Distributions: Topology and Sequential Compactness

Distributions: Topology and Sequential Compactness.

May 1, 2014

Contents

1 Introduction 2

2 General deﬁnitions, and methods for Topological Vector Spaces. 4

3 The topology of D(Ω) 9 3.1 The spaces DK ...... 10 3.2 LF-spaces ...... 12 3.3 D(Ω) is an LF-space ...... 14 3.4 Semi-norms deﬁning the topology on D(Ω) ...... 15

4 Properties of the Space D(Ω) 15 4.1 Completeness of D(Ω) ...... 15 4.2 Non-metrizability of D(Ω) ...... 18 4.3 D(Ω) is a Montel Spaces ...... 19 4.4 Separability ...... 19

5 The space D0(Ω) 20 5.1 Positive distributions ...... 21 5.2 Topologies on D0(Ω) ...... 21 5.3 Non-metrizability ...... 22 5.4 These Spaces are Hausdorﬀ ...... 24 5.5 The Banach-Steinhaus Theorem & Barrelled Spaces ...... 24 5.6 The Banach-Alaoglu-Bourbaki Theorem ...... 25 5.7 Sequential Compactness ...... 27 5.8 More Montel Spaces & Equivalence of the Topologies on Bounded sets ...... 28 5.9 The Weak topology on D(Ω) ...... 29 5.10 Reﬂexitivity ...... 30 5.11 Completeness ...... 31 5.12 The space D as a subset of D0 and Separability...... 32

6 Multiplying Distributions 33 6.1 Multiplication between subspaces of functions and D0 ...... 33 6.2 The Impossibility of Multiplying two Distributions in General ...... 34 6.3 Division by Analytic Functions in the Case of One Real Variable...... 35 6.3.1 Division by x ...... 36 6.3.2 Division by Polynomials...... 36

1 6.3.3 Division by Analytic Functions with Zeroes of Finite Order...... 36 6.4 Applications of Multiplication...... 38

7 Rapidly Decreasing Functions. 38 7.1 Relation between S and D(Sn)...... 40 7.2 The Fourier transform on S ...... 40

8 Tempered Distributions 41 8.1 Relation between S 0 and D0(Sn)...... 42 8.2 The Fourier Transform on S ’...... 42

9 The space E0. 43

10 Comparison and Relations between the spaces D, S and E and their duals. 43

11 Tensor Product of two Distributions 44 11.1 The Tensor Product of two Vector Spaces ...... 44 11.2 The Tensor Product of Function Spaces ...... 44 11.3 Tensor Products of Distributions ...... 45 0 11.3.1 The Space L(D(Ω1), D (Ω2)) ...... 48 11.4 Topological Tensor Products ...... 48 11.4.1 The π-topology ...... 48 11.4.2 The -topology ...... 49

12 Kernels and the kernels theorem. 49 12.1 Regular Kernels ...... 52

13 Nuclear Locally Convex Spaces 55 13.1 Nuclear Mappings ...... 55 13.2 Nuclear Spaces ...... 56 13.3 Nuclear Spaces and the Kernels Theorem...... 57

A Appendix: Completions 57

1 Introduction

This essay largely concerns the topological properties of the space of distributions as developed by Laurent Schwartz between 1945 and 1950. We begin this introduction with some discussion of the history and uses of distributions and will then move on to an outline of the main topics of the essay. Much of the historical details are taken from [6]. Examples of generalized functions pre-date the rigorous development of distributions. A particularly prevalent example is Heaviside’s operational calculus in [12] where he, unrigorously, uses algebraic manipulation of differential operators to treat many problems, particularly in electrodynamics. Schwartz was unaware of Heaviside’s calculus when he initially created the theory of distributions. He was told of it by electrical engineers who were able to put the theory of distributions to great use. In [6] Lützensays that the theory of distributions was well received by physicists as it “allowed them to use improper functions in good conscience”. Though it was in fact Sobolev who first defined distributions as the continuous linear forms on a space of test functions. Schwartz did not know of this work when he began to work on the

2 generalized solutions to partial diﬀerential equations. He originally worked by introducing the convolution operators T : D → E which satisfy

T (φ ∗ ψ) = T (φ) ∗ ψ.

Here, for example δ is the identity operator. This theory along with many theorems was made by Schwartz in October 1944. It was originally very fruitful and Schwartz was able to prove the analogies of many of the theorems which hold in the current theory of distributions. However, it became more challenging when dealing with Fourier transforms. Consequently, in 1945 Schwartz switched to studying D0. Lützenreports in [6] that Schwartz, himself, felt that there were two factors which made D0 a good place to work on his theories of generalized solutions. Firstly, that functional analysis was at that time making great developments and Schwartz’s own private work on the dual spaces of Fréchet spaces allowed him to manipulate these dual spaces without having a ‘concrete representation’. Secondly, he already knew that δ as a measure could already be represented as a functional. Distribution spaces have many applications, particularly in PDE, which are largely not addressed in this essay, in particular the work done by Lars Hörmander. Furthermore, the space of tempered distributions allows for many generalizations and extensions of theories using Fourier analysis. The distributions generalize the way in which locally summable functions and measures can act as the kernels of integral operators on an appropriate space of test functions. This, after appropriate checks and with restrictions, allows us to extend the weak formulation of PDEs outside function spaces. This essay begins by introducing the space of test functions and studying its topology, especially results about compactness and metrizability. This space is the smooth functions of compact support. The topology on this space was developed by Schwartz in a method which was extended by Schwartz along with Dieudonnéto a general class of topological vector space called LF-spaces. We work by first introducing the LF-space in general and seeing how D can be viewed as a particular case. The Theory of Distributions is a theory of duality. We define the space of distributions to be the analytic dual of the test functions. We can also define many operations on the space of distributions; multiplication, differentiation, Fourier transform, in terms of the adjoint map of the maps on the space of test functions where they are defined in the classical sense. We can, therefore, discover many topological properties of the space of distributions by looking at theorems about dual topologies. For example, the Banach-Alaoglu-Bourbaki Theorem which shows us that weak-* bounded, weak-* closed sets in the dual of a Topological Vector Space are weak-* compact. Again in exploring the dual topologies we will focus on compactness and metrizablity results. In fact for both the distributions and the test functions we can show that, although neither space is metrizable, the closed, bounded sets are compact and metrizable (so also sequentially compact). This will lead to the equivalence of the weak-* and strong dual topology on bounded sets and the reflexivity of all spaces. We then explore the multiplication and division of distributions. Multiplication, can be defined by duality between the space E of smooth functions and the distributions. We will explore the continuity properties of this mapping and show that it cannot be extended to a multiplication between two distribution spaces, though here we only look at the case of one variable. We then go on to look at the case where the smooth function we multiply is fixed and examine the invertibility of this operation. This shows another example of how distributions differ from functions. Division by an analytic function is possible in general even when the function has zeros (though these have to be of finite order), the result of division is, however, not uniquely defined. We finish the section on multiplication by very briefly showing how these

3 results make clear and rigorous how to pose linear PDE with smooth coefficients in the space of distributions. We will go on to look at some other spaces of distributions which are found by enlarging the space of test functions. These are the space S 0 of tempered distributions which is defined as the dual of S the space of rapidly decreasing function. This space is particularly useful for Fourier analysis, which we will look at extremely briefly. We also look at the space E0 which are the distributions of compact support. The topology of all these spaces are very similar and we will not reprove all the results that we gave for the space D0. The next section looks at tensor products of distributions which allows us to develop the theory of kernel operators and examines the way in which an element of D0(Ω × Ω) can act as a mapping from D(Ω) to D0(Ω). We also look at the topology on tensor products as developed by Grothendiek in an example of how the theory of distributions fed back into functional analysis. We will demonstrate Schwartz’s Kernels Theorem which gives that all linear mappings form D(Ω) to D0(Ω) can be given in this way. This was proved by Schwartz for example in [11] or [10] using the theory of nuclear locally convex spaces which was again work by Grothendieck when he was Schwartz’s student. This theory is very beautiful but in this section we show how the algebraic, not the topological isomorphism, can be found from a much simpler proof using Fourier transforms. We then look at fundamental kernels and parametrices for differential operators and we relate their regularity properties to the hypo-ellipticity of the operator. We finish by briefly giving some notions of Nuclear spaces including the ‘idea’ of Schwartz’s proof of the kernels theorem and of a very elegant similar proof given in [4]. Nuclear spaces also aided Schwartz in creating the theory of vector valued distributions in [10].

Acknowledgements

I would like to thank Prof. Cl´ement Mouhot for supervising this essay, and also Tim Talbot for helping me understand some of the material on topological vector spaces viewed through their neighbourhoods and particularly for pointing out that not all neighbourhoods are open which is a piece of information without which nothing makes any sense! I also used to a great extent my notes from the part III courses ‘Functional Analysis’ taught by Dr Andras Zsak, and ‘Distribution Theory and Applications’ taught by Dr Anthony Ashton.

2 General deﬁnitions, and methods for Topological Vector Spaces.

As the theory of distributions concerns topological vector spaces, there are several definitions and Theorems which we will use repeatedly throughout so I will list them here. Definition. A Topological Vector Space (TVS) is a set V related to a field F which carries the normal algebraic structure of a vector space with a topology which has the following properties. (1) (x, y) 7→ x + y : V × V → V is continuous. (2) (λ, x) 7→ λx : F × V is continuous. where both V × V and F × V carry the normal product topology. Because of these properties it is possible to recover the whole of the topology of a vector space V by looking at the neighbourhood base around 0. In particular addition and multiplication are separately continuous since λ 7→ (λ, x)

4 for x fixed is continuous and x 7→ (x, y) is continuous for y fixed. Definition. A neighbourhood of x is a set N such that there exists U ∈ τ with

x ∈ U ⊂ N

Definition. A local neighbourhood base of 0 is a collection of sets B s.t. N is a neighbourhood of 0 iff ∃B ∈ B with B ⊂ N. Suppose we are given a collection of sets B in a vector space V then for each B ∈ B construct the set B0 = {x ∈ B|∃A ∈ B, x + A ⊂ B} we hope these will be the open neighbourhoods of 0. Let B0 be the collection of such B0 for each B ∈ B then the set: A = {x + B0|x ∈ V,B0 ∈ B0} we hope will form a base for a topological vector space topology on V provided that the sets in B satisfy some conditions to make them consistent with the axioms of a topological vector space. We will find this conditions but first need to make some definitions. Definition. A set, N, in V is called balanced if for all |λ| ≤ 1 we have that λN ⊂ N. Definition. A set A in V is called absorbing if for every φ there exists a scalar λ such that φ ∈ λA. To find what conditions we need the set B to satisfy we will look at the neighbourhoods of 0 in some Topological Vector Space V . Firstly it is obvious that they all must contain 0. We know that addition is continuous from V × V to V . So if N is an open neighbourhood of 0 in V then there must be some open neighbourhoods of 0 M,L such that M + L ⊂ N. (Since the topology on V ×V has a base of the form M ×L for M,L open in V .) If we take M 0 = M ∩L then M 0 + M 0 ⊂ N. Since every neighbourhood of 0 in V contains an open neighbourhood this means we have the following property:

(A) If N is a neighbourhood of 0 there will exist a neighbourhood of 0, M, such that M + M ⊂ N.

Since, multiplication is continuous given an open neighbourhood of 0, N, we have t > 0 and M a neighbourhood of 0 such that if |λ| < t and x ∈ M we have λx ∈ N. Therefore [ M 0 = λM ⊂ N |λ|

(B) If N is a neighbourhood of 0 then there will exist M ⊂ N such that M is balanced.

We also know that multiplication is continuous V → V when λ is ﬁxed (and supposed not to be 0) so if N is an open neighbourhood of 0 there exists M, an open neighbourhood of 0, 1 such that λ M ⊂ N. Hence M ⊂ λN so we have the following property:

5 (C) N is a neighbourhood of 0 and λ 6= 0 means that λN is a neighbourhood of 0.

We also know that multiplication is continuous if x ∈ V is ﬁxed. So given N an open 1 neighbourhood of 0 there exists t > 0 such that |λ| < t means that λx ∈ N hence x ∈ λ N. So we have property

(D) Every neighbourhood of 0 is absorbing.

Therefore, B is going to be a base of neighbourhoods for we need that collectively the sets that can be generated by arbitrary unions and ﬁnite intersections of elements of B will have the properties above.

Suppose that B satisﬁes the following properties:

(A’) If N ∈ B then there exists M ∈ B such that

M + M ⊂ N

(B’) If N ∈ B there exists B a balanced set such that there is M ∈ B with

M ⊂ B ⊂ N

(C’) If λ is a scalar, N ∈ B then there exists M ∈ B such that M ⊂ λN

(D’) If N ∈ B and x ∈ V then there exists λ such that

x ∈ λN

(E’) If N,N 0 ∈ B then there exists M ∈ B such that

M ⊂ N ∩ N 0

We now claim that A as defined above will define a vector space topology on V . First we show that it does indeed define a topology.

y ∈ (x + A) ∩ (x0 + A0) we have 0 ∈ (x − y + A) ∩ (x0 − y + A0) and since 0 ∈ x − y + A we know that y − x ∈ A so by the deﬁnition of B0 there exists BB0 such that y − x + B ∈ A and similarly we can ﬁnd B0 such that y − x0 + B0 ∈ A0 so if we take C = B ∩ B0 then there is C0 ∈ B0 such that C0 ⊂ C then

y + C0 ⊂ (x + A) ∩ (x0 + A0) which shows that A forms a base for the topology. We would like it to be a topological vector space topology so we need that addition is continuous. This means that if we are given x, y, z ∈ V with

y + z = x.

6 and also we are given A ∈ B0 we would like to ﬁnd B,C ∈ B0 such that

(y + B) + (z + C) ⊂ x + A we know that we have D ∈ B0 such that

D + D ⊂ A then we can take B = C = D to get

(y + B) + (z + C) = x + D + D ⊂ x + A

We also need to show that multiplication is continuous. This means that if we are given x, y ∈ V and λ ∈ F such that λ.y = x and we are also given A ∈ B0 then we would like to ﬁnd > 0 and B ∈ B0 such that if |t| < then (λ + t).(y + B) ⊂ x + A this will be ⇔ λy + ty + (λ + t)B ⊂ x + A ⇔ ty + (λ + t)B ⊂ A we know that we can ﬁnd C a balanced open neighbourhood of 0 such that

C + C ⊂ A then since C is absorbing we can choose such that

ty ∈ C, ∀ |t| < and since C is balanced and λC will be an open neighbourhood of 0 for all non-zero λ we can ﬁnd B such that (λ + t)B ⊂ C ∀ |t| < by letting B be a shrunken C. Consequently we have

(λ + t)(y + B) = x + ty + (λ + t)B ⊂ x + C + C ⊂ x + A

Proposition. If a vector space V carries two topological vector space topologies τ, τ 0 which prescribe the same neighbourhoods of 0 then τ = τ 0. This shows the topology constructed above is the unique one admitting this neighbourhood base so we have an identiﬁcation between topologies for TVS and neighbourhood bases.

Proof. In this case let B be the collection of neighbourhoods of 0 in both τ and τ 0. . Further, suppose U is an open set in τ then if x ∈ U, ∃V ∈ B0 with x + V ⊂ U then since V is a neighbourhood of 0 in τ 0 there will be a V 0 ∈ τ 0 with 0 ∈ V 0,V 0 ⊂ V then x + V 0 ⊂ U so U is open in τ 0 and this argument is symmetric so τ = τ 0.

7 Further, most of the spaces I will use will be locally convex and Hausdorff. (Because of the definition of a TVS any T1 space will be Hausdorff so the Hausdorff condition is necessary that the topology ’sees’ the difference between any two points.) In particular to show that a space is Hausdorff it is sufficient to show that given any x ∈ V there is some N a neighbourhood of 0 with x∈ / N. Definition. A locally convex space is a topological vector space which has a base of convex, balanced neighbourhoods of 0. To each convex, balanced set in the neighbourhood we can associate a semi-norm

µB(x) = inf{λ > 0|x ∈ λB}

This is called the Minkowski functional of the set B. Therefore, we can equivalently deﬁne a locally convex topology by a collection of semi-norms {pi|i ∈ I} where I is some indexing set. This has a base of local neighbourhoods of the form.

V = {x|pi1 (x) < 1, ..., pik (x) < k} where k ∈ N, i1, ..., ik ∈ I, 1, ...k ∈ R. We can see here that we automatically have continuity of addition for any convex, balanced 1 1 possible base as if N is a convex neighbourhood base then 2 N + 2 N ⊂ N. We also have a few helpful little things like for N any convex, balanced neighbourhood of 0 we will have: 1 1 1 1 N − N = N + N ⊂ N, 2 2 2 2 1 1 1 N + N + N ⊂ N, 3 3 3 etc. Consequently, to check a convex, Balanced, collection will form a neighbourhood base we only need that λN is a neighbourhood and that we have the intersection condition. Therefore the topology we described above generated by semi-norms is always a TVS topology. And the topology is defined by any collection of semi-norms whose semi-balls for a sub-local-base. We can also see that a locally convex space is Hausdorff iff \ ker(pi) = {0} i∈I

Definition. A Fréchetspace is a locally convex space which is metrizable with a complete, translation-invariant metric. It will also be useful to have definitions of bounded sets and Cauchy sequences in non-metric spaces. Definition. A subset B of a topological vector space is said to be bounded if for every N a neighbourhood of 0, there exists a scalar λ s.t. B ⊂ λN. So we can see that since every neighbourhood is absorbing that {x} is a bounded set for x ∈ V .

Proposition. In a locally convex space deﬁned by a collection of semi-norms pi, i ∈ I a set B is bounded iﬀ pi(B) is bounded for each i.

8 Proof. Let Ni = {x|pi(x) ≤ 1} and let B be bounded. Then there exists λ with λNi ⊃ B so if x ∈ B then pi(x) ≤ λ. Conversely, suppose that B is a set with x ∈ B ⇒ pi(x) ≤ λi. Then let N be an arbitrary neighbourhood of 0 since the semi-norms deﬁne the topology there is a set

0 N = {x|pi1 (x) ≤ 1, ..., pin (x) ≤ k}

0 0 with N ⊂ N then let µ = max(λi/i, ..., λn/n). Then µN ⊃ B so µN ⊃ B.

Deﬁnition. A sequence (xj) in a topological vector space is Cauchy if for every N a neighbourhood of 0, there exists K ∈ N s.t. if j, k ≥ K, xj − xk ∈ N. Proposition. In a locally convex space all Cauchy sequences are bounded.

Proof. Given a Cauchy sequence {φj} and N a neighbourhood of 0 we wish to show that there exists a µ s.t. φj ∈ µN∀j. As {φj} is Cauchy there exists J s.t. for all j ≥ J

φj − φJ ∈ N

⇒ φj ∈ φJ + N there exists a λ s.t. φJ ∈ λN and since N is convex λN + N ⊂ (λ + 1)N. λ 1 ( This is because λx + y = (λ + 1)( λ+1 x + λ+1 y).

