Functional Analysis/Topological vector spaces

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(May 28, 2008)

A vector space endowed by a topology that makes translations (i.e., $x + y$ ) and dilations (i.e., $α x$ ) 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 $λ$ 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 $λ 1 x + λ 2 y = 1$ .

1 Corollary $(s + t) E = s E + t E$ 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 $f (λ, x) = λ x$ for scalars $λ$ , 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 $λ 1 + λ 2 = 1$

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

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 $δ > 0$ and a neighborhood $W$ of $0$ such that:

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

It follows that the set ${λ; | λ | < δ} W$ 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, $g = λ f$ for some scalar $λ$ . 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 ${x; f (x) < δ}$ over $δ > 0$ 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 $λ 1 + λ 2$ , 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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lp with 0 < p < 1 is not locally convex

Functional Analysis/Topological vector spaces

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