# The Fréchet space of holomorphic functions on the unit disc

Jordan Bell
April 3, 2014

## 1 Introduction

The goal of this note is to develop all the machinery necessary to understand what it means to say that the set $H(D)$ of holomorphic functions on the unit disc is a separable and reflexive Fréchet space that has the Heine-Borel property and is not normable.

## 2 Topological vector spaces

If $X$ is a topological space and $p\in X$, a local basis at $p$ is a set $\mathscr{B}$ of open neighborhoods of $p$ such that if $U$ is an open neighborhood of $p$ then there is some $U_{0}\in\mathscr{B}$ that is contained in $U$. We emphasize that to say that a topological vector space $(X,\tau)$ is normable is to say not just that there is a norm on the vector space $X$, but moreover that the topology $\tau$ is induced by the norm.

A topological vector space over $\mathbb{C}$ is a vector space $X$ over $\mathbb{C}$ that is a topological space such that singletons are closed sets and such that vector addition $X\times X\to X$ and scalar multiplication $\mathbb{C}\times X\to X$ are continuous. It is not true that a topological space in which singletons are closed need be Hausdorff, but one can prove that every topological vector space is a Hausdorff space.11 1 Walter Rudin, Functional Analysis, second ed., p. 11, Theorem 1.12. For any $a\in X$, we check that the map $x\mapsto a+x$ is a homeomorphism. Therefore, a subset $U$ of $X$ is open if and only if $a+U$ is open for all $a\in X$. It follows that if $X$ is a vector space and $\mathscr{B}$ is a set of subsets of $X$ each of which contains $0$, then there is at most one topology for $X$ such that $X$ is a topological vector space for which $\mathscr{B}$ is a local basis at $0$. In other words, the topology of a topological vector space is determined by specifying a local basis at $0$. A topological vector space $X$ is said to be locally convex if there is a local basis at $0$ whose elements are convex sets.

If $X$ is a vector space and $\mathscr{F}$ is a set of seminorms on $X$, we say that $\mathscr{F}$ is a separating family if $x\neq 0$ implies that there is some $m\in\mathscr{F}$ with $m(x)\neq 0$. (Thus, if $m$ is a seminorm on $X$, the singleton $\{m\}$ is a separating family if and only if $m$ is a norm.) The following theorem presents a local basis at $0$ for a topology and shows that there is a topology for which the vector space is a locally convex space and for which this is a local basis at $0$.22 2 Paul Garrett, Seminorms and locally convex spaces, http://www.math.umn.edu/~garrett/m/fun/notes_2012-13/07b_seminorms.pdf We call this topology the seminorm topology induced by $\mathscr{F}$.

###### Theorem 1 (Seminorm topology).

If $X$ is a vector space and $\mathscr{F}$ is a separating family of seminorms on $X$, then there is a topology $\tau$ on $X$ such that $(X,\tau)$ is a locally convex space and the collection $\mathscr{B}$ of finite intersections of sets of the form

 $B_{m,\epsilon}=\{x\in X:m(x)<\epsilon\},\qquad m\in\mathscr{F},\epsilon>0$

is a local basis at $0$.

###### Proof.

We define $\tau$ to be those subsets $U$ of $X$ such that for all $x\in U$ there is some $N\in\mathscr{B}$ satisfying $x+N\subseteq U$. If $\mathscr{U}$ is a subset of $\tau$ and $x\in\bigcup_{U\in\mathscr{U}}U$, then there is some $U_{0}\in\mathscr{U}$ with $x\in U_{0}$, and there is some $N_{0}\in\mathscr{B}$ satisfying $x+N_{0}\subseteq U_{0}$. We have

 $x+N_{0}\subseteq U_{0}\subseteq\bigcup_{U\in\mathscr{U}}U,$

which tells us that $\bigcup_{U\in\mathscr{U}}U\in\tau$. If $U_{1},\ldots,U_{n}\in\tau$ and $x\in\bigcap_{k=1}^{n}U_{k}$, then there are $N_{1},\ldots,N_{n}\in\mathscr{B}$ satisfying $x+N_{k}\in U_{k}$ for $1\leq k\leq n$. But the intersection of finitely many elements of $\mathscr{B}$ is itself an element of $\mathscr{B}$, so $N=\bigcap_{k=1}^{n}N_{k}\in\mathscr{B}$, and

 $x+N\subseteq\bigcap_{k=1}^{n}U_{k},$

showing that $\bigcap_{k=1}^{n}U_{k}\in\tau$. Therefore, $\tau$ is a topology.

Suppose that $x\in X$. For $y\neq x$, let $m_{y}\in\mathscr{F}$ with $\epsilon_{y}=m_{y}(x-y)\neq 0$; there is such a seminorm because $\mathscr{F}$ is a separating family. Then $U_{y}=y+B_{m_{y},\epsilon_{y}}$ is an open set that contains $y$ and does not contain $x$. Therefore $X\setminus U_{y}$ is a closed set that contains $x$ and does not contain $y$, and

 $\bigcap_{y\neq x}X\setminus U_{y}=\{x\}$

is a closed set, showing that singletons are closed.

Let $x,y\in X$ and $N\in\mathscr{B}$. There are $m_{k}\in\mathscr{F}$ and $\epsilon_{k}>0$, $1\leq k\leq n$, such that $N=\bigcap_{k=1}^{n}B_{m_{k},\epsilon_{k}}$. Let $U=\bigcap_{k=1}^{n}B_{m_{k},\epsilon_{k}/2}$. If $v\in(x+U)+(y+U)$ and $1\leq k\leq n$, then there are $x_{k}\in B_{m_{k},\epsilon_{k}/2}$ and $y_{k}\in B_{m_{k},\epsilon_{k}/2}$ such that $v=x+x_{k}+y+y_{k}$, and

 $m_{k}(v-(x+y))=m_{k}(x_{k}+y_{k})\leq m_{k}(x_{k})+m_{k}(y_{k})<\frac{\epsilon% _{k}}{2}+\frac{\epsilon_{k}}{2}=\epsilon_{k},$

so $v\in x+y+B_{m_{k},\epsilon_{k}}$. This is true for each $k$, $1\leq k\leq n$, so $v\in x+y+N$. Hence

 $(x+U)+(y+U)\subseteq x+y+N,$

showing that vector addition is continuous at $(x,y)\in X\times X$: for every basic open neighborhood $x+y+N$ of the image $x+y$, there is an open neighborhood $(x+U)\times(y+U)$ of $(x,y)$ whose image under vector addition is contained in $x+y+N$.

