′ if and only ifA hasan inverse inA′, that is an elementA−1 ∈ A′ such that

A∗A−1 = A−1 ∗A = δ .

The inverse A−1 of A is unique and the equation has a unique solution. NamelyX = A−1 ∗ B.

This algebraic result is easy to show. One just copies the method of proof fromelementary algebra.

Not everyAhas an inverse in general. IfA is aC∞ functionwith compact support,thenA−1 ∉A′, forA∗A−1 is then C∞, but on the other hand equal to δ. In principlehowever A∗ X = B might have a solution inA′ for several B.

The Poisson equation ΔU = V has been considered in Section 6.5. This is a con-volution equation: ΔU = Δδ∗ U = V with elementary solution X = −1/(4πr) inR3. If V has compact support, then all solutions of ΔU = V are

U = − 14πr

∗ V + harmonic C∞ functions .

Though Δδ and V are inE′(R3), X is not. There is no solution of ΔU = δ in E′(R3)(see later), but there is one in D′(R3). The latter space is however no convolutionalgebra. There are even infinitely many solutions of ΔU = V inD′(R3).

We continue with convolution equations inD′+.

Theorem 6.25. Consider the ordinary differential operator

D = dm

dxm+ a1

dm−1

dxm−1+ · · · + am−1

ddx

+ am

with complex constants a1, . . . , am. Then Dδ is invertible in D′+ and its inverse is of

the form YZ, with Y the Heaviside function and Z the solution of DZ = 0 with initialconditions

Z(0) = Z′(0) = · · · = Z(m−2)(0) = 0, Z(m−1)(0) = 1 .

Proof. Any solution Z ofDZ = 0 is an analytic function, so C∞. We have to show thatD(YZ) = δ if Z satisfies the initial conditions of the theorem. Applying the formulae


of jumps from Section 3.2, Example 3, we obtain

(YZ)′ = YZ′ + Z(0)δ(YZ)′′ = YZ′′ + Z′(0) δ+ Z(0) δ′

...(YZ)(m−1) = YZ(m−1) + Z(m−2)(0)δ+ · · · + Z(0)δ(m−2)

(YZ)(m) = YZ(m) + Z(m−1)(0)δ+ · · · + Z(0)δ(m−1) .

Hence (YZ)(k) = YZ(k) if k ≤m−1 and (YZ)(m) = YZ(m)+δ, thereforeD(YZ) =YDZ + δ = δ. �

Examples

1. LetD = ddx−λ, withλ ∈ C. One hasDδ = δ′−λδ and (δ′−λδ)−1 = Y(x) eλx.

2. ConsiderD =(

d2

dx2+ω2

)withω ∈ R. Then (δ′′ +ω2 δ)−1 = Y(x)sinωx

ω.

3. SetD =(

ddx

− λ)m

. Then (δ′ − λδ)−m = Y(x) xm−1

(m − 1)!eλx.

As application we determine the solution in the sense of function theory of Dz = fifD is as in Theorem 6.25 and f a given function, andwe impose the initial conditionsz(k)(0) = zk (0 ≤ k ≤m − 1). We shall take z ∈ Cm and f continuous.

We first compute Yz. As in the proof of Theorem 6.25, we obtain

D(Yz) = Y(Dz)+m∑k=1

ck δ(k−1)

with ck = am−k z0+am−k−1 z1+· · ·+zm−k. HenceD(Yz) = Yf +∑mk=1 ck δ(k−1),

and therefore Yz = YZ ∗ (Yf +∑mk=1 ck δ(k−1)), or, for x ≥ 0

z(x) =x∫0

Z(x − t) f(t)dt +m∑k=1

ck Z(k−1)(x) .

Similarly we can proceed inD′− with−Y(−x). One obtains the same formula. Verifythat z satisfiesDz = f with the given initial conditions.

Proposition 6.26. IfA1 andA2 are invertible inD′+, thenA1∗A2 is and (A1∗A2)−1 =

A−12 ∗A−1

1 .

This proposition is obvious. It has a nice application. Let D be the above differ-ential operator and let

P(z) = zm + a1zm−1 + · · · + am−1z + am .

Symbolic Calculus of Heaviside 45

Then P(z) can be written as

P(z) = (z − z1)(z − z2) · · · (z − zm) ,

with z1, . . . , zm being the complex roots of P(z) = 0. Hence

D =(

ddx

− z1

)(d

dx− z2

)· · ·

(d

dx− zm

)or

Dδ = (δ′ − z1δ)∗ (δ′ − z2δ

)∗ · · · ∗ (δ′ − zmδ) .Therefore

(Dδ)−1 = Y(x) ez1x ∗ Y(x) ez2x ∗ · · · ∗ Y(x) ezmx

which is in D′+. For example (δ′ − λδ)m; its inverse is Y(x) eλx xm−1/(m− 1)!,

which can easily be proved by induction onm; see also Example 3 above.

6.7 Symbolic Calculus of Heaviside

It is known thatD′+ is a commutative algebra over Cwith unit element δ and without

zero divisors. The latter property says: if S ∗ T = 0 for S, T ∈ D′+, then S = 0 or T =0. For a proof we refer to [14], p. 325, Theorem 152 and [10], p. 173. Therefore, by awell-known theorem from algebra,D′

+ has a quotient field, where convolution equationscan be solved in the natural way by just dividing. Let us apply this procedure, calledsymbolic calculus of Heaviside.

Consider again, for complex numbers ai,

D = dm

dxm+ a1

dm−1

dxm−1+ · · · + am−1

ddx

+ am .

We will determine the inverse ofDδ again. Let

P(z) = zm + a1zm−1 + · · · + am−1z + am

and writeP(z) =

∏j(z − zj)kj

the zj being the mutually different roots of P(z) = 0 with multiplicity kj .Set p = δ′ and write zj for zjδ, 1 for δ and just product for convolution product.

Then we have to determine the inverse of∏j(p − zj)kj in the quotient field of D′

+.By partial fraction decomposition we get

1∏j(z − zj)kj

=∑j

{cj,kj

(z − zj)kj+ · · · + cj,1

(z − zj)

},


the cj,l being complex scalars. The inverse of (p−zj)m is Y(x) ezjx xm−1/(m− 1)!,which is an element ofD′

+ again. Hence (Dδ)−1 is a sum of such expressions

(Dδ)−1 = Y(x)∑j

(cj,kj

xkj−1

(kj − 1)!+ · · · + cj,1

)ezjx .

Let us consider a particular case.LetD = d2/dx2 +ω2 withω ∈ R. Write

1z2 +ω2

= 12iω

(1

z − iω − 1z + iω

).

Thus(Dδ)−1 = 1

2iωY(x)

(eiωx − e−iωx

) = Y(x) sinωxω

.

Another application concerns integral equations. Consider the following integralequation on [0, ∞]:

x∫0

cos(x − t) f(t)dt = g(x) (g given, x ≥ 0) .

Let us assume that f is continuous and g is C1 for x ≥ 0. Replacing f and g by Yfand Yg, respectively, we can interpret this equation in the sense of distributions, andwe actually have to solve

Y(x) cosx ∗ Yf = YginD′

+. We have

Y(x) cosx = Y(x)[

eix + e−ix

2

]= 1

2

[1

p − i +1

p + i

]= pp2 + 1

.

The inverse is therefore

p2 + 1p

= p + 1p= δ′ + Y(x) .

Hence

Yf = Yg∗(δ′+Y(x)) = (Yg)′+Y(x)x∫0

g(t)dt = Yg′+g(0)δ+Y(x)x∫0

g(t)dt .

We conclude that g(0) has to be zero since we want a function solution. But this wasclear a priori from the integral equation.

Volterra Integral Equations of the Second Kind 47

6.8 Volterra Integral Equations of the Second Kind

Consider for x ≥ 0 the equation

f(x)+x∫0

K(x − t) f(t)dt = g(x) . (6.1)

We assume(i) K and g are locally integrable, f locally integrable and unknown,(ii) K(x) = g(x) = f(x) = 0 for x ≤ 0.

Then we can rewrite this equation as a convolution equation inD′+

f + K ∗ f = g or (δ+ K)∗ f = g .

Equation (6.1) is a particular case of a Volterra integral equation of the second kind.The general form of a Volterra integral equation is:

x∫0

K(x, y)f(y)dy = g(x) (first kind),

f(x) +x∫0

K(x, y)f(y)dy = g(x) (second kind).

Here we assume that x ≥ 0 and that K is a function of two variables on 0 ≤ y ≤ x.For an example of a Volterra integral equation of the first kind, see Section 6.7. Volter-ra integral equations of the second kind arise for example when considering linearordinary differential equations with variable coefficients

dmudxm

+ a1(x)dm−1udxm−1

+ · · · + am−1(x)dudx

+ am(x) = g(x)

in which the ai(x) and g(x) are supposed to be continuous.Let us impose the initial conditions

u(0) = u0, u′(0) = u1, . . . , u(m−1)(0) = um−1 .

Then this differential equation together with the initial conditions is equivalent withthe Volterra integral equation of the second kind

v(x)+x∫0

K(x, y)v(y)dy = w(x)


with

v(x) = u(m)(x) ,

K(x,y) = a1(x)+ a2(x) (x−y)+ a3(x)(x−y)2

2!+ · · · + am(x) (x−y)

m−1

(m−1)!,

w(x) = g(x)−m−1∑l=0

m−1∑k=l

uk am−l(x)xk−l

(k− l)! .

In particular, we obtain a convolution Volterra equation of the second kind if allai areconstants, so in case of a constant coefficient differential equation. Of course we haveto restrict to [0, ∞), but a similar treatment can be given on (−∞, 0], by working inD′−.

ExampleConsider d2u/dx2 +ω2u = |x|. Then v = u′′ satisfies

v(x) +ω2

x∫0

(x −y)v(y)dy = |x| −ω2u0 −ω2u1x .

Theorem 6.27. For all locally bounded, locally integrable functions K on [0, ∞), thedistribution A = δ+ K is invertible inD′

+ and A−1 is again of the form δ +H with Ha locally bounded, locally integrable function on [0, ∞).

Proof. Let us write, symbolically, δ = 1 and K = q. Then we have to determine theinverse of 1+ q, first in the quotient field ofD′

+ and then, hopefully, inD′+ itself. We

have1

1+ q =∞∑k=0

(−1)k qk

provided this series converges inD′+. Set

E = δ− K + K∗2 + · · · + (−1)kK∗k + · · · .

Is this series convergent in D′+? Yes, take an interval 0 ≤ x ≤ a and set Ma =ess. sup0≤x≤a |K(x)|. For x in this interval one has

∣∣∣K∗2(x)∣∣∣ = ∣∣∣∣∣

x∫0

K(x − t)K(t)dt

∣∣∣∣∣ ≤ xM2a ,

hence, with induction on k, ∣∣∣K∗k(x)∣∣∣ ≤ xk−1

(k− 1)!Mka ,

and thus |K∗k(x)| ≤ Mkaak−1/(k− 1)! (k = 0,1,2, . . .) for 0 ≤ x ≤ a. On the right-

hand side we see terms of a convergent series, with sumMa eMaa. Therefore the series

Exercises 49

∑∞k=1(−1)k K∗k converges to a locally bounded, locally integrable, function H on

[0, ∞) by Lebesgue’s theorem on dominated convergence, hence the series convergesinD′

+. Thus E = δ+H and E ∈ D′+. Verify that E ∗A = δ. �

Thus the solution of the convolution equation (6.1) is

f = (δ+H)∗ g , or f(x) = g(x)+x∫0

H(x − t)g(t)dt (x ≥ 0) .

We conclude that the solution is given by a formula very much like the original equa-tion (6.1).

6.9 Exercises

Exercise 6.28. Determine the inverses of the following distributions inD′+:

δ′′ − 5δ′ + 6δ; Y + δ′′; Y(x) ex + δ′ .

Exercise 6.29. Solve the following integral equation:

x∫0

(x − t) cos(x − t) f(t)dt = g(x) ,

for g a given function and f an unknown function, both locally integrable and withsupport in [0, ∞).

Exercise 6.30. Let g be a function with support in [0, ∞). Find a distribution f(x)with support in [0, ∞) that satisfies the equation

x∫0

(e−t − sin t

)f(x − t)dt = g(x) .

Which conditions have to be imposed on g in order that the solution is a continuousfunction? Determine the solution in case g = Y , the Heaviside function.

Exercise 6.31. Solve the following integral equation:

f(x) +x∫0

cos(x − t) f(t)dt = g(x) .

Here g is given, f unknown. Furthermore Suppf and Suppg are contained in [0, ∞).


Exercise 6.32. Let f(t) be the solution of the differential equation

u′′′ + 2u′′ +u′ + 2u = −10 cos t ,

with initial conditions

u(0) = 0, u′(0) = 2, u′′(0) = −4 .

Set F(t) = Y(t)f(t). Write the differential equation for F , in the sense of distributions.Determine then F by symbolic calculus inD′+.

6.10 Systems of Convolution Equations*

Consider the system of n equations with n unknowns Xj

A11 ∗X1 +A12 ∗X2 + · · · +A1n ∗Xn = B1

...An1 ∗X1 +An2 ∗X2 + · · · +Ann ∗Xn = Bn ,

where Aij ∈ A′, Xj ∈A′, Bj ∈ A′.The following considerations hold for any system of linear equations over an ar-

bitrary commutative ring with unit element (for exampleA′).Let us writeA = (Aij), B = (Bj), X = (Xj). Then we have to solve A∗X = B.

Theorem 6.33. The equationA∗X = B has a solution for all B if and only ifΔ = detAhas an inverse inA′. The solution is then unique, X = A−1 ∗ B.

Proof.(i) SupposeA∗X = B has a solution for all B. Choose B such that Bi = δ, Bj = 0 for

j �= i, and call the solution in this case Xi = (Cij). Let i vary between 1 and n.Then

n∑k=1

Aik ∗ Ckj ={

0 if i �= jδ if i = j ,

or A∗ C = δ I with C = (Cij). Then detA∗ detC = δ, hence Δ has an inverseinA′.

(ii) Assume that Δ−1 exists in A′. Let aij be the minor of A = (Aij) equal to(−1)i+j× the determinant of the matrix obtained fromA by deleting the jth rowand the ith column. SetCij = aij∗Δ−1. ThenA∗C = C∗A = δ I, soA∗X = Bis solvable for any B, X = C ∗ B.

The remaining statements are easy to prove and left to the reader. �

Exercises 51

6.11 Exercises

Exercise 6.34. Solve inD′+ the following system of convolution equations:{δ′′ ∗X1 + δ′ ∗X2 = δδ′ ∗X1 + δ′′ ∗X2 = 0 .

Further Reading

Rigorous proofs of the existence of the tensor product and convolution product arein [5], [10] and also in [15]. For the impact on the theory of differential equations,see [7].

7 The Fourier TransformSummaryFourier transformation is an important tool in mathematics, physics, computer sci-ence, chemistry and themedical andpharmaceutical sciences. It is used for analyzingand interpreting of images and sounds. In this chapter we lay the fundamentals of thetheory. We define the Fourier transform of a function and of a distribution. Unfortu-nately not every distribution has a Fourier transform, we have to restrict to so-calledtempered distributions. As an example we determine the tempered solutions of theheat or diffusion equation.

Learning Targets� A thorough knowledge of the Fourier transform of an integrable function.� Understanding the definition and properties of tempered distributions.� How to determine a tempered solution of the heat equation.

7.1 Fourier Transform of a Function on R

For f ∈ L1(R) we define

f (y) =∞∫−∞f(x) e−2πixydx (y ∈ R) .

