Topological Vector Spaces, Distributions and Kernels

By François Treves

This text for upper-level undergraduates and graduate students focuses on key notions and results in functional analysis. Extending beyond the boundaries of Hilbert and Banach space theory, it explores aspects of analysis relevant to the solution of partial differential equations.
The three-part treatment begins with topological vector spaces and spaces of functions, progressing to duality and spaces of distribution, and concluding with tensor products and kernels. The archetypes of linear partial differential equations (Laplace's, the wave, and the heat equations) and the traditional problems (Dirichlet's and Cauchy's) are this volume's main focus. Most of the basic classical results appear here. There are 390 exercises, several of which contain detailed information that will enable readers to reconstruct the proofs of some important results.

• Language: English
• Publisher: Dover Publications
• Release date: Sep 3, 2013
• ISBN: 9780486318103

Book preview

theorem.

1

Filters. Topological Spaces. Continuous Mappings

A topological vector space E is, roughly speaking, a set which carries two structures: a structure of topological space; a structure of vector space. Furthermore, some kind of compatibility condition must relate these two structures on E. We begin by recalling briefly what each one of them is, in the absence of any relation between the two.

One defines usually a topology on a set E by specifying what the open subsets of E are going to be. However, in dealing with topological vector spaces, as we are going to do in this book, it is more convenient to define a topology by specifying what the neighborhoods of each point are going to be. It is well known that the two approaches are equivalent: an open set will be a set which, whenever it contains a point, contains a neighborhood of this point; one can also say that an open set is a set which is a neighborhood of each one of its points; on the other hand, a neighborhood of a point x of E is simply a set which contains some open set containing x.

In order to define a topology by the system of the neighborhoods of the points, it is convenient to use the notion of filter. This is a very primitive notion, and the student should find it easy to become familiar with it, and to learn how to use filters, just as he learned how to use sequences. The notion of filter is perfectly independent of topology. A filter is given on a set which need not carry any other structure. Let E is a family of subsets of E, submitted to three conditions:

The simplest example of a filter on a set E is the family of all subsets of E which contain a given subset A, provided the latter is nonempty. With every infinite sequence of points of E is associated a filter. Let x1, x2 ,… be the sequence under consideration. The associated filter is the family of all subsets of E which have the following property:

of subsets of E is a basis of a filter on E if the following two conditions are satisfied:

A familiar example of a basis of filter on the straight line is given by the family of all intervals (—a, a) with a > be the filter associated with a sequence S = {x1, x2 ,…, xn ,…}. For each n = 1, 2,…, let us set

and view Sn as a subset of E. Then the sequence of subsets S = S1 ⊃ S2 ⊃ … Sn .

be some family of subsets of our set E. , if it exists, is completely and uniquely determined: it is the family of subsets of E , of subsets of E. :

. Thus we may state: a basis of filter on E is a family of nonempty subsets of E satisfying Condition (BF). The filter generated by the basis is uniquely determined: by Condition (BF2).

Next step: comparison of filters. We want to be able to say: this filter is finer than this other filter. Keep in mind that filters are sets of sets, or rather of subsets. In other words, filters are subsets of the set of subsets of E(E(E′ are two filters on the same set E. It means that every subset of E ′ is finer ′) be the family of all subsets of E which contain a given subset A (resp. A′) of E; ′ if and only if A′ ⊂ A.

A topology on the set E is the assignment, to each point x of E(x) on E, with the additional requirement that the following two conditions be satisfied:

When these conditions are satisfied we say that we have a topology on E (x) the filter of neighborhoods of the point x. At first sight Condition (N2) may seem involved. It expresses, however, a very intuitive fact. Roughly speaking, it says that given any point z near x (i.e., z is a generic element of U), if a third point y lies sufficiently near to x (the sufficiently near is made precise by the neighborhood of x, V, of which y is an element), then z lies near to y (i.e., z ∈ U (y)). In the language of open sets, (N2) becomes evident: since U is a neighborhood of x, U contains an open set containing x; let V be such an open set. Since V is open, and V ⊂ U, U is obviously a neighborhood of each point of V. (x) is called a basis of neighborhoods of x. This simple notion will play an important role in the forthcoming definitions.

Once we have the notion of filter of neighborhoods of a point, hence of neighborhood of a point (any subset of E belonging to the filter of neighborhoods), we can review quickly the concepts that are used to describe a topology. As we have already said, an open set is a set which is a neighborhood of each one of its points. A subset of E is closed if its complement is open. The closure of a set A ⊂ E is the smallest closed set containing A. It will be denoted by Ā. The following is easy to check: a point belongs to Ā if and only if everyone of its neighborhoods meets A (that is to say, has a nonempty intersection with A). The interior of a set is the largest open set contained in it; if A is the set, its interior will be denoted by Å.

