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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) 93.1 The spaces DK . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103.2 LF-spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123.3 D(Ω) is an LF-space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143.4 Semi-norms defining the topology on D(Ω) . . . . . . . . . . . . . . . . . . . . . . 15
4 Properties of the Space D(Ω) 154.1 Completeness of D(Ω) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154.2 Non-metrizability of D(Ω) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184.3 D(Ω) is a Montel Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194.4 Separability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
5 The space D′(Ω) 205.1 Positive distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215.2 Topologies on D′(Ω) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215.3 Non-metrizability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225.4 These Spaces are Hausdorff . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 245.5 The Banach-Steinhaus Theorem & Barrelled Spaces . . . . . . . . . . . . . . . . 245.6 The Banach-Alaoglu-Bourbaki Theorem . . . . . . . . . . . . . . . . . . . . . . . 255.7 Sequential Compactness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 275.8 More Montel Spaces & Equivalence of the Topologies on Bounded sets . . . . . . 285.9 The Weak topology on D(Ω) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 295.10 Reflexitivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 305.11 Completeness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315.12 The space D as a subset of D′ and Separability. . . . . . . . . . . . . . . . . . . . 32
6 Multiplying Distributions 336.1 Multiplication between subspaces of functions and D′ . . . . . . . . . . . . . . . 336.2 The Impossibility of Multiplying two Distributions in General . . . . . . . . . . . 346.3 Division by Analytic Functions in the Case of One Real Variable. . . . . . . . . . 35
6.3.1 Division by x . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 366.3.2 Division by Polynomials. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
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6.3.3 Division by Analytic Functions with Zeroes of Finite Order. . . . . . . . . 366.4 Applications of Multiplication. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
7 Rapidly Decreasing Functions. 387.1 Relation between S and D(Sn) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 407.2 The Fourier transform on S . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
8 Tempered Distributions 418.1 Relation between S ′ and D′(Sn) . . . . . . . . . . . . . . . . . . . . . . . . . . . 428.2 The Fourier Transform on S ’ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
9 The space E ′. 43
10 Comparison and Relations between the spaces D,S and E and their duals. 43
11 Tensor Product of two Distributions 4411.1 The Tensor Product of two Vector Spaces . . . . . . . . . . . . . . . . . . . . . . 4411.2 The Tensor Product of Function Spaces . . . . . . . . . . . . . . . . . . . . . . . 4411.3 Tensor Products of Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
11.3.1 The Space L(D(Ω1),D′(Ω2)) . . . . . . . . . . . . . . . . . . . . . . . . . 4811.4 Topological Tensor Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
11.4.1 The π-topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4811.4.2 The ε-topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
12 Kernels and the kernels theorem. 4912.1 Regular Kernels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
13 Nuclear Locally Convex Spaces 5513.1 Nuclear Mappings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5513.2 Nuclear Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5613.3 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 developedby Laurent Schwartz between 1945 and 1950. We begin this introduction with some discussionof the history and uses of distributions and will then move on to an outline of the main topicsof the essay. Much of the historical details are taken from [6].
Examples of generalized functions pre-date the rigorous development of distributions. Aparticularly prevalent example is Heaviside’s operational calculus in [12] where he, unrigorously,uses algebraic manipulation of differential operators to treat many problems, particularly inelectrodynamics. Schwartz was unaware of Heaviside’s calculus when he initially created thetheory of distributions. He was told of it by electrical engineers who were able to put the theoryof distributions to great use. In [6] Lutzen says that the theory of distributions was well receivedby 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 formson a space of test functions. Schwartz did not know of this work when he began to work on the
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generalized solutions to partial differential equations. He originally worked by introducing theconvolution 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 madeby Schwartz in October 1944. It was originally very fruitful and Schwartz was able to prove theanalogies of many of the theorems which hold in the current theory of distributions. However, itbecame more challenging when dealing with Fourier transforms. Consequently, in 1945 Schwartzswitched to studying D′. Lutzen reports in [6] that Schwartz, himself, felt that there were twofactors which made D′ 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 privatework on the dual spaces of Frechet spaces allowed him to manipulate these dual spaces withouthaving a ‘concrete representation’. Secondly, he already knew that δ as a measure could alreadybe represented as a functional.
Distribution spaces have many applications, particularly in PDE, which are largely notaddressed in this essay, in particular the work done by Lars Hormander. Furthermore, thespace of tempered distributions allows for many generalizations and extensions of theories usingFourier analysis.
The distributions generalize the way in which locally summable functions and measurescan 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 ofPDEs outside function spaces. This essay begins by introducing the space of test functions andstudying its topology, especially results about compactness and metrizability. This space is thesmooth functions of compact support. The topology on this space was developed by Schwartz ina method which was extended by Schwartz along with Dieudonne to a general class of topologicalvector space called LF-spaces. We work by first introducing the LF-space in general and seeinghow D can be viewed as a particular case.
The Theory of Distributions is a theory of duality. We define the space of distributions tobe the analytic dual of the test functions. We can also define many operations on the spaceof distributions; multiplication, differentiation, Fourier transform, in terms of the adjoint mapof the maps on the space of test functions where they are defined in the classical sense. Wecan, therefore, discover many topological properties of the space of distributions by looking attheorems about dual topologies. For example, the Banach-Alaoglu-Bourbaki Theorem whichshows us that weak-* bounded, weak-* closed sets in the dual of a Topological Vector Spaceare weak-* compact. Again in exploring the dual topologies we will focus on compactness andmetrizablity 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 (soalso sequentially compact). This will lead to the equivalence of the weak-* and strong dualtopology on bounded sets and the reflexivity of all spaces.
We then explore the multiplication and division of distributions. Multiplication, can bedefined by duality between the space E of smooth functions and the distributions. We willexplore the continuity properties of this mapping and show that it cannot be extended to amultiplication between two distribution spaces, though here we only look at the case of onevariable. We then go on to look at the case where the smooth function we multiply is fixed andexamine the invertibility of this operation. This shows another example of how distributionsdiffer from functions. Division by an analytic function is possible in general even when thefunction 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
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results make clear and rigorous how to pose linear PDE with smooth coefficients in the space ofdistributions.
We will go on to look at some other spaces of distributions which are found by enlargingthe space of test functions. These are the space S ′ of tempered distributions which is definedas the dual of S the space of rapidly decreasing function. This space is particularly useful forFourier analysis, which we will look at extremely briefly. We also look at the space E ′ which arethe distributions of compact support. The topology of all these spaces are very similar and wewill not reprove all the results that we gave for the space D′.
The next section looks at tensor products of distributions which allows us to develop thetheory of kernel operators and examines the way in which an element of D′(Ω×Ω) can act as amapping from D(Ω) to D′(Ω). We also look at the topology on tensor products as developed byGrothendiek 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 formD(Ω) to D′(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 Grothendieckwhen he was Schwartz’s student. This theory is very beautiful but in this section we showhow the algebraic, not the topological isomorphism, can be found from a much simpler proofusing Fourier transforms. We then look at fundamental kernels and parametrices for differentialoperators 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’sproof of the kernels theorem and of a very elegant similar proof given in [4]. Nuclear spaces alsoaided Schwartz in creating the theory of vector valued distributions in [10].
Acknowledgements
I would like to thank Prof. Clement Mouhot for supervising this essay, and also Tim Talbotfor helping me understand some of the material on topological vector spaces viewed throughtheir neighbourhoods and particularly for pointing out that not all neighbourhoods are openwhich is a piece of information without which nothing makes any sense! I also used to a greatextent 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 VectorSpaces.
As the theory of distributions concerns topological vector spaces, there are several definitionsand 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 thenormal 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 spaceV by looking at the neighbourhood base around 0. In particular addition and multiplicationare separately continuous since
λ 7→ (λ, x)
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for x fixed is continuous andx 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 neighbourhoodof 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 constructthe set
B′ = x ∈ B|∃A ∈ B, x+A ⊂ B
we hope these will be the open neighbourhoods of 0. Let B′ be the collection of such B′ for eachB ∈ B then the set:
A = x+B′|x ∈ V,B′ ∈ B′
we hope will form a base for a topological vector space topology on V provided that the setsin B satisfy some conditions to make them consistent with the axioms of a topological vectorspace. 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 thatφ ∈ λA.
To find what conditions we need the set B to satisfy we will look at the neighbourhoods of0 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 of0 in V then there must be some open neighbourhoods of 0 M,L such that M + L ⊂ N . (Sincethe topology on V ×V has a base of the form M×L for M,L open in V .) If we take M ′ = M ∩Lthen M ′ +M ′ ⊂ N . Since every neighbourhood of 0 in V contains an open neighbourhood thismeans we have the following property:
(A) If N is a neighbourhood of 0 there will exist a neighbourhood of 0, M , such thatM +M ⊂ N .
Since, multiplication is continuous given an open neighbourhood of 0, N , we have t > 0 andM a neighbourhood of 0 such that if |λ| < t and x ∈M we have λx ∈ N . Therefore
M ′ =⋃|λ|<t
λM ⊂ N
and by scaling M we can ensure that t = 1 so that M ′ is balanced. So we have
(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 notto be 0) so if N is an open neighbourhood of 0 there exists M , an open neighbourhood of 0,such that 1
λM ⊂ N . Hence M ⊂ λN so we have the following property:
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(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 openneighbourhood of 0 there exists t > 0 such that |λ| < t means that λx ∈ N hence x ∈ 1
λN . Sowe 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 setsthat can be generated by arbitrary unions and finite intersections of elements of B will have theproperties 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 ′ ∈ B then there exists M ∈ B such that
M ⊂ N ∩N ′
We now claim that A as defined above will define a vector space topology on V . First weshow that it does indeed define a topology.
y ∈ (x+A) ∩ (x′ +A′)
we have0 ∈ (x− y +A) ∩ (x′ − y +A′)
and since 0 ∈ x − y + A we know that y − x ∈ A so by the definition of B′ there exists BB′such that y − x+ B ∈ A and similarly we can find B′ such that y − x′ + B′ ∈ A′ so if we takeC = B ∩B′ then there is C ′ ∈ B′ such that C ′ ⊂ C then
y + C ′ ⊂ (x+A) ∩ (x′ +A′)
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.
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and also we are given A ∈ B′ we would like to find B,C ∈ B′ such that
(y +B) + (z + C) ⊂ x+A
we know that we have D ∈ B′ 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 givenx, y ∈ V and λ ∈ F such that
λ.y = x
and we are also given A ∈ B′ then we would like to find ε > 0 and B ∈ B′ such that if |t| < εthen
(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 ε such that
ty ∈ C, ∀ |t| < ε
and since C is balanced and λC will be an open neighbourhood of 0 for all non-zero λ we canfind B such that
(λ+ t)B ⊂ C ∀ |t| < ε
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 ′ whichprescribe the same neighbourhoods of 0 then τ = τ ′. This shows the topology constructed above isthe unique one admitting this neighbourhood base so we have an identification between topologiesfor TVS and neighbourhood bases.
Proof. In this case let B be the collection of neighbourhoods of 0 in both τ and τ ′. . Further,suppose U is an open set in τ then if x ∈ U, ∃V ∈ B′ with x + V ⊂ U then since V is aneighbourhood of 0 in τ ′ there will be a V ′ ∈ τ ′ with 0 ∈ V ′, V ′ ⊂ V then x + V ′ ⊂ U so U isopen in τ ′ and this argument is symmetric so τ = τ ′.
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Further, most of the spaces I will use will be locally convex and Hausdorff. (Because of thedefinition of a TVS any T1 space will be Hausdorff so the Hausdorff condition is necessary thatthe topology ’sees’ the difference between any two points.) In particular to show that a spaceis Hausdorff it is sufficient to show that given any x ∈ V there is some N a neighbourhood of 0with 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) = infλ > 0|x ∈ λB
This is called the Minkowski functional of the set B. Therefore, we can equivalently define alocally 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) < ε1, ..., pik(x) < εk
where k ∈ N, i1, ..., ik ∈ I, ε1, ...εk ∈ R.We can see here that we automatically have continuity of addition for any convex, balanced
possible base as if N is a convex neighbourhood base then 12N + 1
2N ⊂ N .We also have a few helpful little things like for N any convex, balanced neighbourhood of 0
we will have:1
2N − 1
2N =
1
2N +
1
2N ⊂ N,
1
3N +
1
3N +
1
3N ⊂ N,
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. Thereforethe topology we described above generated by semi-norms is always a TVS topology. And thetopology 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⋂i∈I
ker(pi) = 0
Definition. A Frechet space 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-metricspaces.
Definition. A subset B of a topological vector space is said to be bounded if for every N aneighbourhood 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 forx ∈ V .
Proposition. In a locally convex space defined by a collection of semi-norms pi, i ∈ I a set Bis bounded iff pi(B) is bounded for each i.
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Proof. Let Ni = x|pi(x) ≤ 1 and let B be bounded. Then there exists λ with λNi ⊃ B so ifx ∈ B then pi(x) ≤ λ.
Conversely, suppose that B is a set with x ∈ B ⇒ pi(x) ≤ λi. Then let N be an arbitraryneighbourhood of 0 since the semi-norms define the topology there is a set
N ′ = x|pi1(x) ≤ ε1, ..., pin(x) ≤ εk
with N ′ ⊂ N then let µ = max(λi/εi, ..., λn/εn). Then µN ′ ⊃ B so µN ⊃ B.
