the colombeau theory of generalized functions · 2020-07-13 · the colombeau theory of generalized...

The Colombeau theory of generalizedfunctions

Ta. Ngo.c TrıMathematics Master thesis specialized in Analysis 1

supervised byProf. Dr. Tom H. Koornwinder

KdV Institute, Faculty of ScienceUniversity of Amsterdam

The Netherlands

2005

1Key words: Schwartz distributions, tempered distributions, Fourier trans-form, convolution, multiplication of distributions, Colombeau generalized func-tions, generalized complex numbers, tempered generalized functions.

Abstract

In this report we discuss the nonlinear theory of generalized functions pro-posed by J. F. Colombeau in the 1980’s in two research monographs. Somemotivations coming from the famous linear theory of L. Schwartz will alsobe discussed. Finally, results by the author will be presented about the classof L1(Rn)-functions as a subset of the class of generalized functions.

Contents

Introduction 3

1 Impossibility and Degeneracy Results in Schwartz Distribu-tion Theory 41.1 The Schwartz distributions . . . . . . . . . . . . . . . . . . . . 4

1.1.1 Test function spaces . . . . . . . . . . . . . . . . . . . 41.1.2 Schwartz distributions . . . . . . . . . . . . . . . . . . 81.1.3 Differentiation . . . . . . . . . . . . . . . . . . . . . . 91.1.4 Product of a distribution and a smooth function . . . 111.1.5 Convolution and Fourier transform . . . . . . . . . . . 12

1.2 Product of distributions and related problems . . . . . . . . 18

2 The Colombeau theory of generalized functions 262.1 Preliminaries and notations . . . . . . . . . . . . . . . . . . . 262.2 Definition of Colombeau generalized functions . . . . . . . . . 282.3 Properties of the differential algebra G(Rn) . . . . . . . . . . 292.4 Nonlinear properties of G(Rn) . . . . . . . . . . . . . . . . . . 332.5 Generalized complex numbers . . . . . . . . . . . . . . . . . . 342.6 Point values of generalized functions . . . . . . . . . . . . . . 362.7 Integrals of generalized functions . . . . . . . . . . . . . . . . 382.8 Weak concepts of equality in G(Ω) . . . . . . . . . . . . . . . 422.9 The tempered generalized functions and their Fourier transform 49

3 L1(R) embedded in the Colombeau generalized functions 56

Discussion and Conclusions 64

Acknowledgments 65

References 66

2

Master Thesis – Ta. Ngo.c Trı 3

IntroductionThe theory of distributions, initiated by L. Schwartz, has opened the doorfor important developments in modern Mathematics, especially in PartialDifferential Equations (see [Sch01], [Hor83]). For this theory, L. Schwartzwon the Fields Medal in 1950. That prestigious prize is the recognition bythe mathematical world for his extremely important work.

Soon after the introduction of his own theory, L. Schwartz published apaper in which he showed an impossibility result (see [Sch54]) about theproduct of two arbitrary distributions. However, in some applications thereis need for such a product. Various mathematicians looked for a way aroundthe Schwartz impossibility result (see [Obe92], [Obe01], [Ros87]). They triedto find methods to define the product of two arbitrary distributions. Someof them partly solved that issue (see [Mik66], [Tys81]). But the need for afull solution remained.

In the 1980’s such a theory was proposed by J. F. Colombeau. In twosuccessive monographs ([Col84] and [Col85]) he proposed the theory of gen-eralized functions. In this new theory the distributions are a subset, andwe can multiply two arbitrary generalized functions. After the appearanceof the Colombeau theory, some mathematicians have relied on it, and gotsome interesting results about the solutions of nonlinear partial differentialequations occurring in nature (see [Obe92], [Obe01], [Gko01]). These resultsshow the importance of the new theory.

In this thesis, the author presents some aspects of the issues above. In thefirst chapter we discuss the Schwartz theory of distributions and the issuesto occur. Another part of the thesis is devoted to the main contents of theColombeau theory of generalized functions. Some questions which came upin the study of this theory are discussed and answered in the last chapter.

This is the graduation thesis for the 2 year Master of Mathematics pro-gram at Korteweg de Vries Institute for Mathematics (KdV), University ofAmsterdam (UvA), The Netherlands.

Chapter 1

Impossibility and DegeneracyResults in SchwartzDistribution Theory

1.1 The Schwartz distributions

To solve the issue of non-existence of derivatives for many ordinary functions,L. Schwartz proposed the distributions in the 1940’s. This theory has con-tributed to the development of some important fields in Mathematics sincethen, such as the development of Partial Differential Equations. I would liketo sketch some main ideas of this distribution theory as well as some “issues”which arise in the process of using it in Mathematics.

1.1.1 Test function spaces

We start with some terminologies that will be used later. A multi-index (or,to be precise, an n-multi-index) is an n-tuples α = (α1, α2, . . . , αn) of nonneg-ative integers; its length(or order) is |α| = α1+α2+· · ·+αn. With each multi-index α is associated the differential operator ∂α = ∂α1

1 ∂α22 . . . ∂αn

n , where∂j = ∂

∂xj, or Dα = Dα1

1 · · ·Dαnn where Dj = ∂

i∂xj, j := 1, 2, . . . , n and i =

√−1;

their order is |α| = α1 + α2 + · · ·+ αn.

Let be given a non-empty open set Ω ⊆ Rn. A complex-valued functionf defined on Ω is said to belong to C∞(Ω) if ∂αf exists and is continuousfor every multi-index α. The support of a continuous function f : Ω −→ C,denoted by supp f, is the closure in Ω of the set x ∈ Ω : f(x) 6= 0. Wenotice that supp f is a closed subset of Ω. If K is a compact set in Rn, then

4


we denote by DK the set f ∈ C∞(Rn) : supp f ⊆ K.

A topological vector space over K, (K := R, or C) is a vector space Xover K, equipped with a topology that is compatible with the vector spacestructure, i.e., such that the mapping (x, y) 7−→ x + y and (λ, x) 7−→ λx arecontinuous.

In a topological vector space X, a subset E is bounded if for every neigh-borhood V of 0, there is a number s > 0 such that E ⊂ tV, for all t > s. If 0has a bounded neighborhood, then X is said to be locally bounded.

A subset E of a topological vector space X is said to be absorbing if, forall x ∈ X, there is t = t(x) 6= 0 such that x ∈ tE. If for all α ∈ C, |α| ≤ 1,we have αE ⊂ E, then E is called a balanced subset of X.

A topological vector space X (with topology τ) is called a locally convexif there is a local base for τ whose members are convex. A locally convexspace is called a Frechet space if it is also induced by a complete metric dsatisfying d(x + z, y + z) = d(x, y) (d is translation invariant).

A topological vector space X is said to have the Heine–Borel property ifevery closed and bounded subset of X is compact. And we can see that:

1.1.1.1 Proposition

C∞(Ω) is a Frechet space with the Heine-Borel property, and DK is a closedsubspace of C∞(Ω), whenever K ⊂ Ω.

Proof. Please refer to [Rud85], pp.32–35 in detail. Here, we only sketchsome main points.

Choose compact sets Kj, j = 1, 2, . . . , such that Kj lies in the interior ofKj+1 (denoted by Int Kj+1) and Ω = ∪∞j=1Kj. The family of seminorms pN

for N = 1, 2, . . . , defined by setting

pN(f) = max|∂αf(x)| : x ∈ KN , |α| ≤ Nhas the properties: separating points of C∞(Ω) and generating a topologywith a countable local base. So, a compatible translation-invariant metric dcan be defined in terms of pN , N = 1, 2, . . . as follows:

d(f, g) =∞∑

N=1

2−NpN(f, g)

1 + pN(f, g).

6 The Colombeau theory of generalized functions

It is not difficult to show that this metric is complete, so C∞(Ω) is a Frechetspace.

For each x ∈ Ω, the functional Fx : f 7−→ f(x) is continuous in thetopology induced by the family of pN for N = 1, 2, . . .. We also notice that

DK =⋂

x∈Ω\Kker Fx,

so DK is a closed subspace of C∞(Ω) for arbitrary compact subset K ofΩ. For the proof of the Heine-Borel property of C∞(Ω), please refer to[Rud85] for details. Lastly, we remark that if a sequence (fn) converges to fin the topology defined above on C∞(Ω), then the sequence (fn) convergesuniformly to f on compact sets in Ω.

Note. We would like to comment that the concepts of an increasing se-quence of compact sets in [Fri98], p.35; or, monotone increasing sequence ofcompact subsets in [Yos74], p.27, for Kj, j = 1, 2, . . . mean Kj ⊂ Int Kj+1, j =1, 2, . . . . The following example shown by Prof. Tom Koornwinder to me willexplain the importance of this condition.

Consider Ω = (−1, 1) ⊂ R, Kn = [−1+1

n,− 1

n]∪[

1

n, 1− 1

n]∪0, n = 1, 2, . . . .

We can easily verify that

Kj ⊂ Kj+1, j = 1, 2, . . . ,∪∞n=1Kn = Ω.

However, Kn is not a subset of Int Km whenever n < m. We could take

the sequence (fn) in C∞(Ω) such that fn(x) = 0 on Kn and fn(1

2n) = 1,

for n = 1, 2, . . . . We can verify that the sequence of fn converges to 0 inthe topology on C∞(Ω), obtained by the K ′

ns, but fn does not converge to 0uniformly in the usual sense!

We also notice that C∞(Ω) is not locally bounded (with the topologyinduced by the family of pN , N = 1, 2, . . .), since if it was, it would havefinite dimension and this is absurd (See [Rud85], Theorem 1.23), so it followsfrom Theorem 1.39 in [Rud85] that C∞(Ω) is not normable. We also get thesame result for DK whenever K has nonempty interior.

1.1.1.2 Definition

The union of the spaces DK , as K ranges over all compact subsets of Ω iscalled the test function space, denoted by D(Ω).


It is clear that D(Ω) is a vector space, equipped with the usual definitionsof addition and scalar multiplication of complex-valued functions. It is alsoobvious that φ ∈ D(Ω) if and only if φ ∈ C∞(Ω) and supp φ is a compactsubset of Ω. To construct a locally convex topology τ on D(Ω) in which allCauchy sequences do converge, we take for τ the collection of all unions ofsets of the form φ + W, with φ ∈ D(Ω) and W ∈ β, where β is the collectionof all convex balanced sets W ⊂ D(Ω) : DK ∩ W ∈ τK , for every compactK of Ω, with the topology τK which has been defined in DK . Please refer to[Rud85], pp.137–138 for details. Here, we only list some important propertiesof this topology.

1.1.1.3 Proposition

• τ is a topology in D(Ω), and β is a local base for τ

• τ makes D(Ω) into a locally convex topological vector space.

We also remark that τ has the following further properties:

• τK coincides with the subspace topology DK that inherits from D(Ω)

• D(Ω) has the Heine-Borel property

• If (φj) is a Cauchy sequence in D(Ω), then (φj) ⊂ DK for some compactK ⊂ Ω, and

limk,j→∞

sup|∂αφk(x)−∂αφj(x)| : x ∈ Ω; |α| ≤ N = 0,where N = 0, 1, 2, . . . .

• If φj → 0 as j → ∞ in τ , then there is a compact K ⊂ Ω whichcontains the support of every φj, j = 1, 2, . . . , and ∂αφj → 0 uniformlyas j →∞, for every multi-index α

• In D(Ω), every Cauchy sequence converges.

We would like to mention one special kind of topological vector space calledMontel space in the following definition. Please refer to [Tre67], p.356 forfurther details.


1.1.1.4 Definition

A topological vector space X is called a Montel space if X is locally con-vex Hausdorff such that every absorbing, covex, balanced and closed subsetof X is a neighborhood of zero in X, and X also has the Heine-Borel property.

Montel spaces share a special property which we will use later: a Montelspace is always reflexive. Also to end this section, we recall the followingresult (All the details are in [Tre67], p.357).

1.1.1.5 Proposition

C∞(Ω) and D(Ω) are Montel spaces!

1.1.2 Schwartz distributions

We start with the definition of Schwartz distributions in [Fri98], p.7 asfollows.

1.1.2.1 Definition

A linear form u : D(Ω) −→ C is called a distribution (or a Schwartz distri-bution) if, for every compact K ⊂ Ω, there is a real number c ≥ 0 and anonnegative integer N such that

|〈u, φ〉| ≤ c∑

|α|≤N

sup |∂αφ|,

for all φ ∈ D(Ω) with supp φ ⊂ K. The vector space of distributions on Ω isdenoted by D′(Ω).

Here, the vector space operations on the space of distributions are theusual ones on linear forms.

We also see that all ordinary functions we often meet such as contin-uous functions, or functions in Lp(Ω), 1 ≤ p ≤ ∞ are distributions. Forinstance, if f is continuous on Ω, then f : D(Ω) −→ C such that 〈f, φ〉 =∫

Ωf(x)φ(x) dx, φ ∈ D(Ω), is a distribution. We can easily verify this con-

clusion. Here, we only mention a newcomer: the Dirac distribution. Weconsider δ : D(Rn) −→ C such that 〈δ, φ〉 = φ(0), for all φ ∈ D(Rn). Ob-viously, we get |〈δ, φ〉| = |φ(0)| ≤ 1 sup |φ(x)|, for all φ ∈ D(Rn), such thatsupp φ ⊂ K, K a compact subset of Rn. Therefore, δ is a distribution and


we call it the Dirac distribution. In fact, this strange function appeared inPhysics at the beginning of the 20th century, and it was one of motivationsfor the appearance of distributions!

We can describe distributions in another way as follows (refer [Fri98],p.9).

1.1.2.2 Proposition

A linear form u on D(Ω) is a distribution if and only if limj→∞〈u, φj〉 = 0for every sequence (φj) which converges to zero in D(Ω) as j →∞.

We will consider a corollary of Proposition 1.1.2.2: a locally integrablefunction f on Ω yields a distribution defined by φ 7−→ 〈f, φ〉 :=

∫Ω

f(x)φ(x) dx.Here, a measurable f such that

∫K|f(x)| dx < ∞ for all compact subsets

K ⊂ Ω is called a locally integrable one. The set of all locally integrable func-tions on Ω is denoted by L1

loc(Ω). Now we turn to the proof that the above lin-ear form is a distribution. Indeed, if (φj) → 0 as j →∞ inD(Ω), then there isa compact subset K ⊂ Ω such that supp φj ⊂ K, j = 1, 2, . . . . So, we can de-fine

∫Ω

f(x)φj(x) dx =∫

Kf(x)φj(x) dx. Moreover, from |fφj| ≤ |f | supK |φj|,

and the uniform convergence of (φj) on K, we can apply the dominated con-vergence Lebesgue theorem, and we get limj→∞〈f, φj〉 = 0.

1.1.3 Differentiation

One of the motivations for distribution theory was to get a space whichcontains ordinary functions and in which every element has a derivative,which coincides with ordinary derivative if the element is a differentiablefunction. D′(Ω) is such a space in view of the following result:

1.1.3.1 Proposition

If u ∈ D′(Ω), then the following linear form, denoted by ∂αu (α a multi-index)

〈∂αu, φ〉 = (−1)|α|〈u, ∂αφ〉, φ ∈ D(Ω),

is also a distribution.

The proof is straight forward, see, for instance [Yos74], p.49.

The motivation for this definition of ∂αu is that it coincides with ordinarydifferentiation if u is a differentiable function. Indeed, if f ∈ C1(Ω), the


space of functions which have continuous derivatives up to order 1 on Ω,then we can remark that

∫Ω

∂j(fφ) dxj = 0 for all j = 1, 2, . . . , n. It followsfrom the Fubini’s rule that

∫Ω

∂j(fφ) dx1 · · · dxn = 0. Hence∫

Ω∂jfφ dx =

− ∫Ω

f∂jφ dx. Or 〈∂jf, g〉 = −〈f, ∂jg〉. So, if we identify f ∈ C1(Ω) withthe distribution f ∈ D′(Ω), 〈f, φ〉 =

∫Ω

fφ dx, the above formula shows themotivation of differentiation we have mentioned.

1.1.3.2 Examples

a. If

H(x) =

1 if x ≥ 0,

0 if x < 0

(called Heaviside function), then ∂H = δ. Indeed, we consider Ω = R, andwe have

〈∂H, φ〉 = (−1)1〈H, ∂φ〉 = −∫ ∞

0

∂φ(x) dx = −φ(x)

∣∣∣∣∣

∞

0

= φ(0) = 〈δ, φ〉,

for all φ ∈ D′(R). Therefore, ∂H = δ.

b. If f(x) = log |x|, then we can verify that f is locally integrable on R.Therefore, the corresponding distribution defined by f is

〈f, φ〉 =

∫ ∞

−∞log |x|φ(x) dx.