A very important notion is the one of a set A dense in another set B; both A and B are subsets of the same topological space E. Then, one says that A is dense in B if the closure Ā of A contains B. In particular, A is said to be dense in E (or everywhere dense) if Ā — E. To say that A is dense in B means that, given any neighborhood of any point x of B, U(x), there is a point y of A which belongs to U(x), i.e., A ∩ U(x) ≠ 0. A standard example of a set everywhere dense is the set of rational numbers Q, when regarded as a subset of the real line R (with the usual topology); note that the complement R — Q of Q is also dense in R. Examples of sets which are dense and open are given by the complement of a straight line in the plane or in space, by the complements of a plane in space, etc. Easy to check are the basic intersection and union properties about open or closed sets: that the intersection of a finite number of open sets is open (this follows immediately from the fact, itself obvious in virtue of Axiom (F2), that the intersection of a finite number of neighborhoods of a point is again a neighborhood of that point); that the union of any number of open sets, be that number finite or infinite, is open (this follows from the fact that the union of a neighborhood of a point with an arbitrary set is a neighborhood of the same point: Axiom (F3)). By going to the complements, one concludes that finite unions of closed sets are closed, arbitrary intersections of closed sets are also closed, etc.

Observe that a set E may very well carry several different topologies. When dealing with topological vector spaces, we shall very often encounter this situation of a set, in fact a vector space, carrying several topologies (all compatible with the linear structure, in a sense that is going to be specified soon). For instance, any set may carry the following two topologies (which, in practice, are almost never used):

′ be two topologies on the same set E. is finer than ′ if every subset of E , or equivalent, if every subset of E (x′(x))) be the filter of neighborhoods of an arbitrary point x of E ′, if, for every x ∈ E(x′(x). Given two topologies on the same set, it may very well happen that none is finer than the other. If one is finer than the other, one says sometimes that they are comparable. The discrete topology is finer, on a set E, than any other topology on E; the trivial topology is less fine than all the others. Topologies on a set form thus a partially ordered set, having a maximal and a minimal element, respectively the discrete and the trivial topology.

The notion of a topology has been introduced in order to provide a solid ground for the notions of convergence and of continuity. Of course, the latter were correctly manipulated (or most of the time, at least) well before anybody thought of topology. We proceed now to give their general definition.

Convergenceon a topological space E is a family of subsets of E. is to converge to a point x of E, which, we repeat again, are subsets of E, get smaller and smaller about x, and that the points of these subsets get nearer and nearer to x. This can be made precise in terms of the neighborhoods of x, which we have at our disposal, since E is a topological space: we must express the fact that, however small a neighborhood of x is, it should contain some subset of E which are contained in that particular one. But in view of Axiom (F3), this means that the neighborhood of x . The phrase "however small a neighborhood of x is has to be made mathematically meaningful: it simply means whatever is the neighborhood of x." In brief, we see that the filter converges to the point x if every neighborhood of x belongs to , in other words if is finer than the filter of neighborhoods of x, (x). This is what the convergence to a point of a filter means.

We recall how the convergence of a sequence to a point is defined. Let S = {x1 , x2 ,…} be the sequence. We say that S converges to x if, given an arbitrary neighborhood U of x, there is an integer n(U) such that n n(U) implies xn ∈ U. Let S = S1 ⊃ S2 ⊃ ··· ⊃ Sn ··· be the subsequences introduced on p. 7: S converges to x if to every U (x) there is an integer n(U) such that Sn(U) ⊂ U. As the subsets Sn of E form a basis of the filter associated with the sequence S, we see immediately that a sequence S converges to x if and only if the associated filter converges to x.

Note that a filter may converge to several different points. Suppose, for instance, that E carries the trivial topology (p. 9): then every filter on E converges to every point of E. Note also that a filter may not converge: for instance, if it is the filter associated with some sequence and if this sequence does not converge. Another example is given by a filter on E which is not the filter of all subsets of E which contain a given point x—when E carries the discrete topology: in this topology, the only converging filters are the filters of neighborhoods of the points. So much for convergence in general topological spaces.

