# Functional Analysis/Topological vector spaces

A vector space endowed by a topology that makes translations (i.e., ) and dilations (i.e., ) continuous is called *a topological vector space* or *TVS* for short.

A subset of a TVS is said to be:

*bounded*if for every neighborhood of there exist such that for every*balanced*if for every scalar with*convex*if for any and any with .

**1 Corollary** * for any if and only if is convex.*

Proof: Supposing we obtain for all . Conversely, if is convex,

- , or for any .

Since holds in general, the proof is complete.

Define for scalars , vectors . If is a balanced set, for any , by continuity,

- .

Hence, the closure of a balanced set is again balanced. In the similar manner, if is convex, for

- ,

meaning the closure of a convex set is again convex. Here the first equality holds since is injective if . Moreover, the interior of , denoted by , is also convex. Indeed, for with

- ,

and since the left-hand side is open it is contained in . Finally, a *subspace* of a TVS is a subset that is simultaneously a linear subspace and a topological subspace. Let be a subspace of a TVS. Then is a topological subspace, and it is stable under scalar multiplication, as shown by the argument similar to the above. Let . If is a subspace of a TVS, by continuity and linearity,

- .

Hence, is a linear subspace. We conclude that the closure of a subspace is a subspace.

Let be a neighborhood of . By continuity there exists a and a neighborhood of such that:

It follows that the set is a union of open sets, contained in and is balanced. In other words, every TVS admits a local base consisting of balanced sets.

**1 Theorem** *Let be a TVS, and . The following are equivalent.*

*(i) is bounded.**(ii) Every countable subset of is bounded.**(iii) for every balanced neighborhood of there exists a such that .*

Proof: That (i) implies (ii) is clear. If (iii) is false, there exists a balanced neighborhood such that for every . That is, there is a unbounded sequence in . Finally, to show that (iii) implies (i), let be a neighborhood of 0, and be a balanced open set with . Choose so that , using the hypothesis. Then for any , we have:

**1 Corollary** *Every Cauchy sequence and every compact set in a TVS are bounded.*

Proof: If the set is not bounded, it contains a sequence that is not Cauchy and does not have a convergent subsequence.

**1 Lemma** *Let be a linear operator between TVSs. If is bounded for some neighborhood of , then is continuous.*

**6 Theorem** *Let be a linear functional on a TVS .*

*(i) has either closed or dense kernel.**(ii) is continuous if and only if is closed.*

Proof: To show (i), suppose the kernel of is not closed. That means: there is a which is in the closure of but . For any , is in the kernel of . This is to say, every element of is a linear combination of and some other element in . Thus, is dense. (ii) If is continuous, is closed. Conversely, suppose is closed. Since is continuous when is identically zero, suppose there is a point with . Then there is a balanced neighborhood of such that . It then follows that . Indeed, suppose . Then

- if , which is a contradiction.

The continuity of now follows from the lemma.

**6 Theorem** *Let be a TVS and its subspace. Suppose:*

- is dense in implies in .

(Note this is the conclusion of Corollary 2.something) Then every continuous linear function on a subspace of extends to an element of .
Proof: We essentially repeat the proof of Theorem 3.8. So, let be the kernel of , which is closed, and we may assume . Thus, by hypothesis, we can find such that: in , but for some point outside . By Lemma 1.6, for some scalar . Since both and do not vanish at , .

**Lemma** *Let be a sequence of subsets of a a linear space containing such that for every . If and , then .*

Proof: We shall prove the lemma by induction over . The basic case holds since for every . Thus, assume that the lemma has been proven until . First, suppose are not all distinct. By permutation, we may then assume that . It then follows:

- and .

The inductive hypothesis now gives: . Next, suppose are all distinct. Again by permutation, we may assume that . Since no carry-over occurs then and , and so:

- .

Hence, by inductive hypothesis, .

**1 Theorem** *Let be a TVS.*

*(i) If is Hausdorff and has a countable local base, is metrizable with the metric such that*

*and for every*

*(ii) For every neighborhood of , there is a continuous function such that*

*, on and for any .*

Proof: To show (ii), let be a sequence of neighborhoods of satisfying the condition in the lemma and . Define on and for every . To show the triangular inequality, we may assume that and are both , and thus suppose and . Then

Thus, . Taking inf over all such we obtain:

and do the same for the rest we conclude . This proves (ii) since is continuous at and it is then continuous everywhere by the triangular inequality. Now, to show (i), choose a sequence of balanced sets that is a local base, satisfies the condition in the lemma and is such that . As above, define for each . For the same reason as before, the triangular inequality holds. Clearly, . If , then there are such that and . Thus, by the lemma. In particular, if for "every" , then since is Hausdorff. Since are balanced, if ,

- for every with .

That means , and in particular . Defining will complete the proof of (i). In fact, the properties of we have collected shows the function is a metric with the desired properties. The lemma then shows that given any , for some . That is, the sets over forms a local base for the original topology.

The second property of in (i) implies that open ball about the origin in terms of this is balanced, and when has a countable local base consisting of convex sets it can be strengthened to:, which implies open balls about the origin are convex. Indeed, if , and if and with , then

since the sum of convex sets is again convex. This is to say,

and by iteration and continuity it can be shown that for every .

**Corollary** *For every neighborhood of some point , there is a neighborhood of with *

Proof: Since we may assume that , take .

**Corollary** *If every finite set of a TVS is closed, is Hausdorff.*

Proof: Let be given. By the preceding corollary we find an open set containing .

A TVS with a local base consisting of convex sets is said to be *locally convex*. Since in this book we will never study non-Hausdorff locally convex spaces, we shall assume tacitly that every finite subset of every locally convex is closed, hence Hausdorff in view of Theorem something.

**Lemma** *Let be locally convex. The convex hull of a bounded set is bounded.*

Given a sequence of semi-norms, define:

- .

then becomes a metric. In fact, Since for any seminorm , .

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