We call f the Fourier transform of f , also denoted byFf .We list some elementary properties without proof.

a. (c1f1 + c2f2) = c1f1 + c2f2 for c1, c2 ∈ C and f1, f2 ∈ L1(R).b. |f (y)| ≤ ‖f‖1 , f is continuous and lim|y|→∞ f (y) = 0 (Riemann–Lebesgue

lemma) for all f ∈ L1(R). To prove the latter two properties, approximate f bya step function. See also Exercise 4.9 a.

c. (f ∗ g) = f · g for all f , g ∈ L1(R).

d. Let f (x) = f(−x)(x ∈ R). Then (f ) = f for all f ∈ L1(R).e. Let (Ltf)(x) = f(x − t), (Mρf)(x) = f(ρx), ρ > 0. Then

(Ltf)ˆ(y) = e−2πity f (y) ,

[e2πitxf(x

](y) = (Ltf )(y) ,

(Mρf )(y) = 1ρf(yρ

)for all f ∈ L1(R) .

Fourier Transform of a Function on R 53

Examples1. LetA > 0 and ΦA the characteristic function of the closed interval [−A, A].

Then ΦA(y) = sin 2πAyπy

(y �= 0) , ΦA(0) = 2A .

2. Set

Δ(x) = { 1− |x| for |x| ≤ 1

0 for |x| > 1 .

(triangle function)

Then Δ = Φ1/2 ∗ Φ1/2, so Δ(y) =(

sinπyπy

)2

(y �= 0) , Δ(0) = 1 .

3. Let T be the trapezoidal function

T(x) =

⎧⎪⎪⎨⎪⎪⎩1 if |x| ≤ 1

1− |x| if 1 < |x| ≤ 2

0 if |x| ≥ 2 .

Then T = Φ1/2 ∗ Φ3/2, hence T (y) =sin 3πy sinπy

(πy)2(y �= 0) and T (0) = 3.

4. Let f(x) = e−a|x|, a > 0. Then f (y) = 2aa2 + 4π2y2

.

5. Let f(x) = e−ax2 , a > 0. Then we get by complex integration f (y) = √π/a ·

e−π2y2/a. In particularF(e−πx2) = e−πy

2 .We know

∫∞−∞ e−ax

2dx =

√π/a for a > 0. From this result we shall deduce, ap-

plying Cauchy’s theorem from complex analysis, that∫∞−∞ e−a(t+ib)

2dt = √π/a

for all a > 0 and b ∈ R.Consider the closed pathW in the complex plane consisting of the line segmentsfrom −R to R (k1), from R to R + ib (k2), from R + ib to −R + ib (k3) and from−R + ib to −R (k4). Here R is a positive real number.

R

R + ib−R + ib

−R →

↑

←

↓

k1

k2

k3

k4

54 The Fourier Transform

Since e−az2 is analytic in z, one has by Cauchy’s theorem

∮W e−az

2dz = 0. Let

us split this integral into four pieces. One has

∫k1

e−az2

dz =R∫−R

e−ax2dx ,

∫k3

e−az2

dz = −R∫−R

e−a(t+ib)2dt .

Parametrizing k2 with z(t) = R + it, we get∫k2

e−az2

dz = ib∫0

e−a(R+it)2dt .

Since ∣∣∣ e−a(R+it)2∣∣∣ = e−a(R

2−t2) ≤ e−a(R2−b2) (a ≤ t ≤ |b|) ,

we get ∣∣∣∣∣∫k2

e−az2

dz

∣∣∣∣∣ ≤ |b| e−a(R2−b2) , hence limR→∞

∫k2

e−az2

dz = 0 .

In a similar waylimR→∞

∫k4

e−az2

dz = 0 .

Therefore ∞∫−∞

e−a(t+ib)2dt =

∞∫−∞

e−ax2dx =

√πa.

We now can determine f

f (y) =∞∫−∞

e−ax2e−2πixy dx = e−π

2y2/a∞∫−∞

e−a(x+πiy

a

)2

dx =√πa

e−π2y2

a .

7.2 The Inversion Theorem

Theorem 7.1. Let f ∈ L1(R) and x a point where f(x + 0) = limx↓0 f(x) andf(x − 0) = limx↑0 f(x) exist. Then one has

limα↓0

∞∫−∞f (y) e2πixy e−4π2αy2

dy = f(x + 0)+ f(x − 0)2

.

The Inversion Theorem 55

Proof. Step 1. One has∞∫−∞f (y) e2πixy e−4π2αy2

dy =∞∫−∞

∞∫−∞f(t) e2πi(x−t)y e−4π2αy2

dtdy

=∞∫−∞f(t)

[ ∞∫−∞

e−4π2αy2e2πi(x−t)ydy

]dt (by Fubini’s theorem)

= 12√πα

∞∫−∞f(t) e

(x−t)24α dt = 1

2√πα

∞∫−∞f(x + t) e−

t24α dt .

Step 2. We now get∞∫−∞f (y) e2πixy e−4π2αy2

dy −{f(x + 0)+ f(x − 0)

2

}

= 12√πα

∞∫0

[{f(x + t)− f(x − 0)

}+ {f(x − t)− f(x − 0)}]

e−t2/4αdt

= 12√πα

δ∫0

· · · + 12√πα

∞∫δ

· · · = I1 + I2 ,

where δ > 0 still has to be chosen.

Step 3. Set φ(t) = |f(x + t) − f(x + 0) + f(x − t) − f(x − 0)|. Let ε > 0 begiven. Choose δ such that φ(t) < ε/2 for |t| < δ, t > 0. This is possible, sincelimt↓0φ(t) = 0. Then we have |I1| ≤ ε/2.Step 4. We now estimate I2. We have

|I2| ≤1

2√πα

∞∫δ

φ(t) e−t2/4αdt ≤ 2

√α√π

∞∫δ

φ(t)t2

dt .

Now∫∞δ φ(t)/t2dt is bounded, because

∞∫δ

φ(t)t2

dt ≤∞∫δ

|f(x + t)|t2

dt +∞∫δ

|f(x − t)|t2

dt +∞∫δ

|f(x + 0)− f(x − 0)|t2

dt

≤ 2δ2‖f‖1 + 1

δ{|f(x + 0)− f(x − 0)|} .

Hence limα↓0 I2 = 0, so |I2| < ε/2 as soon asα is small, say 0 < α < η.

Step 5. Summarizing: if 0 < α < η, then∣∣∣∣∣∞∫−∞f (y) e2πixy e−4π2αy2

dy − f(x + 0)+ f(x − 0)2

∣∣∣∣∣ < ε . �


Corollary 7.2. If both f and f belong to L1(R) and f is continuous in x, then

f(x) =∞∫−∞f (y) e2πixy dy .

Corollary 7.3. Let f ∈ L1(R) and f ≥ 0. If f is continuous in x = 0, then one hasf ∈ L1(R) and therefore

f(0) =∞∫−∞f (t)dt .

Proof. Corollary 7.2 is easily proved by applying Lebesgue’s theorem on dominatedconvergence. Corollary 7.3 is a consequence of Fatou’s lemma. �

Applications1. Let Pa(x) = (1/π)[a/(x2+a2)] (a > 0), see Section 6.2. Then we know by Sec-

tion 7.1, Example 4, that Pa(x) = F (e−2πa|y|)(x). Therefore, by the inversiontheorem, e−2πa|y| = F (Pa)(y). Hence

F (Pa ∗ Pb)(y) = e−2πa|y| e−2πb|y| = e−2π(a+b)|y| = F (Pa+b)(y) .

Again by the inversion theorem we get Pa ∗ Pb = Pa+b.2. If f and g are in L1(R), and if both functions are continuous, then f = g implies

f = g. Later on we shall see that this is true without the continuity assumption,or, otherwise formulated, thatF is a one-to-one linear map.

7.3 Plancherel’s Theorem

Theorem 7.4. If f ∈ L1 ∩ L2, then f ∈ L2 and ‖f‖2 = ‖f‖2.

Proof. Let f be a function in L1∩L2. Then f∗f is in L1 and is a continuous function.Moreover (f∗f ) = |f |2. According to Corollary 7.3 we have |f |2 ∈ L1, hence f ∈ L2.Furthermore

f ∗ f (0) =∞∫−∞

∣∣f(x)∣∣2dx =

∞∫−∞

∣∣∣f (y)∣∣∣2du .

So ‖f‖2 = ‖f‖2. �

The Fourier transform is therefore an isometric linear mapping, defined on thedense linear subspace L1∩L2 of L2. Let us extend this mappingF to L2, for instanceby

Ff(x) = limk→∞

k∫−kf(t)e−2πixtdt . (L2-convergence)

Differentiability Properties 57

We then have:

Theorem 7.5 (Plancherel’s theorem).The Fourier transform is a unitary operator on L2.

Proof. For any f , g ∈ L1 ∩ L2 one has∞∫−∞f(x)g(x)dx =

∞∫−∞f (x)g(x)dx .

Since both∫∞−∞ f(x)(Fg)(x)dx and

∫∞−∞(Ff)(x)g(x)dx are defined forf , g ∈ L2,

and because ∣∣∣∣∣∞∫−∞f(x)(Fg)(x)dx

∣∣∣∣∣ ≤ ‖f‖2 ‖Fg‖2 = ‖f‖2 ‖g‖2 ,

∣∣∣∣∣∞∫−∞(Ff)(x)g(x)dx

∣∣∣∣∣ ≤ ‖Ff‖2 ‖g‖2 = ‖f‖2 ‖g‖2 ,

we have for any f , g ∈ L2

∞∫−∞f(x)(Fg)(x)dx =

∞∫−∞(Ff)(x)g(x)dx .

Notice thatF(L2)⊂L2 is a closed linear subspace. We have to show thatF(L2)=L2.Let g ∈ L2, g orthogonal to F(L2), so

∫∞−∞(Ff)(x)g(x)dx = 0 for all f ∈ L2.

Then, by the above arguments,∫∞−∞ f(x)(Fg)(x)dx = 0 for all f ∈ L2, hence

F(g) = 0. Since ‖g‖2 = ‖Fg‖2, we get g = 0. SoF(L2) = L2. �

7.4 Differentiability Properties

Theorem 7.6. Let f ∈ L1(R) be continuously differentiable and let f ′ be in L1(R) too.Then f ′(y) = (2πiy) f (y).

Proof. By partial integration and for y �= 0 we have

A∫−Af(x) e−2πixy dx = e−2πixy

−2πiyf(x)

∣∣∣x=Ax=−A +

12πiy

A∫−Af ′(x) e−2πixy dx .

Notice that lim|x|→∞ f(x) exists, since f(x) = f(0) +∫ x0 f ′(t)dt and f ′ ∈ L1(R).

Hence, because f ∈ L1(R), lim|x|→∞ f(x) = 0. Then we get∞∫−∞f(x) e−2πixy dx = 1

2πiy

∞∫−∞f ′(x) e−2πixy dx ,


or (2πiy) f (y) = f ′(y) for y �= 0. Since both sides are continuous in y, thetheorem follows. �

Corollary 7.7. Let f be a Cm function and let f (k) ∈ L1(R) for all k with 0 ≤ k ≤m.Then

f (m)(y) = (2πiy)m f (y) .

Notice that for f as above, f (y) = o(|y|−m).

Theorem 7.8. Let f ∈ L1(R) and let (−2πix)f(x) = g(x) be in L1(R) too. Then fis continuously differentiable and (f )′ = g.

Proof. By definition

(f )′(y0) = limh→0

∞∫−∞f(x) e−2πixy0

[e−2πixh − 1

h

]dx .

We shall show that this limit exists and that wemay interchange limit and integration.This is a consequence of Lebesgue’s theorem on dominated convergence. Indeed,∣∣∣∣∣ f(x) e−2πixy0

[e−2πixh − 1

h

]∣∣∣∣∣ ≤|f(x)| |2πx| | sin 2πxθ1(h, x)− i cos 2πxθ2(h, x)|

with 0 ≤ θi(h, x) ≤ 1 (i = 1,2), hence bounded by 2|g(x)|. Since g ∈ L1(R) weare done.

�

Corollary 7.9. Let f ∈ L1(R), gm(x) = (−2πix)m f(x) and gm ∈ L1(R) too. Thenf is Cm and f (m) = gm.

Both Corollary 7.7 and Corollary 7.9 are easily shown by induction onm.

7.5 The Schwartz Space S(R)Let f be in L1(R). Then one obviously has, applying Fubini’s theorem, 〈f , ϕ〉 =〈f , ϕ〉 for allϕ ∈ D(R). If T is an arbitrary distribution, then a “natural” definitionof T would be 〈T , ϕ〉 = 〈T , ϕ〉(ϕ ∈ D(R)). The spaceD(R) is however not closedunder taking Fourier transforms. Indeed, if ϕ ∈ D(R), then ϕ can be continuedto a complex entire analytic function ϕ(z). This is seen as follows. Let Suppϕ ⊂[−A, A]. Then

ϕ(z) =A∫−Aϕ(y) e−2πiyz dy =

∞∑k=0

(−2πiz)k

k!

A∫−Aϕ(y)yk dy

The Schwartz Space S(R) 59

and this is an everywhere convergent power series. Therefore, if ϕ were in D(R),thenϕ = 0.

Thus we will replaceD = D(R) with a new space, closed under taking Fouriertransforms, the Schwartz space S = S(R).

Definition 7.10. The Schwartz space S consists of C∞ functions ϕ such that for allk, l ≥ 0, the functions xl ϕ(k)(x) are bounded onR.

Elements of S are called Schwartz functions, also functions whose derivatives are“rapidly decreasing” (à décroissance rapide). Observe that D ⊂ S. An example ofa new Schwartz function isϕ(x) = e−ax

2(a > 0).

Properties of S1. S is a complex vector space.2. Ifϕ ∈ S, then xl ϕ(k)(x) and [xl ϕ(x)](k) belong to S as well, for all k, l ≥ 0.3. S ⊂ Lp for all p satisfying 1 ≤ p ≤ ∞.4. S = S.

The latter property follows from Section 7.4 (Corollaries 7.7 and 7.9).One defines on S a convergence principle as follows: say that ϕj tends to zero

(notationϕj → 0) when for all k, l ≥ 0,

sup∣∣∣xl ϕ(k)

j (x)∣∣∣ → 0

if j tends to infinity.One then has a notion of continuity in S, in a similar way as in the spaces D

and E.

Theorem 7.11. The Fourier transform is an isomorphism of S.

Proof. The linearity of the Fourier transform is clear. Let us show the continuity of theFourier transform. Let {ϕj} be a sequence of functions in S.(i) Notice that (2πiy)l ϕ(k)

j (y) = ψj(y)whereψj(x) = [(−2πix)k ϕj(x)](l).Clearlyψj ∈ S.

(ii) One has |(2πiy)l ϕ(k)j (y)| = |ψj(y)| ≤ ‖ψj‖1.

(iii) The following inequalities hold:

‖ψj‖1 ≤∞∫−∞|ψj(x)|dx =

1∫−1

|ψj(x)|dx +∫

|x|≥1

|ψj(x)|dx

≤ 2 sup |ψj| +∫

|x|≥1

|x2ψj(x)|x2

dx ≤ 2 sup |ψj| + 2 sup |x2ψj(x)| .


Now, if ϕj → 0 in S, then ψj → 0 in S, hence ‖ψj‖1 → 0 and thereforesup |(2πiy)l ϕ(k)

j (y)| → 0 if j → ∞. Hence ϕj → 0 in S. Thus the Fouriertransform is a continuous mapping from S to itself.In a similar way one shows that the inverse Fourier transform is continuous on S.Hence the Fourier transform is an isomorphism. �

Observe that item (iii) in the above proof implies that the injection S ⊂ L1 iscontinuous; the same is true for S ⊂ Lp for all p satisfying 1 ≤ p ≤ ∞.

7.6 The Space of Tempered Distributions S′(R)One has the inclusions D ⊂ S ⊂ E. The injections are continuous, the images aredense. Compare with Section 5.1.

Definition 7.12. A tempered distribution is a distribution that can be extended to a con-tinuous linear form on S.

Let S′ = S′(R) be the space of continuous linear forms on S. Then, by the re-mark at the beginning of this section, S′ can be identified with the space of tempereddistributions.