Continuity. This concerns mappings. In point set topology, a map f: E → F, this is to say a map from a topological space E into another topological space F, is said to be continuous if any one of the following two conditions is satisfied:

The student will easily check the equivalence of (a) and (b). As for the intuitive meaning of these conditions, we may say the following. If the mapping f is to be continuous at the point x, it should mean that if x′ ∈ E "converges to x," then f(x′) should converge to f(x). Note that " f(x′) converges to f(x)" can be made precise in the following way: given an arbitrary neighborhood of f(x), f(x′) should eventually belong to it; and the eventually means here: provided that x′ is sufficiently near to x. Thus given an arbitrary neighborhood V of f(x), if x′ belongs to a sufficiently small neighborhood of x, then f(x′) ∈ V. The sufficiently small can only be determined by the existence of a certain neighborhood U of x, such that, as soon as x′ ∈ U, then f(x′) ∈ V. This is exactly Property (a): to every neighborhood V of f(x) there is a neighborhood U of x such that

It is immediately seen that, if a sequence {x1, x2 ,…} converges in E to a point x, and if f is a continuous function from E into F, then the sequence {f(x1), f(x2),…} converges to f(x) in F. Convergence of filters is also easily related to continuity of mappings. Let

be a mapping from a set E into a set F. be a filter on E. The image f under f is defined as being the filter having the basis

Observe that, in general, (f )0 is not itself a filter; it is always the basis of a filter (the student may check this point as an exercise). Now, if the filter converges to a point x in E and if f is a continuous function, then f converges to f(x) in F. Indeed, the continuity of f implies that f (x(x(xconverges to x). f is finer than f (x(f(x)).

We have only considered continuous functions, which is to say functions defined everywhere and continuous everywhere. Of course, one may prefer to talk about functions continuous at a point. This is defined by the condition (where x is the point under consideration):

or, equivalently,

Let us insist on the fact that all the functions or mapping which will be considered in this book are defined everywhere.

As a last remark, let us consider the case where F is identical with E as a set, but carries a different topology from the one given on E, and where f is the identity mapping of E onto F, I. The following two properties are obviously equivalent:

Exercises

1.1.Let X be a topological space, A, B two subsets of X. Prove that if A is open we have

Give an example of sets A and B such that A is not open and that the preceding inclusion is not true.

1.2.Prove that the image of a dense set under a map, which is continuous and surjective (i.e., onto), is dense.

1.3.Let X, Y be two topological spaces, f: X → Y a continuous function, B a subset of Yits closure. Do we always have

1.4.Give an example of the following situation: X and Y are topological spaces, f: X → y is a continuous mapping, A is a closed subset of X, and f(A) is not closed in Y.

1.5.Consider a straight line L in the plane R². The filter of neighborhoods of L in R² is the filter formed by the sets which contain an open set containing L. Prove that there is no basis of this filter which is countable.

1.6.Let f on the real line, defined by means of the Riemann sums

where 0 = t0 < t1 < ··· < tj < ··· < tr+1 = 1 and tj τj tj+1 for each j. converges to the integral

1.7.Prove that the filter of neighborhoods of the closed unit disk {(x, y) ∈ R²; x² + y1} in the plane has a countable basis (cf. Exercise 1.5).

1.8.Let Xα (α ∈ A) be a family of topological spaces. Consider their product set

Let us denote by pα the projection mapping on the ath coordinate axis Xα:

The product topology on X is defined in the following way: a subset U of X is a neighborhood of one of its points x = (xα) if, for every α, pα(U) is a neighborhood of xα and if, for all α except possibly a finite number of them, pα(U) = Xα. Prove that this is the least fine topology on X such that all the mappings pα be continuous.

1.9be a filter on the product space X of under the projection pα. α of the form pα(M) as M α does.

1.10.Let us say that a set A is predirected if there is a preorder relation a b on A and if, for any pair of elements a, b of A, there is c ∈ A such that a c, b c (the relation a b is a preorder relation if it is reflexive, i.e., a a for all a, and transitive, i.e., a b and b c imply a c; it is an order relation if, furthermore, a b and b a imply a = b). Let Φ be the set of all filters on X, Φ′ the set of all mappings of predirected sets into X. Prove that there is a canonical mapping of Φ′ onto Φ (this mapping is not one-to-one). Under this mapping, the image of a function f on the predirected set of positive integers into X is the filter associated with a sequence in X.