Let $\alpha\in\mathbb{C}$, $x\in X$, and $N\in\mathscr{B}$, say $N=\bigcap_{k=1}^{n}B_{m_{k},\epsilon_{k}}$. Let $\epsilon=\min\{\epsilon_{k}:1\leq k\leq n\}$, let $\delta>0$ be small enough so that $\delta(\delta+|\alpha|+m_{k}(x))<\epsilon$ for each $1\leq k\leq n$, let $\Delta=\{\beta\in\mathbb{C}:|\beta-\alpha|<\delta\}$, and let $U=\bigcap_{k=1}^{n}B_{m_{k},\delta}$. If $(\beta,v)\in\Delta\times(x+U)$ and $1\leq k\leq n$, then

 $\displaystyle m_{k}(\beta v-\alpha x)$ $\displaystyle=m_{k}(\beta v-\beta x+\beta x-\alpha x)$ $\displaystyle\leq m_{k}(\beta(v-x))+m_{k}((\beta-\alpha)x)$ $\displaystyle=|\beta|m_{k}(v-x)+|\beta-\alpha|m_{k}(x)$ $\displaystyle<(\delta+|\alpha|)\delta+\delta m_{k}(x)$ $\displaystyle=\delta(\delta+|\alpha|+m_{k}(x))$ $\displaystyle<\epsilon$ $\displaystyle\leq\epsilon_{k},$

showing that $\beta v\in\alpha x+B_{m_{k},\epsilon_{k}}$. This is true for each $k$, so $\beta v\in N$, which shows that scalar multiplication is continuous at $(\alpha,x)$: for every basic open neighborhood $\alpha x+N$ of the image $\alpha x$, there is an open neighborhood $\Delta\times(x+U)$ of $(\alpha,x)$ whose image under scalar multiplication is contained in $\alpha x+N$.

We have shown that $X$ with the topology $\tau$ is a topological vector space. To show that $X$ is a locally convex space it suffices to prove that each element of the local basis $\mathscr{B}$ is convex. An intersection of convex sets is a convex set, so to prove that each element of $\mathscr{B}$ is convex it suffices to prove that each $B_{m,\epsilon}$ is convex, $m\in\mathscr{F}$ and $\epsilon>0$. If $0\leq t\leq 1$ and $x,y\in B_{m,\epsilon}$, then

 $m(tx+(1-t)y)\leq m(tx)+m((1-t)y)=tm(x)+(1-t)m(y)

showing that $tx+(1-t)y\in B_{m,\epsilon}$ and thus that $B_{m,\epsilon}$ is a convex set. Therefore, $(X,\tau)$ is a locally convex space. ∎

In the other direction, we will now explain how the topology of a locally convex space is induced by a separating family of seminorms. We say that a subset $S$ of a vector space $X$ is absorbing if $x\in X$ implies that there is some $t>0$ such that $x\in tS$. The Minkowski functional $\mu_{S}:X\to[0,\infty)$ of an absorbing set $S$ is defined by

 $\mu_{S}(x)=\inf\{t\geq 0:x\in tS\},\qquad x\in X.$

If $U$ is an open set containing $0$ and $x\in X$, then $0\cdot x=0\in U$, and because scalar multiplication is continuous there is some $t>0$ such that $tx\in U$. Thus an open set containing $0$ is absorbing. We say that a subset $S$ of a vector space $X$ is balanced if $|\alpha|\leq 1$ implies that $\alpha S\subseteq S$. One proves that in a topological vector space, every convex open neighborhood of $0$ contains a balanced convex open neighhborhood of $0$.33 3 Walter Rudin, Functional Analysis, second ed., p. 12, Theorem 1.14. It follows that a locally convex space has a local basis at $0$ whose elements are balanced convex open sets. The following lemma shows that the Minkowski functional of each member of this local basis is a seminorm.

###### Lemma 2.

If $X$ is a topological vector space and $U$ is a balanced convex open neighborhood of $0$, then the Minkowski functional of $U$ is a seminorm on $X$.

###### Proof.

Let $\alpha\in\mathbb{C}$ and $x\in X$. If $\alpha=0$, then

 $\mu_{U}(\alpha x)=\mu_{U}(0)=0=|\alpha|\mu_{U}(x).$

Otherwise, write $\alpha=ru$ with $r>0$ and $|u|=1$. Because $U$ is balanced and $|u^{-1}|=1$, we have

 $\displaystyle\mu_{U}(\alpha x)$ $\displaystyle=\inf\{t\geq 0:\alpha x\in tU\}$ $\displaystyle=\inf\{t\geq 0:rux\in tU\}$ $\displaystyle=\inf\{t\geq 0:x\in r^{-1}tu^{-1}U\}$ $\displaystyle=\inf\{t\geq 0:x\in r^{-1}tU\}$ $\displaystyle=\inf\{rs\geq 0:x\in sU\}$ $\displaystyle=r\inf\{s\geq 0:x\in sU\}$ $\displaystyle=r\mu_{U}(x).$

Therefore, if $\alpha\in\mathbb{C}$ and $x\in X$, then $\mu_{U}(\alpha x)=|\alpha|\mu_{U}(x)$.

Let $x,y\in X$. $U$ is absorbing, so let $s=\mu_{U}(x)$ and $t=\mu_{U}(y)$. If $\epsilon>0$ then $x\in(s+\epsilon)U$ and $y\in(t+\epsilon)U$. We have

 $x+y\in(s+\epsilon)U+(t+\epsilon)U=\{(s+\epsilon)u+(t+\epsilon)v:u,v\in U\},$

and for $u,v\in U$, because $U$ is convex we have

 $s^{\prime}u+t^{\prime}v=(s^{\prime}+t^{\prime})\left(\frac{s^{\prime}}{s^{% \prime}+t^{\prime}}u+\frac{t^{\prime}}{s^{\prime}+t^{\prime}}v\right)\in(s^{% \prime}+t^{\prime})U,$

where $s^{\prime}=s+\epsilon$ and $t^{\prime}=t+\epsilon$, so

 $x+y\in(s+t+2\epsilon)U.$

This is true for every $\epsilon>0$, which means that $\mu_{U}(x+y)\leq s+t$. Therefore

 $\mu_{U}(x+y)\leq s+t=\mu_{U}(x)+\mu_{U}(y),$

showing that $\mu_{U}$ satisfies the triangle inequality and hence that $\mu_{U}$ is a seminorm on $X$. ∎

We proved above that the Minkowski functional of a balanced convex open neighborhood of $0$ is a seminorm. The following lemma shows that the collection of Minkowski functionals corresponding to a balanced convex local basis at $0$ are a separating family.44 4 Walter Rudin, Functional Analysis, second ed., p. 27, Theorem 1.36.

###### Lemma 3.

If $X$ is a topological vector space and $U$ is a balanced convex open neighborhood of $0$, then

 $U=\{x\in X:\mu_{U}(x)<1\}.$

If $\mathscr{B}$ is a local basis at $0$ whose elements are balanced and convex, then

 $\{\mu_{U}:U\in\mathscr{B}\}$

is a separating family of seminorms on $X$.