Examples1. Let f ∈ Lp,1 ≤ p ≤ ∞. Then Tf is tempered since S ⊂ Lq, for q such that

1/p + 1/q = 1, and the injection is continuous.2. Let f be a locally integrable function that is slowly increasing (à croissance lente):

there are A > 0 and an integer k ≥ 0 such that |f(x)| ≤ A |x|k if |x| → ∞.Then Tf is tempered.This follows from the following inequalities:∣∣∣∣∣∞∫−∞f(x)ϕ(x)dx

∣∣∣∣∣ ≤∣∣∣∣∣

1∫−1

f(x)ϕ(x)dx

∣∣∣∣∣+∣∣∣∣∣∫

|x|≥1

f(x)ϕ(x)dx

∣∣∣∣∣≤ sup |ϕ|

1∫−1

∣∣f(x)∣∣dx + sup∣∣∣ϕ(x)xk+2

∣∣∣ ∫|x|≥1

|f(x)||x|k+2

dx .

3. Every distributionT with compact support is tempered. To see this, extend T toE(Theorem 5.2) and then restrict T to S.

4. Sinceϕ → ϕ′ is a continuous linear mapping from S to S, the derivative of a tem-pered distribution is again tempered.

5. If T is tempered and α a polynomial, then αT is tempered, since ϕ → αϕ isa continuous linear map from S into itself.

6. Let f(x) = ex2 . Then Tf is not tempered.

Structure of a Tempered Distribution* 61

Indeed let α ∈ D, α ≥ 0, α(x) = α(−x) and∫∞−∞α(x)dx = 1. Let χj be

the characteristic function of the closed interval [−j, j] and set αj = χj ∗ α. Thenαj ∈ D and αj ϕ → ϕ in S for allϕ ∈ S. Chooseϕ(x) = e−x

2 . Then 〈Tf , αjϕ〉 =∫∞−∞αj(x)dx. But limj→∞

∫∞−∞αj(x)dx = ∞.

Similar to Propositions 2.3 and 5.4 one has:

Proposition 7.13. Let T be a distribution. Then T is tempered if and only if there existsa constant C > 0 and a positive integerm such that

|〈T , ϕ〉| ≤ C∑

k,l≤msup

∣∣∣xkϕ(l)(x)∣∣∣

for allϕ ∈ D.

7.7 Structure of a Tempered Distribution*

This section is devoted to the structure of tempered distributions on R.

Theorem 7.14. Let T be a tempered distribution on R. Then there exists a continuous,slowly increasing, function f on R and a positive integerm such that T = f (m) in thesense of distributions.

Proof. Since T is tempered, there exists a positive integer N such that

|〈T , ϕ〉| ≤ const.∑k,l≤N

sup∣∣∣xkϕ(l)(x)

∣∣∣for allϕ ∈ D. Consider (1+ x2)−N T . We have∣∣∣∣⟨(1+ x2

)−NT , ϕ

�∣∣∣∣ ≤ const.∑k,l≤N

sup

∣∣∣∣∣xk[(

1+ x2)−N

ϕ(x)](l)∣∣∣∣∣

≤ const.∑k≤N

sup∣∣∣ϕ(k)(x)

∣∣∣ 11+ x2

.

By partial integration from −∞ to x, we obtain

|ϕ(k)(x)|1+ x2

≤∞∫−∞

|2xϕ(k)(x)|(1+ x2)2

dx +∞∫−∞

|ϕ(k+1)(x)|1+ x2

dx .

Hence∣∣∣∣⟨(1+ x2

)−NT , ϕ

�∣∣∣∣ ≤ const.∑

k≤N+1

{ ∞∫−∞

∣∣∣ϕ(k)(x)∣∣∣2

dx}1/2

for allϕ ∈ D. LetHN+1 be the (Sobolev) space of all functionsϕ ∈ L2(R) for whichthe distributional derivativesϕ′, . . . ,ϕ(N+1) also belong to L2(R), provided with the


scalar product

(ψ, ϕ) =N+1∑k=0

∞∫−∞ψ(k)(x)ϕ(k)(x)dx .

It is not difficult to show that HN+1 is a Hilbert space (see also Exercise 7.24). There-fore, (1 + x2)−N T can be extended to HN+1 and thus (1 + x2)−N T = g for someg ∈ HN+1, i. e.

⟨(1+ x2)−N T , ϕ⟩ = (g,ϕ) = N+1∑

k=0

∞∫−∞g(k)(x)ϕ(k)(x)dx .

Thus (1+ x2

)−NT =

N+1∑k=0

(−1)k g(2k)

in the sense of distributions. Now write

G(x) =x∫0

g(t)dt .

Then G is continuous, slowly increasing because |G(x)| ≤√|x| ‖g‖2, and G′ = g

in the sense of distributions. Thus(1+ x2

)−NT =

N+1∑k=0

(−1)k G(2k+1)

onD. Hence T is of the form

T =N+1∑k=0

dk Gk

for some constants dk, while eachGk is a linear combination of terms of the form{[(1+ x2)N](l)G}(2k+1−l)

(0 ≤ l ≤ 2k+ 1) .

So, a fortiori, each term of Gk is a slowly increasing continuous function. Integratingsome terms of Gk a number of times and keeping in mind that if h is a continuousslowly increasing function, then x �

∫ x0 h(t)dt is again such a function, we get the

following result:T = f (m) ,

withm = 2N + 3 and f a continuous, slowly increasing, function. This completesthe proof. �

The following corollary turns out to be useful in Chapter 8.

Corollary 7.15. Let T be a tempered distribution onRwith support contained in (0, ∞).There exists a continuous, slowly increasing, function f with Suppf ⊂ (0, ∞) anda positive integerm such that T = f (m) in the sense of distributions.

Fourier Transform of a Tempered Distribution 63

Proof. Let T = f (m) as in Theorem 7.14. Since SuppT ⊂ (0, ∞), we can find β ∈E(R) with Suppβ ⊂ (0, ∞) and β(x) = 1 for x in a neighborhood of SuppT . ThenT = βT . We then get for allϕ ∈ D

〈T , ϕ〉 = 〈T , βϕ〉 =⟨f (m), βϕ

⟩=

m∑k=0

(−1)k(mk

) ⟨[β(k)f

](m−k), ϕ⟩ ,hence T = ∑m

k=0(−1)k(mk

)[β(k)f](m−k). Integrating some terms once more a num-

ber of times gives the desired result. �

7.8 Fourier Transform of a Tempered Distribution

Let T ∈ S′.

Definition 7.16. The Fourier transform T of T is defined by 〈T , ϕ〉 = 〈T , ϕ〉(ϕ ∈ S).

Observe that T ∈ S′. We shall also write T = F T .If

Fϕ(x) =∞∫−∞ϕ(y) e2πixy dy(ϕ ∈ S) ,

then F T can be defined in a similar way for T ∈ S′. One has FF = FF = id in Sand in S′.

Choose the following convergence principle in S′: Tj → 0 if 〈Tj, ϕ〉 → 0 for allϕ ∈ S (j → ∞). Compare this with Section 4.3. As said before, this leads to a notionof continuity in S′. ThenF andF are isomorphisms of S′.

Examples1. δ = 1.2. 1 = δ. This relation is sometimes symbolically written as

∫∞−∞ e−2πixy dy =

δ(x).3. F δ′ = 2πix,F δ(m) = (2πix)m.4. F δ(a) = e2πiax .5. If T ∈ S′, thenF (T (m)) = (2πix)m (F T) andF ((−2πix)m T) = (F T)(m).

One also has– If f ∈ L1, then Tf = Tf . This implies, in particular, using Corollary 6.12, that

f = 0 gives f = 0. See also Section 7.2, Application 2.– If f ∈ L2, then Tf is tempered and, by Plancherel’s theorem, Tf = Tf .– If f ∈ Lp, then f ∈ Lq with q such that 1/p + 1/q = 1, 1 ≤ p ≤ 2 (without

proof; see [11]).


Let T be a distribution with compact support, so T ∈ E′. Then 〈Tx, e−2πixz〉(z ∈ C) is well defined and one has

〈Tx, e−2πixz〉 =⟨Tx,

∞∑k=0

(−2πix)k

k!zk⟩=

∞∑k=0

⟨Tx,

(−2πix)k

k!

⟩zk ,

for every z ∈ C, because the power series for e−2πixz not only converges uniformlyon compact sets (with z fixed), but also in E. The conclusion is: z → 〈Tx, e−2πixz〉is an entire analytic function.

Theorem 7.17. For T ∈ E′, the function y → 〈Tx, e−2πixy〉 is the Fourier transformofT . It is a slowly increasing function that can be extended to an entire analytic functionon the complex plane, given by z → 〈Tx, e−2πixz〉.

Proof. Let α ∈ D be such that α(x) = 1 on a neighborhood of SuppT . Call K =Suppα. By Section 2.2, Proposition 2.3, there is an integer m ≥ 0 and a constantc′ > 0 such that

|〈T , ϕ〉| ≤ c′m∑k=0

sup∣∣∣ϕ(k)(x)

∣∣∣for allϕ ∈ Dwith Suppϕ ⊂ K. Forψ ∈ E setϕ = αψ. Then we get

|〈T , ψ〉| = |〈T , ϕ〉| ≤ cm∑k=0

supx∈K∣∣∣ψ(k)(x)∣∣∣

for some constant c > 0 and allψ ∈ E. Hence,

|〈Tx, e−2πixy〉| ≤ cm∑k=0

supx∈K |(2πiy)m| ≤ const. |y|m ,

if |y| → ∞. So y → 〈Tx, e−2πixy〉 is a slowly increasing function and thus definesa tempered distribution. Moreover, one has for allϕ ∈ D⟨

T , ϕ⟩= ⟨T , ϕ⟩=⟨Tx,

∞∫−∞ϕ(y) e−2πixy dy

⟩

=⟨Tx,

∞∫−∞α(x)ϕ(y)e−2πixy dy

⟩

=⟨Tx ⊗ 1y, α(x)ϕ(y) e−2πixy

⟩

=∞∫−∞ϕ(y)

⟨Tx, e−2πixy

⟩dy . (“Fubini”)

The tempered distributions T and (the distribution defined by) the functionx � 〈Tx, e−2πixy〉 thus coincide. �

Paley–Wiener Theorems onR* 65

Fourier Transform and ConvolutionIf S, T ∈ E′, then S ∗ T ∈ E′ and (S ∗ T) is a C∞ function, which is clearly equal toy � S(y)T (y).

We mention (without proof): if S ∈ E′ and T ∈ S′, then S ∗ T ∈ S′ and(S ∗ T) = S(y) T . Notice that S(y) is a slowly increasing C∞ function. Further-more, we already knew that for f , g ∈ L1, (f ∗ g)(y) = f (y) g(y) for all y ∈ R.

7.9 Paley–Wiener Theorems on R*

The Paley–Wiener theorems characterize the Fourier transforms ofC∞ functions withcompact support and distributions with compact support.

Theorem 7.18.a. Let ϕ ∈ D with Suppϕ ⊂ [−A, A]. Then ϕ is an entire analytic function on C

that satisfies(∗) for every integerm ≥ 0 there is a constant cm > 0 such that for allw ∈ C

|ϕ(w)| ≤ cm (1+ |w|)−m e2πA | Imw| .

b. Let F be an entire analytic function satisfying (∗), for some A > 0. Then there isa functionϕ ∈ D with Suppϕ ⊂ [−A, A] and ϕ = F .

Theorem 7.19.a. Let T ∈ E′ with SuppT ⊂ [−A, A]. Then T is an entire analytic function on C

that satisfies the following:(∗∗) there exists an integerm ≥ 0 and a constant c > 0 such that for allw ∈ C

|T (w)| ≤ c (1+ |w|)m e2πA | Imw| .

b. Let F be an entire analytic function satisfying (∗∗) for some A > 0. Then there isa distribution T ∈ E′ with SuppT ⊂ [−A, A] and T = F .

Proof. (Theorem 7.18)a. Letϕ ∈ D and Suppϕ ⊂ [−A, A]. Setw = u+ iv. Then

|ϕ(w)| ≤A∫−A|ϕ(x)| e2πxv dx ≤ c e2πA|v| .

Let m be a positive integer. Set ψ(x) = 1/(2πi)mϕ(m)(x). Then ψ(w) =wm ϕ(w). Sinceψ ∈ D there is a constant c′m > 0 such that

|ψ(w)| ≤ c′m e2πA |v| ,


hence there is cm > 0 such that

|ϕ(w)| ≤ cm1

(1+ |w|)m e2πA |v| ,

since (1+ |w|)m/|w|m is bounded for |w| → ∞.b. Let f be the restriction of F to R and setϕ = Ff . Since F satisfies (∗),ϕ is C∞

by Corollary 7.9 and

ϕ(x) =∞∫−∞F(t) e2πixt dt .

It suffices to show thatϕ(x) = 0 for |x| > A.Consider the following pathW (with u fixed):

M

M + iu−M + iu

−M →

↑

←

↓

k1

k2

k3

k4

Then∮W F(w) e2πxw dw = 0 for all x ∈ R. We have

∫k1

F(w) e2πxw dw =M∫−Mf(t) e2πixt dt .

−∫k3

F(w) e2πixw dw = e−2πxuM∫−MF(t + iu) e2πixt dt .

Furthermore∣∣∣∣∣∫k2

F(w) e2πixw dw

∣∣∣∣∣ =∣∣∣∣∣

u∫0

F(M + is) e−2πxs ds

∣∣∣∣∣≤

u∫0

cm(1+M)m e2πA|s| e−2πxs ds ≤ const.

cm(1+M)m ,

Paley–Wiener Theorems onR* 67

and this tends to zero ifM tends to infinity. In a similar way we obtain |∫k4| → 0

ifM → ∞. Therefore

ϕ(x) = e−2πxu

( ∞∫−∞F(t + iu) e2πixt dt

)

for all u. Consequently

|ϕ(x)| ≤ e−2πxu∞∫−∞

c2

1+ t2e2πA|u| dt = const. e2π(A|u|−xu) .

If x > A, let u → ∞. Thenϕ(x) = 0. If x < −A, let u → −∞. Thenϕ(x) = 0

as well. �

Proof. (Theorem 7.19)a. Let T ∈ E′ and SuppT ⊂ [−A, A]. By Theorem 7.17, T is an entire analytic

function. From Proposition 5.4 we obtain

|〈T , ψ〉| ≤ ck∑j=1

supx∈[−A,A] |ψ(j)(x)|

for some c>0, some integer k>0 and allψ∈E. Moreover T (w)=〈Tx, e−2πixw〉.From this relation we easily obtain

|T (w)| ≤ c (k+ 1) (2π)k (1+ |w|)k e2πA| Imw| .

b. Let α∈D, α≥0, α(x)=α(−x) and∫∞−∞α(x)dx=1. Set αε(x)=1/ε α(x/ε)

for any ε > 0.Let F be an entire analytic function satisfying (∗∗). Call f the restriction of Fto R. Then f is a slowly increasing function, hence in S′. Set T = Ff , so T ∈ S′as well. We shall show that SuppT ⊂ [−A, A]. Call Tε = αε ∗ T . From theidentity

〈αεT , ϕ〉 = 〈T , αεϕ〉 = 〈T , (αε ∗ϕ)〉 = 〈T , (αε ∗ϕ)ˇ〉 = 〈αε ∗ T , ϕ〉

forϕ∈D and ϕ(x)=ϕ(−x) (x ∈ R), we see that Tε∈S′ and Tε= αε ·T . Fur-thermore, Tε is an entire analytic function, namely Tε = αεF . Since Suppαε ⊂[−ε, ε]we have by Theorem 7.18: for any k there is a ck > 0 such that

|αε(w)| ≤ck

(1+ |w|)k e2πε| Imw| ,

hence|Tε(w)| ≤

c ck(1+ |w|)k−m e2π(A+ε) | Imw| .

By Theorem 7.18 b., we have Tε ∈ D and SuppTε ⊂ [−A − ε, A + ε]. SinceT = limε↓0 Tε we get SuppT ⊂ [−A, A]. �


7.10 Exercises

Exercise 7.20.a. Show that the Fourier transform of an odd (even) distribution is an odd (even) dis-

tribution.b. Determine all odd distributions satisfying xT = 1.c. Deduce from b. the Fourier transform of pv(1/x).