2

Vector Spaces. Linear Mappings

We recall first what a vector space is. The vector spaces we shall consider will be defined only on one of the two classical fields: the field of real numbers, R, or the field of complex numbers, C. As a rule, we shall suppose that the field is C. When we specifically need the field to be R, we shall always say so. In other words, we deal always with complex vector spaces. A vector space E over C is a system of three objects (E, Av , Ms) consisting of a set E and of two mappings:

Of course, there are conditions to be satisfied by these objects. The mapping Av , called vector additioriy must be a commutative group composition law, i.e., it must have the following properties:

Of course, we write x – y instead of x + (– y). The mapping Ms is called scalar multiplication, or multiplication by scalars, and should satisfy the following conditions:

(i) λ(μx) = (λμ)x;

(ii) (λ + μ)x = λx + μx

(iii) 1 · x = x;

(iv) 0 · x = 0;

(v) λ(x + y) = λx + λy.

We do not recall the meaning of such notions as linear independence, basis, vector (or linear) subspace, etc. A mapping f: E → F of a vector space E into another, F, is called linear if for all x, y ∈ E, λ, μ ∈ C,

Let us recall that a linear mapping f : E → F is one-to-one if and only if f(x) = 0 implies x = 0. Indeed, if f is one-to-one, f(x) = 0 must imply x = 0. Conversely, suppose that f(x) = 0 implies x = 0, and let x, y ∈ E be such that f(x) = f(y). This equation can be written f(x – y) = 0, implying then x – y = 0.

Q.E.D.

A notion with which the student may not be so familiar is the one of quotient space. As it will play a crucial role in the sequel, we shall recall its definition.

Let E be a vector space (over C) and M a linear subspace of E. For two arbitrary elements x and y of E, the property

defines an equivalence relation: it is reflexive, since x – x = 0 ∈ M (every linear subspace contains the origin); it is symmetric, since x – y ∈ M implies – (x – y) = y – x ∈ M (if a linear subspace contains an element, it contains its inverse); it is transitive, since

(when a linear subspace contains two vectors, it also contains their sum). Then we may define the quotient set E/M: it is the set of equivalence classes for the relation x – y ∈ M. There is a canonical mapping of E onto E/M: the mapping which, to each x ∈ E, assigns its class modulo the relation x – y ∈ M. It helps the intuition to visualize the class of elements equivalent to x modulo M, that is to say the y’s such that x – y ∈ M, as a linear subvariety: indeed, they constitute the set

which is the translation of M by x. Observe the following, which is easy to check (using the fact that M is a linear subspace):

Thus we define vector addition and scalar multiplication in E/M: if ϕ(x) is the class of x mod M, λϕ(x) = ϕ(λx) and ϕ(x) + ϕ(y) = ϕ(x + y). These definitions are unambiguous by virtue of (2.1) and (2.2); they turn E\M into a vector space, and ϕ : E → E/M, the canonical mapping, is then a linear map. It is, of course, onto.

Now let E, F be two linear spaces (over C), and f a linear map E → F. We define the image of f, and denote it by Im f, as the subset of F:

We define the kernel of f, and denote it by Ker f, as the subset of E:

Both Im f and Ker f are linear subspaces (of F and E resp.). We have then the diagram

where i is the natural injection of Im f into F, that is to say the mapping which to each element y of Im f assigns that same element y, regarded as an element of F; ϕ is the canonical map of E onto its quotient, E/Ker fis defined so as to make the diagram commutative, which means that the image of x ∈ E under f is identical with the image of ϕ(x) (i.e., the class of x modulo Ker fis well defined by the equation

Indeed, if ϕ(x) = ϕ(y), in other words if x – y ∈ Ker f, then f(x) = f(y). It is an immediate consequence of the linearity of f and of the linear structure of the quotient space E/Ker f is a one-to-one linear map of E/Ker f onto Im f. The onto property is evident from the definition of Im f (ϕ(x(ϕ(y)), it means by definition that f(x) = f(y), hence f(x – y) = 0 or x – y ∈ Ker f, which means that ϕ(x) = ϕ(y).

Q.E.D.

Let E be an arbitrary set (not necessarily a vector space) and F (E; F) the set of all mappings of E into F. It can be equipped with a natural structure of vector space. We must first define the sum of two mappings f, g : E → F. It must be a function of the same kind, and we must therefore say what its value should be at an arbitrary point x of E. Naturally, we take this value to be equal to the sum of the values of the factors, f(x), g(x):

Similarly, to define λf, where λ is an arbitrary scalar, we define its value at an arbitrary point x of E. We set (this is again a definition):

When E also is a vector space (over the same scalar field as F, for us the field of complex numbers C), we will be particularly interested in the linear mappings of E into F(E; F.

When F is denoted by E* and called the algebraic dual of E. When E is a topological vector space (see next chapter) we shall be interested in a smaller dual of E, namely the linear subspace of E* consisting of the linear mappings E → C which are continuous; this will be called the dual of E and denoted by E′. One should always be careful to distinguish between E* and E′ (except in exceptional cases, e.g., when E is finite dimensional†). The elements of E* are most of the time referred to as the linear functionals, or the linear forms on E.