###### Proof.

Let $U\in\mathscr{B}$. If $x\in U$, then because $1\cdot x\in U$ and scalar multiplication is continuous, there is some $\delta>0$ and some open neighborhood $N$ of $x$ such that the image of $[1-\delta,1+\delta]\times N$ under scalar multiplication is contained in $U$. In particular, if $(1+\delta)x\in U$ and so $x\in\frac{1}{1+\delta}U$. Thus we have

 $\mu_{U}(x)=\inf\{t\geq 0:x\in tU\}\leq\frac{1}{1+\delta}<1.$

Therefore, if $x\in U$ then $\mu_{U}(x)<1$. On the other hand, if $x\in X$ and $\mu_{U}(x)<1$, then there is some $t<1$ such that $x\in tU$. As $U$ is balanced, we have $x\in U$. Therefore, if $\mu_{U}(x)<1$ then $x\in U$. This establishes that if $U\in\mathscr{B}$ then

 $U=\{x\in X:\mu_{U}(x)<1\}.$

If $x\neq 0$, then because singletons are closed, the set $X\setminus\{x\}$ is open and contains $0$, and thus there is some $U\in\mathscr{B}$ with $U\subseteq X\setminus\{x\}$. Hence $x\not\in U$, which implies by the first claim that $\mu_{U}(x)\geq 1$. In particular, $\mu_{U}(x)\neq 0$, proving the second claim. ∎

If $X$ is a locally convex space then there is a local basis at $0$, call it $\mathscr{B}$, whose elements are balanced and convex, and we have established that $\mathscr{F}=\{\mu_{U}:U\in\mathscr{B}\}$ is a separating family of seminorms on $X$. Therefore by Theorem 1, $X$ with the seminorm topology induced by $\mathscr{F}$ is a locally convex space. The following theorem states that the seminorm topology is equal to the original topology of the space.55 5 Paul Garrett, Seminorms and locally convex spaces, http://www.math.umn.edu/~garrett/m/fun/notes_2012-13/07b_seminorms.pdf

###### Theorem 4.

If $(X,\tau)$ is a locally convex space, then there is a separating family of seminorms on $X$ such that $\tau$ is equal to the seminorm topology.

###### Proof.

Let $\mathscr{B}$ be a local basis at $0$ whose elements are balanced and convex and let $\mathscr{F}=\{\mu_{U}:U\in\mathscr{B}\}$. If $U\in\mathscr{B}$, then $U=\{x\in X:\mu_{U}(x)<1\}$, which is an open neighborhood of $0$ in the seminorm topology induced by $\mathscr{F}$, and this implies that the seminorm topology is at least as fine as $\tau$.

If $U\in\mathscr{B}$ and $\epsilon>0$, then

 $\{x\in X:\mu_{U}(x)<\epsilon\}=\left\{x\in X:\mu_{U}\left(\frac{x}{\epsilon}% \right)<1\right\}=\{\epsilon x\in X:\mu_{U}(x)<1\}=\epsilon U.$

$\epsilon U\in\tau$ and $0\in\epsilon U$, and it follows that $\tau$ is at least as fine as the seminorm topology. Therefore $\tau$ is equal to the seminorm topology induced by $\mathscr{F}$. ∎

We have shown that if $X$ is a vector space and $\mathscr{F}$ is a separating family of seminorms on $X$, then $X$ with the seminorm topology induced by $\mathscr{F}$ is a locally convex space. Furthermore, we have shown that if $X$ is a locally convex space then there is a separating family $\mathscr{F}$ of seminorms on $X$ such that the topology of $X$ is equal to the seminorm topology induced by $\mathscr{F}$. In other words, the topology of any locally convex space is the seminorm topology induced by some separating family of seminorms on the space.

A subset $E$ of a topological vector space $X$ is said to be bounded if for every open neighborhood $N$ of $0$ there is some $s>0$ such that $t>s$ implies that $E\subseteq tN$.

###### Lemma 5.

If $X$ is a locally convex space with the seminorm topology induced by a separating family $\mathscr{F}$ of seminorms on $X$, then a subset $E$ of $X$ is bounded if and only if each $m\in\mathscr{F}$ is a bounded function on $E$.

###### Proof.

Suppose that $E$ is bounded and $m\in\mathscr{F}$. The set $U=\{x\in X:m(x)<1\}$ is an open neighborhood of $0$, so there is some $t>0$ such that $E\subseteq tU$. Hence if $x\in E$ then $m(x), so $m$ is a bounded function on $E$.

Suppose that for each $m\in\mathscr{F}$ there is some $M_{m}$ such that $x\in E$ implies that $m(x)\leq M_{m}$. If $U$ is an open neighborhood of $0$, then there are $m_{1},\ldots,m_{n}\in\mathscr{F}$ and $\epsilon_{1},\ldots,\epsilon_{n}>0$ such that

 $\bigcap_{k=1}^{n}\{x\in X:m_{k}(x)<\epsilon_{k}\}\subseteq U.$

Let $M=\max\left\{\frac{M_{m_{k}}}{\epsilon_{k}}:1\leq k\leq n\right\}$. For $t>M$,

 $\bigcap_{k=1}^{n}\{tx\in X:m_{k}(x)<\epsilon_{k}\}\subseteq tU,$

i.e.,

 $\bigcap_{k=1}^{n}\{x\in X:m_{k}(x)<\epsilon_{k}t\}\subseteq tU.$

But if $x\in E$ and $1\leq k\leq n$ then

 $m_{k}(x)\leq M_{m_{k}}\leq\epsilon_{k}M<\epsilon_{k}t,$

hence $x$ is in the above intersection and thus is in $tU$. Therefore $E\subseteq tU$, showing that $E$ is bounded. ∎

We now prove that if the topology of a locally convex space is induced by a countable separating family of seminorms then the topology is metrizable.

###### Theorem 6.

If $(X,\tau)$ is a locally convex space with the seminorm topology induced by a countable separating family of seminorms $\{m_{n}:n\in\mathbb{N}\}$ and $c_{n}$ is a summable nonincreasing sequence of positive numbers, then

 $d(x,y)=\sum_{n=1}^{\infty}c_{n}\frac{m_{n}(x-y)}{1+m_{n}(x-y)},\qquad x,y\in X,$

is a translation invariant metric on $X$, $\tau$ is equal to the metric topology for $d$, and with this metric the open balls centered at $0$ are balanced.

###### Proof.

For any $x,y\in X$ we have

 $d(x,y)<\sum_{n=1}^{\infty}c_{n}<\infty,$

because the sequence $c_{n}$ is summable. It is apparent that $d(x,y)=d(y,x)$.