Exercise 7.21. Determine the Fourier transform f of

f(x) ={|x| if |x| < 1

0 elsewhere.

Show that f is a C∞ function.

Exercise 7.22. Consider the tempered distribution |x|.a. Compute δ′′ ∗ |x|.b. Deduce thatF(|x|) = A[Pf(1/x2)]+ C δ with constants A and C . Determine A.c. Compute d/dx [pv(1/x)]. DetermineF[Pf (1/x2)] and then the constant C .

Exercise 7.23. If the product of two entire analytic functions is zero, then at least oneof both functions is identically zero. Apply this fact for showing that the convolutionalgebra E′ has no zero divisors.

Exercise 7.24. Let H1 be the space of functions f ∈ L2(R), whose first derivative (inthe sense of distributions) belongs to L2(R) as well. ProvideH1 with the scalar product

(f , g)1 =∞∫−∞f(x) g(x)dx +

∞∫−∞

dfdx

dgdx

dx .

a. Show thatH1 is a Hilbert space with normN1(f) = (f , f)1/21 .b. A tempered distribution f belongs toH1 if and only if√

1+ λ2 f (λ) ∈ L2(R) .

Here f is the Fourier transform of f . Prove it.c. For f ∈ H1 set

‖f‖ =∥∥∥√1+ λ2 f (λ)

∥∥∥2.

Show that ‖f‖ andN1(f) are equivalent norms onH1.

Fourier Transform inRn 69

7.11 Fourier Transform in Rn

If x = (x1, . . . , xn), y = (y1, . . . , yn), then we write 〈x, y〉 = ∑ni=1 xiyi. We shall

also write dx for dx1 . . .dxn.For f ∈ L1(Rn), define the Fourier transform f = Ff of f by

f (y) =∫Rn

f(x) e−2πi〈x,y〉 dx (y ∈ Rn) .

This Fourier transform behaves in the same way as in case n = 1. One has, for exam-ple,

F (e−πr2) = e−πr

2(r 2 = x2

1 + · · · + x2n) .

The Schwartz space S(Rn) consists of all C∞ functionsϕ on Rn such that

supx∈Rn |xl Dkϕ(x)| <∞

for all n-tuples l and k of nonnegative integers. The space S(Rn) carries the conver-gence principle set forward by its defining relations.

Tempered distributions on Rn are elements of S′(Rn), the space of continuouslinear forms on S(Rn). Choose in S′(Rn) the similar convergence principle as incase n = 1, see Section 7.8. The reader may verify that the structure theorem fromSection 7.7 can be generalized to Rn, using the same method of proof.

The Fourier transform of a tempered distribution is defined as in case n = 1.The same kind of theorems hold. Also the Paley–Wiener theorems can be generalizedto Rn.

Here are some formulae:

F[∂δ∂xk

]= 2πiyk , F [−2πixk] = ∂δ

∂yk(1 ≤ k ≤ n) .

We proceed now with a detailed study of the Fourier transforms of so-called radialfunctions f on Rn: f(x1, . . . , xn) depends only on r , so f(x1, . . . , xn) = Φ(r).Notice that the requirement f ∈ L1(Rn) is equivalent with rn−1Φ(r) ∈ L1(0, ∞).

Let f ∈ L1(Rn) be a radial function. We claim that f is also radial. This is seenas follows. Observe that f being radial is the same as saying that f(Ax) = f(x)almost everywhere for all orthogonal transformations A. So we have to show thatf (Ay) = f (y) for all y ∈ Rn and all A. This is however clear from the definition off and the properties of A

f(Ay) =∫Rn

f(x) e−2πi〈x,Ay〉 dx =∫Rn

f(Ax) e−2πi〈x,y〉 dx

=∫Rn

f(x) e−2πi〈x,y〉 dx = f (y) .


Let f (y) = Ψ(ρ) with ρ = ‖y‖ = √y21 + · · · + y2

n. We try to determine Ψ in termsof Φ, and conversely. One has for n ≥ 2

Ψ(ρ) = f (0,0, . . . ,0, ρ) = ∫Rn

e−2πixnρ f(x1, . . . , xn)dx1 · · ·dxn

= Sn−2

∞∫0

π∫0

e−2πirρ cosθ Φ(r) rn−1 sinn−2 θ dθ dr ,

applying spherical coordinates and setting

Sn−2 =2π

n−12Γ (n−1

3

) .We observe that some special function occurs in this integral formula, a Bessel

function.The Bessel function of the first kind of order ν has the following integral represen-

tation:

Jν(x) =(x/2)ν√

π Γ (ν + 1/2)

π∫0

e−ix cosθ sin 2ν dθ(Re(ν) > − 1

2

).

For ν = (n− 2)/2, x = 2πρr one obtains

Jn−22(2πρr) = π

n−22 ρ

n−22 r

n−22

√π Γ (n−1

2 )

π∫0

e−2πirρ cosθ sinn−2 θ dθ

= πn−32ρn−2

2 rn−2

2Γ (n−12 )

π∫0

e−2πirρ cosθ sin n−2 θ dθ .

Therefore

Ψ(ρ) = 2π

ρn−2

2

∞∫0

rn2 Jn−2

2(2πρr)Φ(r)dr (n ≥ 2) . (7.1)

One also has, if ρn−1 Ψ(ρ) ∈ L1(0, ∞) and Φ is continuous, then

Φ(r) = 2π

rn−2

2

∞∫0

ρn2 Jn−2

2(2πρr)Ψ(ρ)dρ (n ≥ 2) . (7.2)

One calls the transform in equation (7.1) (and equation (7.2)) Fourier–Bessel transformor Hankel transform of order (n− 2)/2. It is defined on L1((0, ∞), rn−1 dr).

We leave the casen = 1 to the reader.There are many books on Bessel functions, Fourier–Bessel transforms and Han-

kel transforms, see for example [16]. Because of its importance for mathematicalphysics, it is useful to study them in some detail.

The Heat or Diffusion Equation in One Dimension 71

7.12 The Heat or Diffusion Equation in One Dimension

The heat equation was first studied by Fourier (1768–1830) in his famous publication“Théorie analytique de la chaleur”. It might be considered as the origin of, what isnow called, Fourier theory.

|x = 0

Let us consider a bar of length infinity, thin, homogeneous and heat conducting.Denote by U(x, t) the temperature of the bar at place x and time t. Let c be its heatcapacity. This is defined as follows: the heat quantity in 1 cm of the bar with tem-perature U is equal to c U . Let γ denote the heat conducting coefficient: the heatquantity that flows from left to right at place x in one second is equal to −γ ∂U/∂x,with−∂U/∂x the temperature decay at place x.

Let us assume, in addition, that there are some heat sources along the bar withheat density ρ(x, t). This means that between t and t + dt the piece (x, x + dx)of the bar is supplied with ρ(x, t)dx dt calories of heat. We assume furthermorethat there are no other kinds of sources in play for heat supply (neither positive nornegative).

Let us determine how U behaves as a function of x and t. The increase of heat in(x, x + dx) between t and t + dt is

1.[γ∂U∂x(x + dx, t)− γ ∂U

∂x(x, t)

]dt ∼ γ ∂

2U∂x2

dx dt (by conduction),

2. ρ(x, t)dx dt (by sources).

Together:[γ∂2U∂x2

+ ρ(x, t)]

dxdt.

On the other hand this is equal to c (∂U/∂t) dxdt. Thus one obtains the heat equa-tion

c∂U∂t

− γ ∂2U∂x2

= ρ .

Let us assume that the heat equation is valid for distributions ρ too, e. g. for– ρ(x, t) = δ(x): each second only at place x = 0 one unit of heat is supplied to

the bar:– ρ(x, t) = δ(x)δ(t): only at t = 0 and only at place x = 0 one unit of heat is

supplied to the bar, during one second.

We shall solve the following Cauchy problem: find a functionU(x, t) for t ≥ 0, whichsatisfies the heat equationwith initial conditionU(x,0) = U0(x), withU0 aC2 func-tion.


We will consider the heat equation as a distribution equation and therefore set

U(x, t) ={U(x, t) if t ≥ 0 ,0 if t < 0 ,

and we assume thatU is C2 as a function of x and C1 as a function of t for t ≥ 0. Wehave

∂2U∂x2

={∂2U∂x2

},∂U∂t

={∂U∂t

}+ U0(x)δ(t) ,

so that U satisfies

c∂U∂t

− γ ∂2U∂x2

= ρ (x, t)+ c U0(x)δ(t). (7.3)

We assume, in addition,– x � ρ(x, t) is slowly increasing for all t,– x � U(x, t) is slowly increasing for all t,– x � U0(x) is slowly increasing.

Set V (y, t) = Fx U(x, t), σ (y, t) = Fx ρ(x, t), V0(y) = Fx U0(x). Equation (7.3)becomes, after taking the Fourier transform with respect to x,

c∂V (y, t)∂t

+ 4π2γy2 V (y, t) = σ (y, t)+ c V0(y)δ(t) .

Now fixy (that is: consider the distributionϕ � 〈 . , ϕ(t)⊗ψ(y)〉 for fixedψ ∈ D).We obtain a convolution equation inD′

+, namely

(c δ′ + 4π2γy2 δ)∗ V (y, t) = σ (y, t)+ c V0(y)δ(t) .

CallA = c δ′ + 4π2γy2 δ. ThenA−1 = (Y(t)/c) e−4π2γy2t . We obtain the solution

V (y, t) = σ (y, t)∗t Y (t)c e−4π2(γ/c)y2t + V0(y)Y(t) e−4π2(γ/c)y2t .

Observe that U(x, t) = Fy V (y, t).Important special case: U0(x) = 0, ρ (x, t) = δ(x)δ(t). Then one asks in fact

for an elementary solution of equation (7.3). We have σ (y, t) = δ(t), V0(y) = 0,hence

V (y, t) = Y(t)c

e−4π2(γ/c)y2t .

An elementary solution of the heat equation is thus given by

U(x, t) = Y(t)c

1√(γ/c)πt

e−x2

4(γ/c)t .

The Heat or Diffusion Equation in One Dimension 73

We leave it to the reader to check that U indeed satisfies

c∂U∂t

− γ ∂2U∂x2

= δ(x)δ(t) .

If there are no heat sources, then U(x, t) is given by

U(x, t) = c U0(x)∗x E(x, t) if t > 0 ,

withE(x, t) = 1

2c√(γ/c)πt

e−x2

4(γ/c)t (t > 0) .

We leave it as an exercise to solve the heat equation in three dimensions

c∂U∂t

− γΔU = ρ .Further Reading

More on the Fourier transform of functions is in [4] and the references therein. Moreon tempered solutions of differential equations one finds for example in [7]. Noticethat the Poisson equation has a tempered elementary solution. Is this a general phe-nomenon for differential equations with constant coefficients? Another, more mod-ern, method for analyzing images and sounds is the wavelet transform. For an intro-duction, see [3].

8 The Laplace TransformSummaryThe Laplace transform is a useful tool to find nonnecessarily tempered solutions ofdifferential equations. We introduce it here for the real line. It turns out that there isan intimate connection with the symbolic calculus of Heaviside.

Learning Targets� Learning about how the Laplace transform works.

8.1 Laplace Transform of a Function

For any function f on R, that is locally integrable on [0, ∞), we define the Laplacetransform L by

Lf(p) =∞∫0

f(t) e−pt dt (p ∈ C) .

Existence1. If for t ≥ 0 the function e−ξ0t |f(t)| is in L1([0, ∞)) for some ξ0 ∈ R, then

clearlyLf exists as an absolutely convergent integral for all p ∈ Cwith Re(p) ≥ξ0. Furthermore, Lf is analytic for Re(p) > ξ0 and, in that region, we have forall positive integersm

(Lf)(m)(p) =∞∫0

(−t)m f(t) e−pt dt .

Observe that it suffices to show the latter statements for ξ0 = 0 andm = 1. Inthat case we have for p with Re(p) > 0 the inequalities

|f(t)|∣∣∣∣∣e−(p+h)t − e−pt

h

∣∣∣∣∣ ≤ |f(t)| e−ξt∣∣∣∣∣e−ht − 1

h

∣∣∣∣∣ ≤ |f(t)| e−ξt t et|h|

≤ |f(t)| t e−(ξ−ε)t

if |h| < ε and ξ = Re(p). Because ξ > 0, one has ξ − ε > 0 for ε small. Setξ − ε = δ. Then there is a constant C > 0 such that t ≤ C e

12δt . We thus may

conclude that the above expressions are bounded by C |f(t)| provided |h| < ε.Now apply Lebesgue’s theorem on dominated convergence.

2. If f(t) = O (ekt)(t → ∞), then Lf exists for Re(p) > k. In particular, slowlyincreasing functions have a Laplace transform for Re(p) > 0. If f has compactsupport, then Lf is an entire analytic function.

3. If f(t) = e−t2 , then Lf is entire. If f(t) = et

2 , then Lf does not exist.

Laplace Transform of a Distribution 75

Examples1. L1 = 1/p, L(tm) =m!/pm+1 (Re(p) > 0). By abuse of notation we frequently

omit the argument p in the left-hand side of the equations.2. Let f(t) = tα−1, α being a complex number with Re(α) > 0. Then Lf(p) =Γ (α)/pα for Re(p) > 0.3. Set for c > 0,

Hc(t) ={

1 if t ≥ c ,0 elsewhere.

Then L(Hc)(p) = e−cp/p for all p with Re(p) > 0.4. Let f(t) = e−αt (α complex). Then Lf(p) = 1/(p +α) for Re(p) > −Re(α).

In particular, for a real,– L (sinat) = a

p2 + a2for Re(p) > 0 ,

– L (cosat) = pp2 + a2

for Re(p) > 0 .

5. Set h(t) = e−at f(t) with a ∈ R and assume that Lf exists for Re(p) > ξ0.Then Lh exists for Re(p) > ξ0 − a and Lh(p) = Lf(p + a).In particular,

– L (e−at sinbt) = b(p2 + a2)+ b2

for Re(p) > −a ,

– L (e−at cosbt) = p + a(p2 + a2)+ b2

for Re(p) > −a ,

– L(tα−1 eλtΓ (α)

)= 1(p − λ)α for Re(p) > Re(λ) , and Re(α) > 0 .

6. Let h(t) = f(λt)(λ > 0) and let Lf exist for Re(p) > ξ0. ThenLh(p) exists forRe(p) > λξ0 and Lh(p) = (1/λ)Lf(p/λ).

8.2 Laplace Transform of a Distribution

Let T ∈ D′+, thus the support of T is contained in [0, ∞). Assume that there exists

ξ0 ∈ R such that e−ξ0t Tt ∈ S′(R). Then evidently e−pt Tt ∈ S′(R) as well for allp ∈ C with Re(p) ≥ ξ0. Let now α be a C∞ function with Suppα bounded from theleft and α(t) = 1 in a neighborhood of [0, ∞). Then α(t) ept ∈ S(R) for all p ∈ Cwith Re(p) < 0. Therefore, ⟨

e−ξ0t Tt, α(t) e(−p+ξ0)t⟩

exists, commonly abbreviated by 〈Tt, e−pt〉 (Re(p) > ξ0). We thus define for T ∈ D′+

as above

76 The Laplace Transform

Definition 8.1.LT(p) =

⟨Tt, e−pt

⟩(Re(p) > ξ0) .

Comments1. The definition does not depend on the special choice of α.2. LT is analytic for Re(p) > ξ0 (without proof; a proof can be given by using

a slight modification of Corollary 7.15, performing a translation of the interval).3. The definition coincides with the one given for functions in Section 8.1 in case of

a regular distribution.4. For any positive integerm, (LT)(m)(p) = L((−t)mT)(p) for Re(p) > ξ0.5. L(λT + μS) = λL(T) + μL(S) for Re(p) large, provided both S and T admit

a Laplace transform.

Examples1. Lδ = 1,Lδ(m) = pm,Lδ(a) = e−ap.2. If T ∈ E′ then LT is an entire analytic function.3. Based on the formulae Lδ(m) = pm,L(tl) = l!/pl+1 and L(e−at T)(p) =

LT(p + a), one can determine the inverse Laplace transform of any rationalfunction P(p)/Q(p) (P , Q polynomials) as a distribution in D′+. One just ap-plies partial fraction decomposition.