If E, F, G are three vector spaces over C, and u : E → F, v :F → G two linear mappings, it is clear that the compose v u, defined by

is a linear map of E into G. If G = C, v is a linear functional on F, i.e., v is an element x* of the algebraic dual F* of F; the compose x* u is a linear functional on E. We obtain thus a mapping x* x* u of F* into E* for each given u ∈ . This mapping is obviously linear. It is called the algebraic transpose of u; we shall denote it by u*. As is readily seen, u u* .

Exercises

2.1.Give an example of a linear space E and of two linear mappings u, v of E into itself with the following properties:

2.2.Let E be a vector space, M a linear subspace of E, and u a linear map of M into a vector space F. Prove that there is a linear map v : E → F which extends u, i.e., such that u(x) = v(x) for all x ∈ M.

2.3.Let E, F be two vector spaces, u : E → F a linear map, and u* : F* → E* the algebraic transpose of u. Prove that the following properties are equivalent:

(a)u is surjective;

(b)u* is injective.

2.4.Let E, F be two vector spaces, and u : E → F a linear map. Let M (resp. N) be a linear subspace of E (resp. F), and ϕ (resp. Ψ) the canonical mapping of E (resp. F) onto E/M (resp. F/N). Prove the equivalence of the following two properties:

(a)u(M) ⊂ N;

(b)there exists a linear map v such that the following diagram is commutative:

Prove that, if (a) holds, the mapping v above is unique.

2.5.Let M be a linear subspace of E, j the natural injection of M into E, and ϕ the canonical map of E onto E/M. Let us set

2.6.Let (Eα) (α ∈ A) be an arbitrary family of vector spaces over the complex numbers. Consider the product set

it carries a vector space structure where vector addition and scalar multiplication are performed componentwise. The direct sum of the Eα is the linear subspace of E consisting of those elements x = (xα)α ∈ A which all the components xα are equal to zero with the possible exception of a finite number of them; we shall denote by E0 this direct sum. Prove that there is a canonical isomorphism between the algebraic dual of the direct sum E0 and the product of the algebraic duals E*α of the Eα .

2.7.Let us keep the notation of Exercise 2.6. Let α ∈ A. We denote by jα the (linear) mapping of Eα into E defined as follows: if z ∈ Eα , jα(z) is the element x = (xβ)β ∈ A of E such that xβ = 0 if β ≠ α and xα = z. It is evident that jα is one-to-one; if pα is the coordinate projection x = (xβxα , we have pα jα = identity of Eα . Let j*α : E* → E*α be the algebraic transpose of jα . We may define the following linear map of E*

Prove the equivalence of the following properties:

(a)the set of indices A is finite;

(b)j* is one-to-one;

(c)j* is ah isomorphism of E* .

2.8.Prove that every vector space E is isomorphic to the direct sum of a family of one-dimensional vector spaces. Then, by making use of the results stated in Exercises 2.6 and 2.7, prove that the following properties are equivalent:

---

† Also Hausdorff; see Chapter 9.

3

Topological Vector Spaces. Definition

Let E be a vector space over the field of complex numbers C (in short, a vector space). Let

be the vector addition and the scalar multiplication in E. in E is said to be compatible with the linear structure of E if Av and Ms are continuous when we provide E , E × E , and C × E is the usual topology in the complex plane C. We recall the meaning of a product topology.. Consider two topological spaces E, F. In order to say what the product topology on E × F is, it suffices to exhibit a basis of the filter of neighborhoods of each point (x, y) of E × F. Such a basis is provided by the rectangles

where U (resp. V) is a neighborhood of x (resp. y) in E (resp. F). That these rectangles form a basis of filter is trivial; they obviously do not form a filter (except in trivial cases), since a set which contains a rectangle does not have to be a rectangle. It remains to check that the filters thus defined, for each pair (x, y), indeed can be taken as filters of neighborhoods of (x, y) in a topology on E × F assigns to each point λ of the complex plane a remarkable basis of neighborhoods, the disks, open or closed, with center at this point (and with positive radius p). When provided with a topology compatible with its linear structure, E becomes a topological vector space, which we shall abbreviate into TVS.