If $m$ is any seminorm on $X$, then

 $\frac{m(x)+m(y)}{1+m(x)+m(y)}-\frac{m(x+y)}{1+m(x+y)}=\frac{m(x)+m(y)-m(x+y)}{% (1+m(x)+m(y))(1+m(x+y))}\geq 0,$

so

 $\frac{m(x+y)}{1+m(x+y)}\leq\frac{m(x)+m(y)}{1+m(x)+m(y)}.$

Also, it is straightforward to check that the function $f:[0,\infty)\to[0,\infty)$ defined by $f(a)=\frac{a}{1+a}$ satisfies $f(a+b)\leq f(a)+f(b)$. Define $d_{0}(x)=d(x,0)$. If $x,y\in X$, then

 $\displaystyle d_{0}(x+y)$ $\displaystyle=\sum_{n=1}^{\infty}c_{n}\frac{m_{n}(x+y)}{1+m_{n}(x+y)}$ $\displaystyle\leq\sum_{n=1}^{\infty}c_{n}\frac{m_{n}(x)+m_{n}(y)}{1+m_{n}(x)+m% _{n}(y)}$ $\displaystyle\leq\sum_{n=1}^{\infty}c_{n}\frac{m_{n}(x)}{1+m_{n}(x)}+c_{n}% \frac{m_{n}(y)}{1+m_{n}(y)}$ $\displaystyle=d_{0}(x)+d_{0}(y).$

Hence, for $x,y\in X$,

 $d(x,z)=d_{0}(x-y+y-z)\leq d_{0}(x-y)+d_{0}(y-z)=d(x,y)+d(y,z),$

showing that $d$ satisfies the triangle inequality.

If $d(x,y)=0$, then

 $\sum_{n=1}^{\infty}c_{n}\frac{m_{n}(x-y)}{1+m_{n}(x-y)}=0.$

As each term is nonnegative, each term must be equal to $0$. As each $c_{n}$ is positive, this implies that each $m_{n}(x-y)$ is equal to $0$. But $\{m_{n}:n\in\mathbb{N}\}$ is a separating family so if $x-y\neq 0$ then there is some $m_{n}$ with $m_{n}(x-y)\neq 0$, and this shows that $x-y=0$, i.e. $x=y$. Therefore $d$ is a metric on $X$.

If $x_{0}\in X$, then $d(x+x_{0},y+x_{0})=d(x,y)$: the metric $d$ is translation invariant.

If $|\alpha|\leq 1$ and $x\in X$, then

 $\displaystyle d_{0}(\alpha x)$ $\displaystyle=\sum_{n=1}^{\infty}c_{n}\frac{m_{n}(\alpha x)}{1+m_{n}(\alpha x)}$ $\displaystyle=\sum_{n=1}^{\infty}c_{n}\frac{|\alpha|m_{n}(x)}{1+|\alpha|m_{n}(% x)}$ $\displaystyle=\sum_{n=1}^{\infty}c_{n}\frac{m_{n}(x)}{\frac{1}{|\alpha|}+m_{n}% (x)}$ $\displaystyle\leq\sum_{n=1}^{\infty}c_{n}\frac{m_{n}(x)}{1+m_{n}(x)}$ $\displaystyle=d_{0}(x).$

Thus, if $d(x,0)<\epsilon$ and $|\alpha|\leq 1$ then $d(\alpha x,0)<\epsilon$, so the open ball

 $\{x\in X:d(x,0)<\epsilon\}$

is balanced.

$(X,\tau)$ has a local basis at $0$ whose elements are finite intersections of sets of the form $\{x\in X:m_{n}(x)<\epsilon\}$. Suppose that $\epsilon>0$, let $N$ be large enough so that $\sum_{n=N+1}^{\infty}c_{n}<\frac{\epsilon}{2}$, and let $M$ be large enough so that $\frac{1}{M}\sum_{n=1}^{N}c_{n}<\frac{1}{2}$. If $x\in\bigcap_{n=1}^{N}\{y\in X:m_{n}(y)<\frac{\epsilon}{M}\}$, then

 $\displaystyle d(x,0)$ $\displaystyle=\sum_{n=1}^{N}c_{n}\frac{m_{n}(x)}{1+m_{n}(x)}+\sum_{n=N+1}^{% \infty}c_{n}\frac{m_{n}(x)}{1+m_{n}(x)}$ $\displaystyle<\sum_{n=1}^{N}c_{n}m_{n}(x)+\sum_{n=N+1}^{\infty}c_{n}$ $\displaystyle<\sum_{n=1}^{N}c_{n}\frac{\epsilon}{M}+\frac{\epsilon}{2}$ $\displaystyle<\frac{\epsilon}{2}+\frac{\epsilon}{2}$ $\displaystyle=\epsilon.$

This shows that

 $\bigcap_{n=1}^{N}\left\{x\in X:m_{n}(x)<\frac{\epsilon}{M}\right\}\subseteq\{x% \in X:d(x,0)<\epsilon\},$

and this entails that $\tau$ is at least as fine as the metric topology induced by $d$.

Suppose that $0<\epsilon<\frac{1}{2}$ and $N\in\mathbb{N}$. If $d(x,0), then of course for each $n$ we have

 $c_{n}\frac{m_{n}(x)}{1+m_{n}(x)}

and hence if $1\leq n\leq N$ then

 $\frac{m_{n}(x)}{1+m_{n}(x)}<\frac{c_{N}}{c_{n}}\epsilon\leq\epsilon,$

and hence if $1\leq n\leq N$ then

 $m_{n}(x)<\frac{\epsilon}{1-\epsilon}<2\epsilon.$

Therefore,

 $\{x\in X:d(x,0)

It follows from this that the metric topology induced by $d$ is at least as fine as $\tau$. ∎

If a locally convex space is metrizable with a complete metric, then it is called a Fréchet space.

We now prove conditions under which a topological vector space is normable.

###### Theorem 7.

A topological vector space $(X,\tau)$ is normable if and only if there is a convex bounded open neighborhood of the origin.

###### Proof.

Suppose that $V$ is a convex bounded open neighborhood of $0$. $V$ contains a balanced convex open neighborhood $U$ of $0$,66 6 Walter Rudin, Functional Analysis, second ed., p. 12, Theorem 1.14. and because $V$ is bounded so is $U$. We define $\left\|x\right\|=\mu_{U}(x)$, where $\mu_{U}$ is the Minkowski functional of $U$. If $x\neq 0$, then because $N=X\setminus\{x\}$ is an open neighborhood of $0$ and $U$ is bounded, there is some $t>0$ such that $U\subseteq tN$. Hence $x\not\in\frac{1}{t}U$, i.e., $tx\not\in U$. As $U$ is balanced, by Lemma 3 we get $\mu_{U}(tx)\geq 1$. $\mu_{U}$ is a seminorm, so $\mu_{U}(x)\geq\frac{1}{t}>0$, showing that if $x\neq 0$ then $\mu_{U}(x)>0$, and hence that $\left\|\cdot\right\|$ is a norm on $X$. Also, we check that

 $\{x\in X:\left\|x\right\|

Because $U$ is bounded, for any open neighborhood $N$ of $0$ there is some $t>0$ such that $U\subseteq tN$, hence

 $\left\{x\in X:\left\|x\right\|<\frac{1}{t}\right\}\subseteq N.$

This implies that the norm topology for $\left\|\cdot\right\|$ is at least as fine as $\tau$. And $\{x\in X:\left\|x\right\| is an open set because scalar multiplication is continuous, so $\tau$ is at least as fine as the norm topology for $\left\|\cdot\right\|$. Therefore that $(X,\tau)$ is normable with the norm $\left\|\cdot\right\|$.

