Functional Analysis/Topological vector spaces

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.'' 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:
 * (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 $$E \subset t V$$.
 * $$E \subset t V = s {t \over s} V \subset s V \subset s U$$

$$\square$$

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 $$\mathcal{X}$$. 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
 * (i) $$f$$ has either closed or dense kernel.
 * ''(ii) $$f$$ is continuous if and only if $$\operatorname{ker}f$$ is closed.
 * $$y - {x \over f(x)} \in \operatorname{ker}(f) \cap (y + V)$$ 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 $$p$$, $$d(x, y) \le d(x, z) + d(z, y)$$.