Suppose that E is a TVS. Then its topology is translation invariant, which, roughly speaking, means that, topologically, E looks about any point as it does about any other point. More precisely: the filter of neighborhoods (x) of the point x is the family of sets V + x, where V varies over the filter of neighborhoods of the neutral element(0). Proof of this statement: Let U be an arbitrary neighborhood of x. As the mapping y y + x from E into (as a matter of fact, onto) itself is continuous, which follows immediately from the continuity of the mapping Av: (x, yx + y, the inverse image of U under this mapping must be a neighborhood of the preimage of x under this mapping; this preimage is obviously the neutral element 0. Let V be the inverse image of U. We have U = V + x. Conversely, given an arbitrary neighborhood V of 0, V + x is a neighborhood of x by virtue of symmetry, or by virtue of the continuity of the mappings y y — x. Thus: in order to study the topology of a topological vector space E, it suffices to study the filter of neighborhoods of the origin.

In practice, one always begins by giving the filter of neighborhoods of the origin, or (more frequently) a basis of this filter. It follows from there that we need some criteria on a filter which would insure that it is the filter of neighborhoods of the origin in a topology compatible with the linear structure of E.

THEOREM 3.1. A filter on a vector space E is the filter of neighborhoods of the origin in a topology compatible with the linear structure of E if and only if it has the following properties:

We have used two words, absorbing and balanced, which have not yet been defined.

Definition 3.1. A subset A of a vector space E is said to be absorbing if to every x ∈ E there is a number cx > 0 such that, for all λ ∈ C, | λ cx , we have λx ∈ A.

In more colorful but less precise language, we may say that A is absorbing if it can be made, by dilation, to swallow any single point of the space.

Definition 3.2. A subset A of a vector space E is said to be balanced if for every x ∈ A and every λ ∈ C, | λ 1, we have λx ∈ A.

The only balanced subsets of the complex plane are the open or the closed disks centered at the origin.

Proof of Theorem 3.1. We begin by proving the necessity of Properties (3.1)–(3.5). The necessity of (3.1) goes without saying.

Necessity of (3.2). By V + V we mean the set of points x + y, where x and y run over V. Let U be an arbitrary neighborhood of the origin. Its preimage under the mapping (x, yx + y of U must be a neighborhood of 0, therefore must contain a rectangular neighborhood W × W′, where W and W′ are neighborhoods of 0 in E. But then it contains a square, namely (W ∩ W′) × (W ∩ W′). If we take V = W ∩ W′, this means precisely that V + V ⊂ U.

Necessity of (3.3). By λU we mean the set of vectors λx, where x varies over U. Because of the continuity of the mapping (λ, x) → λx from C × E into E, if we fix λ ∈ C, λ ≠ 0, the map x → λ–1x of E into itself must be continuous. The preimage of any neighborhood U of the origin in E must be such a neighborhood; this preimage is obviously λU.

Necessity of (3.4). Again we use the continuity of the mapping (λ, x) → λx, this time at a point (0, x) where x is an arbitrary point of E. The preimage of a neighborhood U of 0 in E must be a neighborhood of (0, x), since (0, x) is mapped into 0. Hence that preimage must contain a rectangle N × W where N (resp. W) is a neighborhood of 0 (resp. of x) in C (resp. in E). By definition of the topology of a TVS, W is of the form W′ + x, where W′ is a neighborhood of 0 in E (see p. 21). On the other hand, N contains a disk of the complex plane, centered at the origin, Dρ = {λ ∈ C; | λ ρ}, ρ > 0. Thus we see that, for all y ∈ W′ + x and all complex numbers λ such that | λ ρ, we have λy ∈ U. In particular, we may take y = x.

Necessity of (3.5). We duplicate the proof of the necessity of (3.4) but taking this time x = 0, hence W′ = W. We have seen that the preimage of U contains a rectangle Dρ × W, which means that the set

is contained in U. This set V is obviously balanced. It is a neighborhood of zero, since each λW, λ ≠ 0, is one (in view of (3.3)).

Sufficiency of Conditions (3.1)–(3.5). We must first of all show that, if we define the filter of the neighborhoods of an arbitrary point x of E under the translation y x + y, we have indeed a topology on the set E. Once we have proved this, we must show that this topology is compatible with the linear structure of E. (xunder y x + y, that is to say the family of subsets U + x, where U . Since 0 ∈ U, x belongs to U + x. Thus Axiom (N1), p. 8, is satisfied. Let V be such that V + V ∈ U. Take an arbitrary point y of the set V + x; then U + x contains V + (V + x), hence V + y. But V + y (y), and therefore so does U + x. Thus Axiom (N2), p. 8, is satisfied when we take U + x and V + x, respectively, in the place of U and V in the statement of p. 8. We conclude that we have indeed a topology on E. The last two steps consist in proving that the mappings Av and Ms are continuous. The continuity of Av follows immediately from (3.2). Indeed, let (x, y) be an arbitrary element of E × E; let W be a neighborhood of its image, x + y. We know that W = U + x + y, U ∈ . Choose V ∈ such that V + V ∈ U. Then (V + x) + (V + y) ∈ W, which means that the image of the neighborhood of (x, y),

is contained in W. Then the preimage of W contains that same neighborhood of (x, y) and, consequently, is a neighborhood of (x, y).