φj ∈ λN + N ⊂ (λ + 1)N

0 0 0 and there exists λ s.t. k ≤ J − 1 ⇒ φk ∈ λ N so set µ = max(λ, λ ). The Hanh-Banach theorem will also be used frequently so we recall it here.

The Hahn-Banach Theorem. If X is a real or complex vector space, p : X → R a positively homogeneous, subadditive funtional. Let Y be a closed subspace of X and f : Y → R a linear map such that |f(y)| ≤ p(y) for all y ∈ Y . Then f extends to f˜ a linear map on X and furthermore |f˜(x)| ≤ p(x), ∀x ∈ X . We usually take p to be a semi-norm defining a locally convex space topology. We also claim that a linear functional u : V → F is continuous with respect to an lcs topology defined by pi, i ∈ I iff there exists a finite collection i1, ..., in and a positive real, C, such that

|u(x)| ≤ C max pi (x). k k

3 The topology of D(Ω)

We will construct a topology on D(Ω) the space of continuous functions of compact support n inside Ω ⊂ R via a limiting process on spaces of the form DK which are deﬁned below. This topology is called an LF-space topology. Also following Schwartz in [3] we will give an explicit collections of semi-norms which deﬁne the same topology.

9 3.1 The spaces DK

If K is a compact subset of Ω the we can define a linear subspace of D(Ω), DK to be the set ∞ of functions whose support is contained inside K. This is not the same as the space Cc (K) = C∞(K) as the functions must be able to be smoothly extended to the whole of Ω. Then we have [ [ D(Ω) = DK = DKn K n n where ∪nKn = Ω. For example in Ω = R we can take Kn = {x | |x| ≤ n}. We can define a Fréchet space topology on each DK via the semi-norms

α pn,K (φ) = Σ|α|≤n sup |∂ φ(x)| x∈K

We now show that this does indeed deﬁne a Fr´echet space topology: (1) DK is metrizable:

Since the semi-norms make DK into a locally convex, Hausdorﬀ space and there are countably many of them we have the standard construction of a metric.

−n pn(ψ − φ) d(φ, ψ) = Σn2 1 + pn(ψ − φ)

(noticing that p1 ≤ p2 ≤ p3 ≤ ....)

Proposition. This is a metric deﬁning the topology on DK

Proof. d(φ, ψ) = 0 iﬀ pn(φ − ψ) = 0, ∀n which is iﬀ φ = ψ. We know wish to show the triangle inequality. This will follow from:

p (a + b) p (a) p (b) n ≤ n + n 1 + pn(a + b) 1 + pn(a) 1 + pn(b) by summing both sides. This in turn follows from

pn(a + b) ≤ pn(a) + pn(b) 1 1 ⇒ ≤ pn(a) + pn(b) pn(a + b) 1 1 ⇒ 1 + ≤ 1 + pn(a) + pn(b) pn(a + b) 1 1 ⇒ ≤ 1 + 1 1 + 1 pn(a+b) pn(a)+pn(b) p (a + b) p (a) p (b) ⇒ n ≤ n + n pn(a + b) + 1 1 + pn(a) + pn(b) 1 + pn(a) + pn(b) p (a + b) p (a) p (b) ⇒ n ≤ n + n 1 + pn(a + b) 1 + pn(a) 1 + pn(b) So this deﬁnes a metric.

10 Now we want to show that this metric deﬁnes an equivalent topology. We do this by showing both topologies have the same neighbourhoods of 0. Look at the set

{φ|d(0, φ) ≤ }.

Then there exists an N such that

−n Σn≥N+12 ≤ /2.

Then there exists λn, n = 1, ..., N such that

−n pn(φ) pn(φ) ≤ λn ⇒ 2 ≤ /2N 1 + pn(φ) then the set {φ|pn(φ) ≤ λn, n = 1, ..., N} is contained in the ball so the epsilon ball is a neighbourhood of 0 in the semi-norm topology. Now since pn(φ) ≤ ⇒ pn−1(φ) ≤ it is suﬃcient to look at the sets {φ|pn(φ) ≤ λ} then there exists such that

−n pn(φ) 2 ≤ ⇒ pn(φ) ≤ λ 1 + pn(φ) so the ball in the metric topology is contained in this set. So this set is a neighbourhood of 0 in the metric topology.

There are two interesting points which can be taken from this proof. The ﬁrst is that the balls are not bounded in the sense of the locally convex space topology as none of the basic open neighbourhoods are bounded for the lcs topology. It might at ﬁrst seem that since B 1 n form a neighbourhood base of 0 that the −balls would be bounded via B = nB 1 but this n metric is not positively homogeneous i.e.

d(0, tφ) 6= t.d(0, φ) so tB 6= Bt. in general. This shows that a metrizable space is not necessarily locally bounded. Secondly, this construction works in any locally convex space which can be defined by a non- decreasing, countable sequence of semi-norms. It is also the case that any locally convex space that can be defined by a countable sequence of semi-norms can be defined by a non-decreasing sequence of semi-norms (by listing these q1, q2, q3, ... then looking at pn(x) = maxk≤n(qn(x))). This shows that any first countable, locally convex space is metrizable.

(2) DK is complete with this metric:

It can be seen that if a sequence φj is d-Cauchy then for each n, pn(φj −φk) → 0 as j, k → ∞. α α Therefore, ∂ φj is a Cauchy sequence with the uniform norm for each α. So ∂ φj → φα

11 α uniformly and φα = ∂ φ0 and φ0 ∈ DK since the boundary values of every partial derivative will be 0. This is the same as saying φj → φ0 in the DK topology. α Proof that φα = ∂ φ0: We do this iteratively by showing it for ﬁrst partial derivatives.

+|φj(x + h1) − φj(x) − ∂x1 φj(x)h1| + |h1||∂x1 φj(x) − φ(1,0,...,0)(x)|

We may as well assume that |h1| ≤ 1 so we can choose j such that φj is uniformly less than

away from φ0 and ∂x1 φj is uniformly less than away from φ(1,0,...,0) so that we have:

φ0(x + h1) − φ0(x) − φ(1,0,...,0)(x)h1| ≤ 3 + |φj(x + h1) − φj(x) − ∂x1 φj(x)h1| and for this j we can choose h1 so that RHS is ≤ 4.

3.2 LF-spaces In order to introduce the topology on D(Ω) it is first helpful to introduce in general a new kind of TVS topology. This will be a generalization of the kind of topology we will place on D(Ω). Definition. An LF-space is also called the strict inductive limit of Fréchetspaces. If we have a sequence F1 ⊂ F2 ⊂ F3... of nested vector spaces each with a Fréchetspace topology. We also ask that the subspace topology S induced by Fn+1 on Fn is the original topology on Fn. Then if we let F = n Fn we can define the LF-space topology on F via a local base of neighbourhoods of zero.

B = {V |0 ∈ V,V is convex, balanced ,V ∩ Fn is a neighborhood of 0 in Fn∀n} This topology is locally convex since the base we have just constructed is of convex, balanced sets. Fn is called a sequence of deﬁnition for F .

Proposition. The subspace topology induced by F on Fn is the original topology on Fn. Proof. Here we want to show this by proving that the basic neighbourhoods in one topology are neighbourhoods in the other. If V is a neighbourhood of 0 in F then V contains a convex, balanced neighbourhood of 0 W ∈ B as B is a neighbourhood base of 0 in F . W ∩ Fn is a neighbourhood of 0 in Fn by the deﬁnition of B. So V ∩ Fn ⊃ W ∩ Fn is a neighbourhood of 0 in the original topology.

Now we want to prove that if W is a neighbourhood of 0 in the original topology on Fn then there exists a V which is a neighbourhood of 0 in F s.t. V ∩ Fn = W . It is suﬃcient to do this for convex neighbourhoods only. As Fn+1 induces the original topology on Fn there is a neighbourhood Wn+1 of 0 in Fn+1 s.t.

Wn+1 ∩ Fn = Wn.

We will show below that Wn+1 can also be taken to be convex. Iteratively we have Wn+k+1 a convex neighbourhood of 0 in Fn+k+1 s.t. Wn+k+1 ∩ Fn+k = Wn+k. Let [ V = Wn n this has the required properties, and it is a neighbourhood of 0 as it is convex and its intersection with each Fn is a neighbourhood of 0 in Fn.

12 Proposition. An LF-space is Hausdorﬀ.

Proof. This follows [4] book fairly closely. This is proved similarly to the second part of the above proof. We want to show that if x ∈ F, x 6= 0 then there exists some En with x ∈ En. Since En is Hausdorﬀ there exists a convex neighbourhood Nn of 0 in En such that x∈ / Nn. We know there is a neighbourhood of 0 Wn+1 in En+1 such that

Wn+1 ∩ En = Nn.

We would like to reduce Wn+1 to a convex Nn+1 such that

x∈ / Nn+1 and Nn+1 ∩ En = Nn. S Then if we take N = n Nn we have that N is convex so by the deﬁnition of an LF-space topology N is a neighbourhood of 0 in F and x∈ / N so F is Hausdorﬀ.

Now we need to show that if Nn is a convex neighbourhood of 0 in En not containing x then there exists a convex neighbourhood of 0 Nn+1 in En+1 not containing x such that Nn+1 ∩ En = Nn. This will complete both the above proofs. Since, En+1 induces the topology on En there exists a neighbourhood M of 0 in En+1 such that M ∩ En = Nn then M contains some convex neighbourhood C with

x∈ / C since En+1 is Hausdorﬀ and locally convex. Then look at the convex hull of C and Nn this contains C so is a neighbourhood in En+1 and contains Nn so its restriction to En contains Nn. Suppose it contains x then there exists y ∈ C and z ∈ Nn such that

x = ty + (1 − t)z which means that ty = x − (1 − t)z ∈ En so y ∈ C ∩ En ⊂ Nn which means that x ∈ Nn as Nn is convex. This gives a contradiction so

x∈ / conv(Nn ∪ C) and letting Nn+1 = conv(Nn ∪ C) will give all the required properties.

13 3.3 D(Ω) is an LF-space S If K1 ⊂ K2 ⊂ K3... is a nested sequence of compact sets s.t. n Kn = Ω then D(Ω) can be made into an LF-space with sequence of deﬁnition DKn , n = 1, 2, 3, .... It can often make more sense to choose Kn = ωn where ωn is a relatively compact, open set. This is because if K is an arbitrary compact set then for any x ∈ K \ int(K) we will have φ(x) = 0 for every φ ∈ DK . Firstly, we notice that the topologies we have deﬁned on each DK space are compatible with each other. That is if K ⊂ K0 then the subspace topology induced on DK by DK0 is the initial topology on DK . This is because

pn,K0 |K = pn,K so the semi-norms deﬁning both topologies are the same.

Proposition. This topology is independent of the choice of Kn. 0 0 0 0 Proof. Suppose K1 ⊂ K2 ⊂ K3 ⊂ ... and K1 ⊂ K2 ⊂ K3 ⊂ ... are two such sequences. Let τ, τ be the corresponding topologies. Suppose N is a convex neighbourhood of 0 in τ then we want 0 to look at N ∩ D 0 . There exists an m such that K ⊃ K so Kn m n

N ∩ D 0 = (N ∩ D ) ∩ D 0 Kn Km Kn and we already know that N ∩ DKm is a neighbourhood of 0 in DKm . We know that DKm induces the right subspace topology on D 0 so N ∩ D 0 is a neighbourhood of 0 in D 0 so Kn Kn Kn N is a neighbourhood of 0 in D(Ω). This argument is symmetric which is suﬃcient for our conclusion.

From above it is clear that this will induce the right topology on DK either by choosing it to be part of the sequence of definition of noting that in any such sequence there must be an n s.t. Kn ⊃ K and DKn will induce the right subspace topology on DK . In this case we can explicitly find neighbourhoods of 0 n D(Ω) which match the basic neigh- 0 bourhoods of 0 in DK . If N = {φ|pn(φ) ≤ C} then N = {φ| ||φ||n ≤ C} is a neighbourhood of 0 α 0 in D(Ω) s.t. N ∩ DK = N. Where ||φ||n = Σ|α|≤n supx∈Ω |∂ φ(x)|. There is an important consequence of the definition of this topology for working out whether a linear functional on D(Ω) is a distribution. Proposition. If E is a locally convex space and u : D(Ω) → E is a linear function. Then u is continuous iff u|DK is continuous for every K. Proof. Suppose U is a convex, balanced neighbourhood of 0 in E then u−1(U) is a convex, balanced set in D(Ω) and −1 −1 u (U) ∩ DK = (u|DK ) (U) so if U is a convex, balanced neighbourhood of 0 u−1(U) is convex so is open iff u|−1 (U) is open DK for every K which gives the result.

14 3.4 Semi-norms deﬁning the topology on D(Ω) The topology on D(Ω) is locally convex so we can deﬁne it by a collection of semi-norms. We start by looking at a neighborhood base which is given explicitely in Schwartz’s book. If

∅ = U1 ⊂ U¯1 ⊂ U2 ⊂ U¯2 ⊂ U3 ⊂ U¯3 ⊂ ... S is a sequence of relatively compact open sets s.t. n Un = Ω, and m1 < m2 < m3 < ... is a strictly increasing sequences of integers, 1 ≥ 2 ≥ 3 ≥ ... a sequence of real numbers decreasing to 0. Then

α V (U, m, ) = {φ | Σ|α|≤mi sup |∂ φ(x)| ≤ i for each n = 1, 2, 3, ...} x/∈Ui letting m, range over all possible such sequences gives a neighbourhood base of 0. Proposition. These two deﬁntions give the same topology on D(Ω). Proof. Here we use the fact that if the neighborhood base of 0 in each topology are neighbourhoods in the other topology then the two topologies are the same. Let Kn = U¯n. α V (U, m, ) ∩ DK = {φ | Σ|α|≤mi sup |∂ φ(x)| ≤ i, i = 1, 2, ..., n} x/∈Ui so this is a neighbourhood of 0 in DKn as it contains the set

{φ|pmn (φ) ≤ n} ∩ {φ|pmn−1 (φ) ≤ n−1} ∩ ... ∩ {φ|pm0 (φ) ≤ 0}.

Now suppose that V ∈ B then for each Kn,V ∩ DKn is a neighbourhood of 0 in DKn so contains a set of the form {φ|pmn (φ) ≤ n} as the semi-norms are non-decreasing. Therefore V (U, m, ) ⊂ V so these two topologies coincide. This topology can then be deﬁned by the semi-norms

α N(m, )(φ) = sup( sup |∂ φ(x)|/n). n |α|≤mn,x/∈Un

4 Properties of the Space D(Ω)

Here we will explore some results relating to compactness, completeness, separability and metrizability of the space D(Ω). These will have several implications fro the topological properties of the dual.

4.1 Completeness of D(Ω) We will see later that D(Ω) is not metrizable. However, we can still define a notion of topological completeness. Since D(Ω) is not first countable it is not sufficient just to look at sequences. This also gives an opportunity to introduce filters which generalize the notion of sequences to non- metric spaces and will characterize continuity and completeness in this spaces in the same way sequences do for metric spaces. Definition. A filter on a set E is a family F of subsets of E. (1) ∅ ∈ F (2) A, B ∈ F ⇒ A ∩ B ∈ F (3) A ∈ F,B ⊃ A ⇒ B ∈ F.

15 Definition. A filter F converges to a point x if for every neighbourhood N of x, ∃A ∈ F s.t. A ⊂ N i.e. N ⊂ F. Lemma. In a Hausdorff TVS limits are unique. Proof. Let F be a filter and suppose F → x and F → y. Then since the space is Hausdorff there exists U, V neighbourhoods of x and y respectively such that U ∩ V is disjoint. However since F → x, U ∈ F and since F → y, V ∈ F so U ∩ V ∈ F which is a contradiction.

We can relate this to the concept of sequences. To every sequence x1, x2, x3, ... ⊂ E the associated filter F is the subsets of E which contain all but finitely many elements of the sequence. A sequence converges iff the associated filter converges. Definition. A Cauchy filter is a filter F s.t. if N is a neighbourhood of 0 there exists A ∈ F s.t. A − A ⊂ N. A TVS is said to be complete if every Cauchy filter converges. A metrizable TVS is topologically complete then it is complete in the old sense. Definition. B is a base for a filter F if ∀A, B ∈ B there is some C in B such that C ⊂ A ∩ B and then F is the collection of sets N such that there is some B ∈ B with B ⊂ N. Theorem. Every LF-space, F , is complete. Proof. This follows [4] reasonably closely. Let F be a Cauchy filter and U be the filter of all neighbourhoods of 0. Let F 0 be the smallest filter all the sets of the form U + V,U ∈ U,V ∈ F. This is a base for a filter since

(U + V ) ∩ (U 0 + V 0) ⊃ (U ∩ U 0) + (V ∩ V 0) .

Suppose F 0 converges to x then we claim that F converges to x. This is because if F 0 converges to x, then for every N a neighbourhood of x there exists B ∈ F 0 with B ⊂ N then we have some U a neighbourhood of 0 and V ∈ F with U + V ⊂ B. So we have V ⊂ B which means that V ⊂ N so F converges to x. Also, F 0 is a Cauchy ﬁlter. If N is a neighbourhood of 0 then there exists M a balanced neighbourhood of 0 such that +M +M +M ⊂ N. Then there exists A ∈ F such that A−A ⊂ M then we will have that

(A + M) − (A + M) = A − A + M − M = (A − A) + M + M ⊂ N

0 0 Then let Fn = {U ∈ En | U = V ∩ En, for some V ∈ F }. We claim that if for every V ∈ F that V ∩ En is non-empty then Fn is a Cauchy filter on En. This is because Fn will satisfy all the axioms of a filter except, possibly, that it may contain ∅, this shows it is a filter. If N is a neighbourhood of 0 in En then it contains a convex, balanced neighbourhood M. Then there 0 0 0 exists a neighbourhood of 0 M in E such that M ∩ En = M. Then since F is a Cauchy filter there will be some A ∈ F 0 such that A − A ⊂ M 0 then

A ∩ En − A ∩ En ⊂ M which shows that Fn is a Cauchy ﬁlter.

Next we show that there is some n such that Fn does not contain ∅. Suppose that for every 0 n there is a set Un + Vn ∈ F where Un is a convex neighbourhood of 0 and V ∈ F such that

(Un + Vn) ∩ En = ∅

16 then let [ [ U = Un,V = Vn n n

. Since Un is open for each n this means U will be open so

U + V ∈ F 0 and [ U + V = (x + U) x∈V so is open. Since the subspace topology induced by E on EN is the original topology then

(U + V ) ∩ EN is open in EN . However, [ [ (U + V ) ∩ E1 = (Un + Vn) ∩ E1 ⊂ (Un + Vn) ∩ En = ∅ n n which is a contradiction to (U + V ) ∩ E1 being open.

