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., x + y) and dilations (i.e., \alpha x) continuous is called a topological vector space or TVS for short.

A subset E of a TVS is said to be:

  • bounded if for every neighborhood V of 0 there exist s > 0 such that E \subset t V for every t > s
  • balanced if \lambda E \subset E for every scalar \lambda with |\lambda| \le 1
  • convex if \lambda_1 x + \lambda_2 y \in E for any x, y \in E and any \lambda_1, \lambda_2 \ge 0 with \lambda_1 x + \lambda_2 y = 1.

1 Corollary (s + t)E = sE + tE for any s, t > 0 if and only if E is convex.
Proof: Supposing s + t = 1 we obtain s x + t y \in E for all x, y \in E. Conversely, if E is convex,

{s \over s + t} x + {t \over s + t} y \in E, or sx + sy \in (s+t)E for any x, y \in E.

Since (s + t)E \subset sE + tE holds in general, the proof is complete.\square

Define f(\lambda, x) = \lambda x for scalars \lambda, vectors x. If E is a balanced set, for any |\lambda| \le 1, by continuity,

f(\lambda, \overline{E}) \subset \overline{f(\lambda, E)} \subset \overline{E}.

Hence, the closure of a balanced set is again balanced. In the similar manner, if E is convex, for s, t > 0

f((s + t), \overline{E}) = \overline{f((s+t), E)} = \overline{sE + tE},

meaning the closure of a convex set is again convex. Here the first equality holds since f(\lambda, \cdot) is injective if \lambda \ne 0. Moreover, the interior of E, denoted by E^\circ, is also convex. Indeed, for \lambda_1, \lambda_2 \ge 0 with \lambda_1 + \lambda_2 = 1

\lambda_1 E^\circ + \lambda_2 E^\circ \subset E,

and since the left-hand side is open it is contained in E^\circ. Finally, a subspace of a TVS is a subset that is simultaneously a linear subspace and a topological subspace. Let \mathcal{M} be a subspace of a TVS. Then \overline{\mathcal{M}} is a topological subspace, and it is stable under scalar multiplication, as shown by the argument similar to the above. Let g(x, y) = x + y. If \mathcal{M} is a subspace of a TVS, by continuity and linearity,

g(\overline{\mathcal{M}}, \overline{\mathcal{M}}) \subset \overline{g(\mathcal{M}, \mathcal{M})} = \overline{\mathcal{M}}.

Hence, \overline{\mathcal{M}} is a linear subspace. We conclude that the closure of a subspace is a subspace.

Let V be a neighborhood of 0. By continuity there exists a \delta > 0 and a neighborhood W of 0 such that:

f(\{ \lambda; |\lambda| < \delta \}, W) \subset V

It follows that the set \{ \lambda; |\lambda| < \delta \}W is a union of open sets, contained in V and is balanced. In other words, every TVS admits a local base consisting of balanced sets.

1 Theorem Let \mathcal{X} be a TVS, and E \subset \mathcal{X}. The following are equivalent.

  • (i) E is bounded.
  • (ii) Every countable subset of E is bounded.
  • (iii) for every balanced neighborhood V of 0 there exists a t > 0 such that E \subset t V.

Proof: That (i) implies (ii) is clear. If (iii) is false, there exists a balanced neighborhood V such that E \not\subset nV for every n = 1, 2, .... That is, there is a unbounded sequence x_1, x_2, ... in E. Finally, to show that (iii) implies (i), let U be a neighborhood of 0, and V be a balanced open set with 0 \in V \subset U. Choose t so that E \subset t V, using the hypothesis. Then for any s > t, we have:

E \subset t V = s {t \over s} V \subset s V \subset s U

\square

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. \square

1 Lemma Let f be a linear operator between TVSs. If f(V) is bounded for some neighborhood V of 0, then f is continuous.
\square

6 Theorem Let f be a linear functional on a TVS \mathcal{X}.

  • (i) f has either closed or dense kernel.
  • (ii) f is continuous if and only if \operatorname{ker}f is closed.