8.3 Laplace Transform and Convolution

Let S and T be two distributions in D′+ such that e−ξt S and e−ξt T are tempered

distributions for all ξ ≥ ξ0. Then according to [10], Chapter VIII (or, alternatively,using again the slight modification of Corollary 7.15) we have

e−ξt (S ∗ T) ∈ S′ for ξ > ξ0 .

MoreoverL(S ∗ T)(p) =

⟨St ⊗ Tu, e−p(t+u)

⟩=⟨St, e−pt

⟩⟨Tu, e−pu

⟩= LS(p) · LT(p) for Re(p) > ξ0 .

Therefore,L(S ∗ T) = LS · LT .

As a corollary we get:

L(T (m))(p) = L(δ(m) ∗ T)(p) = pmLT(p) for Re(p) > max(ξ0,0) .

Laplace Transform and Convolution 77

Let now f be a function on Rwith f(t) = 0 for t < 0. Assume that e−ξt |f(t)| isintegrable for ξ > ξ0, and, in addition, that f is C1 for t > 0, that limt↓0 f(t) = f(0)and that limt↓0 f ′(t) is finite. We know then that

{f}′ = {f ′} + f(0) δ ,

in which {f}′ is the derivative of the distribution {f}. Then Lf ′ is clearly definedwhere L{f}′ is defined, hence where Lf is defined. Furthermore,

Lf ′(p) = p (Lf)(p)− f(0) (Re(p) > ξ0) .

With induction we obtain the following generalization. Let f bem times continuous-ly differentiable on (0, ∞), and assume that limt↓0 fk(t) exists for all k ≤ m. Setlimt↓0 f (k)(t) = f (k)(0)(0 ≤ k ≤m − 1). Then

Lf (m)(p) = pm Lf(p)− {f (m−1(0)+ p f(m−2(0)+ · · · + pm−1 f(0)}(Re(p) > ξ0

).

Examples1. Let f = Y be the Heaviside function. Then f ′ = 0 and f(0) = 1. Thus, as we

know already, L(Y)(p) = 1/p.2. Y(t) sinat has as Laplace transforma/(p2 + a2) forRe(p) > 0. Differentiation

gives: aY(t) cosat has Laplace transform (ap)/(p2 + a2) for Re(p) > 0. Seealso Section 8.1.

3. Recall that

Y(t) eλttα−1

Γ (α) ∗ Y(t) eλttβ−1

Γ (β) = Y(t) eλttα+β−1

Γ (α+ β)for Re(α) > 0, Re(β) > 0, see Section 6.2. Laplace transformation gives

1(p − λ)α ·

1(p − λ)β =

1(p − λ)α+β (Re(p) > Re(λ)) .

The second relation is easier than the first one, of course. The second relationimplies the first one, because of

Theorem 8.2. Let T ∈ D′+ satisfy e−ξt Tt ∈ S′(R) for ξ > ξ0. If LT(p) = 0 forRe(p) > ξ0, then T = 0.

Proof. One has 〈e−ξt T , α(t) e−εt e2πiηt〉 = 0 for all η ∈ R and all ε > 0. Hereα(t) is chosen as in the definition of LT and ξ is fixed, ξ > ξ0. Let ϕ ∈ S.Then 〈e−ξt T , α(t) e−εt ϕ(η) e2πiηt〉 = 0 for all η ∈ R and all ε > 0. ApplyingFubini’s theorem in S′(R) gives

0 =∞∫−∞

⟨e−ξt T , α(t) e−εt ϕ(η) e2πiηt

⟩dη =

⟨e−ξtT , α(t)ϕ(t) e−εt

⟩


for all ϕ ∈ S and all ε > 0. Hence 〈e−(ξ+ε)t T , ϕ〉 = 0 for all ϕ ∈ S and allε > 0, thus, forϕ ∈ D, 〈T , ϕ〉 = 〈e−(ξ+ε)t T , e(ξ+ε)t ϕ〉 = 0, so T = 0. �

“Symbolic calculus” is nothing else then taking Laplace transforms: Lmaps thesubalgebra ofD′

+ generated by δ and δ′ onto the algebra of polynomials in p. If thepolynomial P is the image of A: LA = P , and if LB = 1/P for some B ∈ D′

+, then Bis the inverse of A by Theorem 8.2.

The Abel Integral EquationThe origin of this equation is in [N. Abel, Solution de quelques problèmes à l’aided’intégrales définies, Werke (Christiania 1881), I, pp. 11–27]. The equation plays a rolein Fourier analysis on the upper half plane and other symmetric spaces. The generalform of the equation is

t∫0

ϕ(τ)(t − τ)α dτ = f(t) (t ≥ 0, 0 < α < 1) ,

a Volterra integral equation of the first kind. Abel considered only the case α = 1/2.The solution of this equation falls outside the scope of symbolic calculus.

Set ϕ(t) = f(t) = 0 for t < 0. Then the equation changes into a convolutionequation inD′+

Y(t) t−α ∗ϕ = f .Let us assume thatϕ and f admit a Laplace transform. Set Lϕ = Φ, Lf = F . Thenwe obtain Γ (1−α)

p1−α Φ(p) = F(p) ,hence Φ(p) = p F(p)Γ (1−α)pα = Lf ′(p)+ f(0)Γ (1−α)pα .

Therefore

ϕ(t) = 1Γ (1−α) Γ (α){Y(t) tα−1 ∗ f ′(t)+ f(0) Y(t) tα−1

}

= sinπαπ

{ t∫0

f ′(τ)(t − τ)1−α dτ + f(0)

t1−α

}

for t ≥ 0, almost everywhere. We have assumed that f is continuous for t ≥ 0, that f ′exists and is continuous for t > 0, and that limt↓0 f ′(t) is finite. We also applied herethe formula Γ (α) Γ (1−α) = (sinπα)/π for 0 < α < 1. See equation (10.8).

Inversion Formula for the Laplace Transform 79

8.4 Inversion Formula for the Laplace Transform

Let f be a function satisfying f(t) = 0 for t < 0 and e−ξtf(t) ∈ L1 for ξ > ξ0.Set F(p) = Lf(p), thus

F(ξ + iη) =∞∫0

f(t) e−ξt e−iηt dt (ξ > ξ0) .

If η� |F(ξ + iη)| ∈ L1 for some ξ > ξ0, then

f(t) e−ξt = 12π

∞∫−∞F(ξ + iη) eiηt du

at the points where f is continuous, or

f(t) = 12π

∞∫−∞F(ξ + iη) e(ξ+iη)t dη = 1

2πi

ξ+i∞∫ξ−i∞

F(p)ept dp .

The following theorem holds:

Theorem 8.3. An analytic function F coincides with the Laplace transform of a distri-bution T ∈ D′+ in some open right half plane if and only if there exists c ∈ R suchthat F(p) is defined for Re(p) > c and

|F(p)| ≤ polynomial in |p| (Re(p) > c) .

Proof. “Necessary”*. Let T ∈ D′+ be such that e−ctTt ∈ S′ for some c > 0. ThenT = S+U with S ∈ D′

+ of compact support, SuppU ⊂ (0, ∞) and e−ctUt ∈ S′. Thisis evident: choose a function α ∈ D such that α(x) = 1 in a neighborhood of x = 0

and set S = αT, U = (1−α)T . ThenLS(p) is an entire function ofp of polynomialgrowth. To see this, compare with the proof of Theorem 7.17. Furthermore, e−ctUt isof the form f (m) for some continuous, slowly increasing, function f with Suppf ⊂(0, ∞) by Corollary 7.15. Then Ut = ect f (m) = ∑m

k=0(−1)k(mk

)ck (ect f )(m−k).

Moreover e−ξt ect f (t) ∈ L1 for ξ > c, hence L (ect f )(p) exists and is boundedfor Re(p) ≥ c + 1. Therefore

LU(p) =m∑k=0

(−1)k(mk

)ckL (ect f )(m−k)(p)

=m∑k=0

(−1)k(mk

)ck pm−kL (ect f )(p)

satisfies |LU(p)| ≤ polynomial in |p| of degreem, for Re(p) > c + 1. This com-pletes this part of the proof.


“Sufficient”. Let F be analytic and |F(p)| ≤ C/|p|2 for ξ = Re(p) > c > 0,C being a positive constant. Then |F(ξ + iη)| ≤ C/(ξ2 + η2) for ξ > c and all η.Therefore,

f(t) = 12πi

ξ+i∞∫ξ−i∞

F(p)ept dp

exists for ξ > c and all t and does not depend on the choice of ξ > c by Cauchy’stheorem. Let us write

e−ξt f(t) = 12π

∞∫−∞F(ξ + iη) eiηt dη (ξ > c) .

Then we see that f is continuous. Moreover, if e−ξt f(t) ∈ L1 for ξ > c, then F(p) =∫∞−∞ f(t) e−pt dt for ξ = Re(p) > c, by Corollary 7.2 on the inversion of the Fouriertransform. We shall now show(i) f(t) = 0 for t < 0,(ii) e−ξt f(t) ∈ L1 for ξ > c.

It then follows that F = Lf .To show (i), observe that

|f(t)| ≤ C eξt

2πc

∞∫−∞

c dηc2 + η2

= C eξt

2c(ξ > c) .

For t < 0 we thus obtain |f(t)| ≤ limξ→∞ C eξt/(2c) = 0, hence f(t) = 0.

To show (ii), use again |f(t)| ≤ C eξt/(2c) for all ξ > c, hence |f(t)| ≤C ect/(2c). Therefore

e−ξt |f(t)| ≤ C2c

e−(ξ−c)t ∈ L1 for ξ > c and t ≥ 0 .

Let now, more generally, F be analytic and

|F(p)| ≤ polynomial in |p| of degreem ,

for ξ = Re(p) > c > 0.Then |F(p)/pm+2| ≤ C/|p|2 for ξ > c and some constant C. Therefore, there is

a continuous function f such that(i) f(t) = 0 for t < 0,(ii) e−ξt f(t) ∈ L1 for ξ > c, and

F(p) = pm+2Lf(p) (ξ > c) .

Inversion Formula for the Laplace Transform 81

Then F(p) = L(f (m+2))(p) for ξ > c or, otherwise formulated, F = LT with T =dm+2f/dtm+2, the derivative being taken in the sense of distributions. �

Notice that the theorem implies that for any T ∈ D′+ which admits a Laplace

transform there exists a positive integer m such that T is themth derivative (in thesense of distributions) of a continuous function f with support in [0, ∞) and suchthat for some ξ0, e−ξ0tf (t) ∈ L1.

Further Reading

Further reading may be done in the “bible” of the Laplace transform [17].

9 Summable Distributions*SummaryThis chapter might be considered as the completion of the classification of distribu-tions. As indicated, it can be omitted in the first instance.

Learning Targets� Understanding the structure of summable distributions.

9.1 Definition and Main Properties

We recall the definition |k| = k1 + · · · + kn if k = (k1, . . . , kn) is an n-tuple ofnonnegative integers. We also recall from Section 2.2:

Definition 9.1. A summable distribution on Rn is a distribution T with the followingproperty: there exists an integerm ≥ 0 and a constant C satisfying

|〈T , ϕ〉| ≤ C∑

|k|≤msup |Dkϕ| (ϕ ∈ D(Rn)) .

The smallest numberm satisfying this inequality is called the summability-orderof T , abbreviated by sum-order (T).

Observe that locally, i. e. on every open bounded subset of Rn, each distribu-tion T is a summable distribution. Moreover, any distribution with compact supportis summable.

A summable distribution is of finite order in the usual sense, and we haveorder(T) ≤ sum-order(T), with equality if T has compact support. It immediatelyfollows from the definition that the derivative of a summable distribution is summa-ble with

sum-order(∂T∂xi

)≤ sum-order(T)+ 1 .

The space of summable distributions of sum-order 0 coincideswith the spaceMb(Rn)of bounded measures on Rn, that is with the space of measures μ with finite totalmass

∫Rn d|μ|(x). This is a well-known fact. First observe that a distribution of order

0 can be uniquely extended to a continuous linear form on the spaceD0 of continu-ous functions with compact support, provided with its usual topology (see [2], Chap-ter III), thus to a measure, using a sequence of functions {ϕk} as in Section 6.2 anddefining 〈T , ϕ〉 = limk→∞〈T , ϕk ∗ ϕ〉 for ϕ ∈ D0. Next apply for example [2],Chapter IV, § 4, Nr. 7.

If μ ∈Mb then the distributionDkμ is summable with sum-order≤ |k|.We shall show the following structure theorem:

The Iterated Poisson Equation 83

Theorem 9.2 (L. Schwartz [10]).Let T be a distribution. Then the following conditions on T are equivalent:1. T is summable.2. T is a finite sum

∑k Dkμk of derivatives of bounded measures μk.

3. T is a finite sum∑k Dkfk of derivatives of L1 functions fk.

4. For everyα ∈ D,α∗ T belongs toMb(Rn).5. For everyα ∈ D,α∗ T belongs to L1(Rn).

We need some preparations for the proof.

9.2 The Iterated Poisson Equation

In this section we will consider elementary solutions of the iterated Poisson equation

Δl E = δwhere l is a positive integer andΔ the Laplace operator in Rn.

One has, similarly to the case l = 1 (see Section 3.5)

Δl(r 2l−n) = (2l−n)(2l− 2−n) . . . (4−n)(2−n)2l−1 (l− 1)!2π

n2Γ (n2 ) δ .

Thus, if 2l − n < 0 or 2l − n ≥ 0 and n odd, then there exists a constant Bl,n suchthat Δl(Bl,n r 2l−n) = δ .If now 2l−n ≥ 0 andn even, then

Δl(r 2l−n log r) = [[(2l−n)(2l− 2−n) . . . (4−n)(2−n)]]2l−1 (l− 1)!2π

n2Γ (n2 ) δ ,

where in the expression between double square brackets the factor 0 has to be omit-ted. Thus, there exists a constant Al,n such that

Δl(Al,n r 2l−n log r) = δ .

We conclude that for all l andn there exist constants Al,n and Bl,n such that

Δl(r 2l−n (Al,n log r + Bl,n)) = δ .

Let us call this elementary solution of the iterated Poisson equation E = El,n.We now replace E by a parametrix γE, with γ ∈ D(Rn), γ(x) = 1 for x in

a neighborhood of 0 inRn. Then one has for any T ∈ D′(Rn){Δl (γE)− ζ = δ for some ζ ∈ D ,Δl (γE ∗ T)− ζ ∗ T = T .

84 Summable Distributions*

This result, of which the proof is straightforward, implies the following. If we takefor T a bounded measure and take l so large that E is a continuous function, then ev-ery bounded measure is a finite sum of derivatives of functions in L1(Rn) (which arecontinuous and converge to zero at infinity). This proves in particular the equivalenceof 2. and 3. of Theorem 9.2. Moreover, replacing T withα∗T in the above parametrixformula easily implies the equivalence of 4. and 5.

9.3 Proof of the Main Theorem

To prove Theorem 9.2, it is sufficient to show the implications: 2 ⇒ 1⇒ 4⇒ 2.

2⇒ 1. Let T =∑|k|≤mDkμk. Then

|〈T , ϕ〉| =∣∣∣∣ ∑|k|≤m

(−1)|k|⟨μk, Dkϕ

⟩∣∣∣∣ ≤ C∑

|k|≤msup |Dkϕ|

for some constant C and allϕ ∈ D. Hence T is summable.

1 ⇒ 4. Let α ∈ D be fixed and T summable. Then T satisfies a relation of theform

|〈T , ϕ〉| ≤ C∑

|k|≤msup |Dkϕ| (ϕ ∈ D) .

Therefore,

|〈T ∗α, ϕ〉| = |〈T , α∗ϕ〉| ≤ C∑

|k|≤msup |Dkα∗ϕ|

≤ C⎛⎝ ∑|k|≤m

sup‖Dkα‖1

⎞⎠ sup |ϕ| (ϕ ∈ D) .