Consequently, there exists an N such that FN is a Cauchy filter. We want to show that FN converges in En. Since En is metrizable we have a countable neighbourhood base of 0 in 1 terms of the balls of radius around 0. Call these B 1 . Since FN is a Cauchy filter there exists n n An ∈ FN such that An − An ⊂ B 1 . Then for each n we can choose n \ xn ∈ Ak 1≤k≤n so that xn will be a Cauchy sequence. Since En is a Fréchet space it is complete so xn converges to some element x. 1 We then claim that FN converges to x. Let Mn be the ball of radius n around x. Then 1 there exists some K so that m ≥ K means that d(xm, x) ≤ 2n and let m be possibly larger so 0 that Am − Am ⊂ B1/2n then xm ∈ Am so Am ⊂ Mn. It follows from this that F converges to x since for each M a neighbourhood of x in E there is some

Mn ⊂ M ∩ EN

0 and we can find an A ∈ FN with A ⊂ Mn and we claim that we can choose C ∈ F such that C ⊂ M and C ∩ EN = A. 0 0 We find this C by iterating upwards. So we find a A ∈ EN+1 and A ∩ EN = A and 0 A ⊂ M ∩ EN+1 by noticing that FN+1 also doesn’t contain the empty set. Then we can move upward and take the union over all these sets to get C. Since, at the beginning we noticed that if F 0 converges to x then so does F we have shown that F converges to x and therefore that E is complete. We also proved here that any metric space in which every Cauchy sequence converges will also have that any Cauchy filter converges. In the specific case of D(Ω) we also have a quicker way of seeing that all Cauchy sequences converge. We recall that in a locally convex space Cauchy sequences are bounded.

17 Proposition. In D(Ω) if B is a bounded set then there exists a K with B ⊂ DK .

Proof. This follows [2]. Suppose E ⊂ D(Ω) is not in any DK . Then give our sequence of compacts Kn increasing to Ω there must be an xn, φn s.t. xn ∈ Kn \ Kn−1 and φn ∈ E with φn(xn) 6= 0. Then let 1 N = {φ| |φ(x )| < |φ (x )|} n n n n since only ﬁnitely many of the xn are in each Kn N is in B. However φn ∈/ nN for each n so there is not λ with E ⊂ λN. Proposition. In D(Ω) every Cauchy sequence converges. Proof. This is a result of the previous proposition, that Cauchy sequences are bounded and the fact that DK is complete. So since any Cauchy sequence is bounded it will be contained inside some DK and therefore will converge as Cauchy sequences converge in DK .

4.2 Non-metrizability of D(Ω) For this section it is useful to have the following deﬁnition.

Deﬁnition. A topological space X is a Baire space if for every countable sequence Fn of closed S sets with n Fn = X there exists N with intFN 6= ∅. With this deﬁnition the Baire Category theorem states that a complete, metric space is a Baire space. Proposition. D(Ω) is not a Baire space.

0 Proof. DK is nowhere dense since if V ⊂ DK and K ∩ K then V ∩ DK0 = ∅ so V cannot be a neighbourhood of 0. DK is a linear subspace of D(Ω) has empty interior. As for suitably chosen S sequence Kn, D(Ω) = n DKn then D(Ω) is not a Baire space. It is a consequence of this that D(Ω) is not completely metrizable. Since D(Ω) is topologically complete if it were metrizable it would be completely metrizable this shows that D(Ω) is not metrizable. Furthermore, as D(Ω) is locally convex, if it were ﬁrst countable then it would be metrizable. Therefore, the space D(Ω) is not ﬁrst countable. Similarly, we can work from the fact that any topological vector space has a base of neighbourhoods which are balanced. Proposition. If X is a Topological Vector Space with a balanced, countable neighbourhood base Vn in which every Cauchy sequence converges then X is a Baire space. T Proof. We prove that if Un is a countable sequence of open dense sets then Un in X. Let T n W be any open set in X we will prove that ( n Vn) ∩ W 6= ∅. Since V1 is open and dense there exists x1 ∈ W ∩ V1 and W1 and open neighbourhood of 0 such that x1 + W1 ⊂ V1 ∩ W and W1 ⊂ V1. Then (x1 + W1) ∩ U2 has non-empty interior so there exists x2 and W2 ⊂ V2 such that x2 + W2 ⊂ (x1 + W1) ∩ U2. Itteratively, we can produce the sequence x1, x2, x3, ... and W1,W2,W3, ... such that n ! \ xn + Wn ⊂ Un ∩ W k=1

We then claim that x1, x2, x3, ... is a Cauchy sequence. Then let m ≥ n

xm ∈ xn + Wn ⇒ xm − xn ∈ Wn

18 so this is a Cauchy sequence. Therefore there is some x with xn → x so x ∈ xn + Wn for each T n therefore x ∈ ( n Un) ∩ W which gives the result.

4.3 D(Ω) is a Montel Spaces Definition. A Montel space is a topological vector space which is locally convex, Hausdorff and has the Heine-Borel property. (That every closed bounded set is compact.) We wish to show that D(Ω) is a Montel space. Since we know already that every bounded set in D(Ω) is contained inside some DK then if we can show that each DK is a Montel space we will have the result. In order to do this we first recall the Arzèla-AscoliTheorem.

Arz`ela-AscoliTheorem. If K is a compact, Hausdorﬀ topological space. Let C(K) be the Banach space of continuous functions K → R with the supremum norm. If E is a uniformly bounded, equicontinuous set in C(K) then it is relatively compact.

Proposition. DK is a Montel space.

Proof. As DK is metrizable we have compactness iﬀ sequential compactness (for its subsets). Consequently, it is suﬃcient to prove that a bounded sequence in DK has a convergent subsequence. Given this let {φj} be a bounded sequence in DK then φj is uniformly bounded in each α of its partial derivatives which shows that {φj} is equicontinuous. So for each α, {∂ φj} is a uniformly bounded, equicontinuous set and therefore has a convergent subsequence. If we order our set of multi-indexes α(1), α(2), α(3), ... and the α(1)th partial derivatives of this subsequence is a uniformly bounded, equicontinuous set so has a subsequence which converges (2) to some continuous function φα(1) , then we look at the α partial derivatives of this further subsequence and etc. α Then we can form a cantor diagonal sequence φjk s.t. ∂ φjk → φα where φα is a continuous α function which is 0 on the boundary of K. It follows that ∂ φ0 = φα so φ0 is in DK and

φjk → φ0 uniformly in each of its derivatives.

This also shows that D(Ω) is sequentially compact since any bounded sequence will be in some DK which is sequentially compact. It is interesting to notice that the only Banach spaces which are Montel spaces are ﬁnite dimensional which is further proof that the topology on DK is not normable. This is because if a vector space has a bounded, compact neighbourhood of 0 then it must be ﬁnite dimensional.

4.4 Separability We are going to show that D(Ω) is separable. First we need a Lemma.

Lemma. If X is a separable metric space and Y is a subspace then Y is separable.

Proof. Since X is separable it has countable dense subset {xn} and so there exists a subset

{xnj } such that {xnj + B1} covers Y and all the xnj are a distance less than 1 away from Y .

Then we can choose a ynj ∈ Y ∩ (xnj + B1). Then if we repeat this procedure but replacing 1 1 1 with 2 , 3 , ... we will produce a countable dense sequence in Y . Proposition. The space D(Ω) is separable.

19 ∞ Proof. First we prove that the space C (K) (with the same semi-norms as for DK ) is separable ∞ and since DK is a subspace of this and C (K) is a Fr´echet space this will show that DK is separable.

Let T be the map T (φ) = ∂x1 ∂x2 ...∂xn φ. Since K is bounded if

||ψ − T (φ)||∞ ≤ then we can integrate ψ to a function Ψ such that if

max{α1, ..., αn} ≤ 1 then |∂α(Ψ − φ)(x)| ≤ .diam(K).

Fix some φ in C∞(K) we want to show it can be approximated by polynomials. By the n above argument and if we can approximate T φ by a polynomial we can choose pn such that for all α with max{α1, ..., αn} ≤ n then 1 |∂α(p − φ)(x)| ≤ . n n ∞ Then the sequence pn will tend to φ in the topology on C (K). Since the polynomials with rational coeﬃcients are dense in the polynomials and since the Stone-Weierstrass theorem gives that any continuous function can be uniformly approximated by polynomials we have that DK is separable. Since D(Ω) is the union of countably many spaces of the form DK we have that D(Ω) is separable.

5 The space D0(Ω)

We deﬁne this as the analytic dual of D(Ω) the space of continuous linear forms on D(Ω) In this section we will describe two topologies on the space D0(Ω) and discuss the properties of these topologies including issues around compactness, metrizability, separability and completeness. This is the space of distributions. As a result of our criterion for a function from an LF-space to annother LCS to be continuous we have the following criteria for a linear form u to be a distribution. Proposition. A linear functional u on D(Ω) is a distribution iﬀ α (1) For each K there exists a C and an N s.t. |u(φ)| ≤ CΣ|α|≤N supx∈K |∂ φ(x)| (2) For each sequence φj → 0, u(φj) → 0.

Proof. These are both results of the fact that u is continuous iﬀ its restriction to each DK is continuous. As D(Ω) is not metric (2) is not just a result of translation being continuous however since every convergent sequence is in some DK it does hold. The space of distributions does indeed generalise the notion of function. We can identify the 1 0 L functions with a subset of D (Ω) via f ↔ uf where Z uf (φ) = f(x)φ(x)dx. Ω

Similarly, measures can be identiﬁed with distributions via µ ↔ uµ where Z uµ(φ) = φ(x)µ(dx). Ω

20 5.1 Positive distributions Before exploring the topology we look at an interessting way in which D0 is not like a function space. A distribution u, is positive if for every φ ≥ 0, u(φ) ≥ 0. A difference between function spaces and spaces of distributions is that distributions cannot in general be written as the difference of two positive distributions so it is not possible to build up any theory by looking first at positive distributions (for instance like is done for Lebesgue integration). This is because of the following result. First we need the Riesz-Representation Theorem. Riesz Representation Theorem. If K is a compact, Hausdorff topological space then the dual to the Banach space (C(K), || ||∞) is the space of Borel measures on K with the total variation norm.

Proposition. Every positive distribution can be identiﬁed with a Borel Measure. Proof. Suppose u is a positive measure. d For K ⊂ R a compact let ||φ||∞,K = supK |φ|.

Suppose φ ∈ D then ﬁx K compact, ρK ∈ D s.t. ρ = 1 on K. Then

||φ||∞,K ρ − φ ≥ 0

⇒ u(||φ||∞,K ρ − φ) ≥ 0

⇒ ||φ||∞,K u(ρ) ≥ u(φ) so u is continuous with respect to the uniform norm on K and so can be extended continuously to a linear form on C(K).

So it is a consequence of this and Riesz representation that u|DK is a measure on K call it µ|K . 0 Its clear that if K ⊂ K then µK |K0 = µK0 .

Given φ ∈ D∃K s.t. φ ∈ DK then let µ(φ) = µK (φ) this is independent of the choice of K, and so as we can approximate the indicator functions of compact sets in Rd by functions in D, µ deﬁnes a regular Borel measure (not necessarilly ﬁnite) on Rd.

5.2 Topologies on D0(Ω) We will introduce two topologies on D0(Ω). Defintion. The weak-* topology on D0(Ω) is the topology of point-wise convergence it is the weakest topology which makes all the evaluation maps continuous. It is a locally convex topology defined by the semi-norms pφ(u) = |u(φ)| φ ∈ D(Ω) Definition. The strong dual topology is the topology of uniform convergence on the bounded subsets of D(Ω). It is a locally convex topology defined by the semi-norms.

pB(u) = sup |u(φ)| φ∈B

where B ranges over the bounded subsets of D(Ω).

21 5.3 Non-metrizability As we have already seen the strong topology is metrizable iff it is first countable. So we would like to investigate whether the strong dual topology can be defined by a countable collection of semi-norms. If A and B are bounded sets in D(Ω) and A ⊂ B then

pA ≤ pB so pA is continuous with respect to a topology which is generated by a set of semi-norms which include pB. Since every bounded subset of D(Ω) is contained in a bounded set of the form

B = {φ | pn(φ) ≤ Mn}.

We can take our sequence of semi-norms to be associated with bounded sets of the above form. These sets are closed, balanced and convex. Furthermore if we have a countable collection of bounded sets B1,B2,B3, ... we can add countably more sets so that this sequence is closed under finite unions and multiplication by an integer. This will mean that if T : D0 → R is continuous with respect to the topology then |T (u)| ≤ pB(u) for some B. This means that φˆ is continuous with respect to the strong topology since {φ} is bounded. Claim: If we have a sequence as defined above an element of D then φˆ is continuous with re- ˆ ˆ spect to the semi-norms pB1 , pB2 , pB3 , ... only if φ ∈ Bi for some i. We know that φ is continuous with respect to the topology induced by the semi-norms if there exists an i with |u(φ)| ≤ pBi (u) 0 for every u ∈ D (Ω). If φ∈ / Bi then there exists n such that pn(φ) > supψ∈B pn(ψ). So there exists an x0 such that α |Σα≤n∂ φ(x0)| > sup pn(ψ) ψ∈B so the element u with α u(φ) = Σα≤n∂ φ(x0) ˆ has that u(φ) > pBi (φ). Therefore φ is continuous only if there exists an i with φ ∈ Bi. Since {φ} is a bounded set this means that the strong dual topology is defined by the semi- norms pB1 , pB2 , pB3 , ... only if for every φ there is some Bi with φ ∈ Bi. We show below that this is not possible.

Proposition. Let K have non-empty interior. If B1,B2, ... is a collection of bounded sets in S some DK then there exists φ ∈ DK which is not it n Bn

Proof. By possibly making the Bn bigger we may as well take them to be sets of the form

Bj = {φ | pn(φ) ≤ Mn,j}

Then let aj = Mj,j +1. So if φ is an element of DK with pj(φ) ≥ aj then φ∈ / Bj for any j so it is not in the union. So it remains to construct such a function. Since K has non-empty interior we can ﬁnd U1,U2,U3, ... a countable sequence of open balls inside K and there exists a sequence Vn 1 of open balls such that for each n we have diam(Vn) = 2 diam(Un) and we have φn ∈ DK such that φn is identically 1 on Vn and identically 0 outside Un. Then we can produce the function

22 φn(x) sin(anx1) and patch these together to get a function with the required properties. This is because for x ∈ int(Vn) we have

∂α(φ (x) sin(a x )) = ∂α1 sin(a x ) n n 1 x1 n 1 so as without loss of generality we can make an ≥ 1 and large enough so that the sine function does a full wiggle inside Vn then n pn(φn) ≥ an ≥ an.

Since for any φ we have that {φ} is a bounded set and that if B is bounded in D(Ω) then B ∩ DK will be bounded the above proposition shows that the strong dual topology is not first countable. Now we will show that the weak-* topology is not first countable. This is equivalent to saying that it cannot be defined by a countable set of semi-norms of the form pφ for φ ∈ D. For this we need a lemma. ˜ ˜ T Lemma. If f, f1, ..., fn are linear functionals on a vector space V . and ker((f)) ⊃ k≤n ker(fk) ˜ then f ∈ span(f1, ..., fn). n ˜ Proof. Let T : V → scalars be the map x 7→ (f1(x), ..., fn(x)). Then ker(T ) ⊂ ker(f) so there ˜ exist a map g such that f = g ◦ T and g will be of the form g(y) = a1y1 + ... + anyn which means ˜ that f(x) = a1f1(x) + ... + anfn(x).

0 Then suppose that the weak-* topology on D is deﬁned by the semi-norms pφ1 , pφ2 , pφ3 , ... then ﬁnd some other φ we know that φˆ is weak-* continuous. Which means the set

V = {u | |u(φ)| < } contains a set of the form

W = {u | |u(φ1)| < 1, ..., |u(φn)| < n} T ˆ which means that U = k≤n ker(φk) ⊂ V and u ∈ U ⇒ λu ∈ U which means that λu(φ) < for each λ so u(φ) = 0 for each u ∈ U therefore ker(φˆ) ⊃ U which by the lemma above means ˆ ˆ ˆ that φ ∈ span(φ1, ..., φn) which means that

φ ∈ span(φ1, ..., φn) which in turn implies that D(Ω) is countable dimensional. This is a contradiction because if D(Ω) is countable dimensional then so is DK which would mean that DK was the union of countably many ﬁnite dimesional subspace.

span{e1} ⊂ span{e1, e2} ⊂ span{e1, e2, e3} ⊂ ...

These subspaces are closed (as they are ﬁnite dimensional) and nowhere dense since any open neighbourhood of 0 is absorbing. So the weak-* topology is not metrizable.

23 5.4 These Spaces are Hausdorff Now we wish to show that both these topologies are Hausdorff.Since every weak neighbourhood of 0 is a strong neighbourhood of 0 it is sufficient to show that for every u 6= 0 ∈ D0(Ω) there + is some φ ∈ D(Ω) such that u(φ) > 0 then fix 0 < α < u(φ) and the sets Uα = {v | v(φ) > α} − and Uα = {v | v(φ) < α} are open neighbourhoods of u and 0 respectively which are disjoint. 0 Therefore D (Ω1) is Hausdorff for both the strong and weak-* topologies. If u and v are not equal then we can find N,M open neighbourhoods of 0 such that N ∩ (u − v + M) = ∅ then u + M and v + N are disjoint open sets which separate u and v.

5.5 The Banach-Steinhaus Theorem & Barrelled Spaces The Banach-Steinhaus theorem says that if a collection of linear functional on a Banach space X are pointwise (weak-*) bounded then they are uniformly bounded (bounded in the strong dual topology). We wish to extend this result to the spaces D, D0. First we will need the notion of a barrel and a barrelled space. Definition. A set B in a TVS. E is a barrel if it is closed, convex, balanced and absorbing. (Absorbing means if x ∈ E∃λ s.t. x ∈ λB.) Definition. A Barrelled space is a TVS in which every barrel is a neighbourhood of 0. A Fréchet space, F , is barrelled as a consequence of the Baire Category theorem. If B is a barrel then F = S nB so by BCT nB has non-empty interior for some n so B has non- n∈N empty interior. Since B is convex and balanced if N ⊂ B is open then conv(N ∪ (−N)) is a neighbourhood of 0 contained in B. An LF-space, F , with sequence of definition Fn is barrelled since if B is a barrel in F , then B ∩ Fn is a barrel so a neighbourhood of 0 in Fn so since B is convex it is a neighbourhood of 0 in F . Theorem. If a set B is weak-* bounded in D0(Ω) then it is strongly bounded. Proof. First, we show that if B is weak-* bounded then it is equicontinuous at 0. Let > 0. T −1 Then let W = u∈B u (B). We want to show that W is a neighbourhood of 0 in D(Ω). Clearly, W is convex, balanced and closed. We wish to show it is absorbing. Fix φ ∈ D(Ω). Then we have a constant C with: |u(φ)| ≤ C ∀ u ∈ B

Since B is bounded so contained in some DK for some K and so for φ ∈ B we will have α |u(φ)| ≤ CΣ|α|≤n|∂ φ| which must be bounded on B so there is a λ s.t. |u(φ)| ≤ λ ∀ u ∈ B and since u is linear −1 −1 u (λB) = λu (B) ⊃ λW so φ ∈ λW and since D(Ω) is barrelled W is a neighbourhood of 0. Now we show that since B s equicontinuous it is strongly bounded. Let A be a bounded set in D(Ω). Then take , W as before. W is a neighbourhood of 0 in D(Ω) so there exists µ s.t. A ⊂ µW so u(A) ⊂ µB for each u ∈ B so B is strongly bounded.