Proof: To show (i), suppose the kernel of f is not closed. That means: there is a y which is in the closure of \operatorname{ker}f but f(y) \ne 0. For any x \in \mathcal{X}, x - {f(x) \over f(y)}y is in the kernel of f. This is to say, every element of \mathcal{X} is a linear combination of y and some other element in \operatorname{ker}. Thus, \operatorname{ker}f is dense. (ii) If f is continuous, \operatorname{ker}f = f^{-1}(\{0\}) is closed. Conversely, suppose \operatorname{ker}f is closed. Since f is continuous when f is identically zero, suppose there is a point y with f(y) = 1. Then there is a balanced neighborhood V of 0 such that y + V \subset (\operatorname{ker}f)^c. It then follows that \sup_V |f| < 1. Indeed, suppose |f(x)| \ge 1. Then

y - {x \over f(x)} \in \operatorname{ker}(f) \cap (y + V) if x \in V, which is a contradiction.

The continuity of f now follows from the lemma. \square

6 Theorem Let \mathcal{X} be a TVS and \mathcal{M} \subset \mathcal{X} its subspace. Suppose:

\mathcal{M} is dense \Longleftrightarrow z \in \mathcal{X}^* = 0 in \mathcal{M} implies z = 0 in \mathcal{X}.

(Note this is the conclusion of Corollary 2.something) Then every continuous linear function f on a subspace \mathcal{M} of \mathcal{X} extends to an element of \mathcal{X}^*.
Proof: We essentially repeat the proof of Theorem 3.8. So, let \mathcal{M} be the kernel of f, which is closed, and we may assume \mathcal{M} \ne \mathcal{X}. Thus, by hypothesis, we can find g \in \mathcal{X}^* such that:g = 0 in M, but g(p) \ne 0 for some point p outside M. By Lemma 1.6, g = \lambda f for some scalar \lambda. Since both f and g do not vanish at p, \lambda = { g(p) \over f(p) } \ne 0. \square

Lemma Let V_0, V_1, ... be a sequence of subsets of a a linear space containing 0 such that V_{n+1} + V_{n+1} \subset V_n for every n \ge 0. If x \in V_{n_1} + ... + V_{n_k} and 2^{-n_1} + ... + 2^{-n_k} \le 2^{-m}, then x \in V_m.
Proof: We shall prove the lemma by induction over k. The basic case k = 1 holds since V_{n} \subset V_{n} + V_{n} for every n. Thus, assume that the lemma has been proven until k - 1. First, suppose n_1, ..., n_k are not all distinct. By permutation, we may then assume that n_1 = n_2. It then follows:

x \in V_{n_1} + V_{n_2} + ... + V_{n_k} \subset V_{n_2 - 1} + ... + V_{n_k} and 2^{-n_1} + ... 2^{-n_k} = 2^{-(n_2 - 1)} + ... + 2^{-n_k} \le 2^m.

The inductive hypothesis now gives: x \in V_m. Next, suppose n_1, ..., n_k are all distinct. Again by permutation, we may assume that n_1 < n_2 < ... n_k. Since no carry-over occurs then and m < n_1, m+1 < n_2 and so:

2^{-n_2} + ... + 2^{-n_k} \le 2^{-(m+1)}.

Hence, by inductive hypothesis, x \in V_{n_1} + V_{m+1} \subset V_m. \square

1 Theorem Let \mathcal{X} be a TVS.

  • (i) If \mathcal{X} is Hausdorff and has a countable local base, \mathcal{X} is metrizable with the metric d such that
d(x, y) = d(x + z, y + z) and d(\lambda x, 0) \le d(x, 0) for every |\lambda| \le 1
  • (ii) For every neighborhood V \subset \mathcal{X} of 0, there is a continuous function g such that
g(0) = 0, g = 1 on V^c and g(x + y) \le g(x) + g(y) for any x, y.