Definition. A sequence (xj) in a topological vector space is Cauchy if for every N a neighbour-hood 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 thereexists 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 .( This is because λx+ y = (λ+ 1)( λ
λ+1x+ 1λ+1y).
φj ∈ λN +N ⊂ (λ+ 1)N
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 positivelyhomogeneous, subadditive funtional. Let Y be a closed subspace of X and f : Y → R a linearmap such that |f(y)| ≤ p(y) for all y ∈ Y . Then f extends to f a linear map on X andfurthermore |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 maxk
pik(x).
3 The topology of D(Ω)
We will construct a topology on D(Ω) the space of continuous functions of compact supportinside Ω ⊂ Rn via a limiting process on spaces of the form DK which are defined below. Thistopology is called an LF-space topology. Also following Schwartz in [3] we will give an explicitcollections of semi-norms which define the same topology.
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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 setof functions whose support is contained inside K. This is not the same as the space C∞c (K) =C∞(K) as the functions must be able to be smoothly extended to the whole of Ω. Then wehave
D(Ω) =⋃K
DK =⋃n
DKn
where ∪nKn = Ω. For example in Ω = Rn we can take Kn = x | |x| ≤ n.We can define a Frechet space topology on each DK via the semi-norms
pn,K(φ) = Σ|α|≤n supx∈K|∂αφ(x)|
We now show that this does indeed define a Frechet space topology:(1) DK is metrizable:
Since the semi-norms make DK into a locally convex, Hausdorff space and there are countablymany of them we have the standard construction of a metric.
d(φ, ψ) = Σn2−npn(ψ − φ)
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:
pn(a+ b)
1 + pn(a+ b)≤ pn(a)
1 + pn(a)+
pn(b)
1 + pn(b)
by summing both sides. This in turn follows from
pn(a+ b) ≤ pn(a) + pn(b)
⇒ 1
pn(a) + pn(b)≤ 1
pn(a+ b)
⇒ 1 +1
pn(a) + pn(b)≤ 1 +
1
pn(a+ b)
⇒ 1
1 + 1pn(a+b)
≤ 1
1 + 1pn(a)+pn(b)
⇒ pn(a+ b)
pn(a+ b) + 1≤ pn(a)
1 + pn(a) + pn(b)+
pn(b)
1 + pn(a) + pn(b)
⇒ pn(a+ b)
1 + pn(a+ b)≤ pn(a)
1 + pn(a)+
pn(b)
1 + pn(b)
So this defines a metric.
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Now we want to show that this metric defines an equivalent topology. We do this by showingboth topologies have the same neighbourhoods of 0. Look at the set
φ|d(0, φ) ≤ ε.
Then there exists an N such that
Σn≥N+12−n ≤ ε/2.
Then there exists λn, n = 1, ..., N such that
pn(φ) ≤ λn ⇒ 2−npn(φ)
1 + pn(φ)≤ ε/2N
then the setφ|pn(φ) ≤ λn, n = 1, ..., N
is contained in the ε ball so the epsilon ball is a neighbourhood of 0 in the semi-norm topology.Now since
pn(φ) ≤ ε⇒ pn−1(φ) ≤ ε
it is sufficient to look at the setsφ|pn(φ) ≤ λ
then there exists ε such that
2−npn(φ)
1 + pn(φ)≤ ε⇒ pn(φ) ≤ λ
so the ε ball in the metric topology is contained in this set. So this set is a neighbourhood of 0in the metric topology.
There are two interesting points which can be taken from this proof. The first is that theε balls are not bounded in the sense of the locally convex space topology as none of the basicopen 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 ε−balls would be bounded via Bε = nεB 1n
but thismetric is not positively homogeneous i.e.
d(0, tφ) 6= t.d(0, φ)
sotBε 6= Bt.ε
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 spacethat can be defined by a countable sequence of semi-norms can be defined by a non-decreasingsequence 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 →∞.Therefore, ∂αφj is a Cauchy sequence with the uniform norm for each α. So ∂αφj → φα
11
uniformly and φα = ∂αφ0 and φ0 ∈ DK since the boundary values of every partial derivativewill be 0. This is the same as saying φj → φ0 in the DK topology.
Proof that φα = ∂αφ0: We do this iteratively by showing it for first partial derivatives.
|φ0(x+ h1)− φ0(x)− φ(1,0,...,0)(x)h1| ≤ |φ0(x+ h1)− φj(x+ h1)|+ |φ0(x)− φj(x)|
+|φj(x+ h1)− φj(x)− ∂x1φj(x)h1|+ |h1||∂x1φj(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
ε away from φ0 and ∂x1φj is uniformly less than ε away from φ(1,0,...,0) so that we have:
φ0(x+ h1)− φ0(x)− φ(1,0,...,0)(x)h1| ≤ 3ε+ |φj(x+ h1)− φj(x)− ∂x1φj(x)h1|
and for this j we can choose h1 so that RHS is ≤ 4ε.
3.2 LF-spaces
In order to introduce the topology on D(Ω) it is first helpful to introduce in general a new kindof 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 Frechet spaces. If we havea sequence
F1 ⊂ F2 ⊂ F3...
of nested vector spaces each with a Frechet space topology. We also ask that the subspace topologyinduced 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, balancedsets. 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 topologyare 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 aneighbourhood of 0 in Fn by the definition of B. So V ∩ Fn ⊃W ∩ Fn is a neighbourhood of 0in the original topology.
Now we want to prove that if W is a neighbourhood of 0 in the original topology on Fn thenthere exists a V which is a neighbourhood of 0 in F s.t. V ∩ Fn = W . It is sufficient to dothis for convex neighbourhoods only. As Fn+1 induces the original topology on Fn there is aneighbourhood 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 aconvex neighbourhood of 0 in Fn+k+1 s.t. Wn+k+1 ∩ Fn+k = Wn+k. Let
V =⋃n
Wn
this has the required properties, and it is a neighbourhood of 0 as it is convex and its intersectionwith each Fn is a neighbourhood of 0 in Fn.
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Proposition. An LF-space is Hausdorff.
Proof. This follows [4] book fairly closely. This is proved similarly to the second part of theabove 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. Weknow 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
andNn+1 ∩ En = Nn.
Then if we take N =⋃nNn 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 containingx then there exists a convex neighbourhood of 0 Nn+1 in En+1 not containing x such thatNn+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 suchthat
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 thiscontains 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 thatty = x− (1− t)z ∈ En
soy ∈ C ∩ En ⊂ Nn
which means that x ∈ Nn as Nn is convex. This gives a contradiction so
x /∈ conv(Nn ∪ C)
and lettingNn+1 = conv(Nn ∪ C)
will give all the required properties.
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3.3 D(Ω) is an LF-space
If K1 ⊂ K2 ⊂ K3... is a nested sequence of compact sets s.t.⋃nKn = Ω then D(Ω) can be
made into an LF-space with sequence of definition DKn , n = 1, 2, 3, .... It can often make moresense to choose
Kn = ωn
where ωn is a relatively compact, open set. This is because if K is an arbitrary compact setthen for any
x ∈ K \ int(K)
we will haveφ(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 ifK ⊂ K ′
then the subspace topology induced on DK by DK′ is the initial topology on DK . This is because
pn,K′ |K = pn,K
so the semi-norms defining both topologies are the same.
Proposition. This topology is independent of the choice of Kn.
Proof. Suppose K1 ⊂ K2 ⊂ K3 ⊂ ... and K ′1 ⊂ K ′2 ⊂ K ′3 ⊂ ... are two such sequences. Let τ, τ ′
be the corresponding topologies. Suppose N is a convex neighbourhood of 0 in τ then we wantto look at N ∩ DK′n . There exists an m such that Km ⊃ K ′n so
N ∩ DK′n = (N ∩ DKm) ∩ DK′nand we already know that N ∩ DKm is a neighbourhood of 0 in DKm . We know that DKminduces the right subspace topology on DK′n so N ∩ DK′n is a neighbourhood of 0 in DK′n soN is a neighbourhood of 0 in D(Ω). This argument is symmetric which is sufficient for ourconclusion.
From above it is clear that this will induce the right topology on DK either by choosing itto be part of the sequence of definition of noting that in any such sequence there must be an ns.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-bourhoods of 0 in DK . If N = φ|pn(φ) ≤ C then N ′ = φ| ||φ||n ≤ C is a neighbourhood of0 in D(Ω) s.t. N ′ ∩ DK = N . Where ||φ||n = Σ|α|≤n supx∈Ω |∂αφ(x)|.
There is an important consequence of the definition of this topology for working out whethera 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 iscontinuous 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
u−1(U) ∩ DK = (u|DK )−1(U)
so if U is a convex, balanced neighbourhood of 0 u−1(U) is convex so is open iff u|−1DK (U) is open
for every K which gives the result.
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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. Westart by looking at a neighborhood base which is given explicitely in Schwartz’s book. If
∅ = U1 ⊂ U1 ⊂ U2 ⊂ U2 ⊂ U3 ⊂ U3 ⊂ ...
is a sequence of relatively compact open sets s.t.⋃n Un = Ω, and m1 < m2 < m3 < ... is a
strictly increasing sequences of integers, ε1 ≥ ε2 ≥ ε3 ≥ ... a sequence of real numbers decreasingto 0. Then
V (U,m, ε) = φ | Σ|α|≤mi supx/∈Ui
|∂αφ(x)| ≤ εi for each n = 1, 2, 3, ...
letting m, ε 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 neighbour-hoods in the other topology then the two topologies are the same.
Let Kn = Un.
V (U,m, ε) ∩ DK = φ | Σ|α|≤mi supx/∈Ui
|∂αφ(x)| ≤ εi, i = 1, 2, ..., n
so this is a neighbourhood of 0 in DKn as it contains the set
φ|pmn(φ) ≤ εn ∩ φ|pmn−1(φ) ≤ εn−1 ∩ ... ∩ φ|pm0(φ) ≤ ε0.
Now suppose that V ∈ B then for each Kn, V ∩ DKn is a neighbourhood of 0 in DKn socontains a set of the form φ|pmn(φ) ≤ εn as the semi-norms are non-decreasing. ThereforeV (U,m, ε) ⊂ V so these two topologies coincide.
This topology can then be defined by the semi-norms
N(m, ε)(φ) = supn
( sup|α|≤mn,x/∈Un
|∂αφ(x)|/εn).
4 Properties of the Space D(Ω)
Here we will explore some results relating to compactness, completeness, separability and metriz-ability of the space D(Ω). These will have several implications fro the topological properties ofthe 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 topologicalcompleteness. Since D(Ω) is not first countable it is not sufficient just to look at sequences. Thisalso 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 waysequences 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 .
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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 Hausdorffthere exists U, V neighbourhoods of x and y respectively such that U ∩ V is disjoint. Howeversince 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 theassociated filter F is the subsets of E which contain all but finitely many elements of thesequence. 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 ∈ Fs.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∩Band 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 allneighbourhoods of 0. Let F ′ 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 ′ + V ′) ⊃ (U ∩ U ′) + (V ∩ V ′) .
Suppose F ′ converges to x then we claim that F converges to x. This is because if F ′ convergesto x, then for every N a neighbourhood of x there exists B ∈ F ′ with B ⊂ N then we havesome U a neighbourhood of 0 and V ∈ F with U + V ⊂ B. So we have V ⊂ B which meansthat V ⊂ N so F converges to x.
Also, F ′ is a Cauchy filter. If N is a neighbourhood of 0 then there exists M a balancedneighbourhood of 0 such that +M+M+M ⊂ N . Then there exists A ∈ F such that A−A ⊂Mthen we will have that
(A+M)− (A+M) = A−A+M −M = (A−A) +M +M ⊂ N
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 allthe axioms of a filter except, possibly, that it may contain ∅, this shows it is a filter. If N is aneighbourhood of 0 in En then it contains a convex, balanced neighbourhood M . Then thereexists a neighbourhood of 0 M ′ in E such that M ′ ∩En = M . Then since F ′ is a Cauchy filterthere will be some A ∈ F ′ such that A−A ⊂M ′ 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 everyn there is a set Un + Vn ∈ F ′ where Un is a convex neighbourhood of 0 and V ∈ F such that
(Un + Vn) ∩ En = ∅
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then letU =
⋃n
Un, V =⋃n
Vn
. Since Un is open for each n this means U will be open so
U + V ∈ F ′
andU + V =
⋃x∈V
(x+ U)
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 =⋃n
(Un + Vn) ∩ E1 ⊂⋃n
(Un + Vn) ∩ En = ∅
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 thatFN converges in En. Since En is metrizable we have a countable neighbourhood base of 0 interms of the balls of radius 1
n around 0. Call these B 1n
. Since FN is a Cauchy filter there existsAn ∈ FN such that An −An ⊂ B 1
n. Then for each n we can choose
xn ∈⋂
1≤k≤n
Ak
so that xn will be a Cauchy sequence. Since En is a Frechet space it is complete so xn convergesto some element x.