We will calculate ∂f. This distribution is usually denoted by1

x. We have

〈∂(log |x|), φ〉 = −∫ ∞

−∞log |x| ∂φ(x) dx

= −∫ 0

−∞log |x| ∂φ(x) dx−

∫ ∞

0

log |x| ∂φ(x) dx

= − limε→0+

[

∫ −ε

−∞log |x| ∂φ(x) dx +

∫ ∞

ε

log |x| ∂φ(x) dx]

= limε→0+

[φ(ε)− φ(−ε)] log ε +

∫ −ε

−∞

φ(x)

xdx +

∫ ∞

ε

φ(x)

xdx

= limε→0+

[

∫ −ε

−∞

φ(x)

xdx +

∫ ∞

ε

φ(x)

xdx],


since [φ(ε)−φ(−ε)] log ε → 0, as ε → 0+. In fact1

xdoes not belong to L1

loc(R).

So, we cannot represent the distribution1

xin integral form. However, in

the above approach we can define the distribution1

x∈ D′(R) as ∂ log |x|,

and it belongs to D′(R)\L1loc(R). The expression limε→0+ [

∫ −ε

−∞φ(x)

xdx +

∫∞ε

φ(x)

xdx], denoted by p.v.

∫∞−∞

φ(x)

xdx, is called the principal value of

the integral∫∞−∞

φ(x)

xdx.

1.1.4 Product of a distribution and a smooth function

In distribution theory we can get the product of a distribution u ∈ D′(Ω),and a smooth function f ∈ C∞(Ω). This product, denoted by fu, is definedas follows

〈fu, φ〉 = 〈u, fφ〉, for all φ ∈ D(Ω).

It is not difficult to verify that the right-hand side is a distribution, and itjustifies the above definition. For instance, if δ ∈ D′(R), then xδ = 0, since

〈xδ, φ〉 = 〈δ, xφ〉 = (xφ)(0) = 0,∀φ ∈ D(R).

For the case of u =1

x, we can prove that x

1

x= 1 in D′(R). Indeed, we have

〈x1

x, φ〉 = 〈1

x, xφ〉

= limε→0+

[

∫ −ε

−∞

xφ(x)

xdx +

∫ ∞

ε

xφ(x)

xdx]

=

∫ ∞

−∞φ(x) dx

= 〈1, φ〉, ∀φ ∈ D(R).

Leibniz’s rule extends to the product of f ∈ C∞(Ω) and u ∈ D′(Ω) by thefollowing proposition.

1.1.4.1 Theorem

Let f ∈ C∞(Ω) and u ∈ D′(Ω), and let α be an arbitrary multi-index, then

∂α(fu) =∑

β+γ=α

α!

β!γ!∂βf ∂γu,


where α! = α1!α2! . . . αn! if α = (α1, α2, . . . , αn).

Proof. Please refer to [Fri98], p.24 for details.

Next, we will extend the concept of support of an ordinary function to thecase of distributions. Firstly, we say that a distribution u ∈ D′(Ω) vanisheson an open set V of Ω if 〈u, φ〉 = 0 for all φ ∈ D(Ω) with support containedin V . Secondly, we notice the following result, for instance, in [Yos74], p.62.

1.1.4.2 Theorem

If Vjj∈I is a family of open subsets of Ω and if u vanishes on each Vj, j ∈ I,then u vanishes on V = ∪j∈IVj.

Finally, we can say that the support of a distribution u, denoted bysupp u, is the complement of the set x ∈ Ω : u vanishes on a neighborhoodof x. So, as in the case of continuous functions, supp u is a closed subsetof Ω. For instance, supp δ = 0. Indeed, if x 6= 0, so we can select U, anopen neighborhood of x such that 0 /∈ U, then 〈δ, φ〉 = φ(0) = 0. So, wehave supp δ ⊂ 0. On the other hand, for every neighborhood V of 0, thereis a function φ ∈ D(Rn) such that supp φ ⊂ V, φ(0) 6= 0. So, with such φwe have 〈δ, φ〉 = φ(0) 6= 0. Therefore, 0 ⊂ supp δ. In summary, we havesupp δ = 0. We also notice that if f is continuous, or locally integrableon Ω, then supp f in the sense of distributions coincides with supp f inusual sense. For f ∈ L1

loc(Ω), supp f means that x /∈ supp f ⇔ there is aneighborhood U of x in Ω on which f = 0 almost everywhere.

1.1.5 Convolution and Fourier transform

For 1 ≤ p < ∞, let

Lp(Rn) = f defined and measurable on Rn :

∫

Rn

|f(x)|pdx < ∞,

where the integral is in Lebesgue sense and almost everywhere equal func-tions are identified with each other. When equipped with the norm ||f ||Lp :=

(∫Rn |f(x)|p dx)

1p , the space Lp(Rn), 1 ≤ p < ∞ becomes a separable Banach

space (Refer to [Ada75] in detail).

If f, g ∈ L1(Rn), then the convolution of f and g, denoted by f ∗ g, is


defined as follows:

(f ∗ g)(x) :=

∫

Rn

f(y)g(x− y) dy.

It can be shown that (f ∗g)(x) exists almost everywhere, f ∗g ∈ L1(Rn), and||f∗g||L1 ≤ ||f ||L1||g||L1 . This operation makes L1(Rn) into a Banach algebra,but without unit (See [Arv98]). In fact we can also define the convolutionof f and g if f ∈ L1(Rn) and g ∈ Lp(Rn), 1 ≤ p ≤ ∞ (see [Ada75] forthe definition of L∞(Rn)). There are some interesting results related to thatcase, see again [Ada75] for details. Here, we will consider the extension ofthis convolution to distributions.

1.1.5.1 Definition

If u, v ∈ D′(Rn), we call the convolution of u and v, denoted by u ∗ v, thefollowing linear form

〈u ∗ v, φ〉 = 〈u(y), 〈v(x), φ(x + y)〉〉,where φ ∈ D(Rn), whenever 〈u, ψ〉 with ψ(y) := 〈v(x), φ(x + y)〉, is well-defined and the resulting linear form u ∗ v is in D′(Rn).

For instance, we can prove that if at least one of u and v has compactsupport, the above definition will be satisfied and u ∗ v = v ∗ u. Moreover, in[Fri98] there are some other cases discussed where u ∗ v is well-defined.

1.1.5.2 Remarks

a) u ∗ δ = δ ∗ u = u, for all u ∈ D′(Rn). Indeed, δ has compact support and

〈u(y), 〈δ(x), φ(x + y)〉〉 = 〈u(y), φ(y)〉 = 〈u, φ〉b) The above definition of convolution is still valid if f, g ∈ L1(Rn). Indeed,for arbitrary φ ∈ D(Rn), set

h(y) =

∫

Rn

g(x)φ(y + x) dx,

then h ∈ L1(Rn) by 1.1.5. Moreover, we get

|h(y)| ≤∫

Rn

|g(x)φ(y + x)| dx =

∫

Rn

|g(t− y)| |φ(t)| dt

≤ supt∈supp φ

|φ(t)|∫

Rn

|g(t− y)| dt = c||g||L1 ,


for y ∈ Rn. So, since

〈f(y), 〈g(x), φ(y + x)〉〉 = 〈f(y), h(y)〉 =

∫

Rn

f(y)h(y) dy,

and |f(y)h(y)| ≤ c||g||L1|f(y)|, we get the existence of 〈f(y), 〈g(x), φ(y+x)〉〉,so f ∗ g exists. We also have (by Fubini)

〈f ∗ g, φ〉 =

∫

Rn×Rn

f(y)g(x)φ(y + x) dx dy

=

∫

Rn

(

∫

Rn

f(y)g(t− y) dy)φ(t) dt

= 〈∫

Rn

f(y)g(t− y) dy, φ(t)〉

.

So, (f ∗ g)(t) =∫Rn f(y)φ(t− y) dy as usual!

Next, we will get an interesting result related to the convolution of u ∈D′(Rn) and ρ ∈ D(Rn) (See, for instance, [Fri98], p.53) as follows.

1.1.5.3 Theorem

If u ∈ D′(Rn) and ρ ∈ D(Rn), then

(ρ ∗ u)(x) = 〈u(y), ρ(x− y)〉, x ∈ Rn,

and this function is a member of C∞(Rn).

In literature ρ∗u is called a regularization of u, and later we will see thatsome mathematicians such as Mikusinki, Itano, Fisher,...have used regular-izations of u, v ∈ D′(Rn) to define the multiplication uv (Refer to [Col84]p.31 for details).

If f ∈ L1(Rn), the Fourier transform f (sometimes f∧, or Ff) of f is

defined by f(ξ) =∫Rn f(x)e−ix.ξ dx, ξ ∈ Rn, where x.ξ is the usual inner

product on Rn and x.ξ =∑n

j=1 xjξj. We will extend this concept to distribu-tions, or rather to a special subspace of D′(Rn) called the space of tempereddistribution.

1.1.5.4 Definition

We denote by S(Rn) the totality of functions f ∈ C∞(Rn), such that

pα,β(f) := supx∈Rn

|xβDαf(x)| < ∞,


for all multi-indices α, β. Such functions are called rapidly decreasing ones.

We can show that f ∈ D(Rn), then f ∈ S(Rn); or e−|x|2

is also a memberof S(Rn).

The family of pα,β : α, β are multi-indices is the one of separating semi-norms defined on S(Rn), and it induces a locally convex topology. Further-more, this topology is metrizable, and the induced distance is complete. So,S(Rn) is a Frechet space. We also have:

φj −→ 0 in S(Rn) ⇔ ∀α, β; pα,β(φj) −→ 0

We also see that the embedding D(Rn) → S(Rn) continuous, and D(Rn) isdense in S(Rn) for the above topology defined above in S(Rn). One importantresult in S(Rn) we need later is in the following lemma

1.1.5.5 Lemma

The Fourier transform F : S(Rn) −→ S(Rn) is a continuous map.

Now, we will define tempered distributions

1.1.5.6 Definition

S ′(Rn) is the subspace of D′(Rn) consisting of distributions which extendedto be continuous linear forms on S(Rn). A sequence (uj)1≤j≤∞ in S ′(Rn) issaid to converge to u ∈ S ′(Rn) if 〈uj, φ〉 −→ 〈u, φ〉 for all φ ∈ S(Rn) asj −→∞. The members of S ′(Rn) are called tempered distributions.

We can show that u ∈ S ′(Rn) if and only if φj −→ 0 in S(Rn) as j −→∞,implies that 〈u, φj〉 −→ 0 as j −→ 0.

Based on that remark, one can easily verify that L1(Rn) ⊂ S ′(Rn). Wealso get that all distributions of D′(Rn) with compact supports are tempereddistributions, too. For instance, the Dirac function δ ∈ S ′(Rn). Indeed, ifu ∈ D′(Rn) and the compact set K in Rn is supp u, we fix ψ ∈ D(Rn) suchthat ψ = 1 on some open set containing K. We define

〈u, f〉 := 〈u, ψf〉, f ∈ S(Rn)

If fj −→ 0 in S(Rn) as j → ∞, then all Dαfj −→ 0 uniformly on Rn asj →∞. Hence, all Dα(ψfj) −→ 0 uniformly on Rn as j →∞. It follows that


u is a continuous linear form on S(Rn). Since 〈u, φ〉 = 〈u, φ〉 for φ ∈ D(Rn), uis an extension of u. That concludes our arguments.

In fact one can prove that every f ∈ Lp(Rn), 1 ≤ p ≤ ∞; is a tempereddistribution. Furthermore, if P is a polynomial, f ∈ S(Rn), and u is a tem-pered distribution, then the distributions Dαu, Pu,and fu are also tempereddistributions. Please refer to [Rud85] for details.

Now we will define the Fourier transform of a tempered distribution.

1.1.5.7 Definition

The Fourier transform of u ∈ S ′(Rn) is the distribution u ∈ S ′(Rn) definedby

〈u, φ〉 := 〈u, φ〉, φ ∈ S(Rn).

We remark that it follows from Lemma 1.1.5.5 that if φj −→ 0 as j −→ 0

in S(Rn), then φj −→ 0 as j −→ 0 in S(Rn), and it verifies the above setting.There arises a consistency question that ought to be settled. If f ∈ L1(Rn),then f also a member of S ′(Rn), denoted by uf . So, there are two definitions

of the Fourier transform of f , namely f(ξ) =∫Rn f(x)e−ix.ξ dx, ξ ∈ Rn, and

uf . The question is if they agree, i.e., whether the distribution uf corresponds

to the function f . The answer is affirmative, and we can show this as follows

〈uf , φ〉 = 〈uf , φ〉 =

∫

Rn

fφ dx =

∫

Rn

fφ dx = 〈f , φ〉,

for all φ ∈ S(Rn) with the “traditional” Fourier transform and the Fouriertransform of f as a tempered distribution in S ′(Rn) (since L1(Rn) ⊂ S ′(Rn)).

1.1.5.8 Examples

a) δ ∈ S ′(Rn), so we can define δ as follows

〈δ, φ〉 = 〈δ, φ〉 = φ(0) =

∫

Rn

e−ix.0φ(x) dx =

∫

Rn

φ(x) dx = 〈1, φ〉

Hence, we have δ = 1.

b) From the above example, we get 1 ∈ S ′(Rn), so we will consider 1, andwe get

〈1, φ〉 = 〈1, φ〉 =

∫

Rn

φ(ξ) dξ = (2π)n[1

(2π)n

∫

Rn

eiξ.0φ(ξ) dξ]

= (2π)nφ(0) = (2π)n〈δ, φ〉,


where we have used the inverse Fourier transform φ(x) =1

(2π)n

∫Rn eiξ.xφ(ξ) dξ.

Therefore, we have 1 = (2π)nδ.

c) Next, we will show an interesting result related to the Fourier transformof a distribution with compact support. That is, if u is a distribution withcompact support, then u(ξ) = 〈u(x), e−ix.ξ〉, and this u is in C∞(Rn). Even,if we extend the variable ξ to the complex domain, then we also have a verynice result called the Paley-Wiener theorem as follows.

1.1.5.9 Theorem

a) If u is a distribution in S ′(Rn) with compact support, then

u(ξ) = 〈u(x), e−ix.ξ〉, ξ ∈ Cn,

is an analytic function on Cn. The function u(ξ), ξ ∈ Cn is often said to bethe Fourier-Laplace transform of ub) If u is a distribution in D′(Rn) with compact support and supp u ⊂ x :|x| ≤ a, where a is a positive real number, then there are constants c,N ≥ 0such that

|u(ξ)| ≤ c(1 + |ξ|N)ea|Im ξ|, ξ ∈ Cn

c) If u ∈ D(Rn) and supp u ⊂ x : |x| ≤ a, then there are constantscm ≥ 0,m = 0, 1, . . . , such that

|u(ξ)| ≤ cm(1 + |ξ|)−mea|Im ξ|, ξ ∈ Cn,m = 0, 1, . . . .

Proof. Please refer to [Fri98] in detail for a) and b).

Here, we sketch some main points of the proof for c) (see the same refer-ence as above)If u ∈ D(Rn), then

u(ξ) =

∫

Rn

u(x)e−ix.ξ dx.

Also, since u ∈ D(Rn), one can perform repeated partial integration in theabove identity and get

ξαu(ξ) = (−i)|α|∫

Rn

∂αu(x)e−ix.ξ dx, ξ ∈ Cn,


for all multi-indices α, |α| ≥ 0. So one has, if supp u ⊂ x : |x| ≤ a, then

|ξα||u(ξ)| ≤ Va sup|x|≤a

|∂αu| supex|Im ξ| : x ∈ supp u,

where Va is the measure of the ball x : |x| ≤ a. Hence,

|ξα||u(ξ)| ≤ Va sup|x|≤a

|∂αu|ea.|Im ξ|, ξ ∈ Cn,

for all α, |α| ≥ 0, and these inequalities clearly imply c).

Next, we would like to present a theorem called the structure theoremwhich shows the relationship among distributions and ordinary functions,and we will apply this result in Chapter 2. We can see the details of prooffor this result in [Sch66], or [Fri98].

1.1.5.10 Theorem

The restriction of a distribution u ∈ D′(Rn) to a bounded open set Ω ⊂ Rn

is a derivative of finite order of a continuous function.