In the other direction, if $\tau$ is the norm topology for some norm $\left\|\cdot\right\|$ on $X$, then

 $U=\{x\in X:\left\|x\right\|<1\}$

is indeed a convex open neighborhood of the origin. Suppose that $N$ is an open neighborhood of $0$. There is some $r>0$ such that

 $\{x\in X:\left\|x\right\|

and thus such that $U\subseteq\frac{1}{r}N$, and hence $U$ is bounded, showing that there exists a convex bounded open neighborhood of the origin. ∎

A topological vector space is called locally bounded if there is a bounded open neighborhood of the origin. A topological vector space is said to have the Heine-Borel property if every closed and bounded subset of it is compact.

###### Theorem 8.

If $X$ is a topological vector space that is locally bounded and has the Heine-Borel property, then $X$ has finite dimension.

###### Proof.

Let $V$ be a bounded neighborhood of $0$. It is a fact that the closure of a bounded set is itself bounded,77 7 Walter Rudin, Functional Analysis, second ed., p. 11, Theorem 1.13(f). and therefore $\overline{V}$ is compact. For any $x\in X$, the set $x+\overline{V}$ is a compact neighborhood of $x$, hence $X$ is locally compact. But a locally compact topological vector space is finite dimensional,88 8 Walter Rudin, Functional Analysis, second ed., p. 17, Theorem 1.22. so $X$ is finite dimensional. ∎

## 3 Continuous functions on the unit disc

Let $D=\{z\in\mathbb{C}:|z|<1\}$, the open unit disc. Let $C(D)$ be the set of continuous functions $D\to\mathbb{C}$. $C(D)$ is a complex vector space. If $K$ is a compact subset of $D$, define

 $\nu_{K}(f)=\sup\{|f(z)|:z\in K\},\qquad f\in C(D).$

It is straightforward to check that $\nu_{K}$ is a seminorm on $C(D)$. If $f\in C(D)$ is nonzero then there is some $z\in D$ with $f(z)\neq 0$, and hence $\nu_{\{z\}}(f)=|f(z)|>0$, so the set of all $\nu_{K}$ is a separating family of seminorms on $C(D)$. Thus, $C(D)$ with the seminorm topology induced by the set of all $\nu_{K}$ is a locally convex space.

Define $K_{n}=\{z\in\mathbb{C}:|z|\leq 1-\frac{1}{n}\}$, $n\geq 1$. If $K$ is a compact subset of $D$, then there is some $n$ with $K\subseteq K_{n}$, so $\nu_{K}(f)\leq\nu_{K_{n}}(f)$, and hence

 $\{f\in C(D):\nu_{K_{n}}(f)<\epsilon\}\subseteq\{f\in C(D):\nu_{K}(f)<\epsilon\}.$

It follows that the seminorm topology induced by $\{\nu_{K_{n}}:n\in\mathbb{N}\}$ is at least as fine as the seminorm topology induced by $\{\nu_{K}:\textrm{K is compact}\}$, thus the topologies are equal. Because the topology of $C(D)$ is induced by the countable family $\{\nu_{K_{n}}:n\in\mathbb{N}\}$, by Theorem 6 it is metrizable: for any summable nonincreasing sequence of positive real numbers $c_{n}$, the topology is induced by the metric

 $d(f,g)=\sum_{n=1}^{\infty}c_{n}\frac{\nu_{K_{n}}(f-g)}{1+\nu_{K_{n}}(f-g)},% \qquad f,g\in C(D).$ (1)

Suppose that $f_{i}\in C(D)$ is a Cauchy sequence. For $n\in\mathbb{N}$, the fact that $f_{i}$ is a Cauchy sequence in $C(D)$ implies that $\nu_{K_{n}}(f_{i}-f_{j})\to 0$ as $i,j\to\infty$. $C(K_{n})$ is a Banach space with the norm $\nu_{K_{n}}$, and hence there is some $f_{K_{n}}\in C(K_{n})$ satisfying $\nu_{K_{n}}(f_{i}-f_{K_{n}})\to 0$ as $i\to\infty$. We define $f:D\to\mathbb{C}$ to be $f_{K_{n}}(z)$, for $z\in K_{n}$; this makes sense because the restriction of $f_{K_{n}}$ to $K_{m}$ is $f_{K_{m}}$ if $n\geq m$. $f$ is continuous at each point in $D$ because for each point in $D$ there is some $K_{n}$ containing an open neighborhood of the point, and $f_{K_{n}}$ is continuous. Hence $f\in C(D)$. Therefore $C(D)$ with the metric (1) is a complete metric space, which means that it is a Fréchet space.

###### Theorem 9.

The topology of $C(D)$ is not induced by a norm.

###### Proof.

Because the topology of $C(D)$ is the seminorm topology induced by the separating family of seminorms $\{\nu_{K_{n}}:n\in\mathbb{N}\}$, by Lemma 5 a subset $E$ of $C(D)$ is bounded if and only if each $\nu_{K_{n}}$ is a bounded function on $E$, i.e., for each $n\in\mathbb{N}$ there is some $M_{n}$ such that $f\in E$ implies $\nu_{K_{n}}(f)\leq M_{n}$.

Suppose by contradiction that there is a bounded convex open neighborhood $V$ of the origin. Because $\nu_{K_{n}}(f)\leq\nu_{K_{n+1}}(f)$ for any $f\in C(D)$, there is some $N\in\mathbb{N}$ and some $\epsilon>0$ such that

 $U=\{f\in C(D):\nu_{K_{N}}(f)<\epsilon\}\subseteq V.$

$V$ being bounded implies that $U$ is bounded. Let

 $\Delta_{1}=\left\{z\in\mathbb{C}:|z|<1-\frac{1}{N}+\frac{1}{N(N+1)}\right\},% \quad\Delta_{2}=\left\{z\in\mathbb{C}:1-\frac{1}{N}<|z|<1\right\},$

and let $\phi_{1},\phi_{2}$ be a partition of unity subordinate to this open cover of $D$. For any constant $M>0$, the restriction of $M\phi_{2}$ to $K_{N}$ is $0$ and hence belongs to $U$. But $\nu_{K_{N+1}}(M\phi_{2})=M$, so $\nu_{K_{N+1}}$ is not a bounded function on $U$, contradicting that $U$ is bounded. Therefore, there is no bounded convex open neighborhood of $0$. By Theorem 7, this tells us that $C(D)$ is not normable. ∎

For each $n$, the set $C(K_{n})$ is a Banach space with norm $\nu_{K_{n}}$. If $n\geq m$ and $f\in C(K_{n})$, let $r_{n,m}(f)$ be the restriction of $f$ to $K_{m}$. For $n\geq m$, the function $r_{n,m}$ is a continuous linear map $C(K_{n})\to C(K_{m})$, and if $n\geq m\geq l$ then $r_{n,l}=r_{m,l}\circ r_{n,m}$. Thus the Banach spaces $C(K_{n})$ and the maps $r_{n,m}$ are a projective system in the category of locally convex spaces, and it is a fact that any projective system in this category has a projective limit that is unique up to unique isomorphism.

