Functional Analysis/Topological vector spaces

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25% developed  as of May 28, 2008 (May 28, 2008)

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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