1.2 Product of distributions and related prob-

lems

In 1.1.4 we have defined the product of f ∈ C∞(Ω) and u ∈ D′(Ω). Inpractice as well as in mathematics view, we would like to define the prod-uct of two arbitrary distributions, say on Rm. Clearly, we cannot use themethod of 1.1.4, because fφ is not a reasonable test function if f ∈ D′(Rm)and φ ∈ D(Rm). To overcome this, some suggestions came up in literature.However, we will see later that for such ways these still do not lead to thedefinition of the product of two arbitrary distributions. We are also inter-ested in some ways to embed D′(Rm) in an algebra, so that we can get theproduct. But some limitations arise, and we will end this section with sucha result called “Schwartz impossibility result”.

Next, we will discuss the first way of defining the product of two arbitrarydistributions called “method of regularization and passage to the limit” (See[Mik66], or [Col84]).


1.2.1 Definition

A δ-sequence is a sequence (δn), n = 1, 2, . . . of elements of D(Rm) such thata) supp δn ⊂ x ∈ Rm : |x| ≤ εn with limn→∞ εn = 0b)

∫Rm δn(x) dx = 1.

We remark that if S, T ∈ D′(Rm), then from Theorem 1.1.1.5, the convo-lution S ∗δn and T ∗δn are C∞-functions on Rm, and they are regularizationsof S and T. So, we can get (S ∗ δn)(T ∗ δn) in the usual sense, and thismotivates the definition of the distribution product below.

1.2.2 Definition

We say that S and T are multiplicable with product S.T if, for any δ-sequence(δn), n = 1, 2, . . . , there exists limn→∞(S ∗ δn)(T ∗ δn) in D′(Rm) and its limitis independent of the choice of δ-sequence.

Further motivation for the above definition is that limn→∞ δn = δ inD′(Rm), so limn→∞(S ∗ δn) = S and limn→∞(T ∗ δn) = T in D′(Rm). Thisdefinition was proposed by Hirata-Ogata, Mikusinki, Itano, Fisher,. . . (See[Col84]). Some products of distributions can indeed be realized in this way.For instance, we have

1.2.3 Example

In D′(R), we get1

x.δ = −1

2δ′

in the sense of Definition 1.2.2

Proof. We will sketch some main points.

Denote by δ−n (x) the function δn(−x), n = 1, 2, . . . , one gets

〈( 1

x∗ δn).(δ ∗ δn), φ〉 =

= 〈( 1

x∗ δn).δn, φ〉 = 〈( 1

x∗ δn), δn.φ〉 = 〈1

x, δ−n ∗ δnφ〉

for all φ ∈ D(R).


If we expand φ(x) = φ(0) + xφ′(0) + x2ψ(x), then

〈( 1

x∗δn).(δ∗δn), φ〉 = φ(0)〈1

x, δ−n ∗δn〉+φ′(0)〈1

x, δ−n ∗(xδn)〉+〈1

x, δ−n ∗(x2ψ)δn〉

One can prove that the last term in the right hand-side of the above identitytends to 0 as n →∞ (See [Ita76] or [Mik66]. . .) for details).

The first term tends to 0 as n →∞ because δ−n ∗ δn is an even function.

Here, we will show that the second term tends to1

2φ′(0). Indeed, we have

φ′(0)〈1x, δ−n ∗ (xδn)〉 =

∫ ∞

−∞

1

xαn(x) dx,

where αn = δ−n ∗ (xδn), and it follows that

α−n = δn ∗ [(−x)δ−n ] = −x(δn ∗ δ−n ) + (xδn) ∗ δ−n , since

it is obvious that if ψ1 and ψ2 are in L1(R), then x(ψ1 ∗ ψ2) = (xψ1) ∗ ψ2 +ψ1 ∗ (xψ2). Therefore, αn − α−n = x(δn ∗ δ−n ), and

〈1x, αn〉 =

1

2〈1x, αn + α−n 〉+

1

2〈1x, αn − α−n 〉 =

1

2〈1x, αn − α−n 〉,

since αn + α−n is even. It follows that

〈1x, αn〉 =

1

2〈1x, x(δn ∗ δ−n )〉 =

1

2

∫ ∞

−∞(δn ∗ δ−n )(x) dx =

1

2

since δn ∈ D(R), n = 1, 2, . . . and∫R δn(x) dx = 1.

So, we have

limn→∞

〈( 1

x∗ δn)(δ ∗ δn), φ〉 =

1

2φ′(0) = −1

2〈δ′, φ〉,

for all φ ∈ D(R), or

limn→∞

(1

x∗ δn)(δ ∗ δn) = −1

2δ′

in D′(R). Therefore, we have

1

x.δ = −1

2δ′ in D′(R),

in the sense of Definition 1.2.2However, we have:


1.2.4 Proposition

In D′(R), there does not exist δ2 in the sense of Definition 1.2.2.

Note. From this result we can see that this way of defining the product ofdistributions does not meet what we need!

Proof. Conversely, we assume that there exists δ2 ∈ D′(R) in the senseof Definition 1.2.2. Taking an arbitrary δ-sequence (δn), n = 1, 2, . . . , wealways have the existence of

limn→∞

〈δ2n, ψ〉, for all ψ ∈ D(R).

We take ψ ∈ D(R) such that ψ ≡ 1 in a neighborhood of 0. Then,

〈δ2n, ψ〉 =

∫ ∞

−∞δ2nψ dx =

∫ ∞

−∞δ2n dx.

Because of the existence of limn→∞〈δ2n, ψ〉, we get (δn), n = 1, 2, . . . bounded

in L2(R). It is well-known that in L2(R) the closed unit ball is weakly com-pact, so there is a subsequence (δnk

), k = 1, 2, . . . of (δn), n = 1, 2, . . . whichweakly converges to g in L2(R) (we can see this from the result of Problem18 in [Hal67] that the weak topology of closed unit ball in L2(R) is metriz-able). Hence, we have for all ψ ∈ L2(R) that 〈g, ψ〉 = limk→∞〈δnk

, ψ〉, so forψ ∈ D(R), we have limk→∞〈δnk

, ψ〉 = 〈g, ψ〉. It follows that δ = g ∈ L2(R),and this is a contradiction. We conclude that (δ2

n), n = 1, 2, . . . never con-verges in D′(R). These arguments above are based on Prof. Tom Koorn-winder’s remarks on Colombeau theory. They clarify the arguments in[Obe92], p.24.

There are some other examples in which there do not exist products ofdistributions like δ2 above. Please refer to [Obe92] for details.

Now we will show another way of defining the product of distributionsbased on the Fourier transform. We notice that, if u ∈ D′(Rm) with compactsupport, then

u∧(ξ) = 〈u(x), e−ix.ξ〉.Denoting by u∨ its converse Fourier transform, we also have

u∨(x) =1

(2π)m〈u(ξ), eiξ.x〉.

Also denote by M(Rm) all the pairs (u, v) ∈ D′(Rm)×D′(Rm) such that forevery x ∈ Rm, there is a neighborhood of x, denoted by Ωx, such that


1. (ωu)∧(Ψv)∨ is integrable over Rm for all ω, Ψ ∈ D(Ωx),

2.∫Rm(ωu)∧(Ψv)∨ dx =

∫Rm(ωv)∧(Ψu)∨ dx for all ω, Ψ ∈ D(Ωx),

3.∫Rm |(ωu)∧(Ψv)∨| dx depends continuously on ω ∈ D(Ωx) for all Ψ ∈D(Ωx)

For such a pair of u, v above, we can define the product of u and v as follows:

1.2.5 Definition

If (u, v) ∈ M(Rm), the product of u and v in D′(Ωx), denoted by uv, is definedlocally on Ωx, as follows:For any ω ∈ D(Ωx), let Ψ ∈ D(Ωx) with Ψ(x) = 1 on supp ω. Then

〈uv, ω〉 =

∫

Rm

(ωu)∧(Ψv)∨ dx.

We remark that with the first condition above, the right-hand side exists.Moreover, it follows from the second that it is independent on the choice ofΨ. Indeed, If Ψ1, Ψ2 both play the role of Ψ, we have

∫

Rm

(ωu)∧(Ψ1v)∨ dx =

∫

Rn

(ωΨ2u)∧(Ψ1v)∨ dx =

∫

Rm

(ωv)∧(Ψ1Ψ2u)∨ dx

=

∫

Rm

(Ψ1ωu)∧(Ψ2v)∨ dx =

∫

Rm

(ωu)∧(Ψ2v)∨ dx.

Therefore, Definition 1.2.5 is fully defined.

The questions arise what δ2 and1

xδ are in the sense of this definition, and

what is the relationship between the product in the sense of this definitionand the product by “regularization and passage to the limit”. It can beshown in three following results.

1.2.6 Proposition

In D′(R), there does not exist δ2 in the sense of Definition 1.2.5.

Proof. We take x = 0, ω and Ψ ∈ D(Ω0) such that ω(0) = Ψ(0) = 1, andΨ = 1 on supp ω, where Ω0 is a neighborhood of 0. Assume that δ2 ∈ M(R),so (ωδ)∧(Ψδ)∨ is integrable in R. However, we have (ωδ)∧ = 1, and (Ψδ)∨ =1

2π. It follows that

1

2πis integrable in R. That is an absurdity. So, δ2 /∈ M(R).

And this concludes our proposition.


1.2.7 Proposition

In D′(R), there does not exist1

xδ in the sense of Definition 1.2.5, either.

Proof. We will use the argument of contradiction as follows.

If there is1

xδ, then we take Ω0, a neighborhood of 0, and ω, Ψ ∈ D(Ω0) such

that ω(0) = Ψ(0) = 1, and Ψ = 1 on supp ω. We notice that if u ∈ D′(R)arbitrarily, then one has

(u, δ) ∈ M(R) =⇒∫

R|(ωu)∧(Ψδ)∨| dx < ∞,

or

∫

R|ωu|∧ dx < ∞, since (Ψδ)∨ =

1

2π.

Hence, based on the property of the Fourier transform, one has ωu iscontinuous on R. Choosing ω = 1 on a neighborhood of 0, the restriction ofu to this neighborhood would be a continuous function. However, obviously

this is wrong in our case of u =1

x.

The results of Example 1.2.3 and Proposition 1.2.7 suggest that conditionsin Definition 1.2.5 are stricter than the ones in Definition 1.2.1. We can seethis more clearly by the following result of Tysk in [Tys81], or in [Col84].

1.2.8 Proposition

If u, v ∈ D′(Rm) and there exists uv in the sense of Definition 1.2.5, thenthe product uv also exists in the sense of Definition 1.2.1 for all δ-sequencessuch that δn(x) ≥ 0; and both products are equal.

Proof. Please see the references above.

We have seen how mathematicians have partly solved the problem ofdistribution multiplication. Although such those methods sometimes helpedscientists to solve some specific problems in nature as well as in theory (See[Obe92]). However, it is clear that our issue has not been fully solved up tonow. To look back our issue, we also see that it is easy to arrive at a wrongconclusion if we are not careful with the distribution multiplication, even in

what we have constructed. For instance, we have1

x.x = 1, and x.δ = 0 in

D′(R). If we assume that the “ideal” operation of multiplication in D′(R)


which extends the one in D(R) is associative, we get an absurd conclusionas follows:

δ = (1

x.x).δ =

1

x.(x.δ) = 0(!).

Another example is related to the Heaviside function H(x). It is clear thatHn = H, n = 2, 3, . . . , so H = H2 = H3. A “naive” argument gives

H ′ = 2H.H ′ = 3H2.H ′ = 3H.H ′,

hence H ′ = 2H.H ′ = 0 which contradicts that H ′ = δ in D′(R).

L. Schwartz in his theory of distribution (see [Sch54]) cautioned his read-ers by stating “an impossibility result” about defining distribution multipli-cation. This result “pointed out some further basic difficulties in trying toconstruct in easy ways nonlinear extensions of the distributions” (See [Ros87],p.27). This will be the topic of the next subsection.

1.2.9 Schwartz impossibility result

Let A be an algebra containing the algebra C0(R) of all continuous func-tions on R as a subalgebra. Let us assume that the function 1 ∈ C0(R) isthe unit element in A. Further let us assume that there exists a linear map∂ : A −→ A extending the derivative of continuously differentiable func-tions and satisfying Leibniz’s rule ∂(ab) = ∂a.b + a.∂b, then ∂2(|x|) = 0. Weremark from the conclusion of this theorem that we cannot hope to constructan algebra A containing D′(R) such that Leibniz’s rule is satisfied.

Proof. One gets

∂(x|x|) = ∂x.|x|+ x.∂(|x|) = |x|+ x.∂(|x|).It follows that,

∂2(x|x|) = 2.∂(|x|) + x.∂2(|x|).On the other hand, in C1(R), and that means in A : ∂(x|x|) = 2|x|. There-fore, ∂2(x|x|) = 2.∂|x|. It follows that x.∂2(|x|) = 0. Now, we will use theresult “In A, if xa = 0, then a = 0 ”, so we conclude ∂2(|x|) = 0. We onlyneed to verify this last result:We can see that x(log |x| − 1) and x2(log |x| − 1) ∈ C1(R) by giving 0 as thevalue of these functions at 0. Using Leibniz’s rule in A, we get

∂x(log |x| − 1)x = ∂x(log |x| − 1).x + x(log |x| − 1).


So,

∂2x(log |x| − 1)x = ∂2x(log |x| − 1).x + 2.∂x(log |x| − 1).

Thus,

∂2x(log |x| − 1).x = ∂2x2(log |x| − 1) − 2∂x(log |x| − 1).

But, since ∂ coincides with the usual derivation operator on C1-functions,and x2(log |x| − 1) ∈ C1(R), one gets

∂x2(log |x| − 1) = 2x(log |x| − 1) + x.

Therefore, in A we have

∂2x2(log |x| − 1)] = 2.∂x(log |x| − 1)+ 1.

So,∂2x(log |x| − 1).x = 1.

To simplify the notation we set y = ∂2x(log |x| − 1), then y.x = 1; thusx.a = 0 =⇒ y.(x.a) = 0 =⇒ 1.a = 0 =⇒ a = 0. This concludes the proof.

If there is an algebra A containing D′(R) in which Leibniz’s rule is satis-fied, then δ = 0, since ∂2(|x|) = 2δ in D′(R). That is an absurdity. This alsoexplains to us why this result called Schwartz impossibility result.

Chapter 2

The Colombeau theory ofgeneralized functions

In the previous chapter we have seen the need to construct a new theory of“generalized functions” which should solve some issues raised there. In the1980’s J. F. Colombeau proposed such a theory, and we will concerned withit in this chapter.

2.1 Preliminaries and notations

For q = 1, 2, . . . and N is the set of all nonnegative integers we denote

Aq =

φ ∈ D(Rn) :

∫

Rn

φ(t) dt = 1 and

∫

Rn

tαφ(t) dt = 0 for 1 ≤ |α| ≤ q

,

where t = (t1, . . . , tn) ∈ Rn, α = (α1, . . . , αn) ∈ Nn, and tα = (t1)α1 . . . (tn)αn .

We can remark that A1 ⊃ A2 ⊃ . . . ⊃ Aq ⊃ Aq+1 ⊃ . . . from the defeni-tion of Aqs. Moreover, we also get

2.1.1 Remark

∞⋂q=1

Aq = ∅.

Proof. Indeed, if we assume that there is φ ∈ ⋂∞q=1Aq, then φ is an analytic

function in Cn. We notice that conditions of φ imply that φ(0) = 1 and

∂αφ(0) = 0, |α| ≥ 1. So, φ(x) = 1, ∀x ∈ Rn. On the other hand, since

26


φ ∈ Aq, q = 1, 2, . . . , then φ ∈ D(Rn), and for each m = 1, 2, . . . , there isCm ≥ 0 such that

|φ(ξ)| ≤ Cm(1 + |ξ|)−mea|Im ξ|, where ξ ∈ Cn,

and a is a positive number such that supp φ ⊂ x : |x| ≤ a ( the Paley-Wiener theorem (Theorem 1.1.5.9)). In particular, when ξ = x ∈ Rn

and |ξ| = |x| −→ ∞, we will get φ(x) −→ 0. This is in contradiction to

φ(x) = 1, ∀x ∈ Rn. Therefore, we have⋂∞

q=1Aq = ∅.We can also shorten the arguments for the proof above if we use the propertyof the Fourier transform in S(Rn). Indeed, if φ ∈ ⋂∞

q=1Aq, then φ ∈ S(Rn),

and so does φ, which yields a contradiction with φ(x) = 1, ∀x ∈ Rn. There-fore, we also get

⋂∞q=1Aq = ∅. We also have a very important following result:

Remark 2.1.2

For each q = 1, 2, . . . , we have Aq 6= ∅.