Hence T ∗α ∈Mb.

4⇒ 2. This is the most difficult part. From the relation

〈T ∗ ϕ, α〉 = 〈T ∗α, ϕ〉

for α,ϕ ∈ D, we see, by 4, that

|〈T ∗ ϕ, α〉| ≤ Cα

for all ϕ ∈ D with sup |ϕ| ≤ 1, with Cα a positive constant. By the Banach–Steinhaus theorem (or the principle of uniform boundedness) inD, see Section 10.1,we get: for any open bounded subset K ofRn there is a positive integerm and a con-stant CK such that

|〈T ∗ ϕ, α〉| ≤ CK∑

|k|≤msup |Dkα| (9.1)

Canonical Extension of a Summable Distribution 85

for all α ∈ D with Suppα ⊂ K and all ϕ ∈ D with sup |ϕ| ≤ 1. Let us denotebyDm the space of Cm functions with compact support and byDm(K) the subspaceconsisting of all α ∈ Dm with Suppα ⊂ K. Similarly, defineD(K) as the space offunctions α ∈ Dwith Suppα ⊂ K. Let us provideDm(K) with the norm

‖α‖ =∑

|k|≤msup

∣∣∣Dkα∣∣∣ (α ∈ Dm(K)

).

It is not difficult to show that D(K) is dense inDm(K). Indeed, taking a sequenceof functions ϕk (k = 1,2, . . .) as in Section 6.2, one sees, by applying the relationDl(ϕk∗ϕ) = ϕk∗Dlϕ for anyn-tuple of nonnegative integers lwith |l| ≤m andusing the uniform continuity of Dlϕ, that anyϕ ∈ Dm(K) can be approximated inthe norm topology ofDm(K) by functions of the form ϕk ∗ ϕ, which are inD(K)for k large. From equation (9.1) we then see that every distribution T ∗ ϕ can beuniquely extended to Dm(K), such that equation (9.1) remains to hold. Therefore,for all α ∈ Dm(K), T ∗ α is inMb, since still 〈T ∗ ϕ, α〉 = 〈T ∗ α, ϕ〉(ϕ ∈ D).Now apply the parametrix formula again, takingα = γ E = γ El,n with l large enoughin order that α ∈ Dm.The proof of the theorem is now complete.

9.4 Canonical Extension of a Summable Distribution

The content of this section (and the next one) is based on notes by the late ErikThomas [13].

Let T be a summable distribution of sum-orderm. Denote byBm = Bm(Rn) thespace of Cm functions ϕ on Rn which are bounded as well as its derivatives Dkϕwith |k| ≤m. Let us provideBm with the norm

pm(ϕ) =∑

|k|≤msup |Dkϕ| (ϕ ∈ Bm) .

Then, by the Hahn–Banach theorem, T can be extended to a continuous linear formonBm.

Let B = ⋂∞m=0Bm and provide it with the convergence principle: a sequence

{ϕj} in B converges to zero if, for all m, the scalars pm(ϕj) tend to zero when jtends to infinity. Then the extension of T toBm is also a continuous linear form onB.There exists a variety of extensions in general. We will now define a canonical exten-sion of T toB. To this end wemake use of a representation of T as a sum of derivativesof bounded measures T = ∑

|k|≤mDkμk. The numberm arising in this summationmay be larger than the sum-order of T . With help of this representation we can easilyextend T to a continuous linear form on B, since bounded measures are naturallydefined on B. Indeed, the functions in B are integrable with respect to any bound-ed measure. We shall show that such an extension does not depend on the specific


representation of T as a sum of derivatives of bounded measures, it is a canonical ex-tension. The reason is the following. Observe that any μ ∈Mb is concentrated, up toε > 0, on a compact set (depending on μ). It follows that a canonical extension hasa special property, T has the bounded convergence property.

A continuous linear form on Bm is said to have the bounded convergence prop-erty of orderm if the following holds: if ϕj ∈ Bm, supj pm(ϕj) < ∞ and ϕj →0 in Em, then 〈T , ϕj〉 → 0 when j tends to infinity. A similar definition holdsin B: a continuous linear form on B has the bounded convergence property if ϕj ∈B, supj pm(ϕj) < ∞ for allm, andϕj → 0 in E, then 〈T , ϕj〉 → 0 when j tendsto infinity.

Relying on this property we can easily relate T with its extension to B. Indeed,let α ∈ D be such that α(x) = 1 in a neighborhood of x = 0, and set αj(x) =α(x/j). Then, for anyϕ ∈ B the functionsαjϕ tend toϕ inE and supj pm(ϕj) <∞ for all m. Hence 〈T , ϕ〉 = limj→∞ 〈T , ϕj〉. Therefore the extension of T to Bdoes not depend on the specific representation. We call it the canonical extension. Inparticular one can define the total mass 〈T , 1〉 of T , sometimes denoted by

∫T (dx),

which accounts for the name “summable distribution”. Furthermore, the canonicalextension satisfies again

|〈T , ϕ〉| ≤ C pm(ϕ)

for some constant C and allϕ ∈ B,m being now the sum-order of T .Conversely, any continuous linear form T on B gives, by restriction toD, a sum-

mable distribution. There is however only a one-to-one relation between T and itsrestriction toD if we impose the condition: T has the bounded convergence property.

The above allows one to define several operations on summable distributionwhich are common for bounded measures.

Fourier TransformAny summable distribution T is tempered, so has a Fourier transform. But there ismore: F T is function,

F T(y) = 〈Tx, e−2πi〈x,y〉〉 (y ∈ Rn) .

Here we use the canonical extension of T . Clearly F T is continuous (since any μ ∈Mb is concentrated up to ε > 0 on a compact set) and of polynomial growth.

ConvolutionLet T and S be summable distributions. Then the convolution product S ∗ T exists.Indeed, let us define 〈S∗T ,ϕ〉 = 〈T , ϕ∗S〉(ϕ ∈ D). This is a good definition, sinceϕ ∗ S ∈ B. Clearly S ∗ T is summable again. Due to the canonical representationof T and S as a sum of derivatives of bounded measures, we see that this convolutionproduct is commutative and associative and furthermore,

F (T ∗ S) = F (T) · F (S) .

Rank of a Distribution 87

Let O be the space of functions f ∈ E(Rn) such that f and the derivatives of fhave at most polynomial growth. Then O operates by multiplication on the spaceS(Rn) and on S′(Rn). Moreover, since the functions in O have polynomial growth,they define themselves tempered distributions.

Theorem 9.3. Every T ∈ F (O) is summable. More precisely, if P is a polynomialthen P T is summable. Conversely, if P T is summable for every polynomial P , thenT ∈ F (O).

Proof. If T = F (f) with f ∈ O and α ∈ D, then α ∗ T = F (βf) with β ∈ S theinverse Fourier transform of α. Since βf belongs to S we have α ∗ T ∈ S, a fortioriα ∗ T ∈ L1. Therefore, by Theorem 9.2, T is summable. Similarly, if P is a polyno-mial, then P T = F(Df) for some differential operator with constant coefficients, soP T ∈ F (O). Conversely, if P T is summable for all polynomials P , then DF(T) iscontinuous and of polynomial growth for all differential operators D with constantcoefficients, and soF(T) belongs to O (cf. Exercise 4.4 b.). �

9.5 Rank of a Distribution

What is the precise relation between the sum-order of T and the number of terms ina representation of T as a sum of derivatives of bounded measures from Theorem 9.2.To answer this question, we introduce the notion of rank of T .

For every n-tuple of nonnegative integers k = (k1, . . . , kn) we set |k|∞ =max1≤i≤n ki.

A partial differential operator with constant coefficientsak is said to have rankmif it is of the form

D =∑

|k|∞≤makDk

and not every ak with |k|∞ =m vanishes.

Definition 9.4. A distribution T is said to have finite rank if there exists a positive inte-germ and a constant C > 0 such that

|〈T , ϕ〉| ≤ C∑

|k|∞≤msup |Dkϕ| (ϕ ∈ D) .

The smallest suchm is called the rank of T .

Clearly, |k|∞ ≤ |k| ≤ n|k|∞. So the distributions of finite rank coincide withthe summable distributions. There is however a striking difference between sum-order and rank: the sum-order may grow with the dimension while the rank remainsconstant. For example sum-order(Δδ) = 2n and rank(Δδ) = 2, Δ being the usualLaplacian in Rn.


Definition 9.5. Amollifier of rankm is a boundedmeasureK having the following prop-erties:1. DkK is a bounded measure for all n-tuples k with |k|∞ ≤m.2. There exists a differential operator with constant coefficients D of rank m such

thatDK = δ .

Theorem 9.6. There exists mollifiers of rank m. Let K be a mollifier of rank m andletD be a differential operator of rankm such thatDK = δ. Then if T is any summabledistribution of rank ≤m, the summable distribution μ = K ∗ T is a bounded measureand T = Dμ.

Thus for summable distributions of rank≤mwe get a representation of the form

T =∑

|k|∞≤mDkμk ,

which gives an answer to the question posed at the beginning of this section. It turnsout that the rank of a distribution is an important notion.

LetE(m)(Rn) be the space of functionsϕ such thatDkϕ exists and is continuousfor all k with |k|∞ ≤ m, and let B(m)(Rn) be the subspace of function ϕ ∈ E(m)such that Dkϕ is bounded for all k with |k|∞ ≤ m. Notice that B(m) is a Banachspace with norm

p(m)(ϕ) =∑

|k|∞≤msup |Dkϕ| (ϕ ∈ B(m)) .

Clearly any summable distribution T of rank ≤ m has a canonical extension to thespaceB(m), having the bounded convergence property of rankm: ifϕj → 0 in E(m)and supj p(m)(ϕj) < ∞, then 〈T , ϕj〉 → 0. This follows in the same way as before.

Proof. (Theorem 9.6)For dimension n = 1 and rankm = 1, we set, if a > 0,

L(x) = La(x) =1aY(x) e−x/a .

Here Y is, as usual, the Heaviside function. Furthermore, set

D = 1+ a ddx

.

ThenDL = δ and L and L′ are bounded measures.Now consider the case of rank 2 in dimension n = 1 in more detail. For a > 0 set

Ka = La ∗ La, with La(x) = (1/a)Y(−x) ex/a. Then

K(x) = Ka(x) =1

2ae−|x|/a .

Rank of a Distribution 89

SetD =

(1+ a d

dx

)(1− a d

dx

)= 1− a2 d2

dx2.

Then K,K′ and K′′ are bounded measures, and

DK = δ ,ϕ′ = K′ ∗Dϕ ,

ϕ = ϕ ∗ δ =ϕ ∗DK = K ∗Dϕ ,ϕ′′ = K′′ ∗Dϕ .

Therefore there exists a constant C such that for allϕ ∈ D(R),

sup |ϕ| ≤ C sup |Dϕ|, sup |ϕ′| ≤ C sup |Dϕ|, sup |ϕ′′| ≤ C sup |Dϕ|

and thus, if T has rank≤ 2, we therefore have, for some C,

|〈T , ϕ〉| ≤ C sup |Dϕ|

for allϕ ∈ D(R), but also forϕ ∈ B(m)(R). It follows that

|〈K ∗ T , ϕ〉| = |〈T , K ∗ϕ〉| ≤ C sup |DK ∗ϕ| ≤ C sup |ϕ|

forϕ ∈ D(R), so that K ∗ T is a bounded measure.Let us now consider the case of dimensionn > 1, but stillm = 2. Set in this case

K(x) = Ka(x1)⊗ · · · ⊗ Ka(xn)

and

D =n∏i=1

(1− a2 ∂2

∂x2i

).

Then DkK is a bounded measure for |k|∞ ≤ 2. By the same argument as before, wesee that for some C,

sup |Dkϕ| ≤ C sup |Dϕ| (ϕ ∈ D(Rn))

if |k|∞ ≤ 2, and if T has rank≤ 2, we have again

|〈T , ϕ〉| ≤ C sup |Dϕ| (ϕ ∈ D(Rn)) ,

so that μ = K ∗ T is a bounded measure and T = Dμ.For higher rank m we take H = K ∗ · · · ∗ K (p times) if m = 2p, and

H = K∗· · ·∗K∗L (p times K) ifm is odd,m = 2p+1. Here L(x) = L(x1)⊗· · ·⊗L(xn) = (1/an)Y(x1, . . . , xn) · e−(x1+···+xn)/a. We choose also the correspondingpowers of the differential operators, and find in the same way: if T has rank ≤ m,then μ = H ∗ T is a bounded measure and T = Dμ. �

Further Reading

An application of this chapter is a mathematically correct definition of the Feynmanpath integral, see [13].

10 Appendix

10.1 The Banach–Steinhaus Theorem

Topological spacesWe recall some facts from topology.

Let X be a set. Then X is said to be a topological space if a system of subsets Thas been selected in X with the following three properties:1. ∅ ∈ T , X ∈ T .2. The union of arbitrary many sets ofT belongs to T again.3. The intersection of finitely many sets of T belongs to T again.

The elements of T are called open sets, its complements closed sets. Let x ∈ X. Anopen setU containing x is called a neighborhood ofx. The intersection of two neigh-borhoods of x is again a neighborhood of x. A system of neighborhoods U(x) of xis called a neighborhood basis of x if every neighborhood of x contains an element ofthis system. Suppose that we have selected a neighborhood basis for every elementof X. Then, clearly, the neighborhood basis of a specific element x has the followingproperties:1. x ∈ U(x).2. The intersection U1(x)∩U2(x) of two arbitrary neighborhoods in the basis con-

tains an element of the basis.3. If y ∈ U(x), then there is a neighborhood U(y) in the basis of y with U(y) ⊂

U(x).

If, conversely, for any x ∈ X a system of sets U(x) is given, satisfying the abovethree properties, then we can easily define a topology on X such that these sets forma neighborhood basis for x, for anyx. Indeed, one just calls a set open inX if it is theunion of sets of the form U(x).

Let X be a topological space. Then X is called a Hausdorff space if for any twopoints x and y with x �= y there exist neighborhoods U of x and V of y withU ∩ V = ∅.

Let f be a mapping from a topological spaceX to a topological space Y . One saysthat f is continuous at x ∈ X if for any neighborhood V of f(x) there is a neigh-borhood U of x with f(U) ⊂ V . The mapping is called continuous on X if it iscontinuous at every point ofX. This property is obviously equivalent with saying thatthe inverse image of any open set in Y is open in X.

Let X be again a topological space and Z, a subset of X. Then Z can also be seenas a topological space, a topological subspace of X, with the induced topology: opensets in Z are intersections of open sets in X with Z.

The Banach–Steinhaus Theorem 91

Fréchet spacesLet now X be a complex vector space.

Definition 10.1. A seminorm on X is a function p : X → R satisfying1. p(x) ≥ 0;2. p(λx) = |λ|p(x);3. p(x +y) ≤ p(x)+ p(y) (triangle inequality)for all x,y ∈ X and λ ∈ C.

Observe that a seminorm differs only slightly from a norm: it is allowed thatp(x) = 0 for some x �= 0. Clearly, by (iii), p is a convex function, hence all setsof the form {x ∈ X : p(x) < c}, with c a positive number, are convex.

We will consider vector spaces X provided with a countable set of seminormsp1, p2, . . . . Givenx ∈ X, ε > 0 and a positive integerm, we define the setsV(x,m; ε)as follows:

V(x,m; ε) = {y ∈ X : pk(x −y) < ε for k = 1,2, . . . ,m}.

Then, clearly, the intersection of two of such sets contains again a set of this form.Furthermore, if y ∈ V(x,m; ε) then there exists ε′ > 0 such that V(y,m; ε′) ⊂V(x,m; ε). We can thus provide X with a topology by calling a subset of X open if itis a union of sets of the form V(x,m; ε). This topology has the following properties:the mappings

(x, y)� x +y(λ,x)� λx ,

from X ×X → X and C×X → X, respectively, are continuous. The spaceX is said tobe a topological vector space.

Clearly, any x ∈ X has a neighborhood basis consisting of convex sets. There-fore X is called a locally convex (topological vector) space.