24 It is a consequence of this that if ui is a sequence with ui(φ) → vφ 6= ±∞. Then the linear map u deﬁned by u(φ) = vφ is continuous. This is because the weak-* closure of the sequence is equicontinuous as was shown in the above proof so u will be continuous. Since the strong topology is stronger this shows that the bounded sets are the same for both topologies.

5.6 The Banach-Alaoglu-Bourbaki Theorem In normed spaces the Banach-Alaoglu theorem says that the unit ball of the dual is weak-* compact. We hope to extend this to a form which will be useful in the space of distributions. First we state Tychonoﬀ’s theorem which we will use.

Tychonoﬀ’s Theorem. If Xi, i ∈ I is a collection of compact spaces with some indexing set I Q then i∈I Xi is compact with the product topology. We introduce the notion of a polar. Deﬁnition. Suppose E is a topological vector space and E0 is its analytic dual. Then if A ⊂ E the polar of A is A˚ = { u ∈ E0 | sup |u(φ)| ≤ 1} φ∈A Using this we have

Banach-Alaoglu-Bourbaki. If E is a topological vector space and E0 its dual and if U is a neighbourhood of 0 in E we have that U˚ is weak-* compact. Proof. Fix x ∈ E then there is λ such that x ∈ λU or equivalently 1 x ∈ U λ consequently if f ∈ U˚ we will have 1 |f x | ≤ 1 λ so that |f(x)| ≤ |λ| We notice also that U˚ is weak-* closed since if u∈ / U˚ then there exists a φ ∈ U such that u(φ) > 1 so the set [ V = { u | |u(φ)| > 1} φ∈U is weak-* open since it is the union of weak-* open sets. This set is also the complement of U˚ which means that U˚ is weak-* closed. It is also balanced since if u ∈ U˚ and |λ| ≤ 1 then we have

|λu(φ)| = |λ||u(φ)| ≤ |λ| ≤ 1 for each φ ∈ U which means that λu ∈ U.˚ The polar is convex since if u and v are in U˚ and 0 ≤ t ≤ 1 and φ ∈ U we will have

|tu(φ) + (1 − t)v(φ)| ≤ t|u(φ)| + (1 − t)|v(φ)| ≤ t + (1 − t) = 1

25 which means that tu + (1 − t)v ∈ U.˚ Let Mφ = supu∈B |u(φ)|. Then let Fφ = [−Mφ,Mφ]. Then let Y F = Fφ φ∈D(Ω) with the product topology. Let πφ be the co-ordinate projections. F is compact by Tychonoﬀ’s theorem. Now we construct a map θ : B → F .

θ(u) = (u(φ))φ∈D(Ω)

Now we need to prove that θ is a homeomorphism onto its image and that its image is closed in F . −1 ˆ θ ◦ φ = πφ on the image of θ so θ−1 is weak-* continuous. ˆ θ ◦ πφ = φ so θ is continuous. The image of θ is the subspace of F where fλφ+µψ = λfφ + µfpsi \ Im(θ) = ker(πλφ+µψ − λπφ − µπψ) φ,ψ,λ,µ which is closed. Theorem. Any weak-* bounded, weak-* closed set in the dual space of a barrelled TVS is weak-* compact. Proof. We use the theorem above and show that any weak-* bounded, weak-* closed set is contained in the polar of a neighbourhood of 0 in the space E. Suppose B is such a weak-* bounded, weak-* closed set then let B˜ be formed by

B˜ = { x ∈ E | |u(φ)| ≤ 1, ∀ u ∈ B}

This space is weakly closed in E so is closed in E. It is also balanced and convex by similar arguments as we used to show U˚ is balanced and convex. Also if we ﬁx x ∈ E then since B is weak-* bounded there is some λ such that

sup |u(x)| ≤ λ u∈B which means that 1 x ∈ B˜ λ so that B˜ is absorbing and therefore a barrel. Since our TVS is barrelled this means B˜ is a neighbourhood of 0 and B is contained in the polar of B˜.

26 5.7 Sequential Compactness 0 Proposition. A weak-* bounded sequence un in D (Ω) has a convergent subsequence. For this we need some Lemmas. Lemma 1. Let K be the closure of a simply connected domain. If u ∈ D0 with u(φ) ≤ α 1 CΣα≤N supK |∂ φ| for each φ ∈ DK then there exists a C function f and a multi-index α such that for every φ ∈ DK |α| α u(φ) = (−1) uf (∂ φ) and α is of the form α1 = α2 = α2 = ... = αn = N + 3

Proof. This follows [2]. We look at the map T = ∂x1 ∂x2 ...∂xn and want to show that it is 0 invertible on DK . Lets look at the map S = ∂x1 ∂x2 ...∂xn on DK . Then for φ ∈ DK since φ is 0 on the boundary of K we have: Z φ(y) = S(φ)(x)dx {xi≤yi,i=1,...,n} This shows that S is injective which implies that Sn is injective. It also shows that |φ(y)| is bounded by the L1 norm of Sφ. The mean value theorem gives that

|φ| ≤ diam(K) max ||(∂x φ)||∞ i i which implies that N 0 N+1 ||φ||N ≤ C||S φ||∞ ≤ C ||S φ||L1 So we have 00 N+1 |u(φ)| ≤ C ||φ||N ≤ C3||S φ||L1 since SN+1 is injective we can deﬁne a map u0 on its image so that

u = u0 ◦ SN+1 therefore 0 N+1 |u (ψ)| ≤ C3||ψ||L1 ψ ∈ Im(S ) 00 Using the Hahn-Banach theorem we can extend this to u a functional on DK such that for all φ ∈ DK we have that 00 |u (φ)| ≤ C3||φ||L1 So u00 ∈ (L1(K))0 which can be identiﬁed with L∞(K). This gives us a function g ∈ L∞(K) which we can then integrate twice normally in all the derivatives to produce the required function 2 f. Since K is bounded we will have ||f||∞ ≤ (diam(K)) ||g||∞. This also shows that the supremum bound of f is bounded in terms of C00 and properties of K.

0 Lemma 2. If uj is a pointwise bounded sequence in D (Ω) then there is an N such that for α each j there exists a Cj such that uj(φ) ≤ CjΣα≤N supK |∂ φ|. Proof. Suppose that this is false. Then we can extract a subsequence such that

α un(φn) > nΣ|α|≤n sup |∂ φn| K and possibly by rescaling we can let pn(φn) = 1. This will give that φn is a bounded sequence in 0 DK . However the Banach-Steinhaus theorem says that a weak-* bounded set in D is bounded on a bounded set in D and {un} is not bounded on {φn} which is a contradiction.

27 0 Proof. Proof of the proposition: Suppose uj is a bounded sequence in D (Ω) and K some compact |α| α subset of Ω then by lemma 1 there exists fj for each j such that uj(φ) = (−1) ufj (∂ φ) and by lemma 2 this α is the same for each j.

Now we want to show that fj is a uniformly bounded, equicontinuous set when restricted to K. We have: Z f(x)(−1)|α|∂αφ(x)dx = u(φ) K for every φ ∈ DK We would like to show that fj are uniformly bounded and equicontinuous by showing that fj and their order one partial derivatives are uniformly bounded. The Banach-Steinhaus theorem gives that uj will be bounded uniformly on each bounded subset of DK which shows that the Cj in lemma 2 can all be taken to be the same. We saw in the proof of lemma 1 that the supremum bound of f and its derivatives only depends on this C and properties of K which shows that the fj are uniformly bounded and equicontinuous since they are also uniformly bounded in each of the order one partial derivatives.

Therefore we can apply Arz`ela-Ascolito the sequence fj to ﬁnd a sub-sequence fjk which converges uniformly to a function f. So the sequence ujk |DK converges to the linear form on DK α α N+3 induced by (−1) ∂ f since uj = uf ◦ T . We can do this for each DKn for some sequence of compacts Kn increasing to Ω, by extracting successive subsequences. We can then form a

Cantor diagonal sequence and since every φ is contained in some DKn this will converge in the weak-* sense.

This shows that all Cauchy sequences converge in D0 since we know that Cauchy sequences are bounded and so by the above proposition they will have a convergent subsequence. It is also the case that if a subsequence of a Cauchy sequence converges to a point u then the whole sequence must, therefore in D0 with the weak-* topology all Cauchy sequences converge.

5.8 More Montel Spaces & Equivalence of the Topologies on Bounded sets We recall that the bounded sets are the same for both topologies. In this section we will prove that D0(Ω) has the property that the topologies restricted to bounded sets are the same and are metrizable when restricted to these sets. We also show that the bounded sets are compact. First we prove a lemma. Lemma. If E is an equicontinuous set in D0(Ω). Then the topology of bounded convergence and the weak-* topology are the same on E.

Proof. Fix A closed and bounded in D(Ω) we have already proved that A is weak-* compact thanks to Banach-Alaoglu-Bourbaki. Fix > 0 we wish to ﬁnd a neighbourhood of 0 in E s.t. supφ∈A |u(φ)| ≤ for all u ∈ E. This will show that PA|E is weak-* continuous. As E is equicontinuous there is a weak-* neighbourhood of 0 U ∈ D s.t. |u(φ)| ≤ for all u ∈ E, φ ∈ U. Then since A is compact we have φ1, φ2, φ3, ..., φn ∈ A s.t. φ1+U, φ2+U, ..., φn+U cover A. For each i = 1, ..., n we have Wi weak-* neighbourhood of 0 s.t.

u ∈ Wi ⇒ |u(φi)| ≤

28 T let W = Wi this is also a weak-* neighbourhood of 0. Then for u ∈ W

sup |u(φ)| = max sup |u(φi + ψ)| φ∈K i≤n ψ∈U

≤ max |u(φi)| + sup |u(ψ)| ≤ 2 i≤n ψ∈U

Then, since every bounded set is contained in a closed, bounded set pB is continuous for every B bounded. This shows the strong dual topology is weaker that the weak-* topology on E. As {φ} is bounded for each φ we already know that the weak-* topology is weaker than the strong dual topology so they must be the same.

From this we can show that D0(Ω) is a Montel space. Proposition. D0(Ω) is a Montel space with the strong dual topology.

Proof. We have already that D0 is locally convex and Hausdorﬀ. We would like to exploit the fact that we know that weak-* bounded, weak-* closed sets are weak-* compact after Banach- Alaoglu-Bourbaki. As the strong dual topology is stronger than the weak-* topology we have that B strongly bounded ⇒ B¯w∗ is weakly compact. Also from the proof of Banach-Steinhaus we have B strongly bounded ⇒ B weak-* bounded ⇒ B equicontinuous. So we know that B¯w∗ is equicontinuous so by the above lemma on B¯w∗ the strong and weak-* topologies coincide. So B¯w∗ is strongly compact and B is a strongly closed subset of B¯w∗ so B is strongly compact.

It is in fact a general truth that the duals of Montel Spaces are Montel Spaces.

Proposition. Restricted to equicontinuous sets, the topology of convergence on M a dense subset of D is equal to the strong dual and weak-* topologies. This along with separability of D shows that these topologies restricted to equicontinuous sets are metrizable. Proof. In the proof of the Lemma we can see that if M is a dense set in D(Ω) then we can choose the φ1, ..., φn to be in M. Therefore, the restriction to any bounded set of the topology induced by the strong or the weak topology is the same as the topology which is induced by the evaluation maps of a dense set in D(Ω). Since we know that D(Ω) is separable with countable dense set M we can look at the weakest topology on D0(Ω) which will make the evaluation maps of all the elements of M continuous. This has a countable base neighbourhood base so is metrizable so any of these topologies restricted to a bounded set are metrizable. This gives annother proof that bounded sets in D0 are sequentially compact since sequential compactness is equivalent to compactness in a metrizable space.

5.9 The Weak topology on D(Ω) We work toward showing that the space D is reflexive. We do this by analogy with the theorem for normed spaces which says that a Banach space which is reflexive if and only if its unit ball is weakly compact. Consequently we need to define the weak topology on D.

29 Deﬁnition. The Weak topology is the weakest topology on the space D(Ω) which makes all the 0 elements of D (Ω) continuous. It is a locally convex topology deﬁned by the semi-norms pu where

pu(φ) = |u(φ)| where u ranges over the distributions. As with normed spaces u is continuous with respect to the weak topology on D(Ω) iﬀ u ∈ D0. We already know that u ∈ D0 implies that u is continuous. Suppose u is weakly continuous then V = {φ | |u(φ)| < } is weakly open contains a set of the form

{φ | |ui(φ)| < 1, i = 1, ..., n}

0 for some set of ui ∈ D . Then \ U = ker(un) ⊂ V n if φ ∈ U then λφ ∈ U so |u(λφ)| < for all λ so u(φ) = 0 for φ ∈ U. So ker(u) ⊃ U which 0 means that u ∈ span{u1, ..., un}. Which means that u ∈ D Proposition. Any set B which is closed and bounded in the normal topology on D(Ω) is weakly compact. Proof. If B is closed and bounded in the LF-space topology then we already know it is compact in the LF-space topology. Any weakly open set is also open in the LF-space topology so any weak open cover of B is an open cover of B in the normal topology so has a ﬁnite subcover. Hence B is weakly compact.

5.10 Reflexitivity Definitions. The bidual, F”, of a TVS F is the dual of F’ with its strong dual topology. A space is called reflexive if F” is isomorphic to F via the cannonical mapping x → xˆ. By the lemma in the previous section the strong and weak-* topologies coincide on the 00 equicontinuous sets if f ∈ D then f|E is weak* continuous for E some equicontinuous set. For this section, we need to recall a special case of the separation form of the Hahn-Banach Theorem. Hahn-Banach Separation. If X is a locally convex space and A, B are disjoint convex sets s.t. A is compact and B is closed. Then there exists f ∈ X0 s.t. sup f(a) < inf f(b) a∈A b∈B Now, on the way to proving that D is reflexive we show something similar to Goldstine’s theorem for normed spaces which says that the ball of the bidual is the weak-* closure of the image of the ball under the canonical injection of D into D00 via the evaluation map. Proposition. This canonical mapping D → D00 is an injection. Proof. We want that if u(φ) = u(ψ) for every u ∈ D0 then φ = ψ. Suppose φ 6= 0 then we can define a linear map on span(φ) by lettingu ˜(λφ) = λ then the Hahn-Banach theorem allows us to extend this to a linear functional u ∈ D0 so D0 strongly separates the points of D so the canonical mapping is indeed an injection.

30 Proposition. Let B be the weakly bounded set in D deﬁned by

|u(φ)| ≤ Mu

0 where u ranges over D and Mus are arbitrarily chosen positive reals. Then let

00 00 B = {f ∈ D | |f(u)| ≤ Mu} this is a weak-* bounded set in D00. Let K be the weak-* closure of the image of B under the canonical injection into D00. Then K = B00. Proof. K is weak-* closed and bounded so by Banach-Alaoglu-Bourbaki it is compact. Suppose 00 K 6= B then we know that K ⊂ B so pick some f0 ∈ B \ K then by Hahn-Banach separation there exists u ∈ D0 s.t. sup f(u) < f0(u) f∈K as everything is convex and balanced we may as well have these quantities positive. Since 00 f ∈ B , f0(u) ≤ Mu but sup |f(u)| = sup |u(φ)| = Mu f∈K φ∈B 00 which gives Mu < Mu for this u which is a contradiction. Hence, K = B . Corollary. D(Ω) is reﬂexive. Proof. Let Dˆ be the image of D in D00. The restriction to Dˆ of the weak-* topology on D00 is the weak topology on D. Let B be a closed, bounded set in D and Bˆ its image in D00. Then by ¯ Bˆ is weak-* compact so weak*-closed. By above, Bˆ = Bˆw∗ = B00. So D00 = D.

5.11 Completeness We already know that all Cauchy sequences converge in D0(Ω) but since the space is not metrizable we would like to show that D0(Ω) is complete. Proposition. The space D0(Ω) is complete for the strong dual topology.

Proof. Let F be a Cauchy ﬁlter on D0(Ω) and deﬁne

F (φ) = {A(φ) | A ∈ F } where A(φ) = {u(φ) | u ∈ A} then we claim that F (φ) is a Cauchy filter in R. Indeed given > 0 we know that φˆ is a continuous linear form on D0(Ω) so there is a neighbourhood U of 0 such that u ∈ U means that |u(φ)| < . Then there is some A ∈ F with A−A ⊂ U which means that A(φ)−A(φ) ⊂ (−, ). Since R is complete we know that A(φ) → tφ. We define a function u : D(Ω) → R by u(φ) = tφ. Since it follows from the definition that A(λφ + µψ) = λA(φ) + µA(ψ), u is a linear functional. We now claim that F converges uniformly to u on any bounded set B, of D(Ω). This follows as Bˆ is an equicontinuous set of maps from D0(Ω) to R so their images are uniformly Cauchy. This means that u|B is continuous and bounded. Since u is bounded on all bounded sets and linear u ∈ D0(Ω). Hence F → u.

31 5.12 The space D as a subset of D0 and Separability. We have already seen how the space D embeds into D0 via the same identiﬁcation as for the 1 0 Lloc functions. Here, we state but do not prove that D is sequentially dense in D . This argument relies on mollifying a distribution u by convolving it with a bump function which will produce a smooth function and then multiplying this function by a smooth function which has large compact support. This result has some implications for the space D0. First, however, we would like to investigate continuity properties of the map I : D → D0. This map cannot be a homeomorphism since the I(D) is dense in D0 yet D is complete so would form a closed subspace of D0 if the subspace topology was equivalent to the initial topology. Proposition. The map I : D(Ω) → D0(Ω) which goes via

φ 7→ uφ where Z uφ(ψ) = φ(x)ψ(x)dx Ω is continuous when D0(Ω) carries either the weak-* or strong dual topology. Proof. We prove this by showing that I is sequentially continuous. We have already shown that since D0(Ω) is a locally convex space with either topology that sequential continuity will imply continuity.