Proof. The following proof comes from Prof. Tom Koornwinder’s com-ments on books on Colombeau theory. This shows more clearly and correctlyall the arguments for the above results in [Gko01], p.3.

It is clear that we only need to verify in case of n = 1. Consider thefollowing continuous linear forms on D(R)

L0(ϕ) =

∫

Rn

ϕ(x) dx

Lj(ϕ) =

∫

Rn

xjϕ(x) dx, 1 ≤ j ≤ q

Clearly, (Lj), 0 ≤ j ≤ q are linearly independent, so for each ϕ we can con-sider a linear form defined on the finite dimensional linear subspace generatedby L0, L1, . . . , Lq. Now, we use the Hahn-Banach theorem for extension toa continuous linear functional on D′(R). Applying the result of Proposition1.1.1.5, we have D(R) is a Montel space and D′′(R) = D(R), here D′(R) andD′′(R) are considered with the strong dual topology. Therefore, there areψk ∈ D(R), k = 0, 1, . . . , q such that

Lj(ψk) = δjk for j, k = 0, 1, . . . , q.

Setting ϕ = ψ0, we have Lj(ϕ) = δj0, or∫

Rn

ϕ(x) dx = 1, and

∫

Rn

xjϕ(x) dx = 0, 1 ≤ j ≤ q.


This implies that ϕ ∈ Aq for each q = 1, 2, . . . .We notice that there is at least one another proof for this remark, but itfocuses on technical calculations. Please see [Col85] for details.

We denote by φε(t) the function1

εnφ(

t

ε), where φ ∈ D(Rn) and ε > 0. It is

not difficult to verify that φ ∈ Aq if and only if φε ∈ Aq.We also denote Tx the translation operator, that is the mapping φ 7−→ φ(· −x). We also write φε,x(t) = (Txφε)(t).

2.2 Definition of Colombeau generalized func-

tions

First, we denote by E [Rn] the set of all the functions

R : A1 × Rn −→ C, where (φ, x) 7−→ R(φ, x),

which are C∞-functions in x for each fixed φ. It is easy to see that E [Rn] isan algebra with the pointwise operators.

2.2.1 Definition

We say that an element R ∈ E [Rn] is moderate if for every compact set K ofRn and every derivation operator ∂α (∂α is accepted even for order zero andin that case it is the identity operator), there is an N ∈ N such that for allφ ∈ AN , we have

(∂αR)(φε, x) = O(ε−N)

as ε ↓ 0, uniformly on K. We denote by EM [Rn] the set of all moderate ele-ments of E [Rn].

Note that N = N(α, K). If we have N = N(α, K) for some N, then wemay replace N by any N ′ such that N ′ > N.

We remark that if R1 and R2 are elements of E [Rn], then we have ∂α1R1 ∂α2R2

and ∂α2R2 ∂α1R1 are both O(ε−N1) for some N1. So, applying the Leibniz’srule and noticing that the sum of finitely many elements of orders O(ε−N) isalso of such order, we get that EM [Rn] is a subalgebra of E [Rn].

Denote by

Γ :=

β : N −→ R+ such that if q < r then β(q) ≤ β(r) and lim

q→∞β(q) = ∞

,

and we will show the concept of “null functions” as follows


2.2.2 Definition

We say that an element R of E [Rn] is null, if for every compact set K of Rn

and every derivation operator ∂α (∂α is accepted even for order zero and inthat case it is the identity operator), there is an N ∈ N and β ∈ Γ such thatfor all q ≥ N and all φ ∈ Aq, we have

(∂αR)(φε, x) = O(εβ(q)−N) as ε ↓ 0, uniformly on K.

We denote by I the set of all null elements of E [Rn].

It is obvious that I ⊂ EM [Rn]. It follows also from the definitions ofEM [Rn] and I, and from Leibniz’s rule that: if R1, R2 ∈ EM [Rn], and at leastone of them belongs to I, then R1.R2 ∈ I. Therefore, I is an ideal in EM [Rn].Now we can use the previous definitions in order to define the “generalizedfunctions” on Rn.

2.2.3 Definition

The algebra of generalized functions of Colombeau, denoted by G(Rn)(or G),is the quotient algebra EM [Rn]/I.

We remark that f is a generalized function in G(Rn) iff f = f + I, wheref ∈ EM [Rn] is a representative of f . We also say that f = g in G(Rn) ifff − g ∈ I, where f, g are representatives of f , g respectively.From these remarks we see that G(Rn) is an associative, commutative algebra.It is obvious that ∂αEM [Rn] ⊂ EM [Rn] and ∂αI ⊂ I, ∀α. Therefore, we candefine

∂α : G(Rn) −→ G(Rn)

f 7−→ ∂αf , where ∂αf = ∂αf + I

It follows that ∂α is linear, and satisfies Leibniz’ s rule of product derivatives.Next, we shall show that all classes of ordinary functions such as C∞(Rn), C0(Rn)(continuous functions on Rn), and the Schwartz distributions are still ele-ments of G(Rn).

2.3 Properties of the differential algebra G(Rn)

We will successively study the embeddings of C∞(Rn), C0(Rn) and D′(R)into G(Rn). In each case we will give the representatives in G(Rn) of thecorresponding functions or distributions.


2.3.1 Theorem

An embedding C∞(Rn) → G(Rn) is defined by the mapping

f ∈ C∞(Rn) 7−→ f + I ∈ G(Rn), where

f(φ, x) = f(x), ∀φ ∈ A1 and x ∈ Rn.

Proof. First, we show that the above mapping is well defined. In-deed, since f ∈ C∞(Rn), f(φ, ·) = f(·) ∈ C∞(Rn). On the other hand,

∂αf(φε, x) = ∂αf(x), φ ∈ A1, x ∈ Rn, and for f ∈ C∞(Rn), there is

supK|∂αf(φε, x)| = sup

K|∂αf(x)| = c < ∞ for an arbitrary compact K.

So, we have for each multi-index α,

supK|∂αf(φε, x)| ≤ c ≤ c

ε0, φ ∈ A1 and 0 < ε < 1.

Therefore, f ∈ EM [Rn], and it follows that f +I ∈ G(Rn). We also see that if

f ∈ C∞(Rn) and its image f +I = I, then it follows from the definition of Ithat for each compact K ⊂ Rn we have f(x) = O(ε) as ε ↓ 0 uniformly on K.So, f = 0 on K, and because K is arbitrary, we get f = 0. Therefore, theabove mapping is injective, and we have the embedding as a conclusion.

To conclude this subsection, we remark from the result of the above the-orem that in the C∞(Rn)-case the derivative operators ∂α on G(Rn) coincidewith the usual partial derivatives of functions when we consider them inC∞(Rn).

2.3.2 Theorem

An embedding C0(Rn) → G(Rn) is defined by the mapping

f ∈ C0(Rn) 7−→ f + I ∈ G(Rn), where

f(φ, x) =

∫

Rn

f(x + y)φ(y) dy =

∫

Rn

f(y)φ(y − x) dy, φ ∈ A1, x ∈ Rn.

Proof. We shall show that f ∈ EM [Rn]. Indeed, since φ ∈ A1, we havef(φ, ·) ∈ C∞(Rn) and

∂αf(φ, x) = (−1)|α|∫

Rn

f(y)∂αφ(y − x) dy, where |α| is the order of ∂α


Furthermore, one has

f(φε, x) =

∫

Rn

f(y)φε(y − x) dy

=1

εn

∫

Rn

f(y)φ(y − x

ε) dy =

∫

Rn

f(x + εt)φ(t) dt

It follows that

∂αf(φε, x) =(−1)|α|

ε|α|+n

∫

Rn

f(y)(∂αφ)(y − x

ε) dy =

(−1)|α|

ε|α|

∫

Rn

f(x+εt)∂αφ(t) dt

We notice that for arbitrary compact K and multi-index α, we have

supK|∫

Rn

f(x+εt)∂αφ(t) dt| ≤ supx∈K, y∈supp φ, δ∈[0,1]

|f(x+δy)|∫

Rn

|∂αφ(t)| dt = c < ∞,

and does not depend on ε, 0 < ε < 1 for all φ ∈ A1. Therefore, we havef ∈ EM [Rn].On the other hand, we can show that if f ∈ C0(Rn) and f ∈ I, then f = 0.Indeed, we notice that:

Lemma For all f ∈ C0(Rn), we have limε↓0 f(φε, x) = f(x).

Proof. We can write limε↓0 f(φε, x) = limε↓0∫Rn f(x + εt)φ(t)dt. Now, we

notice that φ ∈ A1, f is continuous. So, we can apply the dominated con-vergence Lebesgue theorem we get the conclusion above.

If f ∈ I, then for all compact K, there exists N ∈ N such that for allφ ∈ AN , we have

f(φε, x) = O(ε) as ε ↓ 0, uniformly on K.

Hence, apply the result of Lemma above, we have f(x) = 0 on K. It followsthat f = 0, and this concludes the proof of the above theorem.

We notice that C0(Rn) is included in EM [Rn] as a linear subspace, not asubalgebra. It follows that C0(Rn) is not a subalgebra of G(Rn), either. Infact, in general we have

∫

Rn

f(x + εt)φ(t) dt.

∫

Rn

g(x + εt)φ(t) dt 6=∫

Rn

f(x + εt)g(x + εt)φ(t) dt.

We also see another issue to arise: if f ∈ C0(Rn), then we can applyboth Theorem 2.3.1 and Theorem 2.3.2, so what happens? Nothing at all!


because we can show that f − f ∈ I, so both f and f are representatives off ∈ C∞(Rn). For convenience, we will show this in case of n = 1. Indeed, wehave

(f − f)(φ, x) = f(x)−∫

Rf(x + y)φ(y) dy.

Therefore, one gets

(f − f)(φε, x) = f(x)−∫

Rf(x + εt)φ(t) dt = −

∫

R[f(x + εt)− f(x)]φ(t) dt.

Since f ∈ C∞(R), we can apply Taylor’s formula up to order q to f at thepoint x, and we get

f(x + εt)− f(x) =

q∑

k=1

(εt)k

k!∂f (k)(x) + εq+1 tq+1

(k + 1)!∂f (q+1)(x + θεt),

where 0 < θ < 1. Hence, for arbitrary compact K, and for all q ∈ N, φ ∈ Aq

we have

(f − f)(φε, x) = O(εq+1) as ε ↓ 0, uniformly on K.

This fits with Definition 2.3.2 for α = 0, N = 0 and β(q) = q+1. It is similarto estimate ∂α(f − f)(φε, x), so f − f ∈ I.

Finally, we talk about the embedding of D′(Rn) into G(Rn).

2.3.3 Theorem

An embedding D′(Rn) → G(Rn) is defined by the mapping

u ∈ D′(Rn) 7−→ u + I ∈ G(Rn), where

u(φ, x) = 〈u(y), φ(y − x)〉, ∀φ ∈ A1 and x ∈ Rn.

Proof. Set φ−(t) = φ(−t) for φ ∈ D(Rn), then u(φ, x) = (u ∗ φ−)(x)(Theorem 1.1.5.3). So, we have u(φ, x) ∈ C∞(Rn). Also it follows from theabove formula that ∂αu(φ, x) = (−1)|α|〈u(y), (∂α

x φ)(y − x)〉. To prove thatu(φ, x) ∈ EM [Rn] and the above mapping is injective, we can refer to twosolutions: the first one based on Theorem 1.1.5.9 in [Col84], pp. 61–62, andthe other one in [Ros87], pp. 60–61, which is based on the compact supportof φ. To conclude we notice that this theorem is an extension of Theorem2.3.2 in case of continuous functions.


It follows that the δ-Dirac distribution has the form fδ + I ∈ G(Rn),where fδ(φ, x) = 〈δ(y), φ(y − x)〉 = φ(−x). Also, we have the representativeof δ2 in G(Rn) is φ2(−x) + I.

To end this subsection we would like to make some historical notes. Infact before Schwartz introduced distributions as they are now, he had triedan approach starting from the convolution operator. He defined an operatorT mapping D(Rn) into the space E(Rn) of C∞-functions with arbitrary sup-port such that for continuous f, the corresponding operator actually mapsφ ∈ D(Rn) to the convolution of f and φ. However, later in the process ofdefining the product of a C∞-function with such an operator T, as well asdefining the Fourier transform, he saw that this method was not very conve-nient. After that he turned to the definition as we know nowadays. However,it is the original realization of a distribution as a convolution operator whichwas extended in Colombeau’s theory (with D(Rn) being replaced by A1).Please refer to [Sch01] for further details.

The following subsection will show the more extended results on the rep-resentatives of elements in G(Rn).

2.4 Nonlinear properties of G(Rn)

We notice that not only multiplication makes sense in G(Rn), but also themore general nonlinear operations can do. To show that, firstly we considera class of slowly increasing functions at infinity in the following definition:

2.4.1 Definition

A function f : Rn −→ C is said to be slowly increasing at infinity if thereare c > 0, N ∈ N such that

|f(x)| ≤ c(1 + |x|)N , ∀ x ∈ Rn.

The set of all C∞(Rn)-functions which together with all their derivatives areslowly decreasing at infinity is denoted by OM(Rn).

Some special properties of functions in this class make them play animportant role when we study the Fourier transform theory in Schwartz dis-tribution theory (See [Fri98], pp. 97–103, for example). Polynomials, or eix

are examples of slowly increasing functions at infinity, but ex is not. Thenext result, see [Col85], pp. 27–31 for details, will show us the issue we wantto concern in this section.


2.4.2 Theorem

If f ∈ OM(R2n), where R2n ' Cn; if G1, G2, . . . , Gn are generalized functionsin G(Ω), where Ω is an open subset of Rm; and R1, R2, . . . , Rn are respec-tive representatives of G1, G2, . . . , Gn, then f(R1, R2, . . . , Rn) is a moderateelement in EM [Ω]. Therefore, f(R1, R2, . . . , Rn) is a representative of a gen-eralized function in G(Ω), denoted by f(G1, G2, . . . , Gn).

We skip the proof of this theorem and we only clarify some issues fromthe above theorem. Here, we have used the concept of generalized functionsdefined in an open subset Ω of Rn. The construction of the space of all gener-alized functions on Ω, denoted by G(Ω), is almost the same as what we havedone in case of Rn. That is why we do not discuss more. We also notice thatEM [Ω], and I(Ω) denote respectively the moderate and null functions in thiscase. Please refer to [Col85], pp. 18–20 for details. It also follows from theresult above that eiδ is an element of G(R) with eiφ(−x) as its representative,...

2.5 Generalized complex numbers

Up to now, we have defined the space G(Rn) of generalized functions in whichthere exists the multiplication of two arbitrary ones as we wish. However,to go further with this space, we need to understand the value of a gener-alized function F at each point x ∈ Rn. This subsection is devoted to thatgoal, and we will start with the construction of generalized complex numbers.

Denoted by E0 the set of all functions from A1 to C. It is clear that E0 isa linear space and an algebra.

2.5.1 Definition

We call R ∈ E0 a moderate element of E0 if there is a positive integer N suchthat for all φ ∈ AN , we have R(φε) = O(ε−N), as ε ↓ 0. Also we denote byEM the set of all moderate elements of E0.

It follows that EM is a linear subspace and also a subalgebra of E0.

2.5.2 Definition

We denote by I0 the set of all elements R in EM satisfying the followingcondition: there are N ∈ N and β ∈ Γ such that for all q, q ≥ N and all


φ ∈ Aq, we have R(φε) = O(εβ(q)−N) as ε ↓ 0.

It follows from the definitions above and the property of the set Γ thatI0 is an ideal of EM . So we can define the quotient algebra as follows:

2.5.3 Definition

We denote by C the quotient algebra EM/I0, and each member of C is calleda generalized complex number.

What is the relationship between C and C? The answer is as follows:

2.5.4 Proposition

The mapping z ∈ C 7−→ z + I0 ∈ C, where z(φ) = z, ∀φ ∈ A1 defines anembedding from C to C.

Proof. Observe: if z ∈ I0, then z = 0, and it implies the Proposition.

2.5.5 Definition

A complex number z ∈ C is said to be associated with a generalized complexnumber Rz = R + I0, R ∈ EM , denoted by Rz ` z, if there is q ∈ N suchthat for all φ ∈ Aq, we have limε↓0 R(φε) = z.

We see that such z above (corresponding to Rz) is unique. Also R+I0 ` 0if R ∈ I0 since we remark that if R ∈ I0, then limε↓0 R(φε) = 0. We denoteby C0 the set of all such Rz above. We see that C0 $ C. Indeed, if we takeRz ∈ C, Rz = R + I0, where R(φ) = φ(0), ∀φ ∈ A1, then R ∈ EM . On the

other hand, we have R(φε) =1

εnφ(0). So, it follows that there do not exist

q ∈ N and z ∈ C, such that for all φ ∈ Aq, limε↓0 R(φε) = z. Or, we haveRz /∈ C0.