We then claim that FN converges to x. Let Mn be the ball of radius 1n around x. Then
there exists some K so that m ≥ K means that d(xm, x) ≤ 12n and let m be possibly larger so
that Am − Am ⊂ B1/2n then xm ∈ Am so Am ⊂ Mn. It follows from this that F ′ converges tox since for each M a neighbourhood of x in E there is some
Mn ⊂M ∩ EN
and we can find an A ∈ FN with A ⊂ Mn and we claim that we can choose C ∈ F ′ such thatC ⊂M and C ∩ EN = A.
We find this C by iterating upwards. So we find a A′ ∈ EN+1 and A′ ∩ EN = A andA′ ⊂M ∩EN+1 by noticing that FN+1 also doesn’t contain the empty set. Then we can moveupward and take the union over all these sets to get C. Since, at the beginning we noticed thatif F ′ converges to x then so does F we have shown that F converges to x and therefore that Eis complete.
We also proved here that any metric space in which every Cauchy sequence converges willalso have that any Cauchy filter converges.
In the specific case of D(Ω) we also have a quicker way of seeing that all Cauchy sequencesconverge.
We recall that in a locally convex space Cauchy sequences are bounded.
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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 ofcompacts Kn increasing to Ω there must be an xn, φn s.t. xn ∈ Kn \Kn−1 and φn ∈ E withφn(xn) 6= 0. Then let
N = φ| |φ(xn)| < 1
n|φn(xn)|
since only finitely many of the xn are in each Kn N is in B. However φn /∈ nN for each n sothere 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 thefact that DK is complete. So since any Cauchy sequence is bounded it will be contained insidesome 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 closedsets with
⋃n Fn = X there exists N with intFN 6= ∅.
With this definition the Baire Category theorem states that a complete, metric space is aBaire space.
Proposition. D(Ω) is not a Baire space.
Proof. DK is nowhere dense since if V ⊂ DK and K ′ ∩K then V ∩ DK′ = ∅ so V cannot be aneighbourhood of 0. DK is a linear subspace of D(Ω) has empty interior. As for suitably chosensequence Kn,D(Ω) =
⋃nDKn 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 topologicallycomplete if it were metrizable it would be completely metrizable this shows that D(Ω) is notmetrizable. Furthermore, as D(Ω) is locally convex, if it were first countable then it would bemetrizable. 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 neigh-bourhoods which are balanced.
Proposition. If X is a Topological Vector Space with a balanced, countable neighbourhood baseVn in which every Cauchy sequence converges then X is a Baire space.
Proof. We prove that if Un is a countable sequence of open dense sets then⋂n Un in X. Let
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 andW1 ⊂ V1. Then (x1 +W1) ∩ U2 has non-empty interior so there exists x2 and W2 ⊂ V2 suchthat x2 + W2 ⊂ (x1 +W1) ∩ U2. Itteratively, we can produce the sequence x1, x2, x3, ... andW1,W2,W3, ... such that
xn +Wn ⊂
(n⋂k=1
Un
)∩W
We then claim that x1, x2, x3, ... is a Cauchy sequence. Then let m ≥ n
xm ∈ xn +Wn ⇒ xm − xn ∈Wn
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so this is a Cauchy sequence. Therefore there is some x with xn → x so x ∈ xn +Wn for eachn 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 andhas 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 boundedset in D(Ω) is contained inside some DK then if we can show that each DK is a Montel spacewe will have the result.
In order to do this we first recall the Arzela-Ascoli Theorem.
Arzela-Ascoli Theorem. If K is a compact, Hausdorff topological space. Let C(K) be theBanach space of continuous functions K → R with the supremum norm. If E is a uniformlybounded, 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 subse-quence.
Given this let φj be a bounded sequence in DK then φj is uniformly bounded in eachof its partial derivatives which shows that φj is equicontinuous. So for each α, ∂αφjis a uniformly bounded, equicontinuous set and therefore has a convergent subsequence. Ifwe order our set of multi-indexes α(1), α(2), α(3), ... and the α(1)th partial derivatives of thissubsequence is a uniformly bounded, equicontinuous set so has a subsequence which convergesto some continuous function φα(1) , then we look at the α(2) partial derivatives of this furthersubsequence and etc.
Then we can form a cantor diagonal sequence φjk s.t. ∂αφjk → φα where φα is a continuousfunction which is 0 on the boundary of K. It follows that ∂αφ0 = φα so φ0 is in DK andφjk → φ0 uniformly in each of its derivatives.
This also shows that D(Ω) is sequentially compact since any bounded sequence will be insome DK which is sequentially compact.
It is interesting to notice that the only Banach spaces which are Montel spaces are finitedimensional which is further proof that the topology on DK is not normable. This is because ifa 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 subsetxnj 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 1with 1
2 ,13 , ... we will produce a countable dense sequence in Y .
Proposition. The space D(Ω) is separable.
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Proof. First we prove that the space C∞(K) (with the same semi-norms as for DK) is separableand since DK is a subspace of this and C∞(K) is a Frechet space this will show that DK isseparable.
Let T be the map T (φ) = ∂x1∂x2
...∂xnφ. Since K is bounded if
||ψ − T (φ)||∞ ≤ ε
then we can integrate ψ to a function Ψ such that if
maxα1, ..., αn ≤ 1
then|∂α(Ψ− φ)(x)| ≤ ε.diam(K).
Fix some φ in C∞(K) we want to show it can be approximated by polynomials. By theabove argument and if we can approximate Tnφ by a polynomial we can choose pn such thatfor all α with maxα1, ..., αn ≤ n then
|∂α(pn − φ)(x)| ≤ 1
n.
Then the sequence pn will tend to φ in the topology on C∞(K). Since the polynomials withrational coefficients are dense in the polynomials and since the Stone-Weierstrass theorem givesthat any continuous function can be uniformly approximated by polynomials we have that DKis separable. Since D(Ω) is the union of countably many spaces of the form DK we have thatD(Ω) is separable.
5 The space D′(Ω)
We define this as the analytic dual of D(Ω) the space of continuous linear forms on D(Ω) In thissection we will describe two topologies on the space D′(Ω) and discuss the properties of thesetopologies 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-spaceto annother LCS to be continuous we have the following criteria for a linear form u to be adistribution.
Proposition. A linear functional u on D(Ω) is a distribution iff(1) For each K there exists a C and an N s.t. |u(φ)| ≤ CΣ|α|≤N supx∈K |∂αφ(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 DKis continuous. As D(Ω) is not metric (2) is not just a result of translation being continuoushowever 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 theL1 functions with a subset of D′(Ω) via f ↔ uf where
uf (φ) =
∫Ω
f(x)φ(x)dx.
Similarly, measures can be identified with distributions via µ↔ uµ where
uµ(φ) =
∫Ω
φ(x)µ(dx).
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5.1 Positive distributions
Before exploring the topology we look at an interessting way in which D′ is not like a functionspace. A distribution u, is positive if for every φ ≥ 0, u(φ) ≥ 0. A difference between functionspaces and spaces of distributions is that distributions cannot in general be written as thedifference of two positive distributions so it is not possible to build up any theory by lookingfirst at positive distributions (for instance like is done for Lebesgue integration). This is becauseof the following result. First we need the Riesz-Representation Theorem.
Riesz Representation Theorem. If K is a compact, Hausdorff topological space then the dualto the Banach space (C(K), || ||∞) is the space of Borel measures on K with the total variationnorm.
Proposition. Every positive distribution can be identified with a Borel Measure.
Proof. Suppose u is a positive measure.
For K ⊂ Rd a compact let ||φ||∞,K = supK |φ|.Suppose φ ∈ D then fix K compact, ρK ∈ D s.t. ρ = 1 on K. Then
||φ||∞,Kρ− φ ≥ 0
⇒ u(||φ||∞,Kρ− φ) ≥ 0
⇒ ||φ||∞,Ku(ρ) ≥ u(φ)
so u is continuous with respect to the uniform norm on K and so can be extended continu-ously 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 itµ|K .
Its clear that if K ′ ⊂ K then µK |K′ = µK′ .
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 D′(Ω)
We will introduce two topologies on D′(Ω).
Defintion. The weak-* topology on D′(Ω) is the topology of point-wise convergence it is theweakest topology which makes all the evaluation maps continuous. It is a locally convex topologydefined by the semi-norms
pφ(u) = |u(φ)| φ ∈ D(Ω)
Definition. The strong dual topology is the topology of uniform convergence on the boundedsubsets of D(Ω). It is a locally convex topology defined by the semi-norms.
pB(u) = supφ∈B|u(φ)|
where B ranges over the bounded subsets of D(Ω).
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5.3 Non-metrizability
As we have already seen the strong topology is metrizable iff it is first countable. So we wouldlike to investigate whether the strong dual topology can be defined by a countable collection ofsemi-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 whichinclude 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 ofbounded sets B1, B2, B3, ... we can add countably more sets so that this sequence is closed underfinite unions and multiplication by an integer. This will mean that if T : D′ → R is continuouswith respect to the topology then
|T (u)| ≤ pB(u)
for some B. This means that φ is continuous with respect to the strong topology since φ isbounded.
Claim: If we have a sequence as defined above an element of D then φ is continuous with re-spect to the semi-norms pB1
, pB2, pB3
, ... only if φ ∈ Bi for some i. We know that φ is continuouswith respect to the topology induced by the semi-norms if there exists an i with |u(φ)| ≤ pBi(u)for every u ∈ D′(Ω). If φ /∈ Bi then there exists n such that pn(φ) > supψ∈B pn(ψ). So thereexists an x0 such that
|Σα≤n∂αφ(x0)| > supψ∈B
pn(ψ)
so the element u withu(φ) = Σα≤n∂
αφ(x0)
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 insome DK then there exists φ ∈ DK which is not it
⋃nBn
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 isnot in the union. So it remains to construct such a function. Since K has non-empty interior wecan find U1, U2, U3, ... a countable sequence of open balls inside K and there exists a sequence Vnof open balls such that for each n we have diam(Vn) = 1
2diam(Un) and we have φn ∈ DK suchthat φn is identically 1 on Vn and identically 0 outside Un. Then we can produce the function
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φn(x) sin(anx1) and patch these together to get a function with the required properties. Thisis because for x ∈ int(Vn) we have
∂α(φn(x) sin(anx1)) = ∂α1x1
sin(anx1)
so as without loss of generality we can make an ≥ 1 and large enough so that the sine functiondoes a full wiggle inside Vn then
pn(φn) ≥ ann ≥ an.
Since for any φ we have that φ is a bounded set and that if B is bounded in D(Ω) thenB ∩DK will be bounded the above proposition shows that the strong dual topology is not firstcountable.
Now we will show that the weak-* topology is not first countable. This is equivalent tosaying that it cannot be defined by a countable set of semi-norms of the form pφ for φ ∈ D. Forthis we need a lemma.
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).
Proof. Let T : V → scalarsn be the map x 7→ (f1(x), ..., fn(x)). Then ker(T ) ⊂ ker(f) so thereexist a map g such that f = g T and g will be of the form g(y) = a1y1 + ...+anyn which meansthat f(x) = a1f1(x) + ...+ anfn(x).
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 φ is weak-* continuous. Which means the set
V = u | |u(φ)| < ε
contains a set of the form
W = u | |u(φ1)| < ε1, ..., |u(φn)| < εn
which means that U =⋂k≤n ker(φk) ⊂ V and u ∈ U ⇒ λu ∈ U which means that λu(φ) < ε
for each λ so u(φ) = 0 for each u ∈ U therefore ker(φ) ⊃ U which by the lemma above means
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 ifD(Ω) is countable dimensional then so is DK which would mean that DK was the union ofcountably many finite dimesional subspace.
spane1 ⊂ spane1, e2 ⊂ spane1, e2, e3 ⊂ ...
These subspaces are closed (as they are finite dimensional) and nowhere dense since any openneighbourhood of 0 is absorbing. So the weak-* topology is not metrizable.
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5.4 These Spaces are Hausdorff
Now we wish to show that both these topologies are Hausdorff.Since every weak neighbourhoodof 0 is a strong neighbourhood of 0 it is sufficient to show that for every u 6= 0 ∈ D′(Ω) thereis some φ ∈ D(Ω) such that u(φ) > 0 then fix 0 < α < u(φ) and the sets U+
α = v | v(φ) > αand U−α = v | v(φ) < α are open neighbourhoods of u and 0 respectively which are disjoint.Therefore D′(Ω1) is Hausdorff for both the strong and weak-* topologies. If u and v are notequal 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 spaceX are pointwise (weak-*) bounded then they are uniformly bounded (bounded in the strongdual topology). We wish to extend this result to the spaces D,D′. First we will need the notionof 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 Frechet space, F , is barrelled as a consequence of the Baire Category theorem. If B isa barrel then F =
⋃n∈N nB so by BCT nB has non-empty interior for some n so B has non-
empty interior. Since B is convex and balanced if N ⊂ B is open then conv(N ∪ (−N)) is aneighbourhood of 0 contained in B.
An LF-space, F , with sequence of definition Fn is barrelled since if B is a barrel in F , thenB ∩ Fn is a barrel so a neighbourhood of 0 in Fn so since B is convex it is a neighbourhood of0 in F .