Consequently, we take some sequence φj → 0 and wish to show that uφj → 0 in either topology. Since the φj converge to zero their supports are all contained in some ﬁxed compact set K. Z

|uφj (ψ)| = | φj(x)ψ(x)dx K Z ≤ sup |φj(x)| |ψ(y)|dy x∈K K

≤ sup |φj(x)| sup |ψ(y)| |K| x∈K y∈K

= |K|p0,K (φj)p0,K (ψ) which will tend to 0 as j → 0. Since p0,K is bounded on bounded subsets of D(Ω) this means that I is sequentially continuous with respect to both the weak-* and strong dual topology. We know that I−1 cannot be continuous. We now have some consequences of the continuity of I and the density of its image. Proposition. D0(Ω) is sequentially separable with either of the topologies. Proof. This is a consequence of the density of I(D(Ω)) in D0(Ω) and the fact that D(Ω) is separable. Also, since the bounded sets in D(Ω) are metrizable then if N is a countable sequentially dense set in D0(Ω) and B is a bounded set in D0(Ω) then we have

N ∩ B ⊃ N ∩ B = B

Which means that N = D0(Ω). We also see that if T is a continuous map from D0(Ω) to some other space then T ◦ I will be a continuous map from D(Ω).

32 6 Multiplying Distributions

First we introduce the space E(Ω) this is the space of smooth function on Ω. This is a ﬁrst countable locally convex space deﬁned by the semi-norms

α pn,K (φ) = Σ|α|≤n sup |∂ φ(x)| x∈K

By choosing a countable sequence of compacts increasing to Ω we can show that this is a Fr´echet space in exactly the same way as for DK . We show that if φj is a Cauchy sequence then it is uniformly Cauchy on each compact in each of its derivatives. Therefore for each compact α K the sequence ∂ φj converges uniformly to a function φα so ordering the multi-indexes and extracting subsequences gives a Cantor diagonal sequence φjk which converges uniformly on K in each partial derivative. Extracting successive sub-sequences for each Kn will give a Cantor diagonal sequence which converges uniformly in each derivative on each compact and therefore in the sense of E. In this section we always consider D0 with the weak-* topology.

6.1 Multiplication between subspaces of functions and D0 It is possible via duality to deﬁne multiplication between E and D0.

ψu(φ) = u(ψφ)

This agrees with our previous notion of multiplying funtions when u can be identiﬁed with an element of L1. This is a bilinear map E × D0 → D0

Proposition. Multiplication is a separately continuous mapping and if φn → 0 in E and un → 0 0 0 in D then φnun → 0 in D

Proof. First ﬁx φ ∈ E then look at u 7→ φu call this map mφ we want to show this is weak-*- to - ˆ ˆ weak-* continuous. Choose ψ ∈ D then φ.ψ ∈ D so ψ ◦ mφ = ψ.φ. Therefore this is continuous.

0 Now fix u ∈ D and look at φ 7→ φu call this map mu we want to show this is continuous, ˆ it is sufficient to show that if ψ ∈ D then ψ ◦ mu is continuous and since E is metrizable it is sufficient to prove this is sequentially continuous. So take some sequence φj → 0 we want to show that u(ψ.φj) → 0 which true since φj → 0 uniformly in each derivative on each compact so ψ.φj → 0 uniformly in each derivative so u(ψ.φj) → 0 as u is continuous.

Now given φj → 0, uj → 0 then for each ψ ∈ D we wish to show that uj(ψ.φj) → 0. The support of ψ is contained in some compact set K and for this K there exists a function ϕ ∈ DK which majorizes ψ.φj for each j (as this seqence tends to 0 uniformly in each derivative). uj(ϕ) → 0 which gives the result.

Proposition. The map M :(φ, u) 7→ φ.u is not a continuous mapping from E × D0 to D0 Proof. We argue by contradiction. Suppose the map were continuous then we would have ψ1, ..., ψr, ξ1, ..., ξr ∈ D and K a compact subset of Ω, n an integer such that.

max |u(φ.ψi)| ≤ pn,K (φ) max |u(ξj)| ∀φ, u i j

33 S then supp(φ.ψi) ⊂ supp(ψ) so without loss of generality we can take supp(φ) ⊂ supp(ψi) and S i K ⊂ i supp(ψi) so we only need to look at u|DK . Fix some x ∈ K such that ψi(x) 6= 0 for (α) some i, and let u = δx for some |α| > n then we have

(α) then maxj |ξ(x)| is just some number and maxi |ψi(x)| 6= 0 so making φ (x) suﬃciently large relative to the lower derivatives will give a contradiction. We now introduce the notion of the order of a distribution. We have already seen for each u ∈ D0 to each compact set K ⊂ Ω we can associate an integer N and a constant C s.t.

α |u(φ)| ≤ CΣ|α|≤n sup |∂ φ(x)| K for all φ ∈ DK If there is an N which works for every compact set then we call the smallest such N the order of u. Otherwise we say u has infinite order. Proposition. It is possible to define multiplication between the subspace of distributions of order less than or equal to N and the N times continuously differentiable functions in Ω Proof. This is because the distributions of order less than or equal to N are in the analytic N N N dual of the space Cc (Ω) and the multiplication between C (Ω) and D(Ω) is into Cc (Ω) so multiplication can be defined by duality as before.

6.2 The Impossibility of Multiplying two Distributions in General In this section we follow Schwartz proof in [5] closely. For this we will need to have the notion of the derivative of a distribution. Deﬁnition. If u in D0 and α is a multi-index we can deﬁne ∂αu via the ’integration by parts formula’ ∂αu(φ) = (−1)|α|u(∂αφ)

This defines a distribution since its restriction to any DK is continuous and linear as differentiation is a continuous, linear map on DK . This definition agrees with the normal derivative on distributions which can be identified with functions with a continuous derivatives up to the |α|th order, thanks to the normal integration by parts formula.

1 Now we want to look at the particular distribution on Ω = R, p.v.( x ) deﬁned by: 1 Z 1 p.v. φ = lim φ(x)dx x →0 |x|> x

1 Lemma. p.v. x is a distribution. Proof. Z 1 Z φ(x)dx = − log()φ() + log()φ(−) − log(|x|)φ0(x)dx |x|> x |x|> and we have − log()φ() + log()φ(−) → 0 so we have a distribution of order 2.

34 1 We notice that if ϕ(x) = x then ϕp.v.( x ) = 1 Now we would like to specify the properties which a good notion of multiplication between two distributions on R would satisfy. (1) We would like it to be associative. (2) We would like it to satisfy the product rule. i.e.

d dv du (u.v) = u + v dx dx dx and consequently the Liebniz rule. (3) We would like it to agree with our earlier deﬁnition of multiplication between E and D0 when E is identiﬁed with a subspace of D0.

d We also notice that since x|x| is continuously diﬀerentiable dx (x|x|) = 2|x| in the distribu- tional sense. If Liebniz rule holds for distributions then we will have: d2 d2 d (x|x|) = x (|x|) + 2 (|x|) dx2 dx2 dx d d2 d ⇒ (2|x|) = x (|x|) + 2 (|x|) dx dx2 dx d2 ⇒ x (|x|) = 0 dx2 d2 However, since dx2 (|x|) = 2δ0 6= 0 if we have the properties above then: 1 d2 1 d2 0 = p.v.( ) x (|x|) = p.v.( )x (|x|) = 2δ x dx2 x dx2 0 Which is a contridiction. Therefore, we cannot have a sense of multiplication satisfying the properties (1), (2), (3) on the distributions on R. This proof only relies on having an element, ϕ of D(Ω) where there exists distributions u, v 6= 0 s.t. u.ϕ = 1 and v.ϕ = 0.

It might seem in addition to the three points above we would like to have a notion of multiplying two distributions which would coincide with the normal notion of multiplication on 1 1 Lloc but this would not be possible as the product of two distributions in Lloc is not necessarily 1 in Lloc and does not necessarily deﬁne a distribution.

6.3 Division by Analytic Functions in the Case of One Real Variable. The goal of this section is when given u ∈ D0(Ω) and f some smooth function to ﬁnd v such that f.v = u

We will work only in the case of one variable and where Ω = R. We will build up by first dividing by the function x 7→ x, then by polynomials and finally by analytic functions whose zeroes have finite order.

35 6.3.1 Division by x We seek to solve the equation x.v = u then (x.v)(φ) = u(φ), vx(xφ(x)) = u(φ). If ψ is a function in D(R) with ψ(0) = 0 then since ψ is diﬀerentiable at 0 we have ψ(x) = xφ(x) for some φ ∈ D(Ω). In this case v(ψ) = u(φ). If we suppose that v exists then if we ﬁx η with η(0) = 1:

φ − φ(0)η v (φ − φ(0)η) = u x which means that φ − φ(0)η v(φ) = u + φ(0)v(η) x so (x.v)(φ) = v(x.φ) = u(φ − φ(0)0.η) + 0.φ(0)v(η) = u(φ). So if we fix η and v(η) this will determine a linear functional. This will be continuous since we have φ − φ(0)η φ 7→ x is continuous with respect to the topology on D(R) and φ 7→ φ(0) is continuous so v is the composition of continuous functions. v is not defined uniquely it depends on the choice of η and of v(η) so there are infinitely many solutions.

6.3.2 Division by Polynomials. We can deﬁne division by (x − α) in the same way as division by x. Following this we can divide by (x − α)n by dividing by (x − α), n times since division by (x − α) is always possible. Then we can use this to divide by any polynomial by successively. This does not depend on the ordering of the factors since multiplication by polynomials does not depend on the ordering of the factors.

6.3.3 Division by Analytic Functions with Zeroes of Finite Order. For this section we need the following Lemmas.

Lemma. Suppose { ωi | i ∈ I } is a locally ﬁnite covering of Ω ( every compact set K intersects 0 with ﬁnitely many of the ωi) and there exists ui ∈ D (ωi) such that

ui|ωi∩ωj = uj|ωi∩ωj for every i, j, then there exists u ∈ D0(Ω) such that

u|ωi = ui

36 Proof. This follows from the fact that there exist smooth functions ρi such that

supp(ρi) ⊂ ωi and Σi∈I ρi = 1. Then the sum u = Σi∈I ρiui deﬁnes a distributions.

Now we show the existence of the ρi. Supposeρ ˜i is a function with supp(˜ρi) ⊂ ωi andρ ˜i > 0 S on Ω \ j6=i ωj. This is possible since [ C1 = Ω \ ωj and C2 = Ω \ ωi j6=i are closed and disjoint and so we have a smooth functionρ ˜i such thatρ ˜i = 0 on C2 andρ ˜i = 1 on C1. (By mollifying the continuous function found in Urysohm’s lemma.) 1 Then 0 < f = Σiρ˜i < ∞ since the covering ωi is locally ﬁnite. Let ρi = f ρ˜i this gives the required result.

We now prove that u = Σiρiui deﬁnes a distribution. Fix K compact then there exists i1, ..., in the ﬁnite collection of sets which intersect K. Then for φ ∈ DK

n n u(φ) = Σk=1ρik uik (φ) = Σk=1uik (ρik φ) which is linear, well deﬁned and continuous on DK .

Also, we have uωi = ui since if supp(φ) ⊂ ωi we have

u(φ) = Σjuj(ρjφ) = Σjuj|ωi (ρjφ) = Σjui(ρjφ)

= ui (Σjρjφ)) = ui(φ).

After this lemma, suppose f is an analytic function with zero set { ξi | i ∈ I } each with ﬁnite order ni. Then, since f is analytic, we can construct { ωi | i ∈ I } a collection of pairwise disjoint open sets such that ξi ∈ ωi. Then we have another open set ω0 such that [ R \ ωi ⊂ ωo. i∈I

We also ask that ξi ∈/ ω0. Then this will form a locally finite covering of Ω and it is, therefore, sufficient to show that we can divide by f on each ωi. i.e. we would like to fine vi such that in ωi

f.vi = u|ωi then we can patch these vi together to ﬁnd v. Also, we notice that

ni f = (x − ξi) fi

1 where fi 6= 0 on ωi. Then ∈ E(ωi) so we can reduce to the case of dividing by a polynomial. fi

37 6.4 Applications of Multiplication. Here we, very briefly, talk about how our definitions of multiplication and differentiation by duality allow us to set PDEs in the space of distributions. This is arguably the primary motivation for developing the space of distributions. Suppose D is a linear differential operator with coefficients in E(Ω) then if u is a distribution we can define α Du = Σ|α|≤ncα∂ u in the sense of distributions. Therefore, we can try and solve the equation

Du = v where u is an unknown distribution and v is a given distribution. This may or may not have a solution. If we look now at the homogeneous case Du = 0 we want

α Σ|α|≤ncα∂ φ0, ∀φ

α ⇔ Σ|α|≤n∂ (cα.φ) = 0, ∀φ

|α| α ⇔ Σ|α|≤nu((−1) ∂ (cα.φ)) = 0, ∀φ α α ⇔ u Σ|α|≤n(−1) ∂ (cα.φ)) = 0, ∀φ. This gives us another diﬀerential operator

0 α D φ = Σ|α|≤n∂ (cα.φ) on D(Ω) which is the formal adjoint of D. H = D 0(D(Ω)) is a linear subspace of D(Ω) so seeking u to solve the homogeneous equation

Du = 0 is equivalent to seeking u in the annihilator of H . While the equation Du = v may not have a solution in the sense of distributions the operator D is still defined. Therefore, in some sense the space of distributions is the largest space in which we can hope to define, and seek existence of solutions, the problem Du = v. However, we have shown that the multiplication product of two distributions is not always possible to define. Hence, the space of distributions is not an obvious space in which to look for the solution to a non-linear problem.

7 Rapidly Decreasing Functions.

Now we introduce another space of possible test functions on the whole of Rn. These are smooth functions which decrease rapidly at infinity in the sense that they decrease faster than the reciprocal of any polynomial. This space is useful in defining the Fourier Transform for a large class of functions and can be used to extend many results of Fourier analysis to a large class of the distributions. We will show a few properties of this space. Many of the topological properties are the same and we restrict ourselves to the detailed proofs of those which are reasonably different to those in the space D0(Ω). They can be made into a locally convex space with the semi-norms:

38 β α pα,β(φ) = sup |x ∂ φ(x)| n x∈R Where α and β run over all the possible multi-indices. Since this is a countable collection of semi-norms the space is metrizable exactly as for DK . This space is denoted S . Proposition. S is a Fréchetspace. Proof. We already know that S is metrizable, as it is a first countable lcs, so it is sufficient to prove that Cauchy sequences converge. Suppose φj is a Cauchy sequence. Given two multi- β α n indices α, β then the sequence x ∂ φj(x) is uniformly Cauchy so converges uniformly on R .This is because it converges uniformly on compacts and limits must agree and it is uniformly small outside compact sets so the convergence is uniform. More precisely if β0 > β then we know that β0 α |x ∂ φj(x)| ≤ M for some M which doesn’t depend on j. This means that β α β−β0 |x ∂ φj(x)| ≤ |x |M and if we fix we can choose |x| = r sufficiently large so that M ≤ , |x| > r . |xβ−β0 | 2 β α Let K = {|x| ≤ r} then we know that x ∂ φj → φα,β uniformly on K So if we choose j large enough so that β α sup |x ∂ φj(x) − φα,β(x)| < x∈K then outside K we will have β α β α |x ∂ φj(x) − φα,β(x)| ≤ |x ∂ φj(x)| + |φα,β(x)| < . β α It converges to a function φα,β. Then φα,β = x ∂ φ0,0 so φj → φ0,0 in the topology of S .

The space D(Rn) can be embedded into S . Let ι : D(Rn) → S be this map. We wish to show that ι is a continuous injection onto its image. First we show ι is continuous by showing that if φj → 0 in D then φj → 0 in S . This is suﬃcient as we only need to show the restriction to each DK is continuous and DK is metrizable. Now we have already shown that φj → 0 implies β that the supports of all the φj are contained in some compact K then for each β, |x | ≤ Mβ on β α α K so x ∂ φj ≤ Mβ∂ φj → 0. The inverse cannot be continuous since S is metrizable and the subspace topology that S induces on D is metrizable where as the LF-space topology on D is not metrizable so the two topologies cannot be the same. Proposition. The space S has the Heine-Borel property and is therefore a Montel Space. Proof. Suppose B is a closed and bounded set in S . Since S is metrizable we have compactness iﬀ sequential compactness. So let φj be a bounded sequence in S and let

K1 ⊂ K2 ⊂ K3 ⊂ ... n β α be a sequence of compact sets whose union is R . Similarly as for in DK the sequence x ∂ φj is a uniformly bounded equicontinuous set on Ki. By the Arz`ela-Ascolitheorem there exists a β α subsequence s.t. x ∂ φjk converges uniformly on Ki then as we have a countable collection of possible multi-indices α, β we can extract further subsequences and then create a cantor diagonal β α sequence φjk such that x ∂ φjk converges uniformly on Ki for every α, β. Then we can keep extracting subsequences of this for each Kn to produce another Cantor diagonal sequence which will converge in the sense of S .

39 7.1 Relation between S and D(Sn) Schwatz also gives a geometric idea of the rapidly decreasing functions in terms of their relation to the set of smooth functions on the sphere. Since the sphere is compact we have that D(Sn) = E(Sn) and will have a Fr´echet space topology. Since we have a smooth bijection with a smooth inverse from Rn to Sn \{ω} where ω is some arbitrary point on the sphere. This map will induce a map from S (Rn) to D(Sn) whose image is in the closed subset of D(Sn) where the functions are 0 in all of their derivatives at ω. This is because if φ ∈ S and P : Rn → Sn \{ω} is a smooth function with smooth inverse we can deﬁne φ˜ on Sn \{ω} by

−1 n φ˜(x) = φ(P (x)) ∈ E(S \{ω}) Now if we look at lim ∂αφ˜ x→ω we can see that this will be 0 as lim ∂α(φ)(y) = 0 |y|→0

So φ˜ extends smoothly to a function on all of Sn. We can use this to show that S is separable as C∞(Sn) is separable since Sn is compact and Hausdorﬀ by the same argument we used to show that C∞(K) was separable for K a compact subset of Rn. Since C∞(Sn) is metric and we know that subspaces of separable, metric spaces are separable this shows that S is separable. Like with te space D(Ω) this will show that bounded sets in S 0 are metrizable.

Since S is a Fr´echet space, it is barelled so the Banach-Steinhaus theorem holds in S . The above also implies weakly bounded sets are weakly compact.