We sum up some properties of the above relation as follows:

2.5.6 Proposition

• If Rz ∈ C, z1, z2 ∈ C such that Rz ` z1, and Rz ` z2, then z1 = z2

• If Rz = z, where z ∈ C ⊂ C, then Rz ` z


• If Rz1 , Rz2 ∈ C; z1, z2 ∈ C and Rz1 ` z1, Rz2 ` z2; then Rz1 + Rz2 `z1 + z2 and Rz1Rz2 ` z1z2

• If Rz ∈ C, z ∈ C and Rz ` z, then −Rz ` (−z).

The proof follows directly from the definition above.

From the results of the proposition above, we can define the mappingRz ∈ C0 7−→ z ∈ C, such that Rz ` z. However, the following example showsthat this mapping is not injective: Take Rz ∈ C, where Rz = R + I0 and

R(φ) =∫R |x|φ(x) dx, then R(φε) =

∫R |x|

1

εφ(

x

ε) dx = εR(φ) for φ ∈ A1 and

limε↓o R(φε) = 0. So, we have Rz ` 0. However, for each q = 1, 2, . . . we canconstruct φ ∈ Aq such that R(φ) 6= 0 (refer to [Ros87], pp. 127–130). Thatmeans Rz 6= 0 in C.

Now we will define the value of a generalized function F ∈ G(Rn) at eachpoint x ∈ Rn.

2.6 Point values of generalized functions

Take F = f + I ∈ G(Rn), for each x ∈ Rn we have f(·, x) ∈ EM , sincef(φ, x) ∈ EM [Rn]. So we can define

F (x) := fx + I0, where fx(φ) = f(φ, x), φ ∈ A1.

This definition is suitable. Indeed, if g + I is another representative of F,then f − g ∈ I. So, for all x ∈ Rn, we have (f − g)x = (f − g)(·, x) ∈ I0. Itshows that F (x) = fx + I0 does not depend on the choice of representativesof F.

2.6.1 Example

What is δ(0)? Here, δ ∈ D′(R) ⊂ G(R).

We have seen that δ is represented by φ(−x) + I, where φ ∈ A1. Itfollows from the above definition that δ(0) = φ(0) + I0. And as we see from2.5.5 that δ(0) is a “really generalized complex number”: it is not associatedwith any ordinary complex number! If x 6= 0, then δ(x) = f(φ, x) + I0,where f(φ, x) = φ(−x). We remark that if for ε > 0 small enough, we have

f(φε, x) =1

εφ(−x

ε) = 0. That means δ(x) = 0 (in C) as we have seen in the

theory of distributions.


2.6.2 Example

In G(R), we have xδ 6= 0, so the contradiction we have met in the distributiontheory does not affect the algebra structure of G(Rn).

We have seen that in D′(R), xδ = 0. And this would yield a contradic-tion if we assume that there exists an associative multiplication in D′(R)(See 1.2.7). Here, we have xδ 6= 0, so this situation will be avoided. Indeed,we have xδ = f + I ∈ G(R), where f(φ, x) = xφ(−x), φ ∈ A1 and x ∈ R.

So, ∂f(φε, 0) =1

εφ(0), φ ∈ A1 and ε > 0. It implies that f /∈ I. Therefore,

xδ 6= 0 in G(R).

However, the value of this generalized function (in G(R)) is always equalto 0 at any point x ∈ R. Indeed, for every x ∈ R, we have

(xδ)(x) = f(φ, x) + I0 = xφ(−x) + I0.

But, f(φε, x) =x

εφ(−x

ε) = 0 for each fixed x and ε > 0 small enough,

since φ has compact support. Therefore, xφ(−x) ∈ I0, or (xδ)(x) = 0 inC, ∀x ∈ R.

The above example shows that it can happen that a generalized functionis not null, but that at every point its value is zero in the set of generalizedcomplex numbers!

2.6.3 Remark

If F = f ∈ C∞(Rn) ⊂ G(Rn), then the value of F at any x ∈ Rn coincideswith the usual value of f at x.

Indeed, we have F = f + I ∈ G(Rn), where f(φ, x) = f(x), ∀φ ∈ A1 andx ∈ Rn. Therefore, we get F (x) = z +I0, where z(φ) = f(φ, x) = f(x). Now,the embedding in 2.5.4 will conclude our remark.

The last part of this subsection deals with the value of the continuousfunctions in G(Rn), and it is given by the following result

2.6.4 Theorem

Let be given a generalized function which corresponds to a continuous func-tion, i.e., F = f ∈ C0(Rn) ⊂ G(Rn) and x ∈ Rn, then F (x) ∈ C0 and


F (x) ` f(x). That means the value of the generalized function F at x isassociated with the usual complex number which is the usual value of the con-tinuous f at x.

Proof. We have F = f + I ∈ G(Rn), f ∈ EM [Rn] and we have

f(φ, x) =

∫

Rn

f(x + y)φ(y) dy, φ ∈ A1, x ∈ Rn.

So, for each x ∈ Rn, we get F (x) = z + I0 ∈ C, with z(φ) = f(φ, x) =∫Rn f(x + y)φ(y) dy. It follows that

z(φε) =

∫

Rn

f(x + y)1

εnφ(

y

ε) dy =

∫

Rn

f(x + εt)φ(t) dt −→ f(x)

as ε ↓ 0, for all φ ∈ A1. Therefore, F (x) ` f(x), at any point x ∈ Rn.

We notice that F (x) is associated with the usual complex number f(x) incase of F = f ∈ C0(Rn), but that not neccesarily F (x) coincides with f(x).See the following example: If F = f = |x| ∈ C0(R), F (0) =

∫R |y|φ(y) dy

is not zero while f(0) = 0 (Please see [Col85] for details). However, F (0) ` 0.

We also remark that if F ∈ G(Ω), where Ω is an open subset of Rn, thenF (x), x ∈ Ω is defined similarly as above (Please refer [Col85], or [Obe98]for details).

2.7 Integrals of generalized functions

Based on the results of subsections 2.5 and 2.6, we can introduce the conceptof integral of the generalized functions F ∈ G(Rn) on a compact set K of Rn

as follows:

2.7.1 Definition

Let be given F = f + I ∈ G(Rn), where f ∈ EM [Rn] and K is a compact setof Rn, and write

h(φ) =

∫

K

f(φ, x) dx, φ ∈ A1.

Then the generalized complex number h + I0 ∈ C is called the integral of Fover K, and it is denoted by

∫K

F (x) dx.


First, we have to verify the correctness of the above definition. Indeed,∫K

f(φ, x) dx is actually defined because f(φ, ·) ∈ C∞(Rn). Furthermore,since f ∈ EM [Rn] and K is compact, then h ∈ EM . In addition, the abovedefinition does not depend on the choice of representatives of F since it isnot difficult to verify that if f ∈ I, then h ∈ I0. Next, we will consider∫

KF (x) dx, where F = f ∈ C∞(Rn) ⊂ G(Rn).

2.7.2 Proposition

If the generalized function F corresponds to a C∞-smooth function, i.e.,F = f ∈ C∞(Rn) ⊂ G(Rn), then

∫K

F (x) dx (K- a compact subset of Rn)coincides with the ordinary integral of f over K.

Proof. We have seen that F = f + I, where f(φ, x) = f(x), ∀φ ∈A1, x ∈ Rn. So,

∫K

f(φ, x) dx =∫

Kf(x) dx ∈ C, ∀φ ∈ A1. In view of the

embedding in 2.5.4, this settles the Proposition.

The next result is about the integral of a generalized function comingfrom a continuous function.

2.7.3 Proposition

If the generalized function F corresponds to a continuous function f, i.e.,F = f ∈ C0(Rn) ⊂ G(Rn), then C0 3

∫K

F (x) dx ` ∫K

f(x) dx.

Proof. We have F = f + I, where f(φ, x) =∫Rn f(x + y)φ(y) dy, φ ∈

A1, x ∈ Rn. Therefore,∫

KF (x) dx = h + I0, where

h(φ) =

∫

K

f(φ, x) dx =

∫

K

∫

Rn

f(x + y)φ(y) dy dx, φ ∈ A1, ε > 0.

We notice that φ ∈ A1 with compact support and∫Rn φ(y) dy = 1, and we

get∫Rn f(x + εy)φ(y) dy −→ f(x) uniformly in K (for x) as ε ↓ 0. Then, we

have

limε↓0

h(φε) =

∫

K

f(x) dx (∈ C).

It means∫

KF (x) dx ∈ C0, and

∫K

F (x) dx ` ∫K

f(x) dx.

2.7.4 Example

If we take δ ∈ G(R) and a < 0 < b, then∫ b

aδ(x) dx ` 1.


Indeed,∫ b

aδ(x) dx = h + I0, where h(φ) =

∫ b

aφ(−x) dx. So,

h(φε) =

∫ b

a

1

εφ(−x

ε) dx = −

∫ −b/ε

−a/ε

φ(t) dt =

=

∫ −a/ε

−b/ε

φ(t) dt −→∫ ∞

−∞φ(t) dt = 1, as ε ↓ 0.

Therefore,∫ b

aδ(x) dx ` 1. Later, we see that

∫ b

aδ(x) dx = 1.

However, if we consider∫ b

0δ(x) dx, then

∫ b

0δ(x) dx ∈ C\C0. Indeed, for

this case

h(φε) =

∫ 0

−b/ε

φ(x) dx −→∫ 0

−∞φ(x) dx as ε ↓ 0.

We remark that if φ ∈ Aq, q = 1, 2, . . . , then∫ 0

−∞ φ(x) dx can be any complexnumber. So, we have our conclusion above. Similarly, we also have: if a < band a ≤ 0 ≤ b, then

∫ b

aδ2(x) dx ∈ C\C0.

Note. We can extend the concept of integral to the case of F ∈ G(Ω) withcompact support. Here, the support of a generalized function F ∈ G(Ω) isthe complement in Ω of the largest open subset of Ω, where F is null. Assumethat the support of F ∈ G(Ω), denoted by supp F, is a compact subset K ofΩ. If K1 ⊂ K2 are two compact subsets of Ω such that K ⊂ Int K1, then∫

K2\Int K1

F (x) dx = 0,

since F is null on a neighborhood of K2\Int K1. Therefore,∫

K1F (x) dx =∫

K2F (x) dx and we can denote this value as

∫Ω

F (x) dx.

We notice that the condition K ⊂ Int K1 is critical. We can see thisin the following instance: If F = δ ∈ G(R), supp δ = 0(:= K) and∫0 δ(x) dx = 0 as we know. So, if we select K1 = K, but K is not a subset

of Int K1 (in this case, Int K1 = ∅), we would go to a wrong conclusion that∫R δ(x) dx ` 0! In fact, we have to select K1 = [a, b], a < 0 < b and it follows

that ∫

Rδ(x) dx =

∫ b

a

δ(x) dx (` 1)

as the note in Example 2.7.4 above.

To end this subsection, we will be concerned with the integral of theproduct ψT, where ψ ∈ D(Rn) and T ∈ D′(Rn) in Rn. It turns out to be thevalue of distribution T at ψ ∈ D(Rn)!


2.7.5 Theorem

If ψ ∈ D(Rn) and T ∈ D′(Rn), then∫

Rn

(ψT )(x) dx = 〈T, ψ〉 ∈ C ⊂ C,

where ψT is the multiplication in G(Rn).

Proof. Take a compact K ⊂ Rn such that supp ψ ⊂ Int K. Applya well-known result for distributions (Theorem 1.1.5.9) in order to obtainf ∈ C0(Rn) and an multi-index α such that

T |Ω= ∂αf |Ω, Ω = Int K.

From the remark above we have∫

Rn

(ψT )(x) dx =

∫

Rn

(ψ∂αf)(x) dx.

In G(Rn) we have ψ. ∂αf = g + I, g ∈ EM [Rn], where for φ ∈ A1, x ∈ Rn,

g(φ, x) = ψ(x)

∫

Rn

(∂αf)(y)φ(y−x) dy = (−1)|α|ψ(x)

∫

Rn

f(y)(∂αφ)(y−x) dy.

It follows that∫Rn(ψT )(x) dx = h + I0 ∈ C, with φ ∈ A1 and

h(φ) =(−1)|α|∫

Rn

∫

Rn

ψ(x) f(y)(∂αφ)(y − x) dy dx

= (−1)|α|∫

Rn

∫

Rn

(∂αψ)(x) f(y)φ(y − x) dy dx

= (−1)|α|∫

Rn

∫

Rn

(∂αψ)(x− y) f(x)φ(y) dx dy.

So, we have

h(φε) = (−1)|α|∫

Rn

∫

Rn

(∂αψ)(x− εy) f(x)φ(y) dx dy, φ ∈ A1, ε > 0.

Therefore, we have

limε↓0

h(φε) = (−1)|α|∫

Rn

(∂αψ)(x)f(x) dx, since φ ∈ A1

Also we can prove that h(φ) − (−1)|α|∫Rn(∂αψ)(x)f(x) dx ∈ I0 (Refer to

[Ros87], p.67 and p.82 for details). That means∫

Rn

(ψT )(x) dx = (−1)|α|∫

Rn

∂αψ(x)f(x) dx = 〈T, ψ〉

This concludes our proof since the right-hand side of the above identity is ausual complex number.


Note. In case of ψ ∈ C∞(Rn), T ∈ D′(Rn) with compact support, we getthe same result as above (See [Col85], p.54). We can illustrate that resultwith this identity

∫R(ψ. δ)(x) dx = ψ(0) for all ψ ∈ C∞(R). Indeed, in view

of Example 1.3.3.2.a., we have ∂H = δ, where H is the Heaviside function,so for all ψ ∈ D(R),

∫

R(ψδ)(x) dx =

∫

R(ψ∂H)(x) dx = −

∫

RH(x)∂ψ(x) dx = −

∫ ∞

0

∂ψ(x) dx = ψ(0).

Another result is: if a < 0 < b, then∫ b

aδ(x) dx =

∫R δ(x) dx = 1.

2.8 Weak concepts of equality in G(Ω)

In view of Schwartz’s impossibility result in 1.2.9 we do not expect thatidentities like xδ = 0, or Hn = H, n = 2, 3, . . . remain true in G(R). In2.6.2 we have seen that xδ 6= 0 in G(R), and Hn 6= H, n = 2, 3, . . . becauseδ 6= 0 in G(R). So, we need to establish other concepts of equality in G(Ω)which are weaker than the usual ones in D(Ω), for instance. To meet theabove requirements, J. F. Colombeau and his collaborators proposed twoweak concepts of equality in G(Rn), which will discuss in this subsection. Wenotice that all the concepts below are still meaningful in case of an arbitraryopen subset Ω of Rn. However, for brevity we only discuss the case of Rn

here.

2.8.1 Definition

A generalized function F ∈ G(Rn) is said to be test null, denoted F ∼ 0, iffor every ψ ∈ D(Rn) we have

∫

Rn

(ψ.F )(x) dx = 0.

Two generalized functions F1, F2 ∈ G(Rn) are test equal if F1 − F2 ∼ 0.

This concept is suggested by the result of Theorem 2.7.5. We notice thatin case of T ∈ D′(Rn) ⊂ G(Rn), we have

∫

Rn

(ψ.T )(x) dx = 〈T, ψ〉.

So, if T ∼ 0 as above, then T = 0 in D′(Rn). However, this “equality”does not hold in some very simple cases. For instance, if f1(x) = x− =


0 if x ≥ 0,

x if x < 0and f2(x) = x+ =

x if x ≥ 0,

0 if x < 0. Then f1f2 is not “test

equal” to 0 in G(R), but f1f2 = 0 in C0(R). To overcome this problem,J. F. Colombeau defined another kind of equality in G(Rn) which can explainsome cases as above example.

2.8.2 Definition

A distribution T ∈ D′(Rn) is said to be associated with a generalized functionF ∈ G(Rn) (or vice versa), denoted F ° T, if, for every ψ ∈ D(Rn), we have

∫

Rn

(ψ.F )(x) dx ` 〈T, ψ〉.

So this means: ∀ψ ∈ D(Rn), ∃q ∈ N, ∀φ ∈ Aq,∫Rn ψ(x)F (φε, x)dx −→

〈T, ψ〉 as ε ↓ 0. And two generalized functions F1, F2 ∈ G(Rn) are associated,denoted by F1 ≈ F2, if F1−F2 is associated with 0 ∈ D′(Rn), i.e., F1−F2 ° 0.