###### Theorem 10.

$C(D)=\varprojlim C(K_{n})$.

###### Proof.

Define $r_{n}:C(D)\to C(K_{n})$ by taking $r_{n}(f)$ to be the restriction of $f$ to $K_{n}$. Each $r_{n}$ is continuous and linear. Certainly, if $n\geq m$ then $r_{m}=r_{n,m}\circ r_{n}$. Suppose the $Y$ is a locally convex space, that $\phi_{n}:Y\to C(K_{n})$ are continuous linear maps, and that if $n\geq m$ then

 $\phi_{m}=r_{n,m}\circ\phi_{n}.$ (2)

If $z\in K_{m}$ and $n\geq m$, then by (2) we have $\phi_{n}(y)(z)=\phi_{m}(y)(z)$. For $z\in D$, eventually $z\in K_{n}$, and define $\phi(y)(z)$ to be $\phi_{n}(y)(z)$ for any $n$ such that $z\in K_{n}$. For each $z\in D$ there is some $n$ such that $z$ is in the interior of $K_{n}$, and the restriction of $\phi(y)$ to $K_{n}$ is equal to $\phi_{n}(y)$, hence $\phi(y)$ is continuous at $z$. Therefore $\phi(y)\in C(D)$, so $\phi:Y\to C(D)$.

Suppose that $y_{1},y_{2}\in Y$ and $\alpha\in\mathbb{C}$. If $z\in D$, then there is some $n$ with $z\in K_{n}$, and because $\phi_{n}$ is linear,

 $\phi(\alpha y_{1}+y_{2})(z)=\phi_{n}(\alpha y_{1}+y_{2})(z)=\alpha\phi_{n}(y_{% 1})(z)+\phi_{n}(y_{2})(z)=\alpha\phi(y_{1})(z)+\phi(y_{2})(z).$

Therefore $\phi$ is linear.

Suppose that $y_{\alpha}\in Y$ is a net with limit $y\in Y$. For $\phi(y_{\alpha})$ to converge to $\phi(y)$ means that for each $n\in\mathbb{N}$ we have $\nu_{K_{n}}(\phi(y_{\alpha})-\phi(y))\to 0$. But

 $\nu_{K_{n}}(\phi(y_{\alpha})-\phi(y))=\nu_{K_{n}}(\phi_{n}(y_{\alpha})-\phi_{n% }(y)),$

and $\phi_{n}(y_{\alpha})\to\phi_{n}(y)$ because $\phi_{n}$ is continuous. Therefore, for each $n\in\mathbb{N}$ we have $\nu_{K_{n}}(\phi(y_{\alpha})-\phi(y))\to 0$, so $\phi$ is continuous. ∎

We proved in the above theorem that the Fréchet space $C(D)$ is the projective limit of the Banach spaces $C(K_{n})$. It is a fact that the projective limit of any projective system of Banach spaces is a Fréchet space.99 9 J. L .Taylor, Notes on locally convex topological vector spaces, http://www.math.utah.edu/~taylor/LCS.pdf, p. 8, Proposition 2.6, and cf. Paul Garret, Functions on circles: Fourier series, I, http://www.math.umn.edu/~garrett/m/fun/notes_2012-13/04_blevi_sobolev.pdf, p. 37, §13.

A topological space is said to be separable if it has a countable subset that is dense.

###### Theorem 11.

$C(D)$ is separable.

###### Proof.

One proves using the Stone-Weierstrass theorem that the Banach space $C(K_{n})$ is separable. The product of at most continuum many separable Hausdorff spaces each with at least two points is itself separable with the product topology.1010 10 Stephen Willard, General Topology, p. 109, Theorem 16.4. Therefore, $\prod_{n=1}^{\infty}C(K_{n})$ is separable. Because each $C(K_{n})$ is a metric space, this countable product $\prod_{n=1}^{\infty}C(K_{n})$ is metrizable, and any subset of a separable metric space is itself separable with the subspace topology. The projective limit of a projective system of topological vector spaces is a closed subspace of the product of the spaces; thus, using merely that the projective limit is a subset of the product $\prod_{n=1}^{\infty}C(K_{n})$ and has the subspace topology inherited from the direct product, we get that $C(D)$ is separable. ∎

## 4 Holomorphic functions on the unit disc

Let $H(D)$ be the set of holomorphic functions $D\to\mathbb{C}$. $H(D)$ is a linear subspace of $C(D)$. Let $H(D)$ have the subspace topology inherited from $C(D)$. One proves that this topology is equal to the seminorm topology induced by $\{\nu_{K_{n}}:n\in\mathbb{N}\}$. Any subset of a separable metric space with the subspace topology is separable. By Theorem 11 the Fréchet space $C(D)$ is separable, and thus $H(D)$ is separable too.

We now prove that $H(D)$ is a closed subspace of $C(D)$.1111 11 Paul Garrett, Holomorphic vector-valued functions, http://www.math.umn.edu/~garrett/m/fun/notes_2012-13/08b_vv_holo.pdf A closed linear subspace of a Fréchet space is itself a Fréchet space, hence this theorem shows that $H(D)$ is a Fréchet space.

###### Theorem 12.

$H(D)$ is a closed subset of $C(D)$.