Theorem. If a set B is weak-* bounded in D′(Ω) then it is strongly bounded.
Proof. First, we show that if B is weak-* bounded then it is equicontinuous at 0. Let ε > 0.Then let W =
⋂u∈B u
−1(Bε). 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
|u(φ)| ≤ CΣ|α|≤n|∂αφ|
which must be bounded on B so there is a λ s.t.
|u(φ)| ≤ λε ∀ u ∈ B
and since u is linearu−1(λBε) = λu−1(Bε) ⊃ 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 ε,W as before. W is a neighbourhood of 0 in D(Ω) so there exists µ s.t.A ⊂ µW so u(A) ⊂ µBε for each u ∈ B so B is strongly bounded.
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It is a consequence of this that if ui is a sequence with ui(φ)→ vφ 6= ±∞. Then the linearmap u defined by u(φ) = vφ is continuous. This is because the weak-* closure of the sequenceis 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 bothtopologies.
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 Ithen
∏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 E′ is its analytic dual. Then if A ⊂ Ethe polar of A is
A = u ∈ E′ | supφ∈A|u(φ)| ≤ 1
Using this we have
Banach-Alaoglu-Bourbaki. If E is a topological vector space and E′ its dual and if U is aneighbourhood 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
λx ∈ U
consequently if f ∈ U we will have
|f(
1
λx
)| ≤ 1
so that|f(x)| ≤ |λ|
We notice also that U is weak-* closed since if u /∈ U then there exists a φ ∈ U such thatu(φ) > 1 so the set
V =⋃φ∈U
u | |u(φ)| > 1
is weak-* open since it is the union of weak-* open sets. This set is also the complement of Uwhich 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
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which means that tu+ (1− t)v ∈ U.Let Mφ = supu∈B |u(φ)|. Then let Fφ = [−Mφ,Mφ]. Then let
F =∏
φ∈D(Ω)
Fφ
with the product topology. Let πφ be the co-ordinate projections. F is compact by Tychonoff’stheorem. 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 closedin F .
θ−1 φ = πφ
on the image of θ so θ−1 is weak-* continuous.
θ πφ = φ
so θ is continuous.The image of θ is the subspace of F where fλφ+µψ = λfφ + µfpsi
Im(θ) =⋂
φ,ψ,λ,µ
ker(πλφ+µψ − λπφ − µπψ)
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 iscontained 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 similararguments as we used to show U is balanced and convex. Also if we fix x ∈ E then since B isweak-* bounded there is some λ such that
supu∈B|u(x)| ≤ λ
which means that1
λx ∈ B
so that B is absorbing and therefore a barrel. Since our TVS is barrelled this means B is aneighbourhood of 0 and B is contained in the polar of B.
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5.7 Sequential Compactness
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 ∈ D′ with u(φ) ≤CΣα≤N supK |∂αφ| for each φ ∈ DK then there exists a C1 function f and a multi-index αsuch that for every φ ∈ DK
u(φ) = (−1)|α|uf (∂αφ)
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 isinvertible on D′K . Lets look at the map S = ∂x1∂x2 ...∂xn on DK . Then for φ ∈ DK since φ is 0on the boundary of K we have:
φ(y) =
∫xi≤yi,i=1,...,n
S(φ)(x)dx
This shows that S is injective which implies that Sn is injective. It also shows that |φ(y)| isbounded by the L1 norm of Sφ. The mean value theorem gives that
|φ| ≤ diam(K) maxi||(∂xiφ)||∞
which implies that||φ||N ≤ C||SNφ||∞ ≤ C ′||SN+1φ||L1
So we have|u(φ)| ≤ C ′′||φ||N ≤ C3||SN+1φ||L1
since SN+1 is injective we can define a map u′ on its image so that
u = u′ SN+1
therefore|u′(ψ)| ≤ C3||ψ||L1 ψ ∈ Im(SN+1)
Using the Hahn-Banach theorem we can extend this to u′′ a functional on DK such that for allφ ∈ DK we have that
|u′′(φ)| ≤ C3||φ||L1
So u′′ ∈ (L1(K))′ 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 functionf . Since K is bounded we will have ||f ||∞ ≤ (diam(K))2||g||∞. This also shows that thesupremum bound of f is bounded in terms of C ′′ and properties of K.
Lemma 2. If uj is a pointwise bounded sequence in D′(Ω) then there is an N such that foreach j there exists a Cj such that uj(φ) ≤ CjΣα≤N supK |∂αφ|.
Proof. Suppose that this is false. Then we can extract a subsequence such that
un(φn) > nΣ|α|≤n supK|∂αφn|
and possibly by rescaling we can let pn(φn) = 1. This will give that φn is a bounded sequence inDK . However the Banach-Steinhaus theorem says that a weak-* bounded set in D′ is boundedon a bounded set in D and un is not bounded on φn which is a contradiction.
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Proof. Proof of the proposition: Suppose uj is a bounded sequence inD′(Ω) andK some compactsubset of Ω then by lemma 1 there exists fj for each j such that uj(φ) = (−1)|α|ufj (∂
αφ) andby 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 toK. We have: ∫
K
f(x)(−1)|α|∂αφ(x)dx = u(φ)
for every φ ∈ DKWe 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 theoremgives that uj will be bounded uniformly on each bounded subset of DK which shows that the Cjin lemma 2 can all be taken to be the same. We saw in the proof of lemma 1 that the supremumbound of f and its derivatives only depends on this C and properties of K which shows that thefj are uniformly bounded and equicontinuous since they are also uniformly bounded in each ofthe order one partial derivatives.
Therefore we can apply Arzela-Ascoli to the sequence fj to find a sub-sequence fjk whichconverges uniformly to a function f . So the sequence ujk |DK converges to the linear form on DKinduced by (−1)α∂αf since uj = uf TN+3. We can do this for each DKn for some sequenceof compacts Kn increasing to Ω, by extracting successive subsequences. We can then form aCantor diagonal sequence and since every φ is contained in some DKn this will converge in theweak-* sense.
This shows that all Cauchy sequences converge in D′ since we know that Cauchy sequencesare bounded and so by the above proposition they will have a convergent subsequence. It isalso the case that if a subsequence of a Cauchy sequence converges to a point u then the wholesequence must, therefore in D′ with the weak-* topology all Cauchy sequences converge.
5.8 More Montel Spaces & Equivalence of the Topologies on Boundedsets
We recall that the bounded sets are the same for both topologies. In this section we will provethat D′(Ω) has the property that the topologies restricted to bounded sets are the same and aremetrizable 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 D′(Ω). Then the topology of bounded convergenceand 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-* compactthanks to Banach-Alaoglu-Bourbaki. Fix ε > 0 we wish to find a neighbourhood of 0 in E s.t.supφ∈A |u(φ)| ≤ ε 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)| ≤ ε for allu ∈ E, φ ∈ U . Then since A is compact we have φ1, φ2, φ3, ..., φn ∈ A s.t. φ1+U, φ2+U, ..., φn+Ucover A. For each i = 1, ..., n we have Wi weak-* neighbourhood of 0 s.t.
u ∈Wi ⇒ |u(φi)| ≤ ε
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let W =⋂Wi this is also a weak-* neighbourhood of 0. Then for u ∈W
supφ∈K|u(φ)| = max
i≤nsupψ∈U|u(φi + ψ)|
≤ maxi≤n|u(φi)|+ sup
ψ∈U|u(ψ)| ≤ 2ε
Then, since every bounded set is contained in a closed, bounded set pB is continuous forevery B bounded. This shows the strong dual topology is weaker that the weak-* topology onE. As φ is bounded for each φ we already know that the weak-* topology is weaker than thestrong dual topology so they must be the same.
From this we can show that D′(Ω) is a Montel space.
Proposition. D′(Ω) is a Montel space with the strong dual topology.
Proof. We have already that D′ is locally convex and Hausdorff. We would like to exploit thefact 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 havethat
B strongly bounded ⇒ Bw∗ is weakly compact. Also from the proof of Banach-Steinhauswe have B strongly bounded ⇒ B weak-* bounded ⇒ B equicontinuous. So we know that Bw∗
is equicontinuous so by the above lemma on Bw∗ the strong and weak-* topologies coincide. SoBw∗ is strongly compact and B is a strongly closed subset of Bw∗ 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 densesubset of D is equal to the strong dual and weak-* topologies. This along with separability of Dshows 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 canchoose the φ1, ..., φn to be in M . Therefore, the restriction to any bounded set of the topologyinduced by the strong or the weak topology is the same as the topology which is induced by theevaluation maps of a dense set in D(Ω). Since we know that D(Ω) is separable with countabledense set M we can look at the weakest topology on D′(Ω) which will make the evaluationmaps of all the elements of M continuous. This has a countable base neighbourhood base so ismetrizable so any of these topologies restricted to a bounded set are metrizable.
This gives annother proof that bounded sets in D′ are sequentially compact since sequentialcompactness 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 theoremfor normed spaces which says that a Banach space which is reflexive if and only if its unit ballis 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 theelements 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 ∈ D′.We already know that u ∈ D′ implies that u is continuous. Suppose u is weakly continuous then
V = φ | |u(φ)| < ε
is weakly open contains a set of the form
φ | |ui(φ)| < 1, i = 1, ..., n
for some set of ui ∈ D′. Then
U =⋂n
ker(un) ⊂ V
if φ ∈ U then λφ ∈ U so |u(λφ)| < ε for all λ so u(φ) = 0 for φ ∈ U . So ker(u) ⊃ U whichmeans that u ∈ spanu1, ..., un. Which means that u ∈ D′
Proposition. Any set B which is closed and bounded in the normal topology on D(Ω) is weaklycompact.
Proof. If B is closed and bounded in the LF-space topology then we already know it is compactin the LF-space topology. Any weakly open set is also open in the LF-space topology so anyweak 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. Aspace is called reflexive if F” is isomorphic to F via the cannonical mapping x→ x.
By the lemma in the previous section the strong and weak-* topologies coincide on theequicontinuous 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-BanachTheorem.
Hahn-Banach Separation. If X is a locally convex space and A,B are disjoint convex setss.t. A is compact and B is closed. Then there exists f ∈ X ′ s.t.
supa∈A
f(a) < infb∈B
f(b)
Now, on the way to proving that D is reflexive we show something similar to Goldstine’stheorem for normed spaces which says that the ball of the bidual is the weak-* closure of theimage of the ball under the canonical injection of D into D′′ via the evaluation map.
Proposition. This canonical mapping D → D′′ is an injection.
Proof. We want that if u(φ) = u(ψ) for every u ∈ D′ then φ = ψ. Suppose φ 6= 0 then wecan define a linear map on span(φ) by letting u(λφ) = λ then the Hahn-Banach theorem allowsus to extend this to a linear functional u ∈ D′ so D′ strongly separates the points of D so thecanonical mapping is indeed an injection.
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Proposition. Let B be the weakly bounded set in D defined by
|u(φ)| ≤Mu
where u ranges over D′ and Mus are arbitrarily chosen positive reals. Then let
B′′ = f ∈ D′′| |f(u)| ≤Mu
this is a weak-* bounded set in D′′. Let K be the weak-* closure of the image of B under thecanonical injection into D′′. Then K = B′′.
Proof. K is weak-* closed and bounded so by Banach-Alaoglu-Bourbaki it is compact. SupposeK 6= B then we know that K ⊂ B so pick some f0 ∈ B′′ \K then by Hahn-Banach separationthere exists u ∈ D′ s.t.
supf∈K
f(u) < f0(u)
as everything is convex and balanced we may as well have these quantities positive. Sincef ∈ B′′, f0(u) ≤Mu but
supf∈K|f(u)| = sup
φ∈B|u(φ)| = Mu
which gives Mu < Mu for this u which is a contradiction. Hence, K = B′′.
Corollary. D(Ω) is reflexive.
Proof. Let D be the image of D in D′′. The restriction to D of the weak-* topology on D′′ isthe weak topology on D. Let B be a closed, bounded set in D and B its image in D′′. Then by
B is weak-* compact so weak*-closed. By above, B =¯Bw∗ = B′′. So D′′ = D.
5.11 Completeness
We already know that all Cauchy sequences converge in D′(Ω) but since the space is not metriz-able we would like to show that D′(Ω) is complete.
Proposition. The space D′(Ω) is complete for the strong dual topology.
Proof. Let F be a Cauchy filter on D′(Ω) and define
F (φ) = A(φ) | A ∈ F
whereA(φ) = u(φ) | u ∈ A
then we claim that F (φ) is a Cauchy filter in R. Indeed given ε > 0 we know that φ is acontinuous linear form on D′(Ω) so there is a neighbourhood U of 0 such that u ∈ U means that|u(φ)| < ε. Then there is some A ∈ F with A−A ⊂ U which means that A(φ)−A(φ) ⊂ (−ε, ε).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 thatA(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 followsas B is an equicontinuous set of maps from D′(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 andlinear u ∈ D′(Ω). Hence F → u.
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5.12 The space D as a subset of D′ and Separability.