We make some remarks about these definitions.

2.8.3 Remarks

We have:a) The relations ∼ and ≈ are obviously equivalence relations on G(Rn).

However, ° ⊂ G(Rn)×D′(Rn) is neither reflexive nor symmetric!

b) If F1, F2 ∈ G(Rn) and F1 ∼ F2, then F1 ≈ F2. This is an easy conse-quence of the definitions above.

c) If T ∈ D′(Rn) and T ∼ 0, or T ≈ 0, then T = 0. Indeed, the caseT ∼ 0 =⇒ T = 0 has already been discussed above. If T ≈ 0, then∫Rn(ψ.T )(x) dx ` 0. However the left-hand side is an ordinary complex num-

ber (Theorem 2.7.5), so∫Rn(ψ.T )(x) dx = 0. It follows that T = 0.

In summary we can say that relations ∼ and ≈ defined on G(Rn) coincidewith the usual equality = when restricted to D′(Rn). Now we will considersome results related to these new concepts of equality.

2.8.4 Proposition

Let be given f1, f2 ∈ C0(Rn) and its product is f1f2 ∈ C0(Rn), too. If F1, F2

in G(Rn) correspond to f1, f2 respectively, then F1F2 ≈ f1f2, i.e., f1f2 is


associated with the generalized function F1F2.

Proof. We know that for j = 1, 2

Fj = fj + I ∈ G(Rn), with fj =

∫

Rn

fj(x + y)φ(y) dy, φ ∈ A1, x ∈ Rn.

Take an arbitrary function ψ ∈ D(Rn), then

ψ.F1F2 = f+I ∈ G(Rn), where f(φ, x) = ψ(x)f1(φ, x)f2(φ, x), φ ∈ A1, x ∈ Rn.

Therefore, we have

∫

Rn

(ψ.F1F2)(x) dx = g + I0 ∈ C, with g(φ) =

∫

Rn

f(φ, x) dx, φ ∈ A1.

If we denote T = f1f2 ∈ C0(R), then ψT = h + I ∈ G(Rn) with

h(φ, x) = ψ(x)

∫

Rn

f1(x + y)f2(x + y)φ(y) dy, φ ∈ A1, x ∈ Rn.

It follows that∫

Rn

(ψ.T )(x) dx = k + I0 ∈ C, where k(φ) =

∫

Rn

h(φ, x) dx, φ ∈ A1.

So for arbitrary ψ ∈ D(Rn); and φ ∈ A1, ε > 0, we get

g(φε)− h(φε) =

∫

Rn

ψ(x)[

∫

Rn

f1(x + εy)φ(y) dy

∫

Rn

f2(x + εz)φ(z) dz−

−∫

Rn

f1(x + εy)f2(x + εy)φ(y) dy] dx.

Noting that ψ ∈ D(Rn), φ ∈ A1 and applying the Lebesgue dominatedconvergence theorem, we get

limε↓0

[g(φε)− h(φε)] = 0.

This concludes our proof.

We also notice that the relation ≈ in the above proposition cannot bereplaced by the relation ∼ . The following example will show why.


2.8.5 Example

If f1(x) = x− =

0 if x ≥ 0,

x if x < 0and f2(x) = x+ =

x if x ≥ 0,

0 if x < 0play the

roles of f1, f2 in the above proposition, then F1F2 f1f2 (= 0).

Indeed, with a similar way of writing as above we have F1F2 = g + I ∈G(Rn) where

g(φ, x) =

∫ −x

−∞(x + y)φ(y) dy

∫ −∞

−x

(x + y)φ(y) dy, φ ∈ A1, x ∈ R.

It follows that∫

R[ψ(F1F2)](x) dx = h + I0 ∈ C, where φ ∈ D(R), and

h(φ) =

∫

Rψ(x)g(φ, x) dx =

=

∫

Rψ(x)[

∫ −x

−∞(x + y)φ(y) dy

∫ −∞

−x

(x + y)φ(y) dy] dx, φ ∈ A1

So, if we take φ ∈ A1, supp φ ⊂ [a, b], then

h(φε) =

∫ −aε

−bε

ψ(x)[

∫ −x/ε

a

(x + εz)φ(z) dz

∫ b

−x/ε

(x + εz)φ(z) dz] dx.

Hence,

h(φε) = ε3

∫ b

a

ψ(−εy)[

∫ y

a

(z − y)φ(z) dz

∫ b

y

(z − y)φ(z) dz] dy.

We notice that by [Ros87], pp. 127–130, we can construct φ ∈ Am, m ≥3, supp φ ⊂ [a, b] such that

∫ b

a

[

∫ y

a

(z − y)φ(z) dz

∫ b

y

(z − y)φ(z) dz] dy 6= 0.

Hence, with such φ for each m, m ≥ 3 and ψ ∈ D(R), ψ(0) 6= 0, we haveh(φε) /∈ I0. It follows that F1F2 0.

If f ∈ C∞(Rn) and T ∈ D′(Rn), we have S = f.T ∈ D′(Rn), and F =f ¯ T ∈ G(Rn). Here we denote ¯ as the multiplication in G(Rn), so thatwe can distinguish with the one in D′(Rn). What is the relationship betweenthese two products? The following result will be concerned with the answer.


2.8.6 Proposition

With above denotions and hypothesis, we have F ∼ S.

Proof. For ψ ∈ D(Rn), applying the result of Theorem 2.7.5 we get∫

Rn

(ψ ¯ S)(x) dx = 〈f.T, ψ〉 = 〈T, ψ.f〉 =

∫

Rn

[(ψ.f)¯ T ](x) dx.

But for ψ, f ∈ C∞(Rn), we have ψ ¯ f = ψ.f. It follows that (ψ.f) ¯ T =(ψ ¯ f)¯ T = ψ ¯ (f ¯ T ). So,∫

Rn

(ψ¯F )(x) dx =

∫

Rn

(ψ.f)¯T )(x) dx =

∫

Rn

(ψ¯S)(x) dx = 〈T, ψ.f〉 ∈ C

Therefore, for all ψ ∈ D(Rn), we have∫

Rn

[ψ ¯ (S − F )](x) dx = 0, or S ∼ F.

2.8.7 Remarks

We also have:

a) If F1, F2 ∈ G(Rn) and F1 ∼ F2 (or F1 ≈ F2), then we cannot implythat F1 = F2.Indeed, applying the result of Proposition 2.8.6, we get xδ ∼ 0 in G(R) andalso xδ ≈ 0 in G(R). However, xδ 6= 0 in G(R) as we have seen in Example2.6.2.

b) If F1, F2 ∈ G(Rn) and α is a multi-index, then F1 ∼ F2 =⇒ ∂αF1 ∼∂αF2, and F1 ≈ F2 =⇒ ∂αF1 ≈ ∂αF1.Indeed, assume that for j = 1, 2 we have Fj = fj + I ∈ G(Rn), fj ∈ EM [Rn].If F1 ∼ F2, then for all ψ ∈ D(Rn), we have

∫

Rn

[ψ(F1 − F2)](x) dx = 0 ∈ C.

Hence, g ∈ I0, where g(φ) =∫Rn ψ(x)[f1(φ, x) − f2(φ, x)] dx. On the other

hand, we have ∂αFj = ∂αfj+I, j = 1, 2. Hence,∫Rn [ψ(∂αF1−∂αF2)](x) dx =

h + I0 ∈ C, where

h(φ) =

∫

Rn

ψ(x)[∂αf1(φ, x)− ∂αf2(φ, x)] dx

= (−1)|α|∫

Rn

∂αψ(x)[f1(φ, x)− f2(φ, x)] dx.


However, ∂αψ ∈ D(Rn) and g(φ) ∈ I0, so h(φ) ∈ I0, too. This implies that∂αF1 ∼ ∂αF2.

If F1 ≈ F2, then for ψ ∈ D(Rn) we have∫Rn [ψ(F1−F2)](x) dx ` 0. It fol-

lows that limε↓0 g(φε) = 0. Similarly, since ∂αψ ∈ D(Rn) and limε↓0 g(φε) = 0,we have limε↓0 h(φε) = 0, or ∂αF1 ≈ ∂αF1.

c)δ2 is not associated with any distribution T ∈ D′(R), so δ2 is really anewcomer in G(R).Indeed, if there is T ∈ D′(R), such that δ2 ≈ T, then we can write

∫R(ψδ2)(x) dx =

g + I0 ∈ C, where g(φ) =∫R ψ(x)φ2(−x) dx, φ ∈ A1. Hence,

g(φε) =

∫

Rψ(x)φ2

ε(−x) dx =1

ε

∫

Rψ(−εx)φ2(x) dx.

Since δ2 ≈ T, limε↓0 g(φε) = 〈T, ψ〉 ∈ C, ∀ψ ∈ D(R). However, for ψ ∈ D(R)and ψ = 1 on a neighborhood of 0, since

∫R φ2(x) dx 6= 0 the left-hand side

cannot converge as ε ↓ 0, while the right-hand side is an ordinary complexnumber. That is an absurdity, and it verifies our conclusion.

d)We have Hm ≈ H, m = 1, 2, . . . in G(R), where H(x) is the Heavisidefunction.

Indeed, we notice that H(x) = ∂x+, where x+ =

x if x ≥ 0

0 if x < 0,and x+ ∈

C0(R). So, H = h + I ∈ G(R), where

h(φ, x) =

∫ ∞

−x

φ(y) dy, φ ∈ A1, x ∈ R.

Hence, Hm = hm + I ∈ G(R). For given φ ∈ A1, we denote∫ x

−∞ φ(y) dyby χ(x). So χ ∈ C∞(R), χ(−∞) = 0, χ(∞) = 1, χ′(x) = φ(x), x ∈R. Moreover, we also see h(φε, x) = 1 − χ(−x

ε), ε > 0, x ∈ R. For each

given ψ ∈ D(R), then∫R[ψ(Hm − H)](x) dx = k + I0 ∈ C, where k(φε) =∫

R ψ(x)ωε(x) dx, ωε(x) = [(1− χ(−x

ε)]m − [(1− χ(−x

ε)] We remark that

limε↓0

ωε(x) =

0 if x 6= 0

[1− χ(0)]m − [1− χ(0)] if x = 0.

It follows that |ωε(x)| ≤ (1 + M)m + (1 + M), where M =∫R |φ(y)| dy < ∞.

This allows us to apply the Lebesgue dominated convergence theorem and


get limε↓0 k(φε) = 0, and this concludes our proof.

e)The relation ∼ is not compatible with the multiplication in G(Rn).Indeed, if we take F1 = F2 = xδ ∈ G(R), then we have seen that F1 ∼ 0 andF2 ∼ 0. However, we will show that F1F2 0. Assume that F1F2 = f + I,where f(φ, x) = x2φ2(−x), φ ∈ A1, x ∈ R.So, for ψ ∈ D(R), we have

∫R(ψ.F1F2)(x) dx = g + I0 ∈ C, where

g(φ) =

∫

Rψ(x).f(φ, x) dx =

∫

Rψ(x).x2φ2(−x) dx, φ ∈ A1.

Hence,

g(φε) = ε

∫

Rψ(−εx).x2φ2(x) dx, φ ∈ A1, ε > 0.

It follows that limε↓0g(φε)

ε= ψ(0)

∫R x2φ2(x) dx, ∀ψ ∈ D(R) and we can see

that g(φε) /∈ I0. Therefore, F1F2 0.

f)The relation ≈ is not compatible with the multiplication in G(Rn), ei-ther.Indeed, we have seen that Hm+1 ≈ H, m = 0, 1, . . . . It follows that (m +

1)Hm.δ ≈ δ, m = 1, 2, . . . . In particular, we get H ≈ H and H.δ ≈ 1

2δ.

However, H.Hδ is not associated with H.1

2δ. Indeed, if it did, we would have

1

3δ ≈ 1

4δ, or δ ≈ 0!! since we always have H2δ ≈ 1

3δ, and

1

2Hδ ≈ 1

4δ.

g) Finally, we will show that if S, T ∈ D′(Rn) and there exists S.T (inthe sense of Definition 1.2.1, or Definition 1.2.5), then S ¯ T ≈ S.T, whereagain we use the notation ¯ as the product in G(Rn)Indeed, because of Proposition 1.2.8, we only need to prove that S¯T ≈ S.T,where S.T is the product in the sense of Definition 1.2.1. Using the result ofTheorem 2.3.3, we have

S ¯ T = P + I, where P (φ, x) = 〈S, φ(·,−x)〉〈T, φ(·,−x)〉, φ ∈ A1.

For ψ ∈ D(Rn), we have

∫

Rn

(ψ¯S¯T )(x) dx = Q+I0 ∈ C, Q(φ) =

∫

Rn

ψ(x)〈S, φ(·,−x)〉〈T, φ(·,−x)〉 dx

Therefore, Q(φε) =∫Rn ψ(x)〈S, φε(·,−x)〉〈T, φε(·,−x)〉 dx. Now, we set ρε(x) =


1

εnφ(−x

ε), then ρε is a δ-sequence and

Q(φε) =

∫

Rn

ψ(x)〈S, ρε(x−·)〉〈T, ρε(x−·)〉 dx = 〈(S∗ρε)(T∗ρε), ψ〉 −→ 〈S.T, ψ〉,

as ε ↓ 0. Therefore, we have S ¯ T ≈ S.T

2.9 The tempered generalized functions and

their Fourier transform

We will now consider the Fourier transform of generalized functions. Sincethe approach is similar to the one in distribution theory, we will start withthe concept of “tempered generalized functions”. Indeed, we can see that thisconcept is the extension of the concept of tempered distributions in 1.1.5.5in [Col85]. Apart from that, we can also consider the concept of integral ina wider sense, compared to the one discussed in 2.7.1.

2.9.1 Notations

We denote by EM,τ [Rn] the set of all R ∈ E [Rn] such that for each multi-indexα, there is N ∈ N such that for all φ ∈ AN , we have

(1 + |x|N)−1(∂αR)(φε, x) = O(ε−N) as ε ↓ 0, uniformly for x ∈ Rn.

It is clear that EM,τ [Rn] ⊂ EM [Rn]. However, we notice that in spite ofC∞(Rn) ⊂ EM [Rn], C∞(Rn) is not contained in EM,τ [Rn]. Indeed, we takef(x) = ex ∈ C∞(R), but f(x) /∈ EM,τ [R]. For the proof, suppose converselythat f(x) = ex ∈ EM,τ [R], then we have

εN .ex

1 + |x|N ≤ c, ∀x ∈ R, 0 < ε < η with η > 0, for c large enough

and some N ∈ N. However, if x =1

ε, then

εNex

1 + |x|N =εNe

1

ε

1 + |ε|−N=

ε2Ne

1

ε

1 + |ε|N ≥ ε2N

εN + 1

1

(2N + 1)!

1

ε2N+1−→∞ as ε ↓ 0.

This is an absurdity.


Next, we denote by Iτ the set of all R ∈ E [Rn] such that, for each multi-index α, there are N ∈ N and β ∈ Γ such that when q ≥ N and φ ∈ Aq, wehave

(1 + |x|N)−1(∂αR)(φε, x) = O(εβ(q)−N), as ε ↓ 0, uniformly for x ∈ Rn.

Using the definition of EM,τ [Rn] and Iτ , we see that Iτ is an ideal of EM,τ [Rn].We also remark that Iτ ⊂ I. However, we will show that I is not containedin EM,τ [Rn]. Indeed, if

R(φε, x) =

∫

Rex+εµφ(µ) dµ− ex, then R ∈ I

since this difference is the one of two different representatives of ex in G(R).Assume that R ∈ EM,τ [R], so it follows from the definition of EM,τ [Rn] that

R(φε, x) = ex

∫

R(eεµ − 1)φ(µ) dµ

≤ c1 + |x|N

εN

for N large enough, for all x ∈ R, c > 0 large enough, and ε small enough.So, we get

∫

R(eεµ − 1)φ(µ) dµ

≤ c1 + |x|N

exεN−→ 0, as x →∞

It follows that∫

R(eεµ − 1)φ(µ) dµ = 0, for all ε > 0 small enough.

Since φ ∈ A1, so we can take derivatives (with ε as variable), and getφ ∈ ⋂∞

q=1Aq. However, it follows from 2.1.1 that⋂∞

q=1Aq = ∅, and it isabsurd. In conclusion, R /∈ EM,τ [R].

Now, we will define the concept of “tempered generalized functions”.

2.9.2 Definition

A tempered generalized function is an element of the quotient algebra EM,τ [Rn]/Iτ .We denote the set of all tempered generalized functions by Gτ (Rn).

