# Functional Analysis/Topological vector spaces

(May 28, 2008)

A vector space endowed by a topology that makes translations (i.e., ${\displaystyle x+y}$) and dilations (i.e., ${\displaystyle \alpha x}$) continuous is called a topological vector space or TVS for short.

A subset ${\displaystyle E}$ of a TVS is said to be:

• bounded if for every neighborhood ${\displaystyle V}$ of ${\displaystyle 0}$ there exist ${\displaystyle s>0}$ such that ${\displaystyle E\subset tV}$ for every ${\displaystyle t>s}$
• balanced if ${\displaystyle \lambda E\subset E}$ for every scalar ${\displaystyle \lambda }$ with ${\displaystyle |\lambda |\leq 1}$
• convex if ${\displaystyle \lambda _{1}x+\lambda _{2}y\in E}$ for any ${\displaystyle x,y\in E}$ and any ${\displaystyle \lambda _{1},\lambda _{2}\geq 0}$ with ${\displaystyle \lambda _{1}x+\lambda _{2}y=1}$.

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

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

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

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

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

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

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

meaning the closure of a convex set is again convex. Here the first equality holds since ${\displaystyle f(\lambda ,\cdot )}$ is injective if ${\displaystyle \lambda \neq 0}$. Moreover, the interior of ${\displaystyle E}$, denoted by ${\displaystyle E^{\circ }}$, is also convex. Indeed, for ${\displaystyle \lambda _{1},\lambda _{2}\geq 0}$ with ${\displaystyle \lambda _{1}+\lambda _{2}=1}$

${\displaystyle \lambda _{1}E^{\circ }+\lambda _{2}E^{\circ }\subset E}$,

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

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

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

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

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

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

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

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

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

${\displaystyle E\subset tV=s{t \over s}V\subset sV\subset sU}$

${\displaystyle \square }$

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. ${\displaystyle \square }$

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

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

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

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

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

The continuity of ${\displaystyle f}$ now follows from the lemma. ${\displaystyle \square }$

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

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

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

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

${\displaystyle x\in V_{n_{1}}+V_{n_{2}}+...+V_{n_{k}}\subset V_{n_{2}-1}+...+V_{n_{k}}}$ and ${\displaystyle 2^{-n_{1}}+...2^{-n_{k}}=2^{-(n_{2}-1)}+...+2^{-n_{k}}\leq 2^{m}}$.

The inductive hypothesis now gives: ${\displaystyle x\in V_{m}}$. Next, suppose ${\displaystyle n_{1},...,n_{k}}$ are all distinct. Again by permutation, we may assume that ${\displaystyle n_{1}. Since no carry-over occurs then and ${\displaystyle m, ${\displaystyle m+1 and so:

${\displaystyle 2^{-n_{2}}+...+2^{-n_{k}}\leq 2^{-(m+1)}}$.

Hence, by inductive hypothesis, ${\displaystyle x\in V_{n_{1}}+V_{m+1}\subset V_{m}}$. ${\displaystyle \square }$

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

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

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

${\displaystyle x+y\in V_{n_{1}}+...+V_{n_{k}}+V_{m_{1}}+...+V_{m_{j}}}$

Thus, ${\displaystyle g(x+y)\leq 2^{-n_{1}}+...+2^{-n_{k}}+2^{-m_{1}}+...+2^{-m_{j}}}$. Taking inf over all such ${\displaystyle n_{1},...,n_{k}}$ we obtain:

${\displaystyle g(x+y)\leq g(x)+2^{-m_{1}}+...+2^{-m_{j}}}$

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

${\displaystyle \lambda x\in V_{n_{1}}+...+V_{n_{k}}}$ for every ${\displaystyle n_{1},...,n_{k}}$ with ${\displaystyle x\in V_{n_{1}}+...+V_{n_{k}}}$.

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

The second property of ${\displaystyle d}$ in (i) implies that open ball about the origin in terms of this ${\displaystyle d}$ is balanced, and when ${\displaystyle {\mathcal {X}}}$ has a countable local base consisting of convex sets it can be strengthened to:${\displaystyle d(\lambda x,y)\leq \lambda d(x,y)}$, which implies open balls about the origin are convex. Indeed, if ${\displaystyle x,y\in V_{n_{1}}+...+V_{n_{k}}}$, and if ${\displaystyle \lambda _{1}\geq 0}$ and ${\displaystyle \lambda _{2}\geq 0}$ with ${\displaystyle \lambda _{1}+\lambda _{2}}$, then

${\displaystyle \lambda _{1}x+\lambda _{2}y\in V_{n_{1}}+...+V_{n_{k}}}$

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

${\displaystyle f(\lambda _{1}x+\lambda _{2}y)\leq \min\{f(x),f(y)\}\leq {f(x)+f(y) \over 2}}$

and by iteration and continuity it can be shown that ${\displaystyle f(\lambda x)\leq \lambda f(x)}$ for every ${\displaystyle |\lambda |\leq 1}$.

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

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

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

Given a sequence ${\displaystyle p_{n}}$ of semi-norms, define:

${\displaystyle d(x,y)=\sum _{n=0}^{\infty }2^{-n}{p_{n}(x-y) \over 1+p_{n}(x-y)}}$.

${\displaystyle d}$ then becomes a metric. In fact, Since ${\displaystyle (1+p(x)+p(y))p(x+y)\leq (p(x)+p(y))(1+p(x+y))}$ for any seminorm ${\displaystyle p}$, ${\displaystyle d(x,y)\leq d(x,z)+d(z,y)}$.