We have already seen how the space D embeds into D′ via the same identification as for theL1loc 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 willproduce a smooth function and then multiplying this function by a smooth function which haslarge compact support. This result has some implications for the space D′. First, however, wewould like to investigate continuity properties of the map I : D → D′. This map cannot be ahomeomorphism since the I(D) is dense in D′ yet D is complete so would form a closed subspaceof D′ if the subspace topology was equivalent to the initial topology.
Proposition. The map I : D(Ω)→ D′(Ω) which goes via
φ 7→ uφ
where
uφ(ψ) =
∫Ω
φ(x)ψ(x)dx
is continuous when D′(Ω) carries either the weak-* or strong dual topology.
Proof. We prove this by showing that I is sequentially continuous. We have already shown thatsince D′(Ω) is a locally convex space with either topology that sequential continuity will implycontinuity.
Consequently, we take some sequence φj → 0 and wish to show that uφj → 0 in eithertopology. Since the φj converge to zero their supports are all contained in some fixed compactset K.
|uφj (ψ)| = |∫K
φj(x)ψ(x)dx
≤ supx∈K|φj(x)|
∫K
|ψ(y)|dy
≤ supx∈K|φj(x)| sup
y∈K|ψ(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 meansthat 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 continuityof I and the density of its image.
Proposition. D′(Ω) is sequentially separable with either of the topologies.
Proof. This is a consequence of the density of I(D(Ω)) in D′(Ω) and the fact that D(Ω) isseparable. Also, since the bounded sets in D(Ω) are metrizable then if N is a countablesequentially dense set in D′(Ω) and B is a bounded set in D′(Ω) then we have
N ∩B ⊃ N ∩B = B
Which means that N = D′(Ω).
We also see that if T is a continuous map from D′(Ω) to some other space then T I will bea continuous map from D(Ω).
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6 Multiplying Distributions
First we introduce the space E(Ω) this is the space of smooth function on Ω. This is a firstcountable locally convex space defined by the semi-norms
pn,K(φ) = Σ|α|≤n supx∈K|∂αφ(x)|
By choosing a countable sequence of compacts increasing to Ω we can show that this is aFrechet space in exactly the same way as for DK . We show that if φj is a Cauchy sequence thenit is uniformly Cauchy on each compact in each of its derivatives. Therefore for each compactK the sequence ∂αφj converges uniformly to a function φα so ordering the multi-indexes andextracting subsequences gives a Cantor diagonal sequence φjk which converges uniformly on Kin each partial derivative. Extracting successive sub-sequences for each Kn will give a Cantordiagonal sequence which converges uniformly in each derivative on each compact and thereforein the sense of E .
In this section we always consider D′ with the weak-* topology.
6.1 Multiplication between subspaces of functions and D′
It is possible via duality to define multiplication between E and D′.
ψu(φ) = u(ψφ)
This agrees with our previous notion of multiplying funtions when u can be identified with anelement of L1.
This is a bilinear map E × D′ → D′
Proposition. Multiplication is a separately continuous mapping and if φn → 0 in E and un → 0in 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 -
weak-* continuous. Choose ψ ∈ D then φ.ψ ∈ D so ψ mφ = ψ.φ. Therefore this is continuous.
Now fix u ∈ D′ and look at φ 7→ φu call this map mu we want to show this is continuous,it is sufficient to show that if ψ ∈ D then ψ mu is continuous and since E is metrizable it issufficient to prove this is sequentially continuous. So take some sequence φj → 0 we want toshow that u(ψ.φj) → 0 which true since φj → 0 uniformly in each derivative on each compactso 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 functionϕ ∈ 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 × D′ to D′
Proof. We argue by contradiction. Suppose the map were continuous then we would haveψ1, ..., ψr, ξ1, ..., ξr ∈ D and K a compact subset of Ω, n an integer such that.
maxi|u(φ.ψi)| ≤ pn,K(φ) max
j|u(ξj)| ∀φ, u
33
then supp(φ.ψi) ⊂ supp(ψ) so without loss of generality we can take supp(φ) ⊂⋃i supp(ψi) and
K ⊂⋃i supp(ψi) so we only need to look at u|DK . Fix some x ∈ K such that ψi(x) 6= 0 for
some i, and let u = δ(α)x for some |α| > n then we have
|δ(α)x (φ.ψi)| = |δx(∂α(φ.ψi))| = |Σβ≤αcα,βφ(β)(x)ψ
(α−β)i (x)|
then maxj |ξ(x)| is just some number and maxi |ψi(x)| 6= 0 so making φ(α)(x) sufficientlylarge 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 eachu ∈ D′ to each compact set K ⊂ Ω we can associate an integer N and a constant C s.t.
|u(φ)| ≤ CΣ|α|≤n supK|∂αφ(x)|
for all φ ∈ DK If there is an N which works for every compact set then we call the smallest suchN the order of u. Otherwise we say u has infinite order.
Proposition. It is possible to define multiplication between the subspace of distributions of orderless 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 analyticdual of the space CNc (Ω) and the multiplication between CN (Ω) and D(Ω) is into CNc (Ω) somultiplication 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 notionof the derivative of a distribution.
Definition. If u in D′ and α is a multi-index we can define ∂αu via the ’integration by partsformula’
∂αu(φ) = (−1)|α|u(∂αφ)
This defines a distribution since its restriction to any DK is continuous and linear as differ-entiation is a continuous, linear map on DK . This definition agrees with the normal derivativeon 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.
Now we want to look at the particular distribution on Ω = R, p.v.( 1x ) defined by:
p.v.
(1
x
)φ = lim
ε→0
∫|x|>ε
1
xφ(x)dx
Lemma. p.v.(
1x
)is a distribution.
Proof. ∫|x|>ε
1
xφ(x)dx = − log(ε)φ(ε) + log(ε)φ(−ε)−
∫|x|>ε
log(|x|)φ′(x)dx
and we have− log(ε)φ(ε) + log(ε)φ(−ε)→ 0
so we have a distribution of order 2.
34
We notice that if ϕ(x) = x then ϕp.v.( 1x ) = 1
Now we would like to specify the properties which a good notion of multiplication betweentwo 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
dx(u.v) = u
dv
dx+du
dxv
and consequently the Liebniz rule.(3) We would like it to agree with our earlier definition of multiplication between E and D′
when E is identified with a subspace of D′.
We also notice that since x|x| is continuously differentiable ddx (x|x|) = 2|x| in the distribu-
tional sense. If Liebniz rule holds for distributions then we will have:
d2
dx2(x|x|) = x
d2
dx2(|x|) + 2
d
dx(|x|)
⇒ d
dx(2|x|) = x
d2
dx2(|x|) + 2
d
dx(|x|)
⇒ xd2
dx2(|x|) = 0
However, since d2
dx2 (|x|) = 2δ0 6= 0 if we have the properties above then:
0 = p.v.(1
x)
(xd2
dx2(|x|)
)=
(p.v.(
1
x)x
)d2
dx2(|x|) = 2δ0
Which is a contridiction. Therefore, we cannot have a sense of multiplication satisfying theproperties (1), (2), (3) on the distributions on R.
This proof only relies on having an element, ϕ of D(Ω) where there exists distributionsu, 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 ofmultiplying two distributions which would coincide with the normal notion of multiplication onL1loc but this would not be possible as the product of two distributions in L1
loc is not necessarilyin L1
loc 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 ∈ D′(Ω) and f some smooth function to find v suchthat
f.v = u
We will work only in the case of one variable and where Ω = R. We will build up by firstdividing by the function x 7→ x, then by polynomials and finally by analytic functions whosezeroes have finite order.
35
6.3.1 Division by x
We seek to solve the equationx.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:
v (φ− φ(0)η) = u
(φ− φ(0)η
x
)which means that
v(φ) = u
(φ− φ(0)η
x
)+ φ(0)v(η)
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 wehave
φ 7→ φ− φ(0)η
x
is continuous with respect to the topology on D(R) and φ 7→ φ(0) is continuous so v is thecomposition of continuous functions. v is not defined uniquely it depends on the choice of η andof 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 candivide 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 theordering of the factors since multiplication by polynomials does not depend on the ordering ofthe 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 intersectswith 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 ∈ D′(Ω) such that
u|ωi = ui
36
Proof. This follows from the fact that there exist smooth functions ρi such that
supp(ρi) ⊂ ωi
andΣi∈Iρi = 1.
Then the sumu = Σi∈Iρiui
defines a distributions.
Now we show the existence of the ρi. Suppose ρi is a function with supp(ρi) ⊂ ωi and ρi > 0on Ω \
⋃j 6=i ωj . This is possible since
C1 = Ω \⋃j 6=i
ωj and C2 = Ω \ ωi
are closed and disjoint and so we have a smooth function ρi such that ρi = 0 on C2 and ρi = 1on C1. (By mollifying the continuous function found in Urysohm’s lemma.)
Then 0 < f = Σiρi < ∞ since the covering ωi is locally finite. Let ρi = 1f ρi this gives the
required result.
We now prove that u = Σiρiui defines a distribution. Fix K compact then there existsi1, ..., in the finite collection of sets which intersect K. Then for φ ∈ DK
u(φ) = Σnk=1ρikuik(φ) = Σnk=1uik(ρikφ)
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 withfinite order ni. Then, since f is analytic, we can construct ωi | i ∈ I a collection of pairwisedisjoint open sets such that ξi ∈ ωi. Then we have another open set ω0 such that
R \⋃i∈I
ωi ⊂ ωo.
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 inωi
f.vi = u|ωithen we can patch these vi together to find v. Also, we notice that
f = (x− ξi)nifi
where fi 6= 0 on ωi. Then 1fi∈ E(ωi) so we can reduce to the case of dividing by a polynomial.
37
6.4 Applications of Multiplication.
Here we, very briefly, talk about how our definitions of multiplication and differentiation by du-ality allow us to set PDEs in the space of distributions. This is arguably the primary motivationfor developing the space of distributions.
Suppose D is a linear differential operator with coefficients in E(Ω) then if u is a distributionwe can define
Du = Σ|α|≤ncα∂αu
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 asolution.
If we look now at the homogeneous case Du = 0 we want
Σ|α|≤ncα∂αφ0, ∀φ
⇔ Σ|α|≤n∂α(cα.φ) = 0, ∀φ
⇔ Σ|α|≤nu((−1)|α|∂α(cα.φ)) = 0, ∀φ
⇔ u(Σ|α|≤n(−1)α∂α(cα.φ))
)= 0, ∀φ.
This gives us another differential operator
D ′φ = Σ|α|≤n∂α(cα.φ)
on D(Ω) which is the formal adjoint of D .H = D ′(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 whichwe 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 alwayspossible to define. Hence, the space of distributions is not an obvious space in which to look forthe 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 aresmooth functions which decrease rapidly at infinity in the sense that they decrease faster thanthe reciprocal of any polynomial. This space is useful in defining the Fourier Transform for alarge class of functions and can be used to extend many results of Fourier analysis to a largeclass of the distributions. We will show a few properties of this space. Many of the topologicalproperties are the same and we restrict ourselves to the detailed proofs of those which arereasonably different to those in the space D′(Ω).
They can be made into a locally convex space with the semi-norms:
38
pα,β(φ) = supx∈Rn
|xβ∂αφ(x)|
Where α and β run over all the possible multi-indices. Since this is a countable collection ofsemi-norms the space is metrizable exactly as for DK . This space is denoted S .
Proposition. S is a Frechet space.
Proof. We already know that S is metrizable, as it is a first countable lcs, so it is sufficientto prove that Cauchy sequences converge. Suppose φj is a Cauchy sequence. Given two multi-indices α, β then the sequence xβ∂αφj(x) is uniformly Cauchy so converges uniformly on Rn.Thisis because it converges uniformly on compacts and limits must agree and it is uniformly smalloutside compact sets so the convergence is uniform. More precisely if β′ > β then we know that|xβ′∂αφj(x)| ≤M for some M which doesn’t depend on j. This means that
|xβ∂αφj(x)| ≤ |xβ−β′|M
and if we fix ε we can choose |x| = rε sufficiently large so that
M
|xβ−β′ |≤ ε
2, |x| > rε.
Let Kε = |x| ≤ rε then we know that xβ∂αφj → φα,β uniformly on Kε So if we choose j largeenough so that
supx∈Kε
|xβ∂αφj(x)− φα,β(x)| < ε
then outside Kε we will have
|xβ∂αφj(x)− φα,β(x)| ≤ |xβ∂αφj(x)|+ |φα,β(x)| < ε.
It converges to a function φα,β . Then φα,β = xβ∂αφ0,0 so φj → φ0,0 in the topology ofS .