Last step: continuity of

Let U′ be a neighborhood of λ0x0; U′ is of the form U + λ0x0 , where U is a neighborhood of zero in E. Let us select another neighborhood of 0, W, such that

Such a neighborhood of zero, W, exists in view of Properties (3.2) and (3.5). In view of (3.4), W is absorbing. In other words, there is a number ρ 1, such that

Let Dρ be the disk centered at the origin, in the complex plane, with radius ρ. Suppose first that λ0 = 0, which implies λ0x0 = 0 and U′ = U. Then we look at the image under Ms of the set Dρ × (W + x0); it is the set

As | λ ρ 1 and as W is balanced, y ∈ W implies λy ∈ W. As | λ ρ, we have also λx0 ∈ W. We conclude that the set (3.6) is contained in W + W, hence in U. Thus the preimage of U contains Dρ × (W + x0), which is a neighborhood of (0, x0); the preimage of U is a neighborhood of (0, x0). Let us suppose finally that λ0 ≠ 0. In this case we look at the image under Ms of the set

where σ = inf(ρ, | λ0 |), the smallest of the two numbers ρ, | λ0 |. The image of (3.7) is the set

Since the complex numbers λ| λ0 |–1, λ0| λ1, and since W is balanced, the sum

belongs to W + W. Since | λ σ ρ, λx0 ∈ W, so that the set (3.8) is contained in

In other words, the preimage of U + λ0x0 contains (3.7) and, therefore, it is a neighborhood of (λ0 , x0).

Q.E.D.

The following property of the filter of neighborhoods of zero in a TVS E is important:

PROPOSITION 3.1. There is a basis of neighborhoods of zero in a TVS E which consists of closed sets.

Proof. It suffices to show that an arbitrary neighborhood of zero U in E contains a closed neighborhood of 0. Let V be another neighborhood of 0 such that V — V ⊂ U⊂ U. Indeed, let x , which means that every neighborhood of x, in particular V + x, meets V. Thus, there are elements y, z ∈ V such that z = x + y—in other words,

COROLLARY. There is a basis of neighborhoods of 0 in E consisting of closed balanced sets.

Indeed, every neighborhood U of 0 in E contains a closed neighborhood of 0, V, which in turn (Theorem 3.1) contains a balanced neighborhood of 0, W. ⊂ V ⊂ U.

Q.E.D.

The student will easily see that, whatever may be the vector space E, the trivial topology (p. 9) is always compatible with the linear structure of Ey and the discrete topology (p. 9) never is—unless E consists of a single element (the origin). We proceed now to discuss a less trivial example.

Example. Let us denote by C[[X]] the ring of formal power series in one variable, X, with complex coefficients. Such a formal power series is written

where the coefficients un are complex numbers. It is the same thing as a power series as encountered in the theory of analytic functions, except that one does not care if it converges or not. Essentially, it is a sequence of complex numbers (u0 , u1 , u2 ,…, un ,…). Addition and multiplication are immediately defined, by just extending what one does with polynomials or with Taylor expansions of analytic functions about the origin. If

we have

Multiplication by scalars is defined in the obvious way:

Addition and multiplication by scalars turn C[[X]] into a vector space; multiplication of formal power series turn it then into an algebra. There is a unit element in this algebra: the formal power series 1, that is to say the series u having all its coefficients un equal to zero if n 1, and such that u0 = 1. The following fact is not difficult to prove:

For a formal power series u to have an inverse, it is necessary and sufficient that its first coefficient, u0 , be different from zero.

Let us denote by 501 the set of elements which do not have an inverse, that is to say the set of formal power series u such that uis an ideal of the algebra C[[X]], which means that

is the largest proper (i.e., different from the whole algebra C[[X]) ideal of C[[Xmust contain an invertible element, hence must contain the series 1, in view of (2), and hence must be identical with the whole algebra.

