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 $E\subset tV$ 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 $sx+ty\in E$ for all $x,y\in E$. Conversely, if $E$ is convex,

$\frac{s}{s+t}x+\frac{t}{s+t}y\in E$, 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.$◻$

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

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

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

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

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 $ℳ$ be a subspace of a TVS. Then $\overline{ℳ}$ 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 $ℳ$ is a subspace of a TVS, by continuity and linearity,

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

Hence, $\overline{ℳ}$ 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 $𝒳$ be a TVS, and $E\subset 𝒳$. 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 $E\subset ̸nV$ 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:

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

$◻$

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. $◻$

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.
$◻$

6 Theorem Let $f$ be a linear functional on a TVS $𝒳$.

• (i) $f$ has either closed or dense kernel.
• (ii) $f$ is continuous if and only if $\ker f$ 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 $\ker f$ but $f(y)\ne 0$. For any $x\in 𝒳$, $x-\frac{f(x)}{f(y)}y$ is in the kernel of $f$. This is to say, every element of $𝒳$ is a linear combination of $y$ and some other element in $\ker$. Thus, $\ker f$ is dense. (ii) If $f$ is continuous, $\ker f=f^{-1}(\{0\})$ is closed. Conversely, suppose $\ker f$ 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 $y+V\subset (\ker f{)}^c$. It then follows that $\underset{V}{\sup }|f|<1$. Indeed, suppose $|f(x)|\ge 1$. Then

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

The continuity of $f$ now follows from the lemma. $◻$

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

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

(Note this is the conclusion of Corollary 2.something) Then every continuous linear function $f$ on a subspace $ℳ$ of $𝒳$ extends to an element of ${𝒳}^{*}$.
Proof: We essentially repeat the proof of Theorem 3.8. So, let $ℳ$ be the kernel of $f$, which is closed, and we may assume $ℳ\ne 𝒳$. Thus, by hypothesis, we can find $g\in {𝒳}^{*}$ 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 $p$, $\lambda =\frac{g(p)}{f(p)}\ne 0$. $◻$

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, $x\in V_{n_1}+V_{m+1}\subset V_m$. $◻$

1 Theorem Let $𝒳$ be a TVS.

• (i) If $𝒳$ is Hausdorff and has a countable local base, $𝒳$ 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 $V\subset 𝒳$ 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 $V_0=𝒳$. As above, define $f(x)=\inf \{2^{-n_1}+...+2^{-n_k};x\in V_{n_1}+...+V_{n_k}\}$ for each $x\in 𝒳$. 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 $𝒳$ 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. $◻$

The second property of $d$ in (i) implies that open ball about the origin in terms of this $d$ is balanced, and when $𝒳$ 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,

$f({\lambda }_{1}x+{\lambda }_{2}y)\le \min \{f(x),f(y)\}\le \frac{f(x)+f(y)}{2}$

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 $\bar{W}\subset V$
Proof: Since we may assume that $x=0$, take $W=\{x;g(x)<2^{-1}\}$. $◻$

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

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 $𝒳$ be locally convex. The convex hull of a bounded set is bounded.

Given a sequence $p_n$ of semi-norms, define:

$d(x,y)=\sum _{n=0}^{\mathrm{\infty }}{2}^{-n}\frac{p_n(x-y)}{1+p_n(x-y)}$.

$d$ 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 $p$, $d(x,y)\le d(x,z)+d(z,y)$.

References

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