The space D(Rn) can be embedded into S . Let ι : D(Rn) → S be this map. We wish toshow that ι is a continuous injection onto its image. First we show ι is continuous by showingthat if φj → 0 in D then φj → 0 in S . This is sufficient as we only need to show the restrictionto each DK is continuous and DK is metrizable. Now we have already shown that φj → 0 impliesthat the supports of all the φj are contained in some compact K then for each β, |xβ | ≤Mβ onK so xβ∂αφj ≤Mβ∂
αφj → 0. The inverse cannot be continuous since S is metrizable and thesubspace topology that S induces on D is metrizable where as the LF-space topology on D isnot 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 compactnessiff sequential compactness. So let φj be a bounded sequence in S and let
K1 ⊂ K2 ⊂ K3 ⊂ ...
be a sequence of compact sets whose union is Rn. Similarly as for in DK the sequence xβ∂αφjis a uniformly bounded equicontinuous set on Ki. By the Arzela-Ascoli theorem there exists asubsequence s.t. xβ∂αφjk converges uniformly on Ki then as we have a countable collection ofpossible multi-indices α, β we can extract further subsequences and then create a cantor diagonalsequence φjk such that xβ∂αφjk converges uniformly on Ki for every α, β. Then we can keepextracting subsequences of this for each Kn to produce another Cantor diagonal sequence whichwill 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 relationto the set of smooth functions on the sphere. Since the sphere is compact we have that D(Sn) =E(Sn) and will have a Frechet space topology. Since we have a smooth bijection with a smoothinverse from Rn to Sn \ω where ω is some arbitrary point on the sphere. This map will inducea map from S (Rn) to D(Sn) whose image is in the closed subset of D(Sn) where the functionsare 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 inversewe can define φ on Sn \ ω by
φ(x) = φ(P−1(x)) ∈ E(Sn \ ω)
Now if we look atlimx→ω
∂αφ
we can see that this will be 0 aslim|y|→0
∂α(φ)(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 andHausdorff by the same argument we used to show that C∞(K) was separable for K a compactsubset of Rn. Since C∞(Sn) is metric and we know that subspaces of separable, metric spacesare separable this shows that S is separable. Like with te space D(Ω) this will show thatbounded sets in S ′ are metrizable.
Since S is a Frechet 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
φ 7→ φ(λ) =
∫Rnφ(x)eλ.xdx, λ ∈ Rn
The Fourier transform is continuous because
λβ∂αφ(λ) = (xα∂βφ(x))
possibly modulo some constants, and
supλ∈Rn
|φ(λ)| ≤ ||φ||L1((Rn)
and if φ ∈ S then
||φ||L1(Rn) ≤∫B1
|φ(x)|dx+
∫Rn\B1
||x|n+1φ(x)||x|−n−1dx
≤ supx∈Rn
|φ(x)| |B1|+ supx∈Rn
|xn+1φ(x)|∫Rn\B1
x−n−1dx.
40
This means shows that the Fourier transform is continuous because
supλ∈Rn
|λβ∂αφ(λ)| = C supλ∈Rn
|(xα∂βφ(x))(λ)|
and we have thatsupλ∈Rn
|(xα∂βφ(x))(λ)| ≤ ||xα∂βφ(x)||L1x
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.
φ(x) =1
(2π)n
∫Rnφ(λ)e−λ.xdλ
8 Tempered Distributions
The tempered distributions are defined as the analytic dual of the space of rapidly decreasingfunctions. Like with the space D′ we can put both the strong dual and the weak-* topologyon this space S ′ and by compactness of bounded sets in S these are equal when restricted toequicontinuous subsets of S ′.
The functions which increase more slowly at infinity than some polynomial can be embeddedinto the space S ′ via the identification f ↔ uf as with L1
loc in the space D′. This does indeeddefine 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:
|∫Rnf(x)φ(x)dx| = |
∫Rn
(|x|−n−kf(x)
) (|x|n+kφ(x)
)dx|
≤∫Rn|x|−n−k|f(x)|dx. sup
x∈Rn(|x|n+k|φ(x)|)
We can see from this that if f is the weak derivative of a function which decreases slower thansome polynomial then it will also define a tempered distribution.
We can prove, as with the earlier spaces that closed, bounded sets in S ′ are compact in theweak-* and strong dual topology and that the space S is reflexive. These properties all followfrom 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 onthe fact that if we have a countable sequence of bounded sets in D then we can find a functionwhich is not in this bounded set. This is also true in the space S since we can forget about thebound on xβ times the functions and choose the bounded sets to be bounded on each of the pnnorms from DK and we can find a function in DK and therefore in S which is not in any ofthese 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 forD′ where we saw that the weak-* topology being metrizable would imply that D was countabledimensional. If the weak-* topology on S ′ were metrizable this would imply that S wascountable dimensional which is a contradiction.
Using the continuous injection of the space D(Rn) we can see that S ′ ⊂ D′ via the dualof this identification. The fact that ι is a continuous injection onto its image shows that theinjection of S ′ into D′ is continuous and an injection onto its image.
41
8.1 Relation between S ′ and D′(Sn)
We can also relate the tempered distributions to a space of distributions on the sphere. Theycan be identified with the space of continuous linear maps on the subspace of C∞(Sn) where afunction 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 D′ to be alsoan element of S ′.
Proposition. A distribution u ∈ D′(Rn) is a tempered distribution iff there exists a distributionu ∈ D′(Sn) such that u|Rn = u.
Proof. The “if” part follows from the fact that S can be identified with a subspace of D(S2).More precisely, there is
T : S → D(Sn)
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 witha closed linear subspace of D(Sn) means that if we have u ∈ S ′ we have a continuous, linearfunctional on a closed linear subspace of D(Sn) so by Hahn-Banach we can extend it to acontinuous linear functional on the whole space.
The u in the above theorem is not unique they can differ by a distribution whose support iscontained in ω. So the space S ′ can be identified with some quotient of the space D′(Sn).
8.2 The Fourier Transform on S ’
We can define the Fourier Transform on S ′ by dual map of the Fourier Transform on S. i.e.
u(φ) = u(φ)
This agrees with the normal notion of Fourier transform in L1 ∩L2 thanks to the Plancherelidentity which states, for φ, ψ ∈ L1 ∩ L2:∫
Rnφ(x)ψ(x)dx =
∫Rnφ(x)ψ(x)dx
This is an isomorphism since the Fourier transform on S is an isomorphism since the dualmap of an isomorphism is an isomorphism.
Proposition. The Fourier inversion theorem holds for S ′. This is in the sense that(1
(2π)n
∫Rnu(λ)e−ix.λdλ
)(φ) =
∫x∈Rn
∫λ∈Rn
1
(2π)nu(λ)φ(x)e−ix.λdλdx
= uλ
(∫x∈Rn
1
(2π)nφ(x)e−ix.λdx
)= u(φ)
where uλ acts on its argument as a function of λ.
Proof. Let F be the Fourier transform operator. Then since we have the Fourier inversiontheorem on S we have that
uλ
(∫x∈Rn
1
(2π)nφ(x)e−ix.λdx
)= uλ(F−1φ) = u(FF−1φ) = u(φ).
42
9 The space E ′.Many of the theorems about the spaces S ′ and D′ also hold for the space E ′. The proofs arevery similar so we will not develop these. This section will be short and mainly focus on thecharacterization of E ′ as the space of distributions of compact support. This will be useful later.
Definition. If u ∈ D′ then we say that u is ‘null’ on an open set ω if for every φ such thatsupp(φ) ⊂ ω then u(φ) = 0. The support of u is the complement of all the union of the sets onwhich u is null.
Proposition. The set E ′ can be considered as a subset of D′ and this is exactly the set ofdistributions 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 ρmaps E(Ω) into D(Ω). So the map composition of multiplication by ρ and u defines a continuouslinear map E which coincides with u on D ⊂ E . Hence, we can consider u as a member of E .
Conversely, suppose that v ∈ E ′. Then suppose the support of v is not compact. In this caseit will contain ω1, ω2, ... a sequence of disjoint open subsets such that v|D(ωn) is not identically0 for any n. Then there exists φn ∈ D(ωn) such that v(φn) > 1 then Σnφn ∈ E but v(Σnφn) >v(Σn≤Nφn) > N which gives a contradiction as N is arbitrary.
10 Comparison and Relations between the spaces D,Sand E and their duals.
We have already seen that D injects continuously into S . We would like also to see that Sinjects continuously into the space E . Since they are both Frechet spaces it is sufficient to showthat if φj → 0 in S then φj → 0 in E . This follows since supRn φ(x) ≥ supK φ(x) so somethingconverging 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
D′ ← S ′ ← E ′
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 dualtopology.)
(7) The spaces are all reflexive.
43
We do, however, have some differences in the question of metrizability of the whole spacesin that D is not metrizable where both S and E are. This, is not so huge a difference as itmight first seem because continuous, linear maps from D can still be characterized entirely bytheir actions on sequences.
There are larger differences in the utility of the different function spaces and distributionspaces. D being the smallest of the function spaces means thatD′ is the largest of the distributionspaces here. In D′ we still have enough structure to define differentiation which will be intothe space, and differential operators whose coefficients are in E it therefore gives a very generalnotion of differentiation and differential operators. However, the space D is too small to beuseful with the Fourier transform. The space S is constructed so that the Fourier transform isan isomorphism of this space which allows us (via duality) to extend the notion of the Fouriertransform to S ′ which is smaller than D′ 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 theproperty that
e1, e2, ..., en
a linearly independent set in V1 andf1, 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 thetensor 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 combi-nations of basis vectors and exploiting the bilinearity of Φ we can write any linear combinationsof elements Φ(u, v) as a finite linear combination of these vectors. This shows that any twotensor 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 otherfunction spaces.
Proposition. If Ω1 ⊂ Rn and Ω2 ⊂ Rm then we can embed D(Ω1) ⊗ D(Ω2) into D(Ω1 × Ω2)where Ω1 × Ω2 is regarded as an open subset of Rn+m.
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Proof. To do this we construct an injection Φ : D(Ω1) × D(Ω2) → D(Ω1 × Ω2) and show thatthe 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 isbilinear and injective since if one of the entries is fixed and its variable this is just a multiplicationmap. 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. Forexample we can use this same map to give an injection
Lp(Ω1)⊗ Lp(Ω2)→ Lp(Ω1 × Ω2)
orCk(K1)⊗ Cl(K2)→ Ck,l(K1 ×K2)
Proposition. If Pn is the space of polynomials in n variables. Then given some open subspaceof Rn this can be regarded as a function space. We can define the tensor product Pn ⊗Pm inthe same way as for function and for this case the injection given above is a bijection.
Proof. Any monomial xαyβ is a tensor product of two polynomials in Pn and Pm and anypolynomial 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 thatC(K1) ⊗ C(K2) is dense in C(K1 × K2). This is because Pn+m is dense in C(K1 × K2) byStone-Weierstrass and also Pn+m = Pn ⊗Pm ⊂ C(K1) ⊗ C(K2). We can use a variationon this argument to show that D(Ω1) ⊗ D(Ω2) is dense in DK1×K2
. Since, D(Ω1 × Ω2) can bewritten as the union of spaces of the form DK1×K2
this will show D(Ω1) ⊗ D(Ω2) is dense inD(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 pis a polynomial.
11.3 Tensor Products of Distributions
We wish to define a tensor product between two distributions which will coincide with out notionof tensor product on functions. Suppose u ∈ D′(Ω1) and v ∈ D′(Ω2) then we can define a linearform 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
|u(φ)||v(ψ)| ≤ C1C2
(Σα≤n1
supx∈K1
|∂αφ(x)|)(
Σβ≤n2supy∈K2
|∂βψ(y)|)
Where Ki is the projection of K onto Ωi and φ ∈ DK1, ψ ∈ DK2
. Therefore we can extend theform u⊗ v by continuity to a form on D(Ω1 × Ω2). We call this the tensor product of u and vand 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 uxregards 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 holdon the whole space.
This is a kind of ‘Fubini’s Theorem’ for distributions and is extremely useful.We can see that if f ∈ L1
loc(Ω1), g ∈ L1loc(Ω2) then uf⊗g = uf ⊗ ug. Since,∫
x∈Ω1
∫y∈Ω2
f(x)g(y)φ(x, y)dxdy =
∫(x,y)∈Ω1×Ω2
(f ⊗ g)(x, y)φ(x, y)dxdy
So the definition is consistent with the earlier tensor product on function spaces.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).
Proposition. The map T : D′(Ω1)×D′(Ω2)→ D′(Ω1×Ω2) given by (u, v) 7→ u⊗v is separatelycontinuous. 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)). Weknow that u(φ(x, y)) ∈ D(Ω2) so vj(u(φ(x, y))) → 0. We can switch u and v in the aboveargument 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 boundedsequence. 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 ontoΩ2 call this K. We can see that (remembering that uj acts on x.)
∂α(uj(φ(x, y))) = uj(∂αy (φ(x, y)))
Since this holds for φ ∈ D(Ω1)⊗D(Ω2) so for all φ by extending by continuity. hence
supy∈K|∂α(uj(φ(x, y)))| = sup
y∈K|uj(∂αy (φ(x, y)))|
and we have
supy∈K|u(φ(x, y))| ≤ sup
y∈K
(CΣβ≤N sup
x∈K′|∂βxφ(x, y)|
)= CΣβ≤N sup
(x,y)∈K′×K|∂βxφ(x, y)|
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 inD′(Ω1 × Ω2) that D′(Ω1)⊗D′(Ω2) is dense in D′(Ω1 × Ω2).
We can give another concrete realisation of the tensor product D′(Ω1)⊗D′(Ω2) as the spaceof continuous bi-linear mappings from D(Ω1) × D(Ω2) to R. We do this by identifying u ⊗ vwith 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 spaceof weakly continuous bi-linear maps B(D(Ω1),D(Ω2)).