The space X is a Hausdorff space if and only if for any nonzero x ∈ X there isa seminorm pk with pk(x) �= 0. From now we shall always assume that this propertyholds.

Let {xn} be a sequence of elements xn ∈ X. One says that {xn} converges tox ∈ X, in notation limn→∞ xn = x or xn → x (n → ∞), if for any neighborhood Vof x there exists a natural numberN such that xn ∈ V for n ≥ N. Alternatively, onemay say that {xn} converges to x if for any ε > 0 and any k ∈ N there is a naturalnumberN = N(k, ε) satisfying pk(xn −x) < ε forn ≥ N. Because X is a Hausdorffspace, the limit x, if it exists, is uniquely determined.

A sequence {xn} is called a Cauchy sequence if for any neighborhood V of x = 0

there is a natural number N such that xn − xm ∈ V for n,m ≥ N. Alternatively,one may say that {xn} is a Cauchy sequence if for any ε > 0 and any k ∈ N there isa natural numberN = N(k, ε) satisfying pk(xn − xm) < ε for n,m ≥ N.

92 Appendix

The space X is said to be complete (also: sequentially complete) if any Cauchysequence is convergent.

Definition 10.2. A complete locally convex Hausdorff topological vector space with atopology defined by a countable set of seminorms is called a Fréchet space.

Examples of Fréchet Spaces1. Let K be an open bounded subset of Rn and letD(K) be the space of functions

ϕ ∈ D(Rn)with Suppϕ ⊂ K. ProvideD(K) with the set of seminorms (actual-ly norms in this case) given by

pm(ϕ) =∑

|k|≤msup |Dkϕ| (ϕ ∈ D(K)) .

Then D(K) is a Fréchet space. Notice that the seminorms form an increasingsequence in this case: pm(ϕ) ≤ pm+1(ϕ) for allm andϕ.This property might be imposed on any Fréchet spaceX by considering the set ofnew seminorms

p′m = p1 + · · · + pm (m ∈ N) .Obviously, these new seminorms define the same topology on X.

2. The spacesE and S. For E we choose the seminorms

pk,K(ϕ) = supx∈K |Dkϕ| ,

k being a n-tuple of nonnegative integers and K a compact subset of Rn. Takingfor K a ball around x = 0 of radius l (l ∈ N), which suffices, we see that weobtain a countable set of seminorms. For S we choose the seminorms

pk,l(ϕ) = sup |xl Dkϕ| .

Here both k and l are n-tuples of nonnegative integers.

The dual spaceLet X be a Fréchet space with an increasing set of seminorms pm(m = 1,2, . . .). De-note byX′ the space of (complex-valued) continuous linear forms x′ onX. The valueof x′ at x is usually denoted by 〈x′, x〉, showing a nice bilinear correspondence be-tween X′ andX

– 〈x′, λx + μ y〉 = λ 〈x′, x〉 + μ 〈x′, y〉 ,– 〈λx′ + μ y ′, x〉 = λ 〈x′, x〉 + μ 〈y ′, x〉 ,

for x,y ∈ X,x′, y ′ ∈ X′, λ, μ ∈ C.The spaceX′ is called the (continuous) dual space of X.


Proposition 10.3. A linear form x′ on X is continuous at x = 0 if and only if there ism ∈ N and a constant c > 0 such that∣∣〈x′, x〉∣∣ ≤ c pm(x)for all x ∈ X. In this case x′ is continuous everywhere on X.

Proof. Let x′ be continuous at x = 0. Then there is a neighborhood U(0) of x = 0

such that |〈x′, x〉| < 1 for x ∈ U(0). Since the set of seminorms is increasing, wemay assume that U(0) is of the form {x : pm(x) ≤ δ} for some δ > 0. Then we getfor any x ∈ X and any ε > 0,

x1 =δx

pm(x)+ ε∈ U(0) ,

thus |〈x′, x1〉| < 1 and

∣∣〈x′, x〉∣∣ = ∣∣〈x′, x1〉∣∣ pm(x)+ ε

δ<pm(x)+ ε

δ.

Hence |〈x′, x〉| ≤ c pm(x) for all x ∈ X with c = 1/δ, since ε was arbitrary.Conversely, any linear form, satisfying a relation∣∣〈x′, x〉∣∣ ≤ c pm(x) (x ∈ X)

for somem ∈ N and some constant c > 0, is clearly continuous at x = 0. But, weget more. If x0 ∈ X is arbitrary, then we obtain∣∣〈x′, x − x0〉

∣∣ ≤ c pm(x − x0) (x ∈ X) ,

hence x′ is continuous at x0. Indeed, if ε > 0 is given, then we get∣∣〈x′, x − x0〉∣∣ = ∣∣〈x′, x〉 − 〈x′, x0〉

∣∣ < εif pm(x − xo) < ε/c and this latter inequality defines a neighborhood of x0. �

We can also define continuity of a linear form x′ in another way.

Proposition 10.4. A linear form x′ on X is continuous on X if one of the following con-ditions is satisfied:(i) x′ is continuous with respect to the topology of X;(ii) there existsm ∈ N and c > 0 such that∣∣〈x′, x〉∣∣ ≤ c pm(x)

for all x ∈ X;(iii) for any sequence {xn} with limn→∞ xn = x one has limn→∞〈x′, xn〉 = 〈x′, x〉.

94 Appendix

Proof. It is sufficient to show the equivalence of (ii) and (iii). Obviously, (ii) implies(iii). The converse implication is proved by contradiction. If condition (ii) is not sat-isfied, then we can find for all m ∈ N and all c > 0 an element xm ∈ X with|〈x′, xm〉| = 1 and pm(xm) < 1/m, taking c =m. Then we have limm→∞ xm = 0,because the sequence of seminorms is increasing. But |〈x′, xm〉| = 1 for allm. Thiscontradicts (iii). �

There exists sufficiently many continuous linear forms on X. Indeed, comparewith the following proposition.

Proposition 10.5. For any x0 �= 0 in X there is a continuous linear form x′ on X with〈x′, x0〉 �= 0.

Proof. Because x0 �= 0, there exists a seminorm pm with pm(x0) > 0. By Hahn–Banach’s theorem, there exists a linear form x′ on X with |〈x′, x〉| ≤ pm(x) for allx ∈ X and 〈x′, x0〉 = pm(x0). By Proposition 10.4 the linear form x′ is continuous.

�

Metric spacesLet X be a set. We recall:

Definition 10.6. A metric on X is a mapping d : X × X → R with the following threeproperties:(i) d(x,y) ≥ 0 for all x,y ∈ X, d(x,y) = 0 if and only if x = y;(ii) d(x,y) = d(y,x) for all x,y ∈ X;(iii) d(x,y) ≤ d(x, z)+ d(z,y) for all x,y, z ∈ X (triangle inequality).

Clearly, the balls B(x, ε) = {y ∈ X : d(x,y) < ε} can serve as a neighborhoodbasis of x ∈ X. Here ε is a positive number. The topology generated by these ballsis called the topology associated with the metric d. The set X, provided with thistopology, is then called ametric space.

Let now X be a Fréchet space with an increasing set p1 ≤ p2 ≤ · · · of semi-norms. We shall define a metric on X in such a way that the topology associatedwith the metric is the same as the original topology, defined by the seminorms. Thespace X is calledmetrizable.

Theorem 10.7. Any Fréchet space is metrizable with a translation invariant metric.

Proof. For the proof we follow [8], § 18, 2. Set

‖x‖ =∞∑k=1

12k

pk(x)1+ pk(x)

and define d by d(x,y) = ‖x − y‖ for x,y ∈ X. Then d is a translation in-variant metric. To see this, observe that ‖x‖ = 0 if and only if pk(x) = 0 for


all k, hence x = 0. Thus the first property of a metric is satisfied. Furthermore,d(x,y) = d(y,x) since ‖x‖ = ‖ − x‖. The triangle inequality follows from the re-lation: if two real numbers a and b satisfy 0 ≤ a ≤ b, then a/(1+ a) ≤ b/(1+ b).Indeed, we then obtain

pk(x +y)1+ pk(x +y)

≤ pk(x)+ pk(y)1+ pk(x)+ pk(y)

≤ pk(x)1+ pk(x)

+ pk(y)1+ pk(y)

for all x,y ∈ X.Next we have to show that themetric defines the same topology as the seminorms.

Therefore it is sufficient to show that every neighborhood of x = 0 in the first topolo-gy, contains a neighborhood of x = 0 in the second topology and conversely.a. The neighborhood given by the inequality ‖x‖ < 1/2m contains the neighbor-

hood given by pm+1(x) < 1/2m+1. Indeed, since pk(x)/[1+ pk(x)] ≤ pk(x),we have for any x satisfying pm+1(x) < 1/2m+1: p1(x) ≤ p2(x) ≤ · · · ≤pm+1(x) < 1/2m+1, hence ‖x‖ < 1/2m+1

∑m+1k=0 1/2k+∑∞

k=m+2 1/2k < 1/2m.b. The neighborhood given by pk(x) < 1/2m contains the neighborhood given by

‖x‖ < 1/2m+k+1. Indeed, for anyx satisfying ‖x‖ < 1/2m+k+1 one has (1/2k)·[pk(x)/(1+ pk(x))] < 1/2m+k+1, hence pk(x)/[1+ pk(x)] < 1/2m+1. Thuspk(x) · (1− 1/2m+1) < 1/2m+1 and therefore pk(x) < 1/2m.

This completes the proof of the theorem. �

Baire’s theoremThe following theorem, due to Baire, plays an important role in our final result.

Theorem 10.8. If a completemetric space is the countable union of closed subsets, thenat least one of these subsets contains a nonempty open subset.

Proof. Let (X, d) be a complete metric space, Sn closed in X (n = 1,2, . . .) and X =⋃∞n=1 Sn. Suppose no Sn contains an open subset. Then S1 �= X and X\S1 is open.

There is x1 ∈ X\S1 and a ball B(x1, ε1) ⊂ X\S1 with 0 < ε1 < 1/2. Now the ballB(x1, ε1) does not belong to S2, hence X\S2 ∩ B(x1, ε1) contains a point x2 anda ball B(x2, ε2) with 0 < ε2 < 1/4. In this way we get a sequence of balls B(xn, εn)with

B(x1, ε1) ⊃ B(x2, ε2) ⊃ · · · ; 0 < εn <1

2n, B(xn, εn)∩ Sn = ∅ .

For n < m one has d(xn, xm) < 1/2n, which tends to zero if n,m → ∞. TheCauchy sequence {xn} has a limit x ∈ X sinceX is complete. But

d(xn, x) ≤ d(xn, xm)+ d(xm, x) < εn + d(xm, x) → εn (m →∞) .So x ∈ B(xn, εn) for all n. Hence x ∉ Sn for all n, so x ∉ X, which is a contradic-tion. �

96 Appendix

The Banach–Steinhaus theoremWe now come to our final result.

Theorem 10.9. LetX be a Fréchet space with an increasing set of seminorms. LetF bea subset of X′. Suppose for each x ∈ X the family of scalars {〈x′, x〉 : x′ ∈ F} isbounded. Then there exists c > 0 and a seminorm pm such that∣∣〈x′, x〉∣∣ ≤ c pm(x)for all x ∈ X and all x′ ∈ F .

Proof. For the proof we stay close to [12], Chapter III, Theorem 9.1. Since X is a com-plete metric space, Baire’s theorem can be applied. For each positive integer n letSn = {x ∈ X : |〈x′, x〉| ≤ n for all x′ ∈ F}. The continuity of each x′ ensuresthat Sn is closed. By hypothesis, each x belongs to some Sn. Thus, by Baire’s theo-rem, at least one of the Sn, say SN , contains a nonempty open set and hence containsa set of the form V(x0,m;ρ) = {x : pm(x − x0) < ρ} for some x0 ∈ SN, ρ > 0

andm ∈ N. That is, |〈x′ x〉| ≤ N for all x with pm(x − x0) < ρ and all x′ ∈ F .Now, if y is an arbitrary element of X with pm(y) < 1, then we have x0 + ρy ∈V(x0,m;ρ) and thus |〈x′, x0 + ρy〉| ≤ N for all x′ ∈ F . It follows that∣∣〈x′, ρy〉∣∣ ≤ ∣∣〈x′, x0 + ρy〉

∣∣+ ∣∣〈x′, x0〉∣∣ ≤ 2N .

Letting c = 2N/ρ, we have for y with pm(y) < 1,

∣∣〈x′, y〉∣∣ = ∣∣〈x′, ρy〉∣∣ 1ρ≤ c ,

and thus,

∣∣〈x′, y〉∣∣ = ∣∣∣∣∣⟨x′,

ypm(y)+ ε

⟩∣∣∣∣∣ (pm(y)+ ε) ≤ c (pm(y)+ ε)for any ε > 0, hence∣∣〈x′, y〉∣∣ ≤ c pm(y) for all y ∈ X and all x′ ∈ F . �

The theorem just proved is known as Banach–Steinhaus theorem, and also as theprinciple of uniform boundedness.

ApplicationWe can now show the following result, announced in Section 4.3.

Theorem 10.10. Let {Tj} be a sequence of distributions such that the limitlimj→∞〈Tj, ϕ〉 exists for allϕ ∈ D. Set 〈T , ϕ〉 = limj→∞ 〈Tj, ϕ〉 (ϕ ∈ D). Then Tis a distribution.

The Beta and Gamma Function 97

Proof. ClearlyT is a linear formonD. We only have to prove the continuity ofT . LetKbe an open bounded subset of Rn and consider the Fréchet spaceD(K) of functionsϕ ∈ D with Suppϕ ⊂ K and seminorms

pm(ϕ) =∑

|k|≤msup |Dkϕ| (ϕ ∈ D(K)) .

Since limj→∞ 〈Tj, ϕ〉 exists, the family of scalars {〈Tj, ϕ〉 : j = 1,2 . . .} is bound-ed for every ϕ ∈ D(K). Therefore, by the Banach–Steinhaus theorem, there exista positive integerm and a constant CK > 0 such that for each j = 1,2, . . .∣∣∣〈Tj, ϕ〉∣∣∣ ≤ CK ∑

|k|≤msup |Dkϕ|

for allϕ ∈ D(K), hence

|〈T , ϕ〉| ≤ CK∑

|k|≤msup |Dkϕ|

for all ϕ ∈ D(K). Since K was arbitrary, it follows from Proposition 2.3 that T isa distribution. �

10.2 The Beta and Gamma Function

We begin with the definition of the gamma function

Γ (λ) = ∞∫0

e−x xλ−1 dx . (10.1)

This integral converges for Reλ > 0. Expanding xλ−λ0 = e(λ−λ0) logx(x > 0) intoa power series, we obtain for any (small) ε > 0 the inequality∣∣∣ xλ−1 − xλ0−1

∣∣∣ ≤ x−ε ∣∣∣ λ− λ0

∣∣∣ ∣∣∣ xλ0−1∣∣∣ ∣∣∣ logx

∣∣∣ (0 < x ≤ 1)

for all λwith |λ−λ0| < ε. A similar inequality holds for x ≥ 1with−ε replaced by εin the power of x. One then easily sees, by using Lebesgue’s theorem on dominatedconvergence, that Γ (λ) is a complex analytic function of λ for Reλ > 0 and

Γ ′(λ) = ∞∫0

e−x xλ−1 logx dx (Reλ > 0) .

Using partial integration we obtain

Γ (λ+ 1) = λ Γ (λ) (Reλ > 0) . (10.2)

Since Γ (1) = 1, we have Γ (n+ 1) = n! for n = 0,1,2, . . ..

98 Appendix

Weare looking for an analytic continuation of Γ . The above relation (10.2) implies,for Reλ > 0, Γ (λ) = Γ (λ+n+ 1)

(λ+n)(λ +n− 1) · · · (λ+ 1)λ. (10.3)

The right-hand side has however a meaning for all λ with Reλ > −n − 1 minusthe points λ = 0,−1, . . . ,−n. We use this expression as definition for the analyticcontinuation: for every λ with λ �= 0,−1,−2, . . . we determine a natural number nwithn > −Reλ−1. Then we define Γ (λ) as in equation (10.3). Using equation (10.2)one easily verifies that this is a good definition: the answer does not depend on thespecial chosen n, provided n > −Reλ− 1.

