# Functional Analysis/Topological vector spaces

(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 $E\subset tV$ for every $t>s$
* balanced if $\lambda E\subset E$ for every scalar $\lambda$ with $|\lambda |\leq 1$
* convex if $\lambda _{1}x+\lambda _{2}y\in E$ for any $x,y\in E$ and any $\lambda _{1},\lambda _{2}\geq 0$ 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 $sx+ty\in E$ 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 $|\lambda |\leq 1$, 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 $\lambda \neq 0$. Moreover, the interior of $E$, denoted by $E^{\circ }$, is also convex. Indeed, for $\lambda _{1},\lambda _{2}\geq 0$ 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 $E\subset tV$.
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 $E\subset tV$, using the hypothesis. Then for any $s>t$, we have:
    $E\subset tV=s{t \over s}V\subset sV\subset sU$
$\square$
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 $f(y)\neq 0$. 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 $|f(x)|\geq 1$. 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 ${\mathcal {M}}\neq {\mathcal {X}}$. Thus, by hypothesis, we can find $g\in {\mathcal {X}}^{*}$ such that:$g=0$ in $M$, but $g(p)\neq 0$ 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 $p$, $\lambda ={g(p) \over f(p)}\neq 0$. $\square$
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 $n\geq 0$. If $x\in V_{n_{1}}+...+V_{n_{k}}$ and $2^{-n_{1}}+...+2^{-n_{k}}\leq 2^{-m}$, 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 $2^{-n_{1}}+...2^{-n_{k}}=2^{-(n_{2}-1)}+...+2^{-n_{k}}\leq 2^{m}$.
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:
    $2^{-n_{2}}+...+2^{-n_{k}}\leq 2^{-(m+1)}$.
Hence, by inductive hypothesis, $x\in V_{n_{1}}+V_{m+1}\subset V_{m}$. $\square$
1 Theorem Let ${\mathcal {X}}$ be a TVS.
       $d(x,y)=d(x+z,y+z)$ and $d(\lambda x,0)\leq d(x,0)$ for every $|\lambda |\leq 1$
       $g(0)=0$, $g=1$ on $V^{c}$ and $g(x+y)\leq g(x)+g(y)$ 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, $g(x+y)\leq 2^{-n_{1}}+...+2^{-n_{k}}+2^{-m_{1}}+...+2^{-m_{j}}$. Taking inf over all such $n_{1},...,n_{k}$ we obtain:
    $g(x+y)\leq g(x)+2^{-m_{1}}+...+2^{-m_{j}}$
and do the same for the rest we conclude $g(x+y)\leq g(x)+g(y)$. 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 $f(x)\leq 2^{-m}$, then there are $n_{1},...,n_{k}$ such that $2^{-n_{1}}+...+2^{-n_{k}}\leq 2^{-m}$ and $x\in V_{n_{1}}+...+V_{n_{k}}$. Thus, $x\in V_{m}$ by the lemma. In particular, if $f(x)\leq 2^{-m}$ for "every" $m$, then $x=0$ since ${\mathcal {X}}$ is Hausdorff. Since $V_{n}$ are balanced, if $|\lambda |\leq 1$,
    $\lambda x\in V_{n_{1}}+...+V_{n_{k}}$ for every $n_{1},...,n_{k}$ with $x\in V_{n_{1}}+...+V_{n_{k}}$.
That means $f(\lambda x)\leq f(x)$, and in particular $f(x)=f(-(-x))\leq f(-x)\leq f(x)$. 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 $\delta \leq 2^{m}$. 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:$d(\lambda x,y)\leq \lambda d(x,y)$, which implies open balls about the origin are convex. Indeed, if $x,y\in V_{n_{1}}+...+V_{n_{k}}$, and if $\lambda _{1}\geq 0$ and $\lambda _{2}\geq 0$ 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,
    $f(\lambda _{1}x+\lambda _{2}y)\leq \min\{f(x),f(y)\}\leq {f(x)+f(y) \over 2}$
and by iteration and continuity it can be shown that $f(\lambda x)\leq \lambda f(x)$ for every $|\lambda |\leq 1$.
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 $(1+p(x)+p(y))p(x+y)\leq (p(x)+p(y))(1+p(x+y))$ for any seminorm $p$, $d(x,y)\leq d(x,z)+d(z,y)$.

## References

* lp with 0 < p < 1 is not locally convex