###### Proof.

Suppose that $f_{j}\in H(D)$ is a net and that $f_{j}\to f\in C(D)$. We shall show that $f\in H(D)$. (In fact it suffices to prove this for a sequence of elements in $H(D)$ because we have shown that $C(D)$ is metrizable, but that will not simplify this argument.) To show this we have to prove that if $z\in D$ then $\frac{f(z+h)-f(z)}{h}$ has a limit as $h\to 0$, $h\in\mathbb{C}$. Let $\gamma$ be a counterclockwise oriented circle contained in $D$ with center $z$, say of radius $r=\frac{1-|z|}{2}>0$. For each $j$ the function $f_{j}$ is holomorphic on $D$, and so Cauchy’s integral formula gives

 $f_{j}(w)=\frac{1}{2\pi i}\int_{\gamma}\frac{f_{j}(\zeta)}{\zeta-w}d\zeta,% \qquad w\in B_{r}(z).$

Therefore

 $\displaystyle f(w)-\frac{1}{2\pi i}\int_{\gamma}\frac{f(\zeta)}{\zeta-w}d\zeta$ $\displaystyle=$ $\displaystyle f(w)-f_{j}(w)+\frac{1}{2\pi i}\int_{\gamma}\frac{f_{j}(\zeta)}{% \zeta-w}d\zeta-\frac{1}{2\pi i}\int_{\gamma}\frac{f(\zeta)}{\zeta-w}d\zeta$ $\displaystyle=$ $\displaystyle f(w)-f_{j}(w)+\frac{1}{2\pi i}\int_{\gamma}\frac{f_{j}(w)-f(w)}{% \zeta-w}d\zeta.$

As $\gamma$ is a compact subset of $D$ this gives us

 $\left|f(w)-\frac{1}{2\pi i}\int_{\gamma}\frac{f(\zeta)}{\zeta-w}d\zeta\right|% \leq|f(w)-f_{j}(w)|+\frac{1}{2\pi}\cdot 2\pi r\cdot\frac{\nu_{\gamma}(f_{j}-f)% }{r-|w-z|}.$

The right-hand side tends to 0, while the left-hand side does not depend on $j$. Hence

 $f(w)=\frac{1}{2\pi i}\int_{\gamma}\frac{f(\zeta)}{\zeta-w}d\zeta,\qquad w\in B% _{r}(z).$ (3)

Applying (3), we have for $0\leq|h|,

 $f(z+h)=\frac{1}{2\pi i}\int_{\gamma}\frac{f(\zeta)}{\zeta-(z+h)}d\zeta,$

hence

 $\displaystyle f(z+h)-f(z)$ $\displaystyle=$ $\displaystyle\frac{1}{2\pi i}\int_{\gamma}\frac{f(\zeta)}{\zeta-(z+h)}d\zeta-% \frac{1}{2\pi i}\int_{\gamma}\frac{f(\zeta)}{\zeta-z}d\zeta$ $\displaystyle=$ $\displaystyle\frac{1}{2\pi i}\int_{\gamma}f(\zeta)\left(\frac{1}{\zeta-(z+h)}-% \frac{1}{\zeta-z}\right)d\zeta$ $\displaystyle=$ $\displaystyle\frac{1}{2\pi i}\int_{\gamma}f(\zeta)\cdot\frac{h}{(\zeta-(z+h))(% \zeta-z)}d\zeta,$

thus

 $\frac{f(z+h)-f(z)}{h}=\frac{1}{2\pi i}\int_{\gamma}\frac{f(\zeta)}{(\zeta-(z+h% ))(\zeta-z)}d\zeta.$

For $\zeta\in\gamma$ we have $\left|\frac{f(\zeta)}{(\zeta-(z+h))(\zeta-z)}\right|\leq\frac{\nu_{\gamma}(f)}% {(r-|h|)r}$, and so by the dominated convergence theorem we get

 $\lim_{h\to 0}\frac{f(z+h)-f(z)}{h}=\frac{1}{2\pi i}\int_{\gamma}\frac{f(\zeta)% }{(\zeta-z)^{2}}d\zeta.$

Thus, for every $z\in D$, the function $f$ is complex differentiable at $z$. Hence $f\in H(D)$, and therefore $H(D)$ is a closed subset of $C(D)$. ∎

We remind ourselves that a topological vector space is said to have the Heine-Borel property if every closed and bounded subset of it is compact. Lemma 5 tells us that a subset $E$ of $H(D)$ is bounded if and only if each seminorm $\nu_{K_{n}}$ is a bounded function on $E$. The following theorem states that $H(D)$ has the Heine-Borel property.1212 12 Henri Cartan, Elementary Theory of Analytic Functions of One or Several Complex Variables, pp. 162–167, chapter V, §4. An equivalent statement is called Montel’s theorem.

###### Theorem 13 (Heine-Borel property).

The Fréchet space $H(D)$ has the Heine-Borel property.

That $H(D)$ has the Heine-Borel property is a useful tool, and lets us prove that the topology of $H(D)$ is not induced by a norm.

###### Theorem 14.

$H(D)$ is not normable.

###### Proof.

If $H(D)$ were normable then by Theorem 7 there would be a convex bounded open neighborhood of the origin. This would imply that $H(D)$ is locally bounded (has a bounded open neighborhood of the origin). But $H(D)$ has the Heine-Borel property, and a topological vector space that is locally bounded and has the Heine-Borel property is finite dimensional by Theorem 8. It is straightforward to check that $H(D)$ is not finite dimensional, and hence $H(D)$ is not normable. ∎

For $f\in H(D)$, let $(df)(z)=\lim_{h\to 0}\frac{f(z+h)-f(z)}{h}$. First, if $f\in H(D)$ then one proves that $df\in H(D)$. Then, the following theorem states that $d:H(D)\to H(D)$ is a morphism in the category of locally convex spaces.1313 13 Henri Cartan, Elementary Theory of Analytic Functions of One or Several Complex Variables, p. 143, chapter V, §1.

###### Theorem 15.

Differentiation $H(D)\to H(D)$ is a continuous linear map.

If $K$ is a compact subset of $D$ and $f\in H(D)$, let $r_{K}(f)$ be the restriction of $f$ to $K$, and let $\overline{H}(K)$ be the closure in $C(K)$ of the set $\{r_{K}(f):f\in H(D)\}$. Each element of $\overline{H}(K)$ is holomorphic on the interior of $K$. $C(K)$ is a Banach space with the norm $\nu_{K}$, and hence $\overline{H}(K)$ is a Banach space with the same norm, because it is indeed a linear subspace. If $n\geq m$ and $f\in\overline{H}(K_{n})$, let $r_{n,m}(f)=r_{K_{m}}(f)\in\overline{H}(K_{m})$. The $r_{n,m}$ are continuous and linear, and if $n\geq m\geq l$ then $r_{n,l}=r_{m,l}\circ r_{n,m}$. Thus the Banach spaces $\overline{H}(K_{n})$ and the continuous linear maps $r_{n,m}$ are a projective system in the category of locally convex spaces, and this projective system has a projective limit $\varprojlim\overline{H}(K_{n})$. The following theorem states that this projective limit is equal to the Fréchet space $H(D)$.1414 14 J. L .Taylor, Notes on locally convex topological vector spaces, http://www.math.utah.edu/~taylor/LCS.pdf, p. 8

###### Theorem 16.

$H(D)=\varprojlim\overline{H}(K_{n})$.