It follows also that the gamma function has simple poles at the points λ = 0,−1,−2, . . . with residue at λ = −k equal to (−1)k/k!.

Substituting x = t2 in equation (10.1) one gets

Γ (λ) = 2

∞∫0

e−t2t2λ−1 dt , (10.4)

hence Γ (1/2) = √π .We continue with the beta function, defined for Reλ > 0, Reμ > 0 by

B(λ, μ) =1∫0

xλ−1 (1− x)μ−1 dx . (10.5)

There is a nice relation with the gamma function. One has for Reλ > 0, Reμ > 0,using equation (10.4),

Γ (λ) Γ (μ) = 4

∞∫0

∞∫0

x2λ−1 t2μ−1 e−(x2+t2) dx dt .

In polar coordinates we may write the integral as∞∫0

π/2∫0

r 2(λ+μ)−1 e−r2(cosϕ)2λ−1 (sinϕ)2μ−1 dϕdr =

12 Γ (λ+ μ)

π/2∫0

(cosϕ2λ−1 (sinϕ)2μ−1 dϕ = 14 B(λ, μ) ,

by substituting u = sin2 ϕ in the latter integral. Hence

B(λ, μ) = Γ (λ) Γ (μ)Γ (λ+ μ) (Reλ > 0, Reμ > 0) . (10.6)

Let us consider two special cases. For λ = μ and Reλ > 0 we obtain

B(λ, λ) = Γ (λ)2Γ (2λ) .


On the other hand,

B(λ, λ) =1∫0

xλ−1 (1− x)λ−1 dx = 12

π/2∫0

(cosϕ)2λ−1 (sinϕ)2λ−1 dϕ

= 2−2λπ/2∫0

(sin 2ϕ)2λ−1 dϕ .

Splitting the path of integration [0, π/2] into [0, π/4]∪ [π/4, π/2], using the re-lation sin 2ϕ = sin 2(1

2π −ϕ) and substituting t = sin2 2ϕ we obtain

2−2λπ/2∫0

(sin 2ϕ)2λ−1 dϕ = 2−2λ+1

1∫0

tλ−1 (1− t)− 12 dt = 21−2λ Γ (λ) Γ (1

2

)Γ(λ+ 1

2

) .Thus Γ (2λ) = 22λ−1π−1/2 Γ (λ) Γ (λ+ 1

2

). (10.7)

This formula is called the duplication formula for the gamma function.Another important relation is the following:

Γ (λ) Γ (1− λ) = B(λ, 1− λ) = πsinπλ

, (10.8)

first for real λwith 0 < λ < 1 and then, by analytic continuation, for all noninteger λin the complex plane. This formula can be shown by integration over paths in thecomplex plane. Consider

B(λ, 1− λ) =1∫0

xλ−1 (1− x)−λ dx , (0 < λ < 1) .

Substituting x = t/(1+ t)we obtain

B(λ, 1− λ) =∞∫0

tλ−1

1+ t dt .

To determine this integral, consider for 0 < r < 1 < ρ the closed path W in thecomplex plane which arises by walking from −ρ to −r along the real axis (k1), nextin negative sense along the circle |z| = r from−r back to−r (k2), then via−k1 to−ρand then in positive sense along the circle |z| = ρ back to −ρ (k3) (see Figure 10.1).

Let us consider the function f(z) = zλ−1/(1− z) = e(λ−1) logz/(1− z). In or-der to define zλ−1 as an analytic function, we consider the region G1, with borderconsisting of k1, the part k21 of k2 between −r and ir , k4 and the part k31 of k3

between iρ and −ρ, and the region G2 with border consisting of −k1, the part k32

of k3 between −ρ and iρ, −k4 and the part k22 of k2 between ir and r . We define

100 Appendix

Figure 10.1. The closed pathW .

zλ−1 now on G1 and G2, respectively, including the border, by means of analytic ex-tensions f1 and f2, respectively, of the, in a neighborhood of z = 1 defined, principalvalue, and in such a way that f1(z) = f2(z) for all z ∈ k4. This can be done forexample by means of the cuts ϕ = −(3/4)π and ϕ = (3/4)π , respectively. Weobtain– onG1: f1(z) = e(λ−1)(log |z|+iarg z), π/2 ≤ argz ≤ π ,– onG2: f2(z) = e(λ−1)(log |z|+iarg z), −π ≤ argz ≤ π/2.

On k4 we have to take argz = π/2 in both cases. We now apply the residuetheorem on G1 with g1(z) = f1(z)/(1 − z) and on G2 with g2(z) = f2(z)/(1− z),respectively. Observe that g1 has no poles in G1 and g2 has a simple pole at z = 1

inG2, with residue−f2(1) = −1. We obtain∮∂G1

g1(z)dz +∮∂G2

g2(z)dz = −2πi .

From the definitions of g1 and g2 follows:∫k4

g1(z)dz +∫−k4

g2(z)dz = 0 ,

∫k1

g1(z)dz =−r∫−ρ

e(λ−1)(log |x|+πi)

1− x dx = e(λ−1)πi

ρ∫r

yλ−1

1+y dy ,

∫−k1

g2(z)dz =−ρ∫−r

e(λ−1)(log |x|−πi)

1− x dx = e−(λ−1)πi

ρ∫r

yλ−1

1+y dy .


Furthermore, for |z| = r : |zλ−1/(z − 1)| ≤ rλ−1/(1− r), hence∣∣∣∣∣∣∣∫k21

g1(z)dz +∫k22

g2(z)dz

∣∣∣∣∣∣∣ ≤ 2πrλ

1− r ,

and, similarly, ∣∣∣∣∣∣∣∫k31

g1(z)dz +∫k32

g2(z)dz

∣∣∣∣∣∣∣ ≤ 2πρλ

ρ − 1.

Taking the limits r ↓ 0 and ρ → ∞, we see that[e(λ−1)πi − e−(λ−1)πi

]I + 2πi = 0

with I = B(λ, 1− λ), hence

B(λ, 1− λ) =∞∫0

tλ−1

1+ t dt = πsinπλ

.

11 Hints to the ExercisesExercise 3.2. Use the formula of jumps (Section 3.2, Example 3). For the last equa-tion, apply induction with respect tom.

Exercise 3.3. Write |x| = Y(x)x − Y(−x)x. Then |x|′ = Y(x) − Y(−x) by theformula of jumps, hence |x|′′ = 2δ. Thus |x|(k) = 2δ(k−2) for k ≥ 2.

Exercise 3.4. Applying the formula of jumps, one could take f such that⎧⎪⎪⎨⎪⎪⎩af ′′ + bf ′ + cf = 0

af(0) = naf ′(0)+ bf(0) =m .

This system has a well-known classical solution. The special cases are easily workedout.

Exercise 3.5. The first expression is easily shown to be equal to the derivative of thedistribution Y(x) logx, which itself is a regular distribution. The second expressionis the derivative of the distribution

pvY(x)x

− δ ,

thus defines a distribution. This is easily worked out, applying partial integration.

Exercise 3.6. This is most easily shown by performing a transformation

x = u+ v, y = u− v .

Then dx dy = 2 dudv, ∂2ϕ/∂x2 − ∂2ϕ/∂y2 = ∂2Φ/∂u∂v forϕ ∈ D(R2), withΦ(u, v) =ϕ(u+ v,u− v). One then obtains⟨∂2T∂x2

− ∂2T∂y2

, ϕ⟩= 2

[ϕ(3,1)−ϕ(2,2)+ϕ(1,1)−ϕ(2,0)]

forϕ ∈ D(R2).

Exercise 3.7. Notice that(∂∂x

+ i ∂∂y

)1

x + iy = 2∂∂z

1z= 0

for z �= 0. Clearly, 1/(x + iy) is a locally integrable function (use polar coordinates).Let χ ∈ D(R), χ ≥ 0, χ(x) = 1 in a neighborhood of x = 0 and Suppχ ⊂ (−1,1).Set χε(x,y) = χ[(x2 +y2)/ε2] for any ε > 0. Then we have forϕ ∈ D(R2)⟨(

∂∂x

+ i ∂∂y

)1

x + iy , ϕ⟩=⟨(

∂∂x

+ i ∂∂y

)1

x + iy , χεϕ⟩

Hints to the Exercises 103

for any ε > 0. Working out the right-hand side in polar coordinates and letting ε tendto zero, gives the result.

Exercise 3.8. Apply Green’s formula, as in Section 3.5, or just compute using polarcoordinates.

Exercise 3.9. Apply Green’s formula, as in Section 3.5.

Exercise 3.10. Show first that ∂E/∂t − ∂2E/∂x2 = 0 for t �= 0. Then just compute,applying partial integration.

Exercise 4.4.a. The main task is to characterize allψ ∈ D of the formψ = ϕ′ for someϕ ∈ D.

Suchψ are easily shown to be the functionsψ inDwith∫∞−∞ψ(x)dx = 0.

b. Write F(x) =∫ x0 g(t)dt and show that F = f + const.

Exercise 4.5. The solutions are pv(1/x)+ const.

Exercise 4.9.a. Use partial integration.

b.∞∫−∞

sinλxx

ϕ(x)dx =A∫−A

sinλx[ϕ(x)−ϕ(0)

x

]dx +ϕ(0)

A∫−A

sinλxx

dx

if Suppϕ ⊂ [−A,A]. The first term tends to zero by Exercise 4.4 a. The secondterm is equal to

ϕ(0)A∫−A

sinλxx

dx = ϕ(0)λA∫−λA

sinyy

dy ,

which tends to πϕ(0) if λ→∞, since∞∫−∞

sinyy

dy = π.

c. limε↓0

∫|x|≥ε

cosλxx

ϕ(x)dx =∞∫−∞

cosλx − 1x

ϕ(x)dx +⟨

pv1x, ϕ

�.

104 Hints to the Exercises

Henceϕ � Pf

∞∫−∞

cosλxx

ϕ(x)dx is a distribution. If Suppϕ ⊂ [−A,A], then

∫|x|≥ε

cosλxx

ϕ(x)dx =∫

A≥|x|≥ε

cosλxx

ϕ(x)dx

=∫

A≥|x|≥ε

cosλxx

[ϕ(x)−ϕ(0)] dx

+ϕ(0)∫

A≥|x|≥ε

cosλxx

dx .

The latter term is equal to zero. Hence

limε↓0

∫|x|≥ε

cosλxx

ϕ(x)dx =A∫−A

cosλx[ϕ(x)−ϕ(0)

x

]dx .

This term tends to zero again if λ→ ∞ by Exercise 4.4 a.

Exercise 4.10. lima↓0

ax2 + a2

= πδ, lima↓0

axx2 + a2

= 0.

Exercise 6.16. (i) e−|x| ∗ e−|x| = (1+ |y|) e−|y|.For (ii) and (iii), observe that xe−ax

2 = − 12a

ddx

e−ax2 = − 1

2aδ′ ∗ e−ax

2.

Exercise 6.17. Observe that f(x) = f(−x), Fm(x) = Fm(−x) and f = δ−1∗Y −δ1 ∗ Y . Then SuppFm ⊂ [−m,m].

Exercise 6.18. LetA be the rotation over π/4 in R2:

A = 1√2

(1 −1

1 1

).

Then χ(x) = Y(Ax) for x ∈ R2. Use this to determine f ∗ g and χ ∗ χ.

Exercise 6.28. Apply Sections 6.6 and 6.7.

Exercise 6.29. Idem.

Exercise 6.30. Condition: g′ has to be a continuous function.

Exercise 6.31. Apply Sections 6.7 and 6.8.

Exercise 6.32. Set, as usual, p = δ′ and write z for zδ.

Exercise 6.34. Apply Section 6.10.

Hints to the Exercises 105

Exercise 7.20. T = pv 1x , T = −πi

[Y(y)− Y(−y)].

Exercise 7.21. f (y) = sin 2πy2πy

+ cos 2πy − 12πy

(y �= 0), f (0) = 1.

f is C∞, also because Tf ∈ E′.

Exercise 7.22. F(|x|) = − 12π2

Pf1x2

.

Exercise 7.23. Take the Fourier or Laplace transform of the distributions.

Exercise 7.24. Observe that, if fm tends to f in L2, then Tfm tends to Tf inD′.

References[1] N. Bourbaki, Espaces Vectoriels Topologiques, Hermann, Paris, 1964.[2] N. Bourbaki, Intégration, Chapitres 1, 2, 3 et 4, Hermann, Paris, 1965.[3] I. Daubechies, Ten Lectures on Wavelets, Society for Industrial and Applied Mathematics,

Philadelphia, Pennsylvania, 1992.[4] G. van Dijk, Introduction to Harmonic Analysis and Generalized Gelfand Pairs, de Gruyter,

Berlin, 2009.[5] F. G. Friedlander andM. Joshi, Introduction to the Theory of Distributions, CambridgeUniversity

Press, Cambridge, 1998.[6] I. M. Gelfand and G. E. Shilov, Generalized Functions, Vol. 1, Academic Press, New York, 1964.[7] L. Hörmander, The Analysis of Linear Partial Differential Operators, Vol. I–IV, Springer-Verlag,

Berlin, 1983–85, Second edition 1990.[8] G. Köthe, Topologische Lineare Räume, Springer-Verlag, Berlin, 1960.[9] L. Schwartz,MéthodesMathématiques pour les Sciences Physiques, Hermann, Paris, 1965.[10] L. Schwartz, Théorie des Distributions, nouvelle édition, Hermann, Paris, 1978.[11] E.M. Stein and G. Weiss, Introduction to Fourier Analysis on Euclidean Spaces, Princeton

University Press, Princeton, 1971.[12] A. E. Taylor and D. C. Lay, Introduction to Functional Analysis, second edition, John Wiley and

Sons, New York, 1980.[13] E. G. F. Thomas, Path Distributions on Sequence Spaces, preprint, 2000.[14] E. C. Titchnarsh, Theory of Fourier Integrals, Oxford University Press, Oxford, 1937.[15] F. Trèves, Topological Vector Spaces, Distributions and Kernels, Academic Press, New York,

1967.[16] E. T. Whittaker and G. N. Watson, A Course of Modern Analysis, Cambridge University Press,

Cambridge, 1980.[17] D. Widder, The Laplace Transform, Princeton University Press, Princeton, 1941.

IndexAbel integral equation 78averaging theorem 41

Banach–Steinhaus theorem 23, 84, 90Bessel function 70bounded– from the left 34– from the right 34bounded convergence property 86

canonical extension 85, 86Cauchy distribution 35Cauchy problem 71Cauchy sequence 91closed set 90complete space 92convergence principle 4, 26, 59convolution algebra 42convolution equation 43convolution of distributions 33

derivative of a distribution 9distribution 4dual space 92duplication formula 99

elementary solution 40

Fourier transform– of a distribution 63– of a function 52Fourier–Bessel transform 70Fréchet space 91

Gauss distribution 35Green’s formula 17

Hankel transform 70harmonic distribution 40harmonic function 17, 40heat equation 71Heaviside function 10

inversion theorem 54iterated Poisson equation 83

Laplace transform– of a distribution 75– of a function 74locally convex space 91locally integrable function 5

metric 94metric space 94metrizable space 94mollifier 88

neighborhood 90neighborhood basis 90Newton potential 40

open set 90order of a distribution 6

Paley–Wiener Theorems 65parametrix 83Partie finie 11Plancherel’s theorem 57Poisson equation 40principal value 11

radial function 69rank of a distribution 87rapidly decreasing function 59Riemann–Lebesgue lemma 52

Schwartz space 59seminorm 91slowly increasing function 60spherical coordinates 18structure of a distribution 28, 61summability order 82summable distribution 6, 82support– of a distribution 8– of a function 3symbolic calculus 45

tempered distribution 60tensor product– of distributions 31– of functions 31test function 3topological space 90topological vector space 91trapezoidal function 53triangle function 53

uniform boundedness 96

Volterra integral equation 47

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