7.2 The Fourier transform on S Since x−(n+1) ∈ L1(Rn) the space S ⊂ L1(Rn) so the Fourier transform can be deﬁned via Z λ.x n φ 7→ φˆ(λ) = φ(x)e dx, λ ∈ R n R The Fourier transform is continuous because

λβ∂αφˆ(λ) = (xα∂βφ(x))ˆ possibly modulo some constants, and ˆ sup |φ(λ)| ≤ ||φ||L1(( n) n R λ∈R and if φ ∈ S then Z Z n+1 −n−1 1 n ||φ||L (R ) ≤ |φ(x)|dx + ||x| φ(x)||x| dx n B1 R \B1 Z n+1 −n−1 ≤ sup |φ(x)| |B1| + sup |x φ(x)| x dx. n n n x∈R x∈R R \B1

40 This means shows that the Fourier transform is continuous because

sup |λβ∂αφ(λ)| = C sup |(xα∂βφ(x))ˆ(λ)| n n λ∈R λ∈R and we have that α β α β sup |(x ∂ φ(x))ˆ(λ)| ≤ ||x ∂ φ(x)|| 1 Lx n λ∈R which we have just shown can be bounded by norms deﬁning the topology on S . This also shows that the Fourier Transform of an element of S is in S (so the Fourier inversion theorem will hold for elements of S ). i.e. Z 1 ˆ −λ.x φ(x) = n φ(λ)e dλ (2π) n R 8 Tempered Distributions

The tempered distributions are defined as the analytic dual of the space of rapidly decreasing functions. Like with the space D0 we can put both the strong dual and the weak-* topology on this space S 0 and by compactness of bounded sets in S these are equal when restricted to equicontinuous subsets of S 0. The functions which increase more slowly at infinity than some polynomial can be embedded 0 1 0 into the space S via the identification f ↔ uf as with Lloc in the space D . This does indeed define a tempered distribution by the definition there will exist a k such that |x|−kf(x) → 0, |x| → ∞ so |x|−k−nf(x) ∈ L1(Rn) so suppose φ ∈ S: Z Z | f(x)φ(x)dx| = | |x|−n−kf(x) |x|n+kφ(x) dx| n n R R Z ≤ |x|−n−k|f(x)|dx. sup (|x|n+k|φ(x)|) n n R x∈R We can see from this that if f is the weak derivative of a function which decreases slower than some polynomial then it will also define a tempered distribution.

We can prove, as with the earlier spaces that closed, bounded sets in S 0 are compact in the weak-* and strong dual topology and that the space S is reflexive. These properties all follow from the Heine-Borel property for the space S or Banach-Alaoglu-Bourbaki. Looking at the proofs that the strong dual topology is not metrizable we see that it relies on the fact that if we have a countable sequence of bounded sets in D then we can find a function which is not in this bounded set. This is also true in the space S since we can forget about the β bound on x times the functions and choose the bounded sets to be bounded on each of the pn norms from DK and we can find a function in DK and therefore in S which is not in any of these sets using the same construction as for DK . We can also show that the weak-* topology is not metrizable in exactly the same way as for D0 where we saw that the weak-* topology being metrizable would imply that D was countable dimensional. If the weak-* topology on S 0 were metrizable this would imply that S was countable dimensional which is a contradiction. Using the continuous injection of the space D(Rn) we can see that S 0 ⊂ D0 via the dual of this identification. The fact that ι is a continuous injection onto its image shows that the injection of S 0 into D0 is continuous and an injection onto its image.

41 8.1 Relation between S 0 and D0(Sn) We can also relate the tempered distributions to a space of distributions on the sphere. They can be identified with the space of continuous linear maps on the subspace of C∞(Sn) where a function and all its derivatives are fixed to be 0 at a point ω. Schwartz uses this to give a necessary and sufficient condition for an element of D0 to be also an element of S 0. Proposition. A distribution u ∈ D0(Rn) is a tempered distribution iff there exists a distribution 0 n n u ∈ D (S ) such that u|R = u. Proof. The “if” part follows from the fact that S can be identified with a subspace of D(S2). More precisely, there is n T : S → D(S ) so if u ∈ D(Sn) then u = u ◦ T is continuous. The “only if” part follows from the Hahn-Banach theorem. Our identification of S with a closed linear subspace of D(Sn) means that if we have u ∈ S 0 we have a continuous, linear functional on a closed linear subspace of D(Sn) so by Hahn-Banach we can extend it to a continuous linear functional on the whole space.

The u in the above theorem is not unique they can diﬀer by a distribution whose support is contained in {ω}. So the space S 0 can be identiﬁed with some quotient of the space D0(Sn).

8.2 The Fourier Transform on S ’ We can deﬁne the Fourier Transform on S 0 by dual map of the Fourier Transform on S. i.e.

uˆ(φ) = u(φˆ) This agrees with the normal notion of Fourier transform in L1 ∩ L2 thanks to the Plancherel identity which states, for φ, ψ ∈ L1 ∩ L2: Z Z φˆ(x)ψ(x)dx = φ(x)ψˆ(x)dx n n R R This is an isomorphism since the Fourier transform on S is an isomorphism since the dual map of an isomorphism is an isomorphism. Proposition. The Fourier inversion theorem holds for S 0. This is in the sense that Z Z Z 1 −ix.λ 1 −ix.λ n uˆ(λ)e dλ (φ) = n uˆ(λ)φ(x)e dλdx (2π) n n n (2π) R x∈R λ∈R Z 1 −ix.λ =u ˆλ n φ(x)e dx = u(φ) n (2π) x∈R where uˆλ acts on its argument as a function of λ. Proof. Let F be the Fourier transform operator. Then since we have the Fourier inversion theorem on S we have that Z 1 −ix.λ −1 −1 uˆλ n φ(x)e dx =u ˆλ(F φ) = u(FF φ) = u(φ). n (2π) x∈R

42 9 The space E0.

Many of the theorems about the spaces S 0 and D0 also hold for the space E0. The proofs are very similar so we will not develop these. This section will be short and mainly focus on the characterization of E0 as the space of distributions of compact support. This will be useful later. Deﬁnition. If u ∈ D0 then we say that u is ‘null’ on an open set ω if for every φ such that supp(φ) ⊂ ω then u(φ) = 0. The support of u is the complement of all the union of the sets on which u is null. Proposition. The set E0 can be considered as a subset of D0 and this is exactly the set of distributions with compact support. Proof. Suppose that u has support in a compact set K. Let ρ ∈ D(Ω) be such that ρ = 1 on K. Then supp(ρφ−φ)∩K = ∅ so u(ρφ−φ) = 0 for every φ. Hence ρ.u = u and multiplication by ρ maps E(Ω) into D(Ω). So the map composition of multiplication by ρ and u deﬁnes a continuous linear map E which coincides with u on D ⊂ E. Hence, we can consider u as a member of E. Conversely, suppose that v ∈ E0. Then suppose the support of v is not compact. In this case it will contain ω1, ω2, ... a sequence of disjoint open subsets such that v|D(ωn) is not identically 0 for any n. Then there exists φn ∈ D(ωn) such that v(φn) > 1 then Σnφn ∈ E but v(Σnφn) > v(Σn≤N φn) > N which gives a contradiction as N is arbitrary.

10 Comparison and Relations between the spaces D, S and E and their duals.

We have already seen that D injects continuously into S . We would like also to see that S injects continuously into the space E. Since they are both Fr´echet spaces it is suﬃcient to show that if φ → 0 in then φ → 0 in E. This follows since sup n φ(x) ≥ sup φ(x) so something j S j R K converging uniformly on the whole space will imply it converges uniformly on every compact. Therefore φj and all its derivatives will converge uniformly on each compact. Therefore we have

D ,→ S ,→ E and

D0 ←- S 0 ←- E0 where ,→ represents a continuous injection. We have several properties that all the spaces share: (1) Cauchy sequences converge. (2) Closed, bounded sets are compact. (3) The topology restricted to bounded sets is metrizable. (4) For the distribution spaces the weak-* topology and the strong dual topology coincide on bounded sets. (5) The function spaces are all separable. (6) None of the distribution spaces are metrizable. (E contains DK as a topological subspace so a countable sequence of bounded sets in E cannot be sufficient to define the strong dual topology.) (7) The spaces are all reflexive.

43 We do, however, have some differences in the question of metrizability of the whole spaces in that D is not metrizable where both S and E are. This, is not so huge a difference as it might first seem because continuous, linear maps from D can still be characterized entirely by their actions on sequences. There are larger differences in the utility of the different function spaces and distribution spaces. D being the smallest of the function spaces means that D0 is the largest of the distribution spaces here. In D0 we still have enough structure to define differentiation which will be into the space, and differential operators whose coefficients are in E it therefore gives a very general notion of differentiation and differential operators. However, the space D is too small to be useful with the Fourier transform. The space S is constructed so that the Fourier transform is an isomorphism of this space which allows us (via duality) to extend the notion of the Fourier transform to S 0 which is smaller than D0 but never-the-less much larger that L2(Rn).

11 Tensor Product of two Distributions 11.1 The Tensor Product of two Vector Spaces

Suppose V1,V2 are two vector spaces, we would like to give a good algebraic deﬁnition of V1 ⊗V2. Suppose E is another vector space and Φ is a bilinear mapping from V1 × V2 to E. With the property that e1, e2, ..., en a linearly independent set in V1 and f1, f2, ..., fm a linearly independent set in V2 then

Φ(ei, fj), i = 1, ..., n, j = 1, ..., m is a linearly independent set in E. And that span(Φ(V1,V2)) = E. Then we can call E the tensor product of V1 and V2. We can see that if eα, α ∈ A is a basis for V1 and fβ, β ∈ B is a basis for V2 then

Φ(eα, fβ), α ∈ A, β ∈ B is a linearly independent set in E. It is also spanning since if x ∈ E we have

x = c1Φ(u1, v1) + ... + ckΦ(uk, vk).

Further we can write Φ(u, v) as a linear combination of Φ(eα, fβ) by writing u, v as linear combinations of basis vectors and exploiting the bilinearity of Φ we can write any linear combinations of elements Φ(u, v) as a ﬁnite linear combination of these vectors. This shows that any two tensor products are isomorphic.

11.2 The Tensor Product of Function Spaces We can now look at the tensor product of spaces of functions, and at their embedding into other function spaces.

n m Proposition. If Ω1 ⊂ R and Ω2 ⊂ R then we can embed D(Ω1) ⊗ D(Ω2) into D(Ω1 × Ω2) n+m where Ω1 × Ω2 is regarded as an open subset of R .

44 Proof. To do this we construct an injection Φ : D(Ω1) × D(Ω2) → D(Ω1 × Ω2) and show that the image of Φ will satisfy the axioms of being a tensor product. We use the map Φ with

Φ(ϕ, ψ)(x, y) = ϕ(x).ψ(y), (x, y) ∈ Ω1 × Ω2 We call Φ(ϕ, ψ) = ϕ ⊗ ψ and we can see that supp(ϕ ⊗ ψ) ⊂ supp(ϕ) × supp(ψ). This map is bilinear and injective since if one of the entries is ﬁxed and its variable this is just a multiplication map. If φ1, ..., φn ∈ D(Ω1) and ψ1, ..., ψm ∈ D(Ω2) then suppose that

Σi,jai,jφi(x)ψj(y) = 0 ∀x, y this means that by linear independence of the ψj(y) we have for each j:

Σiai,jφi(x) = 0 which in turn means that every for every i, j we have that ai,j = 0. We can see that this form of tensor product would hold for any two function space. For example we can use this same map to give an injection p p p L (Ω1) ⊗ L (Ω2) → L (Ω1 × Ω2) or k l k,l C (K1) ⊗ C (K2) → C (K1 × K2)

Proposition. If Pn is the space of polynomials in n variables. Then given some open subspace n of R this can be regarded as a function space. We can deﬁne the tensor product Pn ⊗ Pm in the same way as for function and for this case the injection given above is a bijection. α β Proof. Any monomial x y is a tensor product of two polynomials in Pn and Pm and any polynomial is a ﬁnite linear combination of monomials so Pn+m = Pn ⊗ Pm. We can use the above proposition, along with the Stone-Weierstrass theorem to show that C(K1) ⊗ C(K2) is dense in C(K1 × K2). This is because Pn+m is dense in C(K1 × K2) by Stone-Weierstrass and also Pn+m = Pn ⊗ Pm ⊂ C(K1) ⊗ C(K2). We can use a variation on this argument to show that D(Ω1) ⊗ D(Ω2) is dense in DK1×K2 . Since, D(Ω1 × Ω2) can be written as the union of spaces of the form DK1×K2 this will show D(Ω1) ⊗ D(Ω2) is dense in D(Ω1 × Ω2). This is because if we take φ1 ∈ D(Ω1) with φ1 identically 1 on K1 and φ2 similarly.

Then we can approximate any element of DK1×K2 by functions of the form (φ1 ⊗ φ2)p where p is a polynomial.

11.3 Tensor Products of Distributions We wish to deﬁne a tensor product between two distributions which will coincide with out notion 0 0 of tensor product on functions. Suppose u ∈ D (Ω1) and v ∈ D (Ω2) then we can deﬁne a linear form u ⊗ v on the subspace D(Ω1) ⊗ D(Ω2) ⊂ D(Ω1 × Ω2). We do this via (u ⊗ v)(φ ⊗ ψ) = u(φ)v(ψ)

This is continuous with respect to the subspace topology on D(Ω1) ⊗ D(Ω2) as on α β |u(φ)||v(ψ)| ≤ C1C2 Σα≤n1 sup |∂ φ(x)| Σβ≤n2 sup |∂ ψ(y)| x∈K1 y∈K2

Where Ki is the projection of K onto Ωi and φ ∈ DK1 , ψ ∈ DK2 . Therefore we can extend the form u ⊗ v by continuity to a form on D(Ω1 × Ω2). We call this the tensor product of u and v 0 0 and can easily check that it deﬁnes a tensor product on D (Ω1) ⊗ D (Ω2).

45 Proposition. (u ⊗ v)(φ(x, y)) = ux(vy(φ(x, y))) = vy(ux(φ(x, y))) where we mean that ux regards φ(x, y) as a function of x for ﬁxed y so ux(φ(x, y)) ∈ D(Ω2) as a function in y.

Proof. We see that this holds on the subspace D(Ω1) ⊗ D(Ω2) and so by continuity it will hold on the whole space.

This is a kind of ‘Fubini’s Theorem’ for distributions and is extremely useful. 1 1 We can see that if f ∈ Lloc(Ω1), g ∈ Lloc(Ω2) then uf⊗g = uf ⊗ ug. Since, Z Z Z f(x)g(y)φ(x, y)dxdy = (f ⊗ g)(x, y)φ(x, y)dxdy x∈Ω1 y∈Ω2 (x,y)∈Ω1×Ω2 So the deﬁnition is consistent with the earlier tensor product on function spaces. 0 We can also see from this that if u ∈ D (Ω1) then u generates a linear map from D(Ω1 × Ω2) to D(Ω2) via u ↔ ux where ux(φ(x, y)) will be an element of D(Ω2).

0 0 0 Proposition. The map T : D (Ω1)×D (Ω2) → D (Ω1 ×Ω2) given by (u, v) 7→ u⊗v is separately continuous. Furthermore, if uj → 0, vj → 0 then uj ⊗ vj → 0.

Proof. Let u be ﬁxed. Then by our earlier proposition (u ⊗ vj)(φ(x, y)) = vj(u(φ(x, y)). We know that u(φ(x, y)) ∈ D(Ω2) so vj(u(φ(x, y))) → 0. We can switch u and v in the above argument to show that T is separately continuous.

Now suppose that uj, vj → 0. We claim that uj(φ(x, y)) is a bounded sequence in D(Ω2). If this is true then the weak-* and strong topology will coincide on vj since vj is a bounded sequence. Therefore vj(uj(φ(x, y))) → 0. We know that uj(φ(x, y)) have supports contained in the projection of the support of φ onto Ω2 call this K. We can see that (remembering that uj acts on x.)

α α ∂ (uj(φ(x, y))) = uj(∂y (φ(x, y)))

Since this holds for φ ∈ D(Ω1) ⊗ D(Ω2) so for all φ by extending by continuity. hence

Since uj is a bounded sequence we can take C and N to work for each uj therefore uj(φ(x, y)) is a bounded sequence in D(Ω2).

We can also see that since D(Ω1) ⊗ D(Ω2) is dense in D(Ω1 × Ω2) which in turn is dense in 0 0 0 0 D (Ω1 × Ω2) that D (Ω1) ⊗ D (Ω2) is dense in D (Ω1 × Ω2). 0 0 We can give another concrete realisation of the tensor product D (Ω1) ⊗ D (Ω2) as the space of continuous bi-linear mappings from D(Ω1) × D(Ω2) to R. We do this by identifying u ⊗ v with the map Φ(u, v):(φ, ψ) 7→ u(φ)v(ψ) i.e. Φ(u, v)(φ, ψ) = (u ⊗ v)(φ ⊗ ψ). We call the space of weakly continuous bi-linear maps B(D(Ω1), D(Ω2)). Proposition. The above identiﬁcation is indeed a tensor product.

46 Proof. This follows [4]. First we prove linear independence. If u1, ..., un are linearly independent 0 0 and v1, ..., vm are linearly independent in D (Ω1) and D (Ω2) respectively. Then if

Σi,jai,jΦ(ui, vj) = 0 then by linear independence of the vj we have for each j that

Σiai,jφi = 0 which in turn means that ai,j = 0. Now we prove surjectivity. Suppose that B is a weakly continuous bi-linear form on D(Ω1)× 0 0 D(Ω2) then there exists u1, ..., un ∈ D (Ω1) and v1, ..., vm ∈ D (Ω2) such that if

max |ui(φ) ≤ 1, max |vj(ψ)| ≤ 1 i j then |B(φ, ψ)| ≤ 1. T T Let U = i≤n ker(ui) and V = j≤m kervj. Then if φ ∈ U and maxj |vj(ψ)| ≤ 1 then we have |B(λφ, ψ)| ≤ 1, ∀λ therefore B(φ, ψ) = 0 and since for all ψ ∈ D(Ω2) there is some µ such that maxj |µ.vj(ψ)| ≤ 1 then B(φ, ψ) = 0 for φ ∈ U and any ψ. So we can see that B(φ, ψ) = 0 whenever φ ∈ U or ψ ∈ V . Without loss of generality we can take the ui linearly independent and the vj linearly independent. Consequently U has co-dimension n in D(Ω1) and V has codimension m in D(Ω2). Therefore there exists φ1, ..., φn and ψ1, ..., ψm such that given φ, ψ we have ˜ φ = a1φ1 + ... + anφn + φ and ˜ ψ = b1ψ1 + ... + bmψm + ψ ˜ ˜ where ai, bj are scalars, φ ∈ U and ψ ∈ V . Therefore f we ﬁnd new ui,j and vj,i such that ui,j is 0 on U and the span of φk, k 6= i and ui,j(φi) = B(φi, ψj) which exists by Hahn-Banach. Then have vj,i to be 0 on V and span vk, k 6= j and vj,i(ψj) = 1. Then we look at the element

Σi,j(ui ⊗ vj)(φ ⊗ ψ) = Σi,j(ui ⊗ vj)(Σk,lakbl(φk ⊗ ψl))

= Σi,jaibjB(φi, ψj) = B(φ, ψ)

0 0 We also give another canonical injection of the space D (Ω1) ⊗ D (Ω2) this time into the 0 0 space L(D(Ω1), D (Ω2)), which is the space of continuous linear functions from D(Ω1) to D (Ω2) when both are given the strong topology. We induce this by making a bilinear map from 0 0 D (Ω1) × D(Ω2) to L(D(Ω1), D (Ω2)) and then looking at the span of its image. This map is

(u, v) 7→ (φ 7→ (ψ 7→ u(φ)v(ψ)))

0 i.e. If we call this map L then L (u, v)(φ) = u(φ).v(.) ∈ D (Ω2).