## 5 Dual spaces

The dual of a topological vector space $X$ is the set $X^{*}$ of continuous linear maps $X\to\mathbb{C}$. If $E$ is a bounded subset of $X$ and $\lambda\in X^{*}$, then $\lambda(E)$ is a bounded subset of $\mathbb{C}$ (the image of a bounded set under a continuous linear map is a bounded set). Hence

 $p_{E}(\lambda)=\sup\{|\lambda x|:x\in E\}<\infty.$

The function $p_{E}$ is a seminorm on $X^{*}$, and if $\lambda\neq 0$ then there is some $x\in X$ with $\lambda x\neq 0$, hence $p_{\{x\}}(\lambda)>0$. The strong dual topology on $X^{*}$ is the seminorm topology induced by the separating family

 $\{p_{E}:\textrm{E is a bounded subset of X}\}.$

(To add to our vocabulary: the set of all bounded subsets of a topological vector space is called the bornology of the space. Similar to how one can define a topology as a collection of sets satisfying certain properties, one can also define a bornology on a set without first having the structure of a topological vector space.) We denote by $X_{\beta}^{*}$ the dual space $X^{*}$ with the strong dual topology. $X_{\beta}^{*}$ is a locally convex space. If $X$ is a normed space, one can prove1515 15 K. Yosida, Functional Analysis, sixth ed., p. 111, Theorem 1. that $X_{\beta}^{*}$ is normable with the operator norm

 $\left\|\lambda\right\|=\sup\{|\lambda x|:\left\|x\right\|\leq 1\}.$

We say that a topological vector space $X$ is reflexive if $(X_{\beta}^{*})_{\beta}^{*}=X$; since the strong dual of a topological vector space is locally convex, for a topological vector space to be reflexive it is necessary that it be locally convex.

Let $X$ be a locally convex space. The Hahn-Banach separation theorem1616 16 Walter Rudin, Functional Analysis, second ed., p. 59, Theorem 3.4. yields that $X^{*}$ separates $X$: if $x\neq 0$ then there is some $\lambda\in X^{*}$ with $\lambda x\neq 0$. If $\lambda\in X^{*}$, then $|\lambda|$ is a seminorm on $X$ and $\{|\lambda|:\lambda\in X^{*}\}$ is therefore a separating family of seminorms on $X$. We call the seminorm topology induced by this separating family the weak topology on $X$, and $X$ with the weak topology is a locally convex space. The original topology on $X$ is at least as fine as the weak topology on $X$: any set that is open using the weak topology is open using the original topology.

The following lemma shows that a Fréchet space with the Heine-Borel property is reflexive, and therefore that $H(D)$ is reflexive.

###### Lemma 17.

If a Fréchet space has the Heine-Borel property, then it is reflexive.

###### Proof.

A subset of a locally convex space is called a barrel if it is closed, convex, balanced, and absorbing. A locally convex space is said to be barreled if each barrel is a neighborhood of $0$. It is a fact that every Fréchet space is barreled.1717 17 K. Yosida, Functional Analysis, sixth ed., p. 138, Corollary 1. A locally convex space is reflexive if and only if it is barreled and if every set that is closed, convex, balanced, and bounded is weakly compact.1818 18 K. Yosida, Functional Analysis, sixth ed., p. 140, Theorem 2. Therefore, for a Fréchet space with the Heine-Borel property to be reflexive it is necessary and sufficient that every set that is compact, convex, and balanced be weakly compact. But if a subset of a locally convex space is compact then it is weakly compact, because the original topology is at least as fine as the weak topology and hence any cover of a set by elements of the weak topology is also a cover of the set by elements of the original topology. Therefore, any Fréchet space with the Heine-Borel property is reflexive. ∎

Morphisms in the category of locally convex spaces are continuous linear maps. If $X$ and $Y$ are locally convex spaces and $\phi:X\to Y$ is a morphism, the dual of $\phi$ is the morphism

 $\phi^{*}:Y_{\beta}^{*}\to X_{\beta}^{*}$

defined by

 $\phi^{*}(\lambda)=\lambda\circ\phi,\qquad\lambda\in Y_{\beta}^{*}.$

One verifies that $\phi^{*}$ is in fact a morphism. If the spaces $X_{j}$ and the morphisms $\phi_{i,j}:X_{i}\to X_{j}$, $i\geq j$, are a projective system in the category of locally convex spaces, then the dual spaces $(X_{j})_{\beta}^{*}$ and the morphisms $\phi_{i,j}^{*}:(X_{j})_{\beta}^{*}\to(X_{i})^{*}_{\beta}$, $i\geq j$, are a direct system in this category. It is a fact that the dual of a projective limit of Banach spaces is isomorphic to the direct limit of the duals of the Banach spaces.1919 19 Paul Garrett, Functions on circles: Fourier series, I, http://www.math.umn.edu/~garrett/m/fun/notes_2012-13/04_blevi_sobolev.pdf, p. 15, Theorem 5.1.1. Thus, as $H(D)$ is the projective limit of the Banach spaces $\overline{H}(K_{n})$, its dual space $H^{*}(D)=(H(D))_{\beta}^{*}$ is isomorphic to the direct limit of the duals of these Banach spaces:

 $H^{*}(D)=\varinjlim(\overline{H}(K_{n}))_{\beta}^{*}.$

Cooper2020 20 J. B. Cooper, Functional analysis– spaces of holomorphic functions and their duality, http://www.dynamics-approx.jku.at/lena/Cooper/holloc.pdf, p. 11, §5. shows that $H^{*}(D)$ is isomorphic to the space of germs of functions on the complement of $D$ in the extended complex plane that vanish at infinity. Let $\mathfrak{A}$ be those sequences $a\in\mathbb{C}^{\mathbb{N}}$ satisfying

 $\limsup|a_{n}|^{1/n}\leq 1.$

By Hadamard’s formula for the radius of convergence of a power series, these are precisely the sequences of coefficients of power series with radius of convergence $\geq 1$, and $\mathfrak{A}$ is a complex vector space. The map

 $a\mapsto\sum_{n=0}^{\infty}a_{n}z^{n}$

is linear and has the linear inverse

 $f\mapsto\left(\frac{f^{(n)}(0)}{n!}\right),$

so $H(D)$ and $\mathfrak{A}$ are linearly isomorphic. For $0, define

 $q_{r}(a)=\max\{|a_{n}|r^{n}:n\in\mathbb{N}\}.$

Each $q_{r}$ is a norm, yet we do not give $\mathfrak{A}$ the norm topology. Rather, we give $\mathfrak{A}$ the seminorm topology induced by the family $\{q_{r}:0, and with this topology $\mathfrak{A}$ is a locally convex space. One proves that the above two linear maps are continuous, and hence that $H(D)$ is isomorphic as a locally convex space to $\mathfrak{A}$. Then, one proves that the dual space of $\mathfrak{A}$ are those sequences $b\in\mathbb{C}^{\mathbb{N}}$ such that

 $\limsup|b_{n}|^{1/n}<1,$

and $b$ corresponds to

 $\sum_{n=0}^{\infty}b_{n}\left(\frac{1}{z}\right)^{n+1}.$