Proof: To show (ii), let V_0, V_1, ... be a sequence of neighborhoods of 0 satisfying the condition in the lemma and V = V_0. Define g = 1 on V^c and g(x) = \inf \{2^{-n_1} + ... + 2^{-n_k}; x \in V_{n_1} + ... + V_{n_k} \} for every x \in V. To show the triangular inequality, we may assume that g(x) and g(y) are both < 1, and thus suppose x \in V_{n_1} + ... + V_{n_k} and y \in V_{m_1} + ... + V_{m_j}. Then

x + y \in V_{n_1} + ... + V_{n_k} + V_{m_1} + ... + V_{m_j}

Thus, g(x + y) \le 2^{-n_1} + ... + 2^{-n_k} + 2^{-m_1} + ... + 2^{-m_j}. Taking inf over all such n_1, ..., n_k we obtain:

g(x + y) \le g(x) + 2^{-m_1} + ... + 2^{-m_j}

and do the same for the rest we conclude g(x + y) \le g(x) + g(y). This proves (ii) since g is continuous at 0 and it is then continuous everywhere by the triangular inequality. Now, to show (i), choose a sequence of balanced sets V_0, V_1, ... that is a local base, satisfies the condition in the lemma and is such that V_0 = \mathcal{X}. As above, define f(x) = \inf \{2^{-n_1} + ... + 2^{-n_k}; x \in V_{n_1} + ... + V_{n_k} \} for each x \in \mathcal{X}. For the same reason as before, the triangular inequality holds. Clearly, f(0) = 0. If f(x) \le 2^{-m}, then there are n_1, ..., n_k such that 2^{-n_1} + ... + 2^{-n_k} \le 2^{-m} and x \in V_{n_1} + ... + V_{n_k}. Thus, x \in V_m by the lemma. In particular, if f(x) \le 2^{-m} for "every" m, then x = 0 since \mathcal{X} is Hausdorff. Since V_n are balanced, if |\lambda| \le 1,

\lambda x \in V_{n_1} + ... + V_{n_k} for every n_1, ..., n_k with x \in V_{n_1} + ... + V_{n_k}.

That means f(\lambda x) \le f(x), and in particular f(x) = f(-(-x)) \le f(-x) \le f(x). Defining d(x, y) = f(x - y) will complete the proof of (i). In fact, the properties of f we have collected shows the function d is a metric with the desired properties. The lemma then shows that given any m, \{x; f(x) < \delta\} \subset V_m for some \delta \le 2^m. That is, the sets \{x; f(x) < \delta\} over \delta > 0 forms a local base for the original topology. \square

The second property of d in (i) implies that open ball about the origin in terms of this d is balanced, and when \mathcal{X} has a countable local base consisting of convex sets it can be strengthened to:d(\lambda x, y) \le \lambda d(x, y), which implies open balls about the origin are convex. Indeed, if x, y \in V_{n_1} + ... + V_{n_k}, and if \lambda_1 \ge 0 and \lambda_2 \ge 0 with \lambda_1 + \lambda_2, then

\lambda_1 x + \lambda_2 y \in V_{n_1} + ... + V_{n_k}

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

f(\lambda_1 x + \lambda_2 y) \le \min \{ f(x), f(y) \} \le {f(x) + f(y) \over 2}

and by iteration and continuity it can be shown that f(\lambda x) \le \lambda f(x) for every |\lambda| \le 1.

Corollary For every neighborhood V of some point x, there is a neighborhood of x with \overline{W} \subset V
Proof: Since we may assume that x = 0, take W = \{ x; g(x) < 2^{-1} \}. \square

Corollary If every finite set of a TVS \mathcal{X} is closed, \mathcal{X} is Hausdorff.
Proof: Let x, y be given. By the preceding corollary we find an open set V \subset \overline{V} \subset \{y\}^c containing x. \square

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 \mathcal{X} be locally convex. The convex hull of a bounded set is bounded.

Given a sequence p_n of semi-norms, define:

d(x, y) = \sum_{n=0}^\infty 2^{-n} {p_n(x - y) \over 1 + p_n(x - y)}.

d then becomes a metric. In fact, Since (1 + p(x) + p(y)) p(x + y) \le (p(x) + p(y))(1 + p(x + y)) for any seminorm p, d(x, y) \le d(x, z) + d(z, y).


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