We remark that F ∈ Gτ (Rn) if F = f +Iτ , where f ∈ EM,τ [Rn]. It followsthat ∂αGτ (Rn) ⊂ Gτ (Rn), and the Leibniz rule is still applicable in Gτ (Rn).


2.9.3 Example

Assume that f ∈ Cτ (Rn), where Cτ (Rn) = f ∈ C0(Rn) : ∃c ≥ 0, N ∈N such that |f(x)| ≤ c(1 + |x|N), ∀x ∈ Rn, then f ∈ Gτ (Rn).

Indeed, we can associate f with the map R ∈ EM,τ [Rn] as follows

R(φ, x) =

∫

Rn

f(x + µ)φ(µ) dµ(

=

∫

Rn

f(y)φ(y − x) dy).

Now, we will show why R ∈ EM,τ [Rn]. We have

(∂αR)(φ, x) = (−1)|α|∫

Rn

f(y)(∂αφ)(y − x) dy, so

(∂αR)(φε, x) = (−1)|α|∫

Rn

f(x + εµ)(∂αφ)(µ) dµ. It follows that

|(∂αR)(φε, x)| ≤ c1

∫

Rn

(1 + |x + εµ|)N(∂αφ)(µ) dµ

≤ c1(1 + |x|N)

∫

Rn

(1 + |εµ|N)(∂αφ)(µ) dµ

≤ c(1 + |x|N), ∀x ∈ Rn, for all φ ∈ A1.

It verifies our conclusion.

2.9.4 Remark

If F = f + Iτ ∈ Gτ (Rn), then f + I ∈ G(Rn). It follows that Gτ (Rn) can beconsidered as a subset of G(Rn).

Indeed, it is because of EM,τ [Rn] ⊂ EM [Rn]. However, we also notice thatthere exists R ∈ EM,τ [Rn]

⋂ I, but R /∈ Iτ (Please refer to [Col85], pp. 101–104 for details). That means that the map f + Iτ 7−→ f + I from Gτ (Rn)to G(Rn) is not injective. So, it is not an embedding! However, there is anembedding from Gc(Rn), the space of all generalized functions with compactsupport, to G(Rn). Indeed, if F = R + I ∈ Gc(Rn), and K = supp F, acompact set, then we can take f ∈ D(Rn) such that f ≡ 1 on K, then fR isanother representative of F (since fR−R + I = I). If we take S as anotherrepresentative of F and g ≡ 1 on K, then it follows from

fR− gS = (fR−R) + (R− S) + (S − gS) ∈ I


that fR does not depend on the selections of f and R for F ∈ Gc(Rn). Wenotice that D(Rn) ⊂ EM,τ [Rn] and it follows that fR ∈ EM,τ [Rn], too. So wecan associate F = R + I ∈ Gc(Rn) with fR + Iτ ∈ Gτ (Rn). The mappingR + I 7−→ fR + Iτ is injective since fR is only another representative ofF ∈ Gc(Rn) ⊂ G(Rn).

As far as we know the concept of integral of a generalized function F isconfined to a compact set K, or supp F compact. Now we can extend thisconcept to a class of tempered generalized functions.

2.9.5 Integration of the tempered generalized functions

Let F ∈ Gτ (Rn), F = R + Iτ , where R ∈ EM,τ [Rn]. We remark that forsome N if φ ∈ AN and ε > 0 small enough, then R(φε, ·)φ(·) ∈ S(Rn) since

R ∈ EM,τ [Rn] and φ ∈ D(Rn). So, I(φε) =∫Rn R(φε, x)φε(x) dx makes sense

with such N and ε. To define a generalized complex number we have to definefor all φ ∈ A1. However, we can see at once from the definition of a complexnumber that this generalized complex number does not change if we considerφ ∈ AN with some N, or ε is not small enough. So, we can set I(φε) = 0 whenφ /∈ AN , or ε is not small enough. In summary we will consider a generalizedcomplex number I such that

I(φε) =

∫Rn R(φε, x)φε(x) dx, for φ ∈ AN , for some N and ε small enough

0 otherwise.

We will show that I ∈ EM . We notice that

I(φε) =

∫

Rn

R(φε, x)φ(εx) dx, since φε(x) = φ(εx).

We only need to show I ∈ EM in case of φ ∈ AN , for some N and ε smallenough. In such cases we get

|R(φε, x)| ≤ c1 + |x|N

εN, ∀x ∈ Rn.

On the other hand, since φ ∈ D(Rn), so φ ∈ S(Rn), and for all p we have

|φ(y)| ≤ cp

1 + |y|p , ∀y ∈ Rn.

Selecting p = N + n + 1, we have for all x ∈ Rn

|R(φε, x)||φ(εx)| ≤ c′

εN

1 + |x|NεN+n+1(1 + |x|N+n+1)

.


It follows that

|I(φε)| ≤ C

ε2N+n+1, or I ∈ EM .

On the other hand, if R ∈ Iτ , then

|R(φε, x)| ≤ cR (1 + |x|N) εβ(q)−N , ∀x ∈ Rn, for β ∈ Γ.

Similarly as above, we get

|I(φε)| ≤ CRεγ(q)−N , where γ(q) = β(q)−N − n− 1, or I ∈ I0.

Therefore, I ∈ C does not depend on the choice of representative of F. Andwe call I the integral of F ∈ Gτ (Rn) in Rn and denote it by

∫Rn F (x) dx ∈ C.

If F ∈ Gc(Rn), then we can define∫Rn F (x) dx both in the sense of the note

after Example 2.7.4, and in the sense above. What is the relationship betweenthem? It turns out that the two values define the same complex number inC. Indeed, since F has compact support, so we can choose a representativeof F such that R(φε, x) = 0 if |x| ≥ a, for a > 0 and supp F ⊂ x : |x| ≤ a,where ε > 0 small enough. Based on arguments in Remark 2.9.4, we onlyneed to show that d ∈ I0, where d(φ) =

∫|x|≤a

R(φ, x)[φ(x) − 1] dx. Indeed,

we have

d(φε) =

∫

|x|≤a

R(φε, x)[φ(εx)− 1] dx.

We can see this by expressing φ in the form of Taylor series up to order q +1if φ ∈ Aq. For instance, with n = 1 we can write as follows

|φ(εx)− 1| = |[φ]′(0)εx + [φ]′′(0)(εx)2

2!+ . . . + [φ](q+1)(0)

(εx)q+1

(q + 1)!+ . . . |

= |[φ](q+1)(0)(εx)(q+1)!

q + 1+ . . . | ≤ c.εq+1, uniform for |x| ≤ a,

since [φ]′(0) = [φ]′′(0) = . . . = [φ](q)(0) = 0 (since φ ∈ Aq ). It follows thatd(φε, x) = O(εβ(q)−1), where β(q) = q + 2 ∈ Γ. Hence, d ∈ I0.

Concerning other classes we have the following proposition

2.9.6 Proposition

a) If f ∈ S(Rn), then f can be considered as an element of Gτ (Rn), and theclassical integral

∫Rn f(x) dx coincides with the integral of f as a tempered

generalized function


b) If f ∈ C0(Rn) and there are c > 0 and ν > 0 such that |f(x)| ≤c

1 + |x|n+ν,∀x ∈ Rn, then f can be considered as an element of Gτ (Rn).

Moreover, we have (G)∫Rn f(x) dx ` (O)

∫Rn f(x) dx, where the left and the

right-hand side integrals are in the sense of Gτ (Rn) and in ordinary sense,respectively.

Proof. If f ∈ S(Rn), then we can associate f with Rf + Iτ ∈ Gτ (Rn),where Rf (φ, x) = f(x), ∀x ∈ Rn. It follows from properties of S(Rn) thatRf ∈ EM,τ [Rn]. Denote by d the difference between these two integrals, weneed to show that d ∈ I0. Indeed, we have

d(φε) =

∫

Rn

f(x)[φ(εx)− 1] dx, then for φ ∈ Aq we have

|d(φε)| ≤∫

Rn

|f(x)| |φ(εx)− 1| dx

≤∫

Rn

cq+n+2

1 + |x|q+n+2cq εq+1 |x|q+1 dx

≤ c εq+1, where c =

∫

Rn

cq+n+2 cq |x|q+1

1 + |x|q+n+2dx < ∞.

It follows that d ∈ I0.

b) It is obvious that f ∈ Cτ (Rn), and in Gτ (Rn), f is written as R + Iτ ,where

R(φε, x) =

∫

Rn

f(x + εµ)φ(µ) dµ (see Example 2.9.4).

Therefore (G)∫Rn f(x) dx has its representative I, where

I(φε) =

∫

Rn

( ∫

Rn

f(x + εµ)φ(µ) dµ)φ(εx) dx.

It follows from the properties of φ that

limε↓0

I(φε) =

∫

Rn×Rn

f(x)φ(µ) dxdµ =

∫

Rn

f(x) dx.

So, we have (G)∫Rn f(x) dx ` (O)

∫Rn f(x) dx.

2.9.7 Definition

If F = R + Iτ ∈ Gτ (Rn), where R ∈ EM,τ [Rn], then we define FF (or F ,even F∧), called the Fourier transform of F, to be the following tempered


generalized function

FF = R + Iτ ∈ Gτ (Rn), where

R(φε, x) =

∫

Rn

e−ixyR(φε, y)φ(εy) dy,

for some N ∈ N and ε small enough, otherwise R(φε, x) = 0.

We have to verify that this is a good definition by proving that R ∈EM,τ [Rn], and that this definition does not depend on the choice of represen-tative of F. We can do that by repeating what we have done in 2.9.5 andnotice that |e−ixy| = 1. Please see the details in [Col85].

We also define “the inverse Fourier transform” F−1 as follows

F−1F = F−1R + Iτ ∈ Gτ (Rn), where

(F−1R)(φε, x) = (2π)−n

∫

Rn

eixyR(φε, y)φ(εy) dy,

Concerning the relationship with the usual Fourier transform, we have thefollowing result:

2.9.8 Proposition

If F ∈ S(Rn), then this new Fourier transform coincides with the classicalone.

Proof. We have to prove that d ∈ Iτ , where

d(φε, x) =

∫

Rn

e−ixyF (y)[φ(εy)− 1] dy.

Indeed, we notice that

(∂αd)(φε, x) = (−i)|α|∫

Rn

yαe−ixyF (y)φ(εy) dy.

Now, we can use the Taylor expression as we have done in 2.9.5, and we getd ∈ Iτ .

Chapter 3

L1(R) embedded in theColombeau generalizedfunctions

In this chapter we will be concerned with the study of some issues related toL1(R)-functions as a subset of the class of generalized functions.

3.1 Remarks on L1(R)-functions in G(Rn) and in Gτ (Rn)

We have seen that L1(Rn) ⊂ D′(Rn) in the Schwartz distribution theory, andthat L1(Rn)-functions are tempered distributions. So, to f ∈ L1(Rn), therecorresponds an element in G(Rn), denoted by f+I where f ∈ EM [Rn]. Can weexpress f in term of f more clearly? Yes, and we can do as follows: applyingthe result of Theorem 2.3.3, we have

f(φ, x) = 〈f(y), φ(y − x)〉 =

∫

Rn

f(y)φ(y − x)dy, since f ∈ L1(Rn).

So, we have f + I as the corresponding element of f ∈ L1(R). However, wecannot conclude from this that f + Iτ ∈ Gτ (Rn), see Remark 2.9.4.

When Colombeau started the theory of generalized functions, he usedthe definition of tempered distribution by means of the structure theoremfor S ′(Rn) (see Theorem 8.3 in [Fri98]). That is “G ∈ Gτ (Rn) is a tempereddistribution if there is a continuous function f ∈ Cτ (Rn) and a derivativeD (of any order) such that G = Df in Gτ (Rn)” ( see Definition 4.4.1 in[Col85]). It follows from this definition that all the tempered distributionsin the sense of Definition 1.1.5.6 are the tempered distributions in the sense

56


above. And it is obvious that the definition above is the extension of Defi-nition 1.1.5.6 in the Colombeau theory, and L1(Rn)-functions are elementsof Gτ (Rn). However, in case of L1(R)-functions we can use Theorem 8.17 in[Rud74] to show directly that L1(R)-functions are elements of Gτ (R). We alsoget their representatives in Gτ (R), and from that, their representatives inG(Rn). Indeed, we can do as follows:

Since f ∈ L1(R), so the function g, where g(x) =∫ x

−∞ f(t) dt, x ∈ R iscontinuous in R (Theorem 8.17 in [Rud74]). We also remark that

|g(x)| = |∫ x

−∞f(t) dt| ≤

∫ x

−∞|f(t)| dt ≤

∫

R|f(t)| dt = ‖f‖L1 , for all x ∈ R.

So, g ∈ Cτ (R). Now, it follows from Example 2.9.3 that g belongs to Gτ (R);and in Gτ (R), g is assigned with g + Iτ , where g(φ, x) =

∫R g(x + y)φ(y) dy.

It implies that f belong to Gτ (R), and it is assigned with ∂g + Iτ . So, f isassigned with the element

∂x(

∫

R(

∫ x+y

−∞f(t) dt)φ(y) dy) + Iτ .

Now, we notice that φ ∈ A1 and f ∈ L1(R), so we get

∫ ∞

−∞(

∫ x+y

−∞f(t) dt)φ(y) dy =

∫ ∞

−∞(

∫ x

−∞f(t + y) dt)φ(y) dy

=

∫ x

−∞(

∫ ∞

−∞φ(y)f(t + y) dy) dt,

and the inner integral as the function of t is in L1(R)⋂

C∞(R). It followsthat

∂x(

∫

R(

∫ x+y

−∞f(t) dt)φ(y) dy) =

∫ ∞

−∞φ(y)f(x + y) dy.

Therefore, f is assigned with the element∫R f(x + y)φ(y)dy + Iτ in Gτ (R).

It also shows us that in G(R), the function f ∈ L1(R) is assigned with theelement

∫R f(x + y)φ(y)dy + I.

Now, we will use the results above to study the relationship between theintegral of f ∈ L1(R) in the usual sense and the one in the sense of temperedgeneralized function.


3.2 Relationship between two kinds of integral of f ∈ L1(R)

We have seen that if f ∈ L1(R), then there exists the integral of f in the usualsense, or in Lebesgue sense, denoted by (L)

∫R f(x) dx. On the other hand,

f can be considered as a member of Gτ (R), then there exists the integral off in the sense of tempered generalized functions denoted by (G)

∫R f(x) dx.

Thus, we can consider the relationship between two kinds of integral in C.And we can prove that:

Proposition. In C, the ordinary complex number (L)∫R f(x) dx is as-

sociated with the generalized complex number (G)∫R f(x) dx. Or, we have

(G)∫R f(x) dx ` (L)

∫R f(x) dx.

Proof. Indeed, we have (G)∫R f(x) dx = I + I0, where

I(φε) =

∫

R(

∫

Rf(x + y)φε(y) dy)φε(x) dx

=

∫

R(

∫

Rf(x + εt)φ(t) dt)φ(εx) dx.

We notice that for f ∈ L1(R) and φ ∈ A1 we have

I(φε) =

∫

R(

∫

Rf(x + εt)φ(t) dt)φ(εx) dx

=

∫

R(

∫

Rf(x + εt)φ(εx) dx)φ(t) dt

=

∫

R(

∫

Rf(x)φ(εx− ε2t) dx)φ(t) dt

=

∫

R(

∫

Rφ(εx− ε2t)φ(t) dt)f(x) dx.

So, it follows from functions φ ∈ A1 that

limε↓0

I(φε) =

∫

R(

∫

Rφ(t) dt)f(x) dx = (L)

∫

Rf(x) dx.

This concludes the proof.

3.3 Relationship between Gτ -Fourier transform and the usual Fouriertransform for f ∈ L1(R)

If f ∈ L1(R), then the usual Fourier transform f of f in Chapter 1 is givenby


f(x) =

∫

Re−iyxf(y) dy.

Moreover, since f ∈ L1(R), then f ∈ Gτ (R) and to f is assigned the elementf + Iτ , where f(φ, x) =

∫R f(x + y)φ(y) dy := R(φ, x) for φ ∈ A1 and x ∈ R.

In Gτ (R) we can consider the Gτ -Fourier transform of f, or of f + Iτ as in

Definition 2.9.7 which is defined by R + Iτ , where

R(φε, x) =

∫

Re−iyxR(φε, x)φε(y) dy,

for φ ∈ Aq with q large enough and ε > 0 small enough. We will denote by

(Gτ )f the Fourier transform of f considered in Gτ (R). For the Fourier trans-form in Gτ (Rn), we have some related results. Please refer to [Col85], pp.114–

121 for details. Here, we want to find out some relationship between f and(Gτ )f for f ∈ L1(R). We have:

Theorem. In G(R) we have f ≈ (Gτ )f . Otherwise stated, f is associ-

ated with (Gτ )f in G(R).