Proposition. The above identification is indeed a tensor product.
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Proof. This follows [4]. First we prove linear independence. If u1, ..., un are linearly independentand 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)×
D(Ω2) then there exists u1, ..., un ∈ D′(Ω1) and v1, ..., vm ∈ D′(Ω2) such that if
maxi|ui(φ) ≤ 1, max
j|vj(ψ)| ≤ 1
then |B(φ, ψ)| ≤ 1.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)| ≤ 1then B(φ, ψ) = 0 for φ ∈ U and any ψ. So we can see that B(φ, ψ) = 0 whenever φ ∈ U orψ ∈ V . Without loss of generality we can take the ui linearly independent and the vj linearlyindependent.
Consequently U has co-dimension n in D(Ω1) and V has codimension m in D(Ω2). Thereforethere exists φ1, ..., φn and ψ1, ..., ψm such that given φ, ψ we have
φ = a1φ1 + ...+ anφn + φ
andψ = 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 is0 on U and the span of φk, k 6= i and ui,j(φi) = B(φi, ψj) which exists by Hahn-Banach. Thenhave 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)
We also give another canonical injection of the space D′(Ω1) ⊗ D′(Ω2) this time into thespace 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 fromD′(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)))
i.e. If we call this map L then L (u, v)(φ) = u(φ).v(.) ∈ D′(Ω2).
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11.3.1 The Space L(D(Ω1),D′(Ω2))
In order to make use of our mapping of D′(Ω1) ⊗ D′(Ω2) into the space L(D(Ω1),D′(Ω2)) wewould 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 D′ as the topology of uniform conver-gence on the bounded subsets of D and also that D′ = L(D,R). This motivates our definitionof the topology. Suppose pB is a semi-norm on D′(Ω2) i.e. B is a bounded subset of D(Ω2) and
pB(v) = supψ∈B|v(ψ)|
suppose also that A is a bounded set in D(Ω1) then we define the semi-norm pA;B via
pA;B(T ) = supφ∈A
pB(T (φ))
if we let A range over the bounded sets ofD(Ω1) andB range over the bounded sets ofD(Ω2) thenthis 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 strongtopology on D′.
To show this space is Hausdorff it is sufficient to show that if pA;B(T ) = 0 for every choice ofA 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 thatpφ;ψ(T ) 6= 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 toproving that D′(Ω) is complete. Let F be a Cauchy filter on L and we claim that for each
φ ∈ D(Ω1), F (φ) is a Cauchy filter on D′(Ω2). This is again because φ is a continuous mapfrom 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 thatA − A ⊂ U we will have A(φ) − A(φ) ⊂ V . We claim that if B is a bounded subset of D(Ω1)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, φ ∈ B is an equicontinuous set of mapping on L. Therefore ifF (φ)→ uφ and T is the map T (φ) = uφ then T will be linear and bounded on bounded subsetsof 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 ε and π topology of Grothendiek and look at their specific forms on tensorproducts of distributions and test functions. Suppose E and F are locally convex, Hausdorfftopological vector spaces. Spaces on which the ε and π topology coincide are called nuclearspaces and were developed by Grothendiek. They have many interesting properties and all themain 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 themapping
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 alocal neighbourhood base for E and BF a local neighbourhood base for F we can take U ∈ BEand 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 basesthis will form a neighbourhood base of E ⊗π F .
The π-topology on D′(Ω1)⊗D′(Ω2) is induced by the semi-norms pA ⊗ pB where A and Bare bounded sets in D(Ω1) and D(Ω2) respectively and we define pA ⊗ pB to be the Minkowskifunctional 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 ε-topology
We saw earlier that we can identify D′(Ω1)⊗D′(Ω2) with B(D(Ω1),D(Ω2)) the space of contin-uous bilinear forms, where here the test function spaces carry their weak topology. It is in factgenerally true that we can identify E ⊗ F with the space B(E′σ, F
′σ) in exactly the same way.
The ε-topology is the topology on this space defined by the topology of uniform convergence onthe product of equicontinuous subsets of E′ and F ′. Suppose A, B are such subsets then wehave the semi-norm
pA,B(Φ) = supx∈A
supy∈B|Φ(x, y)|.
For the topology this gives on tensor products of distribution or test function spaces itis interesting to note that we have an identification between the bounded, precompact andequicontinuous sets.
12 Kernels and the kernels theorem.
In this section we will develop some of the basic theory of kernels and the Schwartz kernelstheorem which gives a representation theorem in terms of kernels for linear maps from the spaceof test functions to the space of distributions.
We have already seen that the space L1loc can act on D(Ω) as integral kernels. For f ∈ L1
loc(Ω)we have
φ 7→∫
Ω
f(x)φ(x)dx
and we extended this action to the whole of the space of distributions. Similarly to the tensorproduct if f ∈ L1
loc(Ω1 × Ω2) then we can induce a map from D(Ω1) to D′(Ω2) via
φ 7→(ψ 7→
∫Ω1×Ω2
f(x, y)φ(x)ψ(y)dxdy
)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 haveφ 7→ (ψ 7→ v1(φ)v2(ψ))
so the linear map which corresponds to v1 ⊗ v2 has one dimensional image.
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Suppose that Φ ∈ L(D(Ω1),D′(Ω2)) has one dimensional image λu;λ ∈ R, u ∈ D′(Ω2). Thendefine a linear map v from D(Ω2) to R such that v(φ)u = Φ(φ). Since Φ is continuous it issequentially continuous so if φj → 0 in D(Ω1) then we will have v(φj)u → 0 in D′(Ω2). So inparticular 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 linearmap from D(Ω1) to R is continuous iff it is sequentially continuous this shows that v ∈ D′(Ω1)therefore
Φ = v ⊗ u.
Then if an element of L(D(Ω1),D′(Ω2)) has finite dimensional image it a finite linear combinationof elements with one dimensional image so can be identified with an element of D′(Ω1)⊗D′(Ω2)which in turn can be identified with an element of D′(Ω1 × Ω2). Our goal is to extend this toan identification between the whole of D′(Ω1 × Ω2) and L(D(Ω1),D′(Ω2)).
Schwartz Kernels Theorem. To any element of L(D(Ω1),D′(Ω2)) we can associate a uniqueelement 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.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 thereforeequicontinuous set in D′(Ω2) therefore (Φ(φj))(ψj) tends to 0. Therefore B is sequentiallycontinuous and since B can be considered as a linear functional on D(Ω1) ⊗ D(Ω2) it can beextended to a sequentially continuous linear functional on D(Ω1×Ω2) and here we already knowthat 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
sη(x) = s(x)e−ix.η, tξ = t(y)e−iy.ξ.
Then B(sη, tξ) is a continuous function since if η′ → η then sη′ → sη in D(Ω). (η, ξ) ∈ Rn+m.Also, since pn(e−ix.η) = (1 + ||η||)n we will have some contant c(s, t) such that
|B(sη, tξ)| ≤ c(s, t)(1 + ||η||)n(1 + ||ξ||)n
Now, for intuition we suppose that k(x, y) is a smooth function which generates B. i.e.
B(φ, ψ) =
∫Ω1×Ω2
k(x, y)φ(x)ψ(y)dxdy
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Then we have that
B(sη, tξ) = (2π)n+m(F−1(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 sois in S . Then after Parseval’s theorem we have for φ ∈ DK , ψ ∈ DL
1
(2π)n+m
∫Rn+m
φ(η)ψ(ξ)B(sη, tξ)dηdξ =
∫K×L
φ(x)ψ(y)k(x, y)s(x)t(y)dxdy
=
∫Ω1×Ω2
φ(x)ψ(y)k(x, y)dxdy = B(φ, ψ).
We would like to extend this to the case of a general B. Since D(Ω1 × Ω2) is an LF-spaceand we have a sequence of definition of D(Ω1 × Ω2) of the form
DL1×L′1 ⊂ DL2×L′2 ⊂ DL3×L′3 ⊂ ...
if we have a sequence of maps Kn, n = 1, 2, ... such that Kn ∈ DLn×L′n and that
Kn+1|DLn×L′n = Kn
then this will define a map K ∈ D′(Ω1×Ω2). Take σ ∈ DLn×L′n and s(n), t(n) as before. DefineKn by analogy with the case for smooth functions.
Kn(σ) =1
(2π)n+m
∫Rn+m
σ(η, ξ)B(s(n)η , t
(n)ξ )dηdξ
Since, σ ∈ S and we have that |B(sη, tξ)| ≤ c(s, t)(1+||η||)n(1+||ξ||)n this integral will convergeand therefore define an element of DLn×L′n . It remains to show that Kn will agree with B onDLn⊗DL′n and since this is dense in DLn×L′n this will show that Kn is independent of the choiceof sn, tn and since we could choose sn+1 = sn and tn+1 = tn that we will have the restrictionpropety.
Kn(φ⊗ ψ) =1
(2π)n+m
∫Rn+m
φ(η)ψ(ξ)B(s(n)η , t
(n)ξ )dηdξ
=1
(2π)n+m
∫Rn+m
B(φ(η)s(n)η , ψ(ξ)t
(n)ξ )dηdξ
= B
(1
(2π)n
∫Rnφ(η)s(n)
η dη,1
(2π)m
∫Rm
ψ(ξ)t(n)ξ dξ
)= B
(1
(2π)n
∫Rnφ(η)e−ix.ηdη,
1
(2π)m
∫Rm
ψ(ξ)e−iy.ξdξ
)= B(φ, ψ)
Therefore we have shown existence.
We know need to answer the question of uniqueness. Suppose K and K ′ both representthe linear map Φ. Then
(K −K ′)(φ⊗ ψ) = 0
51
for every φ⊗ ψ ∈ D(Ω1)⊗D(Ω2). Since we know that K −K ′ is continuous on the whole ofD(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
K (σ) = Kx,y
(1
(2π)n+m
∫Rn+m
σ(η, ξ)e−x.ηe−y.ξdηdξ
)
=1
(2π)n+m
∫Rn+m
Kx,y(σe−x.ηe−y.ξ)dηdξ
=1
(2π)n+m
∫Rn+m
σ(η, ξ)Kx,y(e−x.ηe−y.ξ)dηdξ
=1
(2π)n+m
∫Rn+m
σ(η, ξ)(Φx(e−x.η)
)y
(e−y.ξ)dηdξ
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 withcoefficients in E(Ω) then D defines a map from D(Ω) to E(Ω) so consequently a map from D(Ω)to D′(Ω) if we can show this is continuous then there exists K ∈ D′(Ω× Ω) such that∫
Ω
D(φ)(x)ψ(x)dx = K (φ⊗ ψ)“ = ”
∫Ω×Ω
K (x, y)φ(x)ψ(y)dxdy
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 toD′(Ω) generated by K . In this section we will relate the regularity property of kernels to partialdifferential operators particularly hypoelliptic operators. To begin we make some definition.
Definition. A kernel K is left regular if the linear map Φ : D(Ω1)→ D′(Ω2) maps D(Ω1) intoC∞(Ω2) when C∞(Ω2) is considered as a subset of D′(Ω2).
Definition. A kernel K is right regular if the linear map Φt : D(Ω2) → D′(Ω1) maps D(Ω2)into C∞(Ω1). Here Φt 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 regularand is equal to a smooth function away from the diagonal of Ω× Ω.
Proposition. If K is left regular then we can extend Φt to a function from E ′(Ω2) into D′(Ω1).
Proof. Since for any φ ∈ D(Ω1) we have that
Φ(φ) ∈ E(Ω2)
then we know that if u ∈ E ′(Ω2) thatu(Φ(φ))
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makes sense. Then we can define Φt by
(Φt(u))φ = u(Φ(φ))
which clearly extends Φt so it remains to show that Φt is continuous. Since D′(Ω1) is a locally
convex space, it is sufficient to prove that Φt is bounded on bounded sets or equivalently thatits restriction to bounded sets is continuous. Since the bounded sets in E ′(Ω2) are metrizableit is suficient to prove that it is sequentially continuous at 0. So let us fix B a bounded set inD(Ω1) and we want to show that if uj → 0 then
supφ∈B|(Φt(uj))(φ)| → 0
equivalently thatsupφ∈B|uj(Φ(φ))|
but we already know that Φ(φ) is bounded in E(Ω2) and uj → 0 strongly so on bounded sets sowe are done.
If K is a kernel in D′(Ω×Ω) then we can let a differential operator D act on K by lookingat
(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 thatDx(K (x, y) ∗ v(y)) = v(x).
Also, the existence of a left fundamental kernel will give uniqueness since if we have a solutionto the homogeneous kernel.
Dyu(y) = 0
then we will have0 = 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 dΩε = (x, y) ∈ Ω × Ω | ||x − y|| < ε. Then if a linear differential operator Dhas a parametrix which is very regular, it will have a parametrix which is very regular and whosesupport lies inside dΩε.
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Proof. Here we follow [4]. Let η ∈ D(Rn) such that suppη ⊂ Bε and η is identically zero in someneighbourhood of the origin. Suppose K is a parametrix associated to the differential operatorD . 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 originand where x, y are sufficiently close together that η is identically one near by, we have that∂αx 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 andη(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 thatη(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 propertiesof 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. Welook 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 issmooth in some neighbourhood of a that there is a small neighbourhood of a which we shallcall ωa in which u is smooth. We know there exists a parametrix K for D whose support iscontained inside dΩε.