For n n the set of formal power series u such that up = 0 if p < n. n can be written

with u. As the series X (i.e., all coefficients except the one of X¹ are equal to zero, and the coefficient of Xn is the product of n n is an ideal of C[[Xn, as n → + ∞, is obviously the zero power series (i.e., the power series having all its coefficients equal to zero). As the sequence of sets

be the filter it generates: a set U n for large enough n. Let u n + u be the set of formal power series v + u, where v n(un + u, n = satisfies Condition (n being a vector subspace, we have

(u) are the filters of neighborhoods of the points u in a topology on C[[x]]: if v ∈ n + un + u n + v and hence is a neighborhood of v. n + u n + u is open. But (3.9) also implies that the preimage of

under the addition mapping, viewed as a map from the topological space C[[X]) × C[[X]) into the topological space C[[X]), contains

and hence is a neighborhood of the pair (u, v). This proves the continuity of addition.

The continuity of multiplication,

follows from the obvious inclusion

where the left-hand side is the set of products uv where u ∈ p and V ∈ q.

These continuity properties turn C[[X]] into what is called a topological ring.

However they do not turn C[[X]] into a topological vector space. n is not absorbing as soon as n > 0: for there is no complex number λ ≠ 0 such that λ · 1 ∈ n. Thus the multiplication by scalars (λ, uλu, viewed as a map from C × C[[X]] into C[[X]), is not continuous (although it is continuous if we identify λ with the formal power series u such that u0 = λ, up = 0 for p > 0, and if we view the multiplication by scalars as a mapping from [C[X]] × C[[X]) into C[[X])).

n are open. They are also linear subspaces. Now the following is easy to check:

PROPOSITION 3.2. In a TVS E, if a vector subspace M is open, we have M = E.

Indeed, M being open is a neighborhood of each one of its points, in particular of the origin, hence must be absorbing (Property (3.4) of Theorem 3.1). But if λx ∈ M with λ ≠ 0, then x = λ–1(λx) ∈ M.

m (m m consists of sets which are both closed and open (cf. Exercise 3.4).

The topology which we have just described is actually used in algebra. Note that every point has a countable basis of neighborhoods in that topology.

There is another topology which is used on C[[X]], and which is compatible with the linear structure of C[[X]). It is the topology of simple convergence of the coefficients. A formal power series u = is said to converge to another formal power series v if, for each n separately, the complex number un converges to vn. Note that in the first topology described, u did converge to v if the numbers p = 0, 1,…, such that un = vn for n < p n ′.)

As we shall always do in these chapters, we define a topology on a vector space, compatible with its linear structure, by exhibiting a basis of neighborhoods of zero. In our case, the basis will be the collection of the following sets of formal power series:

Here m and n are integers, n = 0, 1,…, m = generated by the basis {Vm,n} (m = 1, 2,…, n = 0, 1,…) satisfies Conditions (3.1)–(3.5) in Theorem 3.1. That the Vm, n indeed form a basis of a filter is an obvious consequence of the fact that

where sup(a, b) means the greatest of the two numbers a, b. Let {u(v)} (v = 1, 2,…) be a sequence of formal power series. It converges to a series u if and only if, to every pair of integers m 1, n 0, there is another v(m, n1 such that

This means, roughly speaking, that u(v) converges to u if more and more coefficients of u(v) get nearer and nearer to the coefficients with the same index of u. n are closed, as is immediately seen (if a formal power series u is a limit of formal power series v such that vp = 0 for p < n, in the sense that the coefficients of u are the limits of the corresponding coefficients of the v’s, we must have up = 0 for p < n). They are not open in view of Proposition 3.2. It should also be noted that in the topology of simple convergence of the coefficients, the origin, and therefore each point, has a countable basis of neighborhoods. This property was also valid for the first topology we have defined on C[[X]].

Exercises

3.1.Let X be a set. Let us assign, to each x ∈ X, a topological vector space Ex ; let us denote by E the disjoint union of the spaces Ex as x varies over X and by Γ(X; E) the set of mappings f of X into E such that, for every x ∈ X, f(x) ∈ Ex . Show that there is a natural structure of vector space over Γ(X; E). Next, consider a finite subset S of X and, for every x ∈ S, a neighborhood of zero, Ux , in Ex . We may then consider the following subset of Γ(X; E):

Show that, when S varies in all possible ways and so do also the neighborhoods of zero Ux in each Ex , the above sets form a basis of a neighborhood of zero in Γ{X; E) for a topology compatible with the linear structure of Γ(X; E) (called topology of pointwise convergence in X).

3.2.Prove that, in a topological vector space E over the field of complex numbers, a set different from ø and from E cannot be both open and closed.

3.3.Let E be a vector space, {Eα} (α ∈ A) a family of topological vector space