Proof. Since f ∈ L1(R), then f ∈ L∞(R) and f(·) is continuous in R(Refer to [Rud74], Theorem 9.6). So, f is also a generalized function and in

G(R) it is assigned with the element f +I, where f(φ) =∫R f(x+ y)φ(y) dy.

We also remark that since φε(x) = φ(εx), so (Gτ )f is the element R(φε, x) +Iτ , where

R(φε, x) =

∫

Re−ixyR(φε, y)φ(εy) dy

in Gτ (R), and so it is in Gτ (R). To prove our theorem we need to show thatfor all ψ ∈ D(R), we have

limε↓0

∫

Rψ(x)A(x, ε) dx = 0, where A(x, ε) = f(φε, x)− R(φε, x).

Now, noticing that φ ∈ A1 and f ∈ L1(R), we apply Fubini’s theorem andget

R(φε, x) =

∫

R(

∫

Re−ixyf(y + εt)φ(εy) dy)φ(t) dt.


It follows that

|R(φε, x)| = |∫

R(

∫

Re−ixyf(y + εt)φ(εy) dy)φ(t) dt|

= |∫

R(

∫

Re−ix(z−εt)f(z)φ(ε(z − εt)) dz)φ(t) dt|

≤ (maxz∈R

|φ(z)|)∫

R

∫

R|f(z)| |φ(t)| dzdt < ∞

So, we can apply the dominated convergence Lebesgue and get

limε↓0

∫

Rψ(x)A(x, ε) dx =

∫

Rψ(x)(

∫

Rf(x)φ(y) dy−

∫

Re−ixy

∫

Rf(y) φ(t)dt φ(0) dy)dx

since φ ∈ A1. It follows that limε↓0∫R ψ(x)A(x, ε)dx = 0, and it concludes

our proof.

3.4 About the existence of the embedding of f ∈ L1(R) into G(R)

We will show that there is an embedding of L1(R) into G(R), where

f ∈ L1(R) 7−→ f + I ∈ G(R).

This makes sense because it follows from arguments in Remark 2.9.4 thatthere is no general embedding from Gτ (R) into G(R). In [Col85], pp.121–126Colombeau defined some other concepts and used another method to provethe existence of the embedding above from the set of Colombeau temperedgeneralized functions to G(Rn). Here, we can use the special property of L1-functions related to concept of the Lebesgue point, and we can get a simplerproof.

We will show that if f ∈ L1(R) and its corresponding element f +I = I,then f = 0 in L1(R). Indeed, we have

f(φε, x) =

∫

Rf(x + y)φε(y) dy

=

∫

Rf(x + εy)φ(t) dt.

If f ∈ I, then for φ ∈ Aq with q large enough we have limε↓0 f(φε, x) = 0uniformly on an arbitrary compact K of R. On the other hand, we notice


that if f ∈ L1(R) then almost every point x ∈ R is the Lebesgue point of f.So, for φ ∈ A1 we have

|f(φε, x)− f(x)| = |∫

R[f(x + εt)φ(t)− f(x)φ(t)] dt|

≤∫

R|f(x + εt)− f(x)| |φ(t)| dt

≤ c1

∫

B

|f(x + εt)− f(x)| dt

≤ c1

2ε

∫

Bε

|f(x + y)− f(x)| dy −→ 0 as ε ↓ 0

on K for almost everywhere x ∈ R. Here, B is a ball such that supp φ ⊂ Band Bε = εB. It follows that f = 0 almost everywhere on K. Since K isarbitrary compact subset of R, we have f = 0 almost everywhere in R andit concludes our proof.

3.5 About the value of a Ck-function at a point

We saw in Remark 2.6.3 that “If F = f ∈ C∞(Rn) ⊂ G(Rn), then the valueof F at any x ∈ Rn coincides with the usual value of f at x.” However, if fis continuous, we cannot get the same result and we have shown a counter-example about that after Theorem 2.6.4. One may wonder whether the resultin Remark 2.6.3 is still right if we ”increase” the level of smoothness of f atx? The answer is “no”, as we state in the following proposition:

Proposition. If we increase the level of smoothness of a function, butnot to be C∞-function, then the value of that function at a point in C stillmay not coincide with the usual one.

Proof. We will take a counter-example to illustrate to result above. In-

deed, if we choose f(x) =

x3 if x ≥ 0,

−x3 if x < 0, then f(0) = 0 as the usual sense

and f ∈ C2(R), but we will show that f(0) 6= 0 in C. First, we know that inG(R), f is represented in the form f + I, where

f(φε, x) =

∫

Rf(x + y)φ(y) dy

=

∫

Rf(y)φ(y − x) dy = −

∫ 0

−∞y3φ(y − x) dy +

∫ ∞

0

y3φ(y − x) dy.


It follows that the value of this function as a generalized function at 0 in Cis the generalized complex number f(φ, 0) + I0, where

f(φ, 0) = −∫ 0

−∞y3φ(y) dy +

∫ ∞

0

y3φ(y) dy =

∫ ∞

0

y3[φ(−y) + φ(y)] dy.

We will prove that f(φ, 0) + I0 6= I0, or the value of f as the generalizedfunction at 0 is not 0 as its usual value. So, we have to prove that f(φ, 0) /∈ I0.We have

f(φε, 0) =

∫ ∞

0

y3[φε(−y) + φε(y)] dy = ε3

∫ ∞

0

x3[φ(−x) + φ(x)] dx.

To prove that f(φ, 0) /∈ I0, we will show that for each q = 1, 2, . . . , wecan find φ ∈ Aq such that

∫∞0

x3[φ(−x) + φ(x)] dx 6= 0. We choose ψ ∈D(R), supp ψ ⊂ (0,∞) and

∫R ψ(x)dx = 1. Setting φ1 = ψ + α1ψ

′ ∈ A1

with α1 =∫R xψ(x)dx. Then, we set φ2 = φ1 + α2ψ

′′ ∈ A2 by choos-

ing α2 = −1

2

∫R x2φ1(x)dx. Now, we choose α3 = −1

6− 1

6

∫R x3φ2(x)dx,

then∫R x3φ3(x)dx = 1, where φ3 = φ2 + α3ψ

(3). We notice that often in

Colombeau’s method this value is 0. Now, we set φ4 = φ3 + α4ψ(4), where

α4 =1

24

∫R x4φ3(x) dx and we get

∫R x4φ4(x) dx = 0. We continue to do up

to q-th step and get φq ∈ D(R), supp φq ⊂ (0,∞), and satisfy∫R φq(x) dx =

1,∫R xpφq(x) dx = 0 for 1 ≤ p ≤ q, p 6= 3 and

∫R x3φq(x) dx = 1. Finally,

we only need to choose φ(x) =1

2[φq(x) + φq(−x)], for each q = 1, 2, . . . , and

we get such a function φ ∈ Aq we need. Indeed, we can easily to verify that∫R φ(x) dx = 1 and

∫R xpφ(x) dx = 0 for 1 ≤ p ≤ q. Moreover, noticing that

supp φq ⊂ (0,∞), we also have

∫ ∞

0

x3[φ(x) + φ(−x)] dx =

∫ ∞

0

x3[φq(x) + φq(−x)] dx =

∫ ∞

0

x3φq(x) dx

=

∫

Rx3φq(x) dx = 1 6= 0.

And it concludes our proof.

We also remark that we can extend the result above in cases of

f(x) =

x2n+1 if x ≥ 0,

−x2n+1 if x < 0,

for an arbitrary positive integer number n.


3.6 About the relations ∼ and ≈ in G(Rn)

As we see in Remark 2.8.7 relations ∼ and ≈ are not compatible with themultiplication in G(Rn). However, we can get this compatibility if we confinethese relations to some special cases as the following results

Proposition. We have

• if f ∈ C∞(Rn), F ∈ G(Rn) and F ∼ 0, then fF ∼ 0

• if f ∈ C∞(Rn), F ∈ G(Rn) and F ≈ 0, then fF ≈ 0.

Proof. The proof of those conclusions is straightforward. For instance, toprove that if f ∈ C∞(Rn), F ∈ G(Rn) and F ∼ 0, then fF ∼ 0, we can doas follows: since F ∼ 0, for all ψ ∈ D(Rn) we have

∫Rn(ψF )(x)dx = 0 in C.

So, there is β ∈ Γ such that for all ψ ∈ D(Rn), we have∫

Kψ(x)R(φε, x)dx =

O(εβ(q)−N), where F = R+I, φ ∈ Aq, q ≥ N for some positive integer N, K,a compact subset of Rn with supp ψ ⊂ K. We remark that if f ∈ C∞(Rn)and ψ ∈ D(Rn), then ψ(x)f(x) is in D(Rn), its support is a subset of K, andit does not depend on φ. So, it can play the role of the function ψ(x) in thearguments above. It follows that

∫Rn(ψfF )(x)dx = 0 in C. That concludes

our arguments.

In addition we remark that if f ∈ D′(Rn), F ∈ G(Rn) and F ∼ 0, wecannot imply that fF ∼ 0. For instance, if f = xδ ∈ D′(R) and F = xδ ∼ 0,but fF = x2δ2 0 as we have seen in Example 2.8.7. We also get a similarconclusion for the relation ≈: if f ∈ D′(Rn), F ∈ G(Rn) and F ≈ 0, wecannot imply that fF ≈ 0.


Discussion and ConclusionsIn Chapter 2 we have met the main points of Colombeau theory of gen-

eralized functions. We saw that in G(Ω) one can multiply two arbitrarygeneralized functions and Leibniz’s rule is valid. Moreover, the multiplica-tion operation is commutative, associative, and the Schwartz distributionsare embedded as a subset of G(Ω) in a way which is compatible with differ-entiation.

The draw-back of multiplication in G(Ω) is that not always f u−fu ∈ I(Ω)if u ∈ D′(Ω), f ∈ C∞(Ω). This is repaired by working with the equivalence∼ . Also, not always f u − fu ∈ I(Ω) if f, g ∈ C0(Ω). This is repaired byworking with ≈ . However, F : F ∼ 0 and F : F ≈ 0 are not idealsin G and ∼ and ≈ do not respect multiplication in G. Already there areu, w ∈ D(Rn), v ∈ C∞(Rn) with vw ∼ 0 but uvw is even not associatedwith 0.

Another way to construct a space of generalized functions so that somelimitations in the Schwartz theory of distributions are taken away is in[Koo84]. In this article the authors Koornwinder and Lodder propose anothermethod to approach the goal above. Their ideas rely on Schwartz’s defin-ition of the Fourier transform of a tempered distribution. They start witha special subspace PC of the space of tempered distributions and introducethe space SPC of “new” generalized functions (They called them generalisedfunctions) which are linear functionals on PC. It is very interesting that inSPC all the usual operations including the product are applicable. Theyalso propose a bigger algebra of generalized functions, denoted by GF whichare the linear functionals on SPC. Then all elements of PC and SPC alsobelong to this new algebra. Please refer to [Koo84], and also [Lod91] in detail.

Lately, in [Gko01] the authors have found some results of application ofColombeau theory in differential geometry and Physics. These results showthe future of the development of Colombeau theory in theory and in appli-cation. The properties of the Fourier transform in Gτ (Rn) for special classessuch as L1(Rn)-functions, the construction of sufficiently “good” topologyon G(Ω), the relationship between the Colombeau theory and the theoryproposed in [Koo84], as well as the applications of the Colombeau theoryof generalized functions in nonlinear analysis are topics for possible furtherexploration.


AcknowledgmentsFirst, the author would like to thank the Dutch government and the

Dutch people. The author could not have come here, to UvA, without theNFP (The Netherlands Fellowship Programmes) scholarship. This beautifulcountry of the Tulip flower and the windmill has given the author a greatopportunity, which always reminds the author that he has to work in orderto deserve that honor!

The author would like to thank Prof. Tom H. Koornwinder for his en-thusiasm, cordial attitude, criticism, and his deep and broad understanding!The author is very honored for studying with such an experienced professorlike Tom H. Koornwinder and to have become one of his students.

The author also expresses his thanks to Dr. J. J. O. O. Wiegerinck atKdV of UvA, Drs. A. J. P. Heck at the Amstel Institute of UvA, Dr. Marcelde Jeu at the Mathematical Institute of Leiden University, library staff ofthe math library of UvA and of the CWI library, lecturers and colleaguesof the Mathematics and Science Education programme at the AMSTEL In-stitute of UvA, officers in the International Office of Faculty of Science ofUvA, colleagues at Math Department of Hanoi University of Pedagogy 2 inVietnam, and Vietnamese friends who are studying here in Amsterdam, fortheir support.

Last, but not least, the author’s deep thanks are devoted to his family:grandparents, parents, Dr. K. Ninh, aunt and uncle, brothers, sisters, niecesT. Nguyen and N. Mai, and nephew V. Anh, especially his wife T. Ha andhis son N. Sang. The author could not have finished this thesis without theirsupport. “Thank you for encouraging me while I am living alone in Amster-dam. I am trying my best to live up to your expectation”.

Amsterdam, June, 2005


References[Ada75] R. A. Adams, Sobolev spaces, Academic Press(1975).

[Arv98] W. Arveson, A Short Course on Spectral Theory, Springer(2002).

[Col84] J. F. Colombeau, New Generalized Functions and Multiplications ofDistributions, North Holland, Math. Studies 84, Amsterdam(1984).

[Col85] J. F. Colombeau, Elementary Introduction to New Generalized Func-tions, North Holland, Math. Studies 113, Amsterdam(1985).

[Fri98] F. G. Friedlander, Introduction to the theory of distributions, Cam-bridge University Press(1998).

[Gko01] M. Grosser, M. Kunzinger, M. Oberguggenberger, R. Steinbauer,Geometric theory of generalized functions with applications to generalrelativity, Mathematics and its Applications 537, Kluwer AcademicPublisher(2001).

[Hal67] P. R. Halmos, A Hilbert space problem book, Van Nostrand(1967).

[Hor83] L. Hormander, The Analysis of Linear Partial Differential Operator I,Springer(1983).

[Ita76] M. Itano, Remarks on the multiplicative products of distributions, Hi-roshima Math. J. 6(1976), pp. 365–375.

[Koo84] T. H. Koornwinder, J. J. Lodder, Generalised functions as linear func-tionals on generalized functions, In P. L. Butzer, R. L. Stens andB. Sz. -Nagy (Eds), Anniversary Volume in Appromation Theory andFunctional Analysis, Birkhauser Verlag, Basel(1984), pp. 151–163.

[Lod91] J. J. Lodder, Towards a symmetrical theory of generalised functions,CWI Tract, 79. Stichting Mathematisch Centrum, Centrum voor Wiskundeen Informatica, Amsterdam(1991).

[Mik66] J. Mikusinski, On the square of the Dirac delta distribution, Bull. Aca.Pol. Sci 14, 9(1966).

[Obe92] M. Oberguggenberger, Multiplication of Distributions and Applicationsto Partial Differential Equations, Longman Sci. & Tech.(1992).

[Obe01] M. Oberguggenberger, Generalized functions in nonlinear models- asurvey, Nonlinear Analysis 47(2001), pp. 5029–5040.


[Ros87] E. E. Rosinger, Generalized Solutions of Nonlinear Partial DifferentialEquations, North-Holland, Math. Studies 146, Amsterdam(1987).

[Rud74] W. Rudin, Real and Complex Analysis, Mc Graw-Hill, Inc., SecondEdition(1974).

[Rud85] W. Rudin, Functional Analysis, Tata Mc Graw-Hill, Inc., New Delhi(1985).

[Sch54] L. Schwartz, Sur l’Impossibilite de la Multiplication des Distributions,Comptes Rendus Acad. Sci. Paris, 239(1954), pp. 847–848.

[Sch66] L. Schwartz, Theorie des distributions, Hermann, Paris(1966).

[Sch01] L. Schwartz, A mathematician grappling with his century, English ver-sion translated by L. Schneps, Birkhauser Verlag(2001).

[Tre67] F. Treves, Topological vector spaces, distributions and kernels, Acad-emic Press, New York and London(1967).

[Tys81] J. Tysk, On the Multiplication of Distributions, Uppsala University(1985).

[Yos74] K. Yosida, Functional Analysis, Fourth Edition, Springer-Verlag(1974).

the colombeau theory of generalized functions · 2020-07-13 · the colombeau theory of generalized...

Documents