D tyK (x, y)− δ(x− y) ∈ E(Ω× Ω)
This shows that the above can be extended to a map from E ′(Ω) to D′(Ω) so if we take φa tobe a function of compact support which is identical to one in some neighbourhood of a. Thenthe above acts on φa.u so we have(
D tyK − δ(x− y)
)yφa(y)u(y) = K (x, y)y(D(φa(y)u(y)))− φa(x)u(x) ∈ E(Ω)
since what we really have is(D tyK − δ(x− y)
)yφa(y)u(y) = (φa.u)y(K − δ(x− y)) ∈ E(Ω)
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when considered as a function of x. So since Du is smooth and we can choose the supportof 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) = φaDuwhich 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 ofa this shows that u is smooth in some small neighbourhood of a. Since a is arbitrary this showsthat u is smooth.
It can be shown that, in a kind of converse to the above theorem, if D and D t are hypoellipticoperator on Ω then for every point, x ∈ Ω, there is a small neighbourhood of x in which D has atwo sided fundamental kernel. Also, due to Malgrange if the map from E(Ω) to itself generatedby D is surjective that we will have a two sided fundamental kernel. However, the proofs ofthese 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 convexspaces which was substantially developed by Grothendieck in his phd thesis. In this section weuse 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 willin general be very brief and focus on a summary of the results in this theory rather than theirproofs. 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 wehave been dealing with are nuclear. It is intended as a ‘flavour’ of the theory of Nuclear spaceswhich 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 dothis by first defining a nuclear mapping between Banach spaces. We then extend this notion tothe 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 topologyof convergence on the bounded subsets of E.
Definition. If E and F are Banach spaces then L1(E,F ), called the nuclear mappings, is theclosure of the space of maps with finite dimensional image.
In order to extend this notion to general locally convex spaces we introduce two ways ofproducing Banach spaces from locally convex spaces. Firstly, suppose N is a convex, balancedneighbourhood of 0 in E. Then we can define a new topology on E by the semi-norm which isthe Minkowski functional of N . This space will not be Hausdorff unless E is normable but wemay quotient by kerµN in order to produce a normed space. We can then complete this spaceto produce a Banach space which we call EN . There is a canonical mapping E → EN givenby the quotient map. This map will be continuous but not always open since the topology onEN does not induce the same topology on the image of this mapping as the quotient topology.
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The second way is to take B a bounded, convex, balanced subset around 0 in F . Let FB bethe span of B and we can put a topology on FB using the Minkowski functional of B. We callB infracomplete if FB is Banach. The inclusion mapping j : FB → F is continuous since Bis bounded it is bounded on every semi-norm. So given a neighbourhood U of the origin in Fthere is some λ such that B ⊂ λU then if x ∈ 1
λB 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 : EN → FBsuch that
u = i u j.Where i : E → EN 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 ofL(E,F ) and that the general condition for nuclearity coincides with that which we have givenfor mappings between Banach spaces. However, we will not do this here. We also state withoutproof the following theorem which is reasonably intuitive particularly when we remember thatif N is a neighbourhood of 0 in E then its polar forms an equicontinuous set in E′.
Proposition. If u is a nuclear map from E to F then there exists sequences λn ∈ l1, x′n ∈E′ equicontinuous and yn ∈ B where B is infracomplete and bounded such that
u(x) = Σnλnx′n(x)yn.
This proposition is illustrative of the properties of nuclear maps. It is very clear here, thoughalso from the definition that the nuclear maps can be approximated by finite dimensional mapsand further restrictions can be given on these maps. This in particular shows that all nuclearmaps are compact. However, it is known that there are compact maps which are not the limitof 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 showthey are equivalent since the proof relies on an understanding of the dual spaces of E⊗εF andE⊗πF . Here we introduce the notation that if p is a seminorm on E with semi-ball Np then
Ep = ENp .
Characterizations of Nuclear Spaces. The following criteria are all equivalent and locallyconvex 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 Ep to
Eq 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 ε topologies are equal on E ⊗ F .
We can see in particular that if E is nuclear then the identity map will be nuclear andconsequently approximable by finite dimensional maps. Since if T : E → F = T I for any T alinear map, F a locally convex space then we hope that T In → T where In → I and are finitedimensional. This gives an indication of the powerful approximation properties that we have innuclear spaces in particular it gives a clue to the fact that the finite dimensional mappings aredense in L(E′, F ) when E is nuclear and under suitable further assumptions on E,F .
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Since the completion of E ⊗ F with either the ε or π topology is the same we now writeE⊗F . We can derive from the above theorem and our knowledge of nuclear maps that we havethe following isomorphisms which we state without proof.
D′(Ω1)⊗D′(Ω2) ∼= Lε(D(Ω1),D′(Ω2)) ∼= L(D(Ω1),D′(Ω2))
where the ε denotes the topology of convergence on the equicontinuous sets of D(Ω1) when it isidentified with the dual of D′(Ω1).
D(Ω)⊗ E ∼= C∞c (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] asthe space of continuous linear forms from D(Ω) to E then we will have
D′(Ω)⊗ E ∼= D′(Ω, E).
This provides a path into studying the vector valued distributions. These isomorphisms all relyon the fact that all the spaces D,D′, E , E ′,S and S ′ 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
D′(Ω1)⊗D′(Ω2) → L(D(Ω1),D′(Ω2))
and thatD′(Ω1)⊗D′(Ω2) → D′(Ω1 × Ω2).
We wish to extend these injections to D′(Ω1)⊗D′(Ω2). We claimed above that D′(Ω1)⊗D′(Ω2) ∼=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 subspacetopology induced by D′(Ω1 × Ω2) on D′(Ω1) ⊗ D′(Ω2) . This result follows from the fact thatevery bounded set in D(Ω1 × Ω2) is contained in the convex hull of a set of the form A ⊗ Bwhere A and B are bounded sets in D(Ω1) and D(Ω2) respectively.
Schwartz’s original proof gives the result originally for the spaces E ′ where there are somesimplifications in the proof of the intermediate results which follow from the fact that E is aFrechet space. He then makes an element of D′(Ω1×Ω2) as an inductive limit of functionals onthe spaces of the form D′(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 V is a completion of V if:(1)V is complete in the sense of Filters,(2)V can be identified with a dense linear subspace of V .
Proposition. Every topologyical vector space V has a completion.
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Proof. We will do this similarly to the metric case by looking at equivalence classes of CauchyFilters. Suppose F and G are two Cauchy filters. We say that they are equivalent if for everyneighbourhood 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 aneighbourhood of 0, then let M be a neighbourhood of 0 such that M +M ⊂ N . Then x+Mis a neighbourhood of x. Therefore there exists C ∈ F such that C ⊂ x + M . Then there isalso 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 thenthere 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) ∩ (A′ +B′) ⊃ (A ∩ A′) + (B ∩B′) we can generate a filter (F + G ) whichcontains B. Then we want to consider the filter
λF = λA | A ∈ F
Now we would like to check that if F ∼ F ′ and G ∼ G ′ then (F + G ) ∼ (F ′ + G ′) andthat λF ∼ λF ′.
To show this let N be a neighbourhood of 0. Let M be a neighbourhood of 0 such thatM + M ⊂ N . Then since F ∼ F ′ there will be A ∈ F , A′ ∈ F ′ such that A − A′ ⊂ M .Similarly there is B ∈ G , B′ ∈ G ′ such that B−B′ ⊂M then we will have (A+B)−(A′+B′) ⊂M + M ⊂ N . Now suppose N is a neighbourhood of 0 then 1
λN is also a neighbourhood of 0so there exists A ∈ F , A′ ∈ F ′ such that A−A′ ⊂ 1
λN . 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 V . LetN be a neighbourhood of 0 in V . Then we have the set
N = α ∈ V | ∃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 V .Now we would like to show that the topology we have just defined is Hausdorff. Suppose
that α 6= 0 is an element of V . Let F ∈ α. Then let O be the filter of all sets of V whichcontain 0, this is an element of 0 in V . Then we know that F is not equivalent to O. So thereexists a neighbourhood N of 0 in V such that A−0 * N for every A ∈ F . Therefore A * Nfor every A ∈ F . Now let M be a neighbourhood of 0 in V such that M +M ⊂ N . Let G ∈ αand suppose that there exists B ∈ G with B ⊂ M . Then there exists A ∈ F and B′ ∈ G suchthat A−B′ ⊂M . Let B′′ = B ∩B′ ∈ G . Then
A+ 0 ⊂ A− (B′′ −B′′) = (A−B′′) +B′′ ⊂M +M ⊂ N
which is a contradiction. So there doesn’t exist a B ∈ G with B ⊂ M . Therefore M is aneighbourhood of 0 not containing α.
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Now we would like to construct a canonical linear map T : V → V . 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 filtersconverging 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 neighbourhoodof x and therefore a member of the filter of neighbourhoods of x. Therefore T (N) ⊂ N . SupposeT (x) ∈ N then N is in a filter converging to x. Therefore for every M a neighbourhood of 0 inV . We have N ∩ (x+M) 6= ∅ therefore x ∈ N therefore N ∩ T (V ) ⊂ T (N). This shows that Tis an isomorphism onto its image.
Now we would like to show that the image of T is dense in V . We would like to do thisby showing that given any open set U that T (V ) ∩ U 6= ∅. Fix α ∈ U then there is some N aneighbourhood of 0 in V such that α+N ⊂ U . Pick some F ∈ α and let M be a neighbourhoodof 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 sayingthat 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 V is complete. We will denote Cauchy filters on V with
hats on top like F to distinguish them from Cauchy filters on V . Suppose F is a Cauchyfilter on V and G is the Cauchy filter of neighbourhoods of 0 in V . We can generate the filterF + G and we showed in the proof that LF-spaces are complete that this filter converges iffF converges. So we would like to show that F + G converges. F + G is a Cauchy filtersince if N is a neighbourhood of 0 in V and M is another neighbourhood of 0 in V such thatM+M−M⊂ N then there exists A ∈ F such that A−A ⊂M hence
(A+M)− (A+M) ⊂M+M+M⊂ N
Now we would like to relate F + G to a Cauchy filter on V . Let A ∈ F + G then let A =T−1(A ∩ T (V )). Let F be the collection of such A. We claim that F is a filter. Suppose
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 V we know that T (V ) ∩ (B + G) 6= ∅ so thatT (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 F + G 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 F + G is a Cauchy filter there exists A ∈ F + G 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 Cauchyfilter. Now let α be the equivalence class of F . We claim that F + G → α. To show this letN be a basic neighbourhood of 0 in V then we want to show that α+N ∈ F + G . We do thisby showing that there is some A ∈ F such that A ⊂ T−1((α+N )∩ t(V )). In this case there is
some A ∈ F + G such that
A = T−1(A ∩ T (V )) ⊂ T−1((α+N ) ∩ T (V ))
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soA ∩ T (V ) = (α+N ) ∩ T (V )
and A ⊃ B + G where B ∈ F and G ∈ G so that B + G will be open in V which means that
B + G ⊂ α+N
since both the above sets are open and T (V ) is dense. This will mean that F + G → α. Itremains to prove that such an A exists.
We know that T (x) ∈ α + N means that T (x) − α ∈ N which means that there existsG ∈ T (x) − α with N ∈ G . We will in fact show that we can take G = F . Choose M aneighbourhood of 0 such that M + M ⊂ N . Since F ∈ α is Cauchy there is A ∈ F such thatA−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.
Proof. We have the maps Ti : V → Vi for each i. So take T−11 T2 : T1(V ) → V2 which is a
homeomorphism from a dense subspace of V1 onto a dense subspace of V2. Therefore we canextend it to a homeomorphism from the whole of V1 to a complete subspace of V2 which containsT2(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, Birkhauser,2010,
[2] Walter Rudin, Functional Analysis, McGraw-Hill, 2005
[3] Laurent Schwartz, Theorie des Distributions, Hermann, 1966
[4] Francois Treves, Topological Vector Spaces, Distributions and Kernels, Academic Press,1967
[5] Laurent Schwartz, “Sur l’impossibilite de la multiplications des distributions”, C.R. Acad.Sci. Paris 239:847-848, 1954
[6] Jesper Lutzen, The Prehistory of the Theory of Distributions, Springer-Verlag, 1982
[7] Gerald B. Folland, Introduction to Partial Differential Equations, Princeton UniversityPress, 1995
[8] Lars Hormander, The Analysis of Linear Partial Differential Operators II, Springer, 1983
[9] Albrecht Pietsch, Nuclear Locally Convex Spaces, Springer-Verlag, 1972
[10] Laurent Schwartz, Theorie des Distributions a Valeurs Vectoriels. I, Annales de l’institutFourier, tome 7, 1957, p. 1-141
[11] Laurent Schwartz, Theorie des Noyaux, Proceedings of the International Congress of Math-ematicians, 1950, vol I, p. 220-230
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[12] O. Heaviside, On Operators in Mathematical Physics, Proc. of the Royal Society, London,52 (1893), p.105-143
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