47 0 11.3.1 The Space L(D(Ω1), D (Ω2)) 0 0 0 In order to make use of our mapping of D (Ω1) ⊗ D (Ω2) into the space L(D(Ω1), D (Ω2)) we would like to define a topology on this space, and show this space is complete with this topology. We recall how we defined the strong dual topology on D0 as the topology of uniform convergence on the bounded subsets of D and also that D0 = L(D, R). This motivates our definition 0 of the topology. Suppose pB is a semi-norm on D (Ω2) i.e. B is a bounded subset of D(Ω2) and

pB(v) = sup |v(ψ)| ψ∈B suppose also that A is a bounded set in D(Ω1) then we deﬁne the semi-norm pA;B via

pA;B(T ) = sup pB(T (φ)) φ∈A if we let A range over the bounded sets of D(Ω1) and B range over the bounded sets of D(Ω2) then 0 this gives a collection of semi-norms which define a locally convex topology on L(D(Ω1), D (Ω2)). This topology is not metrizable, by arguments we have already seen to show that the strong topology on D0. To show this space is Hausdorff it is sufficient to show that if pA;B(T ) = 0 for every choice of A and B then T is identically 0. Suppose T is not identically 0, then there is some φ ∈ D(Ω1) such that T (φ) 6= 0 and so there is some φ ∈ D(Ω2) such that T (φ)(ψ) 6= 0. This means that p{φ};{ψ}(T ) 6= 0.

0 Proposition. With this topology the space L(D(Ω1), D (Ω2)) is complete. Proof. The space is not metrizable so we need to work with filters. We proceed similarly to proving that D0(Ω) is complete. Let F be a Cauchy filter on L and we claim that for each 0 ˆ φ ∈ D(Ω1), F (φ) is a Cauchy filter on D (Ω2). This is again because φ is a continuous map 0 0 from L(D(Ω1), D (Ω2)). Therefore, if V is a neighbourhood of 0 in D (Ω2) then there exists, U, a neighbourhood of 0 in L such that T ∈ U implies that T (φ) ∈ V . Therefore if A is such that A − A ⊂ U we will have A(φ) − A(φ) ⊂ V . We claim that if B is a bounded subset of D(Ω1) 0 0 then the filter F (φ), φ ∈ B converge uniformly on D (Ω2) this follows from the fact that D (Ω2) is complete and the fact that φ,ˆ φ ∈ B is an equicontinuous set of mapping on L. Therefore if F (φ) → uφ and T is the map T (φ) = uφ then T will be linear and bounded on bounded subsets 0 of D(Ω1) so T ∈ L(D(Ω1), D (Ω2)).

11.4 Topological Tensor Products In this section we will look at topologies which we can put on topological tensor products. We introduce the and π topology of Grothendiek and look at their speciﬁc forms on tensor products of distributions and test functions. Suppose E and F are locally convex, Hausdorﬀ topological vector spaces. Spaces on which the and π topology coincide are called nuclear spaces and were developed by Grothendiek. They have many interesting properties and all the main distribution and test function spaces are nuclear.

11.4.1 The π-topology The π- topology is deﬁned to be the strongest locally convex topology on E ⊗ F such that the mapping E × F → E ⊗ F :(x, y) 7→ x ⊗ y

48 is continuous. We can deﬁne a base of neighbourhood for the π-topology by letting BE be a local neighbourhood base for E and BF a local neighbourhood base for F we can take U ∈ BE and V ∈ BF then we can deﬁne

U ⊗ V = { x ⊗ y | x ∈ U, y ∈ V }.

Then if we let W = conv(U ⊗ V ) as U and V range over their appropriate neighbourhood bases this will form a neighbourhood base of E ⊗π F . 0 0 The π-topology on D (Ω1) ⊗ D (Ω2) is induced by the semi-norms pA ⊗ pB where A and B are bounded sets in D(Ω1) and D(Ω2) respectively and we deﬁne pA ⊗ pB to be the Minkowski functional of W as deﬁned above when U is the semi-ball of pA and V is the semi-ball of pB.

11.4.2 The -topology

0 0 We saw earlier that we can identify D (Ω1) ⊗ D (Ω2) with B(D(Ω1), D(Ω2)) the space of continuous bilinear forms, where here the test function spaces carry their weak topology. It is in fact 0 0 generally true that we can identify E ⊗ F with the space B(Eσ,Fσ) in exactly the same way. The -topology is the topology on this space deﬁned by the topology of uniform convergence on the product of equicontinuous subsets of E0 and F 0. Suppose A, B are such subsets then we have the semi-norm pA,B(Φ) = sup sup |Φ(x, y)|. x∈A y∈B For the topology this gives on tensor products of distribution or test function spaces it is interesting to note that we have an identiﬁcation between the bounded, precompact and equicontinuous sets.

12 Kernels and the kernels theorem.

In this section we will develop some of the basic theory of kernels and the Schwartz kernels theorem which gives a representation theorem in terms of kernels for linear maps from the space of test functions to the space of distributions. 1 1 We have already seen that the space Lloc can act on D(Ω) as integral kernels. For f ∈ Lloc(Ω) we have Z φ 7→ f(x)φ(x)dx Ω and we extended this action to the whole of the space of distributions. Similarly to the tensor 1 0 product if f ∈ Lloc(Ω1 × Ω2) then we can induce a map from D(Ω1) to D (Ω2) via Z φ 7→ ψ 7→ f(x, y)φ(x)ψ(y)dxdy Ω1×Ω2

0 0 analogously, we can make an element of D (Ω1 × Ω2) act as a linear map D(Ω1) → D (Ω2).

φ 7→ (ψ 7→ u(φ ⊗ ψ)) if u = v1 ⊗ v2 then we have φ 7→ (ψ 7→ v1(φ)v2(ψ)) so the linear map which corresponds to v1 ⊗ v2 has one dimensional image.

49 0 0 Suppose that Φ ∈ L(D(Ω1), D (Ω2)) has one dimensional image λu; λ ∈ R, u ∈ D (Ω2). Then deﬁne a linear map v from D(Ω2) to R such that v(φ)u = Φ(φ). Since Φ is continuous it is 0 sequentially continuous so if φj → 0 in D(Ω1) then we will have v(φj)u → 0 in D (Ω2). So in particular for every ψ we have v(φj)u(ψ) → 0 since there will exist a ψ such that u(ψ) 6= 0 this implies that

v(φj) → 0 which means that v is a sequentially continuous linear map. Since we showed earlier that a linear 0 map from D(Ω1) to R is continuous iff it is sequentially continuous this shows that v ∈ D (Ω1) therefore Φ = v ⊗ u. 0 Then if an element of L(D(Ω1), D (Ω2)) has finite dimensional image it a finite linear combination 0 0 of elements with one dimensional image so can be identified with an element of D (Ω1) ⊗ D (Ω2) 0 which in turn can be identified with an element of D (Ω1 × Ω2). Our goal is to extend this to 0 0 an identification between the whole of D (Ω1 × Ω2) and L(D(Ω1), D (Ω2)).

0 Schwartz Kernels Theorem. To any element of L(D(Ω1), D (Ω2)) we can associate a unique 0 element of D (Ω1 × Ω2) which we will call the Schwartz kernel. Proof. We follow a very short proof from Duistermaat & Kolk which uses the Fourier transform. 0 We can see that Φ ∈ L(D(Ω1), D (Ω2)) can be related to a bilinear form B on D(Ω1) × D(Ω2) by B(φ, ψ) = (Φ(φ)) (ψ) let suppose φj → 0 and ψj → 0 then since Φ is continuous Φ(φj) is a bounded and therefore 0 equicontinuous set in D (Ω2) therefore (Φ(φj))(ψj) tends to 0. Therefore B is sequentially continuous and since B can be considered as a linear functional on D(Ω1) ⊗ D(Ω2) it can be extended to a sequentially continuous linear functional on D(Ω1 ×Ω2) and here we already know that sequential continuity implies continuity. This means that

|B(φ, ψ)| ≤ CK,Lpn,K (φ)pn,L(ψ) ∀φ ∈ DK , ψ ∈ DL for some n which depends on K and L. Now let us ﬁx some compact sets K and L and let s, t be smooth functions of compact support who are identically one on K and L respectively. Let

−ix.η −iy.ξ sη(x) = s(x)e , tξ = t(y)e .

0 n+m Then B(sη, tξ) is a continuous function since if η → η then sη0 → sη in D(Ω). (η, ξ) ∈ R . −ix.η n Also, since pn(e ) = (1 + ||η||) we will have some contant c(s, t) such that

n n |B(sη, tξ)| ≤ c(s, t)(1 + ||η||) (1 + ||ξ||)

Now, for intuition we suppose that k(x, y) is a smooth function which generates B. i.e. Z B(φ, ψ) = k(x, y)φ(x)ψ(y)dxdy Ω1×Ω2

50 Then we have that

n+m −1 B(sη, tξ) = (2π) F (k(x, y)s(x)t(y)) (η, ξ) and the inverse Fourier transform is well deﬁned since k(x, y)s(x)t(y) has compact support so is in S . Then after Parseval’s theorem we have for φ ∈ DK , ψ ∈ DL Z Z 1 ˆ ˆ n+m φ(η)ψ(ξ)B(sη, tξ)dηdξ = φ(x)ψ(y)k(x, y)s(x)t(y)dxdy (2π) n+m R K×L Z = φ(x)ψ(y)k(x, y)dxdy = B(φ, ψ). Ω1×Ω2

We would like to extend this to the case of a general B. Since D(Ω1 × Ω2) is an LF-space and we have a sequence of deﬁnition of D(Ω1 × Ω2) of the form

D 0 ⊂ D 0 ⊂ D 0 ⊂ ... L1×L1 L2×L2 L3×L3 if we have a sequence of maps , n = 1, 2, ... such that ∈ D 0 and that Kn Kn Ln×Ln

Kn+1|D 0 = Kn Ln×Ln

0 (n) (n) then this will define a map ∈ D (Ω × Ω ). Take σ ∈ D 0 and s , t as before. Define K 1 2 Ln×Ln Kn by analogy with the case for smooth functions. Z 1 (n) (n) Kn(σ) = n+m σˆ(η, ξ)B(sη , tξ )dηdξ (2π) n+m R n n Since,σ ˆ ∈ S and we have that |B(sη, tξ)| ≤ c(s, t)(1+||η||) (1+||ξ||) this integral will converge and therefore define an element of D 0 . It remains to show that will agree with B on Ln×Ln Kn D ⊗D 0 and since this is dense in D 0 this will show that is independent of the choice Ln Ln Ln×Ln Kn of sn, tn and since we could choose sn+1 = sn and tn+1 = tn that we will have the restriction propety.

Z 1 ˆ ˆ (n) (n) Kn(φ ⊗ ψ) = n+m φ(η)ψ(ξ)B(sη , tξ )dηdξ (2π) n+m R Z 1 ˆ (n) ˆ (n) = n+m B(φ(η)sη , ψ(ξ)tξ )dηdξ (2π) n+m R Z Z 1 ˆ (n) 1 ˆ (n) = B n φ(η)sη dη, m ψ(ξ)tξ dξ (2π) n (2π) m R R Z Z 1 ˆ −ix.η 1 ˆ −iy.ξ = B n φ(η)e dη, m ψ(ξ)e dξ (2π) n (2π) m R R = B(φ, ψ) Therefore we have shown existence.

We know need to answer the question of uniqueness. Suppose K and K 0 both represent the linear map Φ. Then (K − K 0)(φ ⊗ ψ) = 0

51 0 for every φ ⊗ ψ ∈ D(Ω1) ⊗ D(Ω2). Since we know that K − K is continuous on the whole of 0 D(Ω1 × Ω2) and that D(Ω1) ⊗ D(Ω2) this means that K − K = 0.

We can also see that we can see that given such a K we will have Z 1 −x.η −y.ξ K (σ) = Kx,y n+m σˆ(η, ξ)e e dηdξ (2π) n+m R Z 1 −x.η −y.ξ = n+m Kx,y(ˆσe e )dηdξ (2π) n+m R Z 1 −x.η −y.ξ = n+m σˆ(η, ξ)Kx,y(e e )dηdξ (2π) n+m R Z 1 −x.η −y.ξ = n+m σˆ(η, ξ) Φx(e ) y (e )dηdξ (2π) n+m R which writes K explicitly in terms of Φ which also gives uniqueness.

An interesting consequence of this is that if D is a linear partial differential operator with coefficients in E(Ω) then D defines a map from D(Ω) to E(Ω) so consequently a map from D(Ω) to D0(Ω) if we can show this is continuous then there exists K ∈ D0(Ω × Ω) such that Z Z D(φ)(x)ψ(x)dx = K (φ ⊗ ψ)“ = ” K (x, y)φ(x)ψ(y)dxdy Ω Ω×Ω which puts integral and differential operators into the same form.

12.1 Regular Kernels The regularity of kernels refer to the regularity of the image of D(Ω) under the linear map to D0(Ω) generated by K . In this section we will relate the regularity property of kernels to partial diﬀerential operators particularly hypoelliptic operators. To begin we make some deﬁnition.

0 Deﬁnition. A kernel K is left regular if the linear map Φ: D(Ω1) → D (Ω2) maps D(Ω1) into ∞ ∞ 0 C (Ω2) when C (Ω2) is considered as a subset of D (Ω2).

t 0 Deﬁnition. A kernel K is right regular if the linear map Φ : D(Ω2) → D (Ω1) maps D(Ω2) ∞ t into C (Ω1). Here Φ is deﬁned by

(Φt(ψ))(φ) = K (φ ⊗ ψ)

Deﬁnition. A kernel K in D(Ω × Ω) is called very regular if K is both left and right regular and is equal to a smooth function away from the diagonal of Ω × Ω.

t 0 0 Proposition. If K is left regular then we can extend Φ to a function from E (Ω2) into D (Ω1).

Proof. Since for any φ ∈ D(Ω1) we have that

Φ(φ) ∈ E(Ω2)

0 then we know that if u ∈ E (Ω2) that u(Φ(φ))

52 makes sense. Then we can deﬁne Φ˜t by

(Φ˜t(u))φ = u(Φ(φ))

t ˜t 0 which clearly extends Φ so it remains to show that Φ is continuous. Since D (Ω1) is a locally convex space, it is sufficient to prove that Φ˜t is bounded on bounded sets or equivalently that 0 its restriction to bounded sets is continuous. Since the bounded sets in E (Ω2) are metrizable it is suficient to prove that it is sequentially continuous at 0. So let us fix B a bounded set in D(Ω1) and we want to show that if uj → 0 then

˜t sup |(Φ (uj))(φ)| → 0 φ∈B equivalently that sup |uj(Φ(φ))| φ∈B but we already know that Φ(φ) is bounded in E(Ω2) and uj → 0 strongly so on bounded sets so we are done.

If K is a kernel in D0(Ω × Ω) then we can let a diﬀerential operator D act on K by looking at (D ⊗ I)K = DxK (x, y) We say that K is a left fundamental kernel for the linear diﬀerential operator D if

DxK (x, y) − δ(x − y) = 0 and K is a parametrix associated to D if

DxK (x, y) − δ(x − y) ∈ E(Ω × Ω).

In general the existence of a right fundamental kernel gives existence via convolution.

DxK (x, y) − δ(x − y) = 0 implies that Dx(K (x, y) ∗ v(y)) = v(x). Also, the existence of a left fundamental kernel will give uniqueness since if we have a solution to the homogeneous kernel. Dyu(y) = 0 then we will have 0 = K (x, y)Dyu(y) = δ(x − y) = u(x) which gives uniqueness, roughly speaking. We can use fundamental kernels and parametrices to study the linear diﬀerential operators to which they are associated. Before this though we make a helpful lemma.

Lemma. Let dΩ = { (x, y) ∈ Ω × Ω | ||x − y|| < }. Then if a linear diﬀerential operator D has a parametrix which is very regular, it will have a parametrix which is very regular and whose support lies inside dΩ.

53 n Proof. Here we follow [4]. Let η ∈ D(R ) such that suppη ⊂ B and η is identically zero in some neighbourhood of the origin. Suppose K is a parametrix associated to the diﬀerential operator D. Then we look at K˜ = η(x − y)K and seek to show that this is also a parametrix for D. Since, K is smooth away from the origin and where x, y are suﬃciently close together that η is identically one near by, we have that α ∂x η(x − y) = 0 in a small neighbourhood of the diagonal. Therefore

Dx(η(x − y)K ) − η(x − y)K ∈ E(Ω × Ω)

We can also see that η(x − y)δ(x − y) = δ(x − y) as the support of δ(x − y) is the diagonal and η(x − y) is one in some neighbourhood of the diagonal. As K is a parametrix for D we know

DxK − δ(x − y) ∈ E(Ω × Ω) so that η(x − y)DxK − δ(x − y) = η(x − y)(DxK − δ(x − y)) ∈ E(Ω × Ω). Therefore K˜ − δ(x − y) = K˜ − η(x − y)K + (η(x − y)K − δ(x − y)) ∈ E(Ω × Ω).

We now introduce the notion of hypoellipticity which we shall relate to regularity properties of the kernels.

Deﬁnition. A diﬀerential operator D is hypoelliptic if

Du ∈ E(Ω) implies that u ∈ E(Ω). We can see that both the Laplacian and the Cauchy-Riemann operator is hypoelliptic. We look at a theorem due to Schwartz which is shown in [4].

Proposition. If D t has a parametrix which is very regular then D is hypoelliptic. Proof. Here again we follow [4]. We ﬁx an arbitrary a ∈ Ω and with to show that if Du is smooth in some neighbourhood of a that there is a small neighbourhood of a which we shall call ωa in which u is smooth. We know there exists a parametrix K for D whose support is contained inside dΩ. t DyK (x, y) − δ(x − y) ∈ E(Ω × Ω) 0 0 This shows that the above can be extended to a map from E (Ω) to D (Ω) so if we take φa to be a function of compact support which is identical to one in some neighbourhood of a. Then the above acts on φa.u so we have

t − δ(x − y) φ (y)u(y) = (x, y) ( (φ (y)u(y))) − φ (x)u(x) ∈ E(Ω) DyK y a K y D a a since what we really have is

t − δ(x − y) φ (y)u(y) = (φ .u) ( − δ(x − y)) ∈ E(Ω) DyK y a a y K

54 when considered as a function of x. So since Du is smooth and we can choose the support of K to be small enough that φa(y) is identically one when x is suﬃciently near to a and (x, y) ∈ supp(K ). Then when x is suﬃciently near a in this sense we will have D(φau) = φaDu which will be smooth by hypothesis. We also supposed that K is left regular so that

K (φaDu) ∈ E(Ω).

Therefore, we have that φau is smooth and since φa is identically one in some neighbourhood of a this shows that u is smooth in some small neighbourhood of a. Since a is arbitrary this shows that u is smooth.

It can be shown that, in a kind of converse to the above theorem, if D and D t are hypoelliptic operator on Ω then for every point, x ∈ Ω, there is a small neighbourhood of x in which D has a two sided fundamental kernel. Also, due to Malgrange if the map from E(Ω) to itself generated by D is surjective that we will have a two sided fundamental kernel. However, the proofs of these statements require the machinery of Sobolev spaces so we will not give them here.

13 Nuclear Locally Convex Spaces

In this section we will briefly look at the theory of nuclear mappings and nuclear locally convex spaces which was substantially developed by Grothendieck in his phd thesis. In this section we use the existence and uniqueness of a completion of a non-metrizable topological vector space. This is given in the first appendix. In this section we mainly follow the exposition given in [4], though some of this is given in a very different way in both [10] and [9]. In this section I will in general be very brief and focus on a summary of the results in this theory rather than their proofs. All the spaces of test functions and distributions which we have seen are nuclear spaces. This section is very light on proofs, in particular I will omit even the proof that the spaces we have been dealing with are nuclear. It is intended as a ‘flavour’ of the theory of Nuclear spaces which grew out of the study of the topology of the distributions spaces as a natural extension.

13.1 Nuclear Mappings In order to define a nuclear space it is necessary to first define a nuclear mapping. We will do this by first defining a nuclear mapping between Banach spaces. We then extend this notion to the nuclear mappings between any two locally convex spaces. We recall that L(E,F ) is the space of continuous linear forms from E to F given the topology of convergence on the bounded subsets of E. Definition. If E and F are Banach spaces then L1(E,F ), called the nuclear mappings, is the closure of the space of maps with finite dimensional image.

In order to extend this notion to general locally convex spaces we introduce two ways of producing Banach spaces from locally convex spaces. Firstly, suppose N is a convex, balanced neighbourhood of 0 in E. Then we can deﬁne a new topology on E by the semi-norm which is the Minkowski functional of N. This space will not be Hausdorﬀ unless E is normable but we may quotient by kerµN in order to produce a normed space. We can then complete this space to produce a Banach space which we call EˆN . There is a canonical mapping E → EˆN given by the quotient map. This map will be continuous but not always open since the topology on EˆN does not induce the same topology on the image of this mapping as the quotient topology.

55 The second way is to take B a bounded, convex, balanced subset around 0 in F . Let FB be the span of B and we can put a topology on FB using the Minkowski functional of B. We call B infracomplete if FB is Banach. The inclusion mapping j : FB → F is continuous since B is bounded it is bounded on every semi-norm. So given a neighbourhood U of the origin in F 1 there is some λ such that B ⊂ λU then if x ∈ λ B we will have x ∈ U. Deﬁnition. We call u a nuclear mapping from E to F if there exists a neighbourhood of 0, N, in E and an infracomplete, bounded set, B in F and further a nuclear mapping u˜ : EˆN → FB such that u = i ◦ u˜ ◦ j.

Where i : E → EˆN and j : FB → F are the canonical mappings. It can be shown that if the set of nuclear mappings E to F form a vector subspace of L(E,F ) and that the general condition for nuclearity coincides with that which we have given for mappings between Banach spaces. However, we will not do this here. We also state without proof the following theorem which is reasonably intuitive particularly when we remember that if N is a neighbourhood of 0 in E then its polar forms an equicontinuous set in E0.

1 0 Proposition. If u is a nuclear map from E to F then there exists sequences {λn} ∈ l , {xn ∈ 0 E } equicontinuous and {yn} ∈ B where B is infracomplete and bounded such that

0 u(x) = Σnλnxn(x)yn. This proposition is illustrative of the properties of nuclear maps. It is very clear here, though also from the definition that the nuclear maps can be approximated by finite dimensional maps and further restrictions can be given on these maps. This in particular shows that all nuclear maps are compact. However, it is known that there are compact maps which are not the limit of finite dimensional mappings or in fact nuclear mappings.

13.2 Nuclear Spaces Here we give the many alternate characterizations of nuclear spaces. However, we will not show they are equivalent since the proof relies on an understanding of the dual spaces of E⊗ˆ F and E⊗ˆ πF . Here we introduce the notation that if p is a seminorm on E with semi-ball Np then ˆ ˆ Ep = ENp . Characterizations of Nuclear Spaces. The following criteria are all equivalent and locally convex spaces E with these criteria are nuclear spaces. For every seminorm p there exists a semi norm q with p ≤ q and there is a map Eˆp to Eˆq which is natural and nuclear. This is the extension by continuity of the map E/ker(p) → E/ker(q) which exists since ker(p) ⊂ ker(q).

For every locally convex space F all the continuous linear maps E → F is nuclear.

For every locally convex space the π and topologies are equal on E ⊗ F . We can see in particular that if E is nuclear then the identity map will be nuclear and consequently approximable by finite dimensional maps. Since if T : E → F = T ◦ I for any T a linear map, F a locally convex space then we hope that T ◦ In → T where In → I and are finite dimensional. This gives an indication of the powerful approximation properties that we have in nuclear spaces in particular it gives a clue to the fact that the finite dimensional mappings are dense in L(E0,F ) when E is nuclear and under suitable further assumptions on E,F .

56 Since the completion of E ⊗ F with either the or π topology is the same we now write E⊗ˆ F . We can derive from the above theorem and our knowledge of nuclear maps that we have the following isomorphisms which we state without proof.

0 0 ∼ 0 ∼ 0 D (Ω1)⊗Dˆ (Ω2) = L(D(Ω1), D (Ω2)) = L(D(Ω1), D (Ω2)) where the denotes the topology of convergence on the equicontinuous sets of D(Ω1) when it is 0 identiﬁed with the dual of D (Ω1).

∼ ∞ D(Ω) ⊗ E = Cc (Ω,E) E(Ω) ⊗ E =∼ C∞(Ω,E) If we deﬁne the distributions with values in a locally convex space E, as is done in [10] as the space of continuous linear forms from D(Ω) to E then we will have

D0(Ω) ⊗ E =∼ D0(Ω,E).

This provides a path into studying the vector valued distributions. These isomorphisms all rely on the fact that all the spaces D, D0, E, E0, S and S 0 are all nuclear.

13.3 Nuclear Spaces and the Kernels Theorem. To ﬁnish, we sketch the proof of the Kernels theorem given in [4]. We have already seen that

0 0 0 D (Ω1) ⊗ D (Ω2) ,→ L(D(Ω1), D (Ω2)) and that 0 0 0 D (Ω1) ⊗ D (Ω2) ,→ D (Ω1 × Ω2). 0 0 0 0 ∼ We wish to extend these injections to D (Ω1)⊗Dˆ (Ω2). We claimed above that D (Ω1)⊗Dˆ (Ω2) = 0 0 0 0 L(D(Ω1), D (Ω2)). We also know that D (Ω1) ⊗ D (Ω2) is dense in D (Ω1 × Ω2). Therefore, admitting our claims about the results in nuclear spaces it is sufficient to show that the subspace 0 0 0 topology induced by D (Ω1 × Ω2) on D (Ω1) ⊗ D (Ω2) . This result follows from the fact that every bounded set in D(Ω1 × Ω2) is contained in the convex hull of a set of the form A ⊗ B where A and B are bounded sets in D(Ω1) and D(Ω2) respectively. Schwartz’s original proof gives the result originally for the spaces E0 where there are some simplifications in the proof of the intermediate results which follow from the fact that E is a 0 Fréchet space. He then makes an element of D (Ω1 × Ω2) as an inductive limit of functionals on the spaces of the form D0(K × L) similarly as was done in the fourier transform proof. He then, separately, proves that the isomorphism given is also topological.

A Appendix: Completions

We hope to prove the existence and uniqueness of the completion of topological vector spaces.

Definition. If V is a Hausdorff topological vector space the Vˆ is a completion of V if: (1)Vˆ is complete in the sense of Filters, (2)V can be identified with a dense linear subspace of Vˆ . Proposition. Every topologyical vector space V has a completion.

57 Proof. We will do this similarly to the metric case by looking at equivalence classes of Cauchy Filters. Suppose F and G are two Cauchy filters. We say that they are equivalent if for every neighbourhood of 0 N. There exists A ∈ F and B ∈ G such that a − B ⊂ N. We claim that if F and G are equivalent and F → x, G → y then x = y. Suppose N is a neighbourhood of 0, then let M be a neighbourhood of 0 such that M + M ⊂ N. Then x + M is a neighbourhood of x. Therefore there exists C ∈ F such that C ⊂ x + M. Then there is also A ∈ F ,B ∈ G such that B − A ⊂ M so B ⊂ A + M so similarly B ⊂ A ∩ C + M ⊂ x + M + M ⊂ x + N. And A ∩ C 6= ∅ since F is a filter. We also claim that if F → x and G → x then F ∼ G . Let U be a neighbourhood of 0 then there exists V a neighbourhood of 0 such that V − V ⊂ U. Then there x + V ∈ F ∩ G and (x + V ) − (x + V ) = V − V ⊂ U. Hence F ∩ G . Now we need to develop the idea of adding two filters and multiplying by a scalar. Let B = {A + B | A ∈ F ,B ∈ G } Then since (A + B) ∩ (A0 + B0) ⊃ (A ∩ A0) + (B ∩ B0) we can generate a filter (F + G ) which contains B. Then we want to consider the filter

λF = {λA | A ∈ F }

Now we would like to check that if F ∼ F 0 and G ∼ G 0 then (F + G ) ∼ (F 0 + G 0) and that λF ∼ λF 0. To show this let N be a neighbourhood of 0. Let M be a neighbourhood of 0 such that M + M ⊂ N. Then since F ∼ F 0 there will be A ∈ F ,A0 ∈ F 0 such that A − A0 ⊂ M. Similarly there is B ∈ G ,B0 ∈ G 0 such that B −B0 ⊂ M then we will have (A+B)−(A0 +B0) ⊂ 1 M + M ⊂ N. Now suppose N is a neighbourhood of 0 then λ N is also a neighbourhood of 0 0 0 0 1 0 so there exists A ∈ F ,A ∈ F such that A − A ⊂ λ N. Then λA − λA ⊂ N. This shows that the equivalence classes of Cauchy ﬁlters form a vector space. Now we need to put a topology on this vector space. We do this by deﬁning neighbourhoods of 0 in Vˆ . Let N be a neighbourhood of 0 in V . Then we have the set

N = {α ∈ Vˆ | ∃F ∈ α with N ∈ F }

Let N,M be two neighbourhoods of 0 in V and produce N and M as above. Then we look at

N ∩ M = {α | ∃F ∈ α with N ∈ F , ∃G ∈ α with M ∈ F }

⊃ {α | ∃F ∈ α with N ∈ F ,M ∈ G } = {α | ∃F ∈ α with N ∩ M ∈ F } which shows these can form a base for neighbourhoods in 0 for a topology on Vˆ . Now we would like to show that the topology we have just defined is Hausdorff. Suppose that α 6= 0 is an element of Vˆ . Let F ∈ α. Then let O be the filter of all sets of V which contain 0, this is an element of 0 in Vˆ . Then we know that F is not equivalent to O. So there exists a neighbourhood N of 0 in V such that A − {0} * N for every A ∈ F . Therefore A * N for every A ∈ F . Now let M be a neighbourhood of 0 in V such that M + M ⊂ N. Let G ∈ α and suppose that there exists B ∈ G with B ⊂ M. Then there exists A ∈ F and B0 ∈ G such that A − B0 ⊂ M. Let B00 = B ∩ B0 ∈ G . Then

A + {0} ⊂ A − (B00 − B00) = (A − B00) + B00 ⊂ M + M ⊂ N which is a contradiction. So there doesn’t exist a B ∈ G with B ⊂ M. Therefore M is a neighbourhood of 0 not containing α.

58 Now we would like to construct a canonical linear map T : V → Vˆ . We do this via T (x) being the equivalence class of the filter of neighbourhoods of x in V . This is the class of all filters converging to x. This is an injection by uniqueness of limits of a filter in a Hausdorff space. Suppose N is an open neighbourhood of 0 in V then for every x ∈ N, N will be a neighbourhood of x and therefore a member of the filter of neighbourhoods of x. Therefore T (N) ⊂ N . Suppose T (x) ∈ N then N is in a filter converging to x. Therefore for every M a neighbourhood of 0 in V . We have N ∩ (x + M) 6= ∅ therefore x ∈ N therefore N ∩ T (V ) ⊂ T (N). This shows that T is an isomorphism onto its image. Now we would like to show that the image of T is dense in Vˆ . We would like to do this by showing that given any open set U that T (V ) ∩ U 6= ∅. Fix α ∈ U then there is some N a neighbourhood of 0 in V such that α+N ⊂ U. Pick some F ∈ α and let M be a neighbourhood of 0 such that M +M ⊂ N. Then since F is a Cauchy filter there exists A ∈ F with A−A ⊂ M. We pick any x ∈ A and would like to claim that T (x) is in α + N . This is equivalent to saying that T (x) − α ∈ N . We want to find some G ∈ T (x) − α with N ∈ G . Now (x + M) − A ⊂ A + M − A ⊂ M + (A − A) ⊂ M + M ⊂ N

Which means that N ∈ F (x) − F ∈ T (x) − α and hence T (x) ∈ α + N . Finally, we would like to show that Vˆ is complete. We will denote Cauchy filters on Vˆ with hats on top like Fˆ to distinguish them from Cauchy filters on V . Suppose Fˆ is a Cauchy filter on Vˆ and Gˆ is the Cauchy filter of neighbourhoods of 0 in Vˆ . We can generate the filter Fˆ + Gˆ and we showed in the proof that LF-spaces are complete that this filter converges iff Fˆ converges. So we would like to show that Fˆ + Gˆ converges. Fˆ + Gˆ is a Cauchy filter since if N is a neighbourhood of 0 in Vˆ and M is another neighbourhood of 0 in Vˆ such that M + M − M ⊂ N then there exists A ∈ Fˆ such that A − A ⊂ M hence (A + M) − (A + M) ⊂ M + M + M ⊂ N

Now we would like to relate Fˆ + Gˆ to a Cauchy filter on V . Let A ∈ Fˆ + Gˆ then let A = T −1(A ∩ T (V )). Let F be the collection of such A. We claim that F is a filter. Suppose ˆ ˆ ˆ ˆ S A ∈ F + G then there is a B ∈ F and G ∈ G such that B + G ⊂ A and B + G = α∈B(α + G) so is open. Therefore since T (V ) is dense in Vˆ we know that T (V ) ∩ (B + G) 6= ∅ so that T (V ) ∩ A= 6 ∅. This shows that ∅ ∈/ F . Suppose that A, B ∈ F then A ∩ B = T −1(T (V ) ∩ A) ∩ T −1(T (V ) ∩ B) = T −1(T (V ) ∩ A ∩ B) and A ∩ B is in Fˆ + Gˆ so A ∩ B ∈ F . Furthermore, we will show that F is a Cauchy filter. To do this we fix N a neighbourhood of 0 in V . Then since Fˆ + Gˆ is a Cauchy filter there exists A ∈ Fˆ + Gˆ such that A − A ⊂ N . We hope that A − A ⊂ N. Since T −1 is linear we have

A − A = T −1(T (V ) ∩ A) − T −1(T (V ) ∩ A) = T −1(T (V ) ∩ (A − A))

⊂ T −1(T (V ) ∩ N ) ⊂ T −1(T (N)) = N since the closed neighbourhoods of 0 form a local base for V this shows that F is a Cauchy ﬁlter. Now let α be the equivalence class of F . We claim that Fˆ + Gˆ → α. To show this let N be a basic neighbourhood of 0 in Vˆ then we want to show that α + N ∈ Fˆ + Gˆ. We do this by showing that there is some A ∈ F such that A ⊂ T −1((α + N ) ∩ t(V )). In this case there is some A ∈ Fˆ + Gˆ such that A = T −1(A ∩ T (V )) ⊂ T −1((α + N ) ∩ T (V ))

59 so A ∩ T (V ) = (α + N ) ∩ T (V ) and A ⊃ B + G where B ∈ Fˆ and G ∈ Gˆ so that B + G will be open in Vˆ which means that B + G ⊂ α + N since both the above sets are open and T (V ) is dense. This will mean that Fˆ + Gˆ → α. It remains to prove that such an A exists. We know that T (x) ∈ α + N means that T (x) − α ∈ N which means that there exists G ∈ T (x) − α with N ∈ G . We will in fact show that we can take G = F . Choose M a neighbourhood of 0 such that M + M ⊂ N. Since F ∈ α is Cauchy there is A ∈ F such that A − A ⊂ M pick some x ∈ A then x + M is in the ﬁlter of neighbourhoods of x. So we have

x + M − A ⊂ A + M − A ⊂ M + M ⊂ N which shows that T (x) ∈ α + N for every x ∈ A. i.e. A ⊂ T −1((α + N ) ∩ T (V )).

Proposition. If V1 and V2 are two completions of V then they are isomorphic. −1 Proof. We have the maps Ti : V → Vi for each i. So take T1 ◦ T2 : T1(V ) → V2 which is a homeomorphism from a dense subspace of V1 onto a dense subspace of V2. Therefore we can extend it to a homeomorphism from the whole of V1 to a complete subspace of V2 which contains T2(V ) and is therefore the whole of V2 hence they are isomorphic.

References

[1] J. J. Duistermaat, J. A. C. Kolk, Distributions: Theory and Applications, Birkhäuser, 2010, [2] Walter Rudin, Functional Analysis, McGraw-Hill, 2005 [3] Laurent Schwartz, Théoriedes Distributions, Hermann, 1966 [4] Fran¸coisTreves, Topological Vector Spaces, Distributions and Kernels, Academic Press, 1967 [5] Laurent Schwartz, “Sur l’impossibilitéde la multiplications des distributions”, C.R. Acad. Sci. Paris 239:847-848, 1954 [6] Jesper Lützen,The Prehistory of the Theory of Distributions, Springer-Verlag, 1982 [7] Gerald B. Folland, Introduction to Partial Differential Equations, Princeton University Press, 1995 [8] Lars Hörmander,The Analysis of Linear Partial Differential Operators II, Springer, 1983 [9] Albrecht Pietsch, Nuclear Locally Convex Spaces, Springer-Verlag, 1972 [10] Laurent Schwartz, Théorie des Distributions àValeurs Vectoriels. I, Annales de l’institut Fourier, tome 7, 1957, p. 1-141 [11] Laurent Schwartz, Théoriedes Noyaux, Proceedings of the International Congress of Math- ematicians, 1950, vol I, p. 220-230

60 [12] O. Heaviside, On Operators in Mathematical Physics, Proc. of the Royal Society, London, 52 (1893), p.105-143 [13] N. Bourbaki, Elements of Mathematics: Topological Vector Spaces, Springer 1987