The Fréchet space of holomorphic functions on the unit disc

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 $ℬ$ of open neighborhoods of $p$ such that if $U$ is an open neighborhood of $p$ then there is some $U_0\in ℬ$ 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 $ℂ$ is a vector space $X$ over $ℂ$ that is a topological space such that singletons are closed sets and such that vector addition $X\times X\to X$ and scalar multiplication $ℂ\times X\to X$ 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. 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 $ℬ$ 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 $ℬ$ 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 $ℱ$ is a set of seminorms on $X$, we say that $ℱ$ is a separating family if $x\ne 0$ implies that there is some $m\in ℱ$ with $m(x)\ne 0$. (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 $0$. We call this topology the seminorm topology induced by $\mathrm{F}$.

Theorem 1 (Seminorm topology).

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

is a local basis at $\mathrm{0}$.

Proof.

We define $\tau$ to be those subsets $U$ of $X$ such that for all $x\in U$ there is some $N\in ℬ$ satisfying $x+N\subseteq U$. If $𝒰$ is a subset of $\tau$ and $x\in {\bigcup }_{U\in 𝒰}U$, then there is some $U_0\in 𝒰$ with $x\in U_0$, and there is some $N_0\in ℬ$ satisfying $x+N_0\subseteq U_0$. We have

$$x+N_0\subseteq U_0\subseteq \bigcup _{U\in 𝒰}U,$$

which tells us that ${\bigcup }_{U\in 𝒰}U\in \tau$. If $U_1,\mathrm{\dots },U_n\in \tau$ and $x\in {\bigcap }_{k=1}^n{U}_k$, then there are $N_1,\mathrm{\dots },N_n\in ℬ$ satisfying $x+N_k\in U_k$ for $1\le k\le n$. But the intersection of finitely many elements of $ℬ$ is itself an element of $ℬ$, so $N={\bigcap }_{k=1}^n{N}_k\in ℬ$, 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 $y\ne x$, let $m_y\in ℱ$ with ${ϵ}_y=m_y(x-y)\ne 0$; there is such a seminorm because $ℱ$ is a separating family. Then $U_y=y+B_{m_y,{ϵ}_y}$ 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

$$\bigcap _{y\ne x}X\setminus U_y=\{x\}$$

is a closed set, showing that singletons are closed.

Let $x,y\in X$ and $N\in ℬ$. There are $m_k\in ℱ$ and ${ϵ}_k>0$, $1\le k\le n$, such that $N={\bigcap }_{k=1}^n{B}_{m_k,{ϵ}_k}$. Let $U={\bigcap }_{k=1}^n{B}_{m_k,{ϵ}_k/2}$. If $v\in (x+U)+(y+U)$ and $1\le k\le n$, then there are $x_k\in B_{m_k,{ϵ}_k/2}$ and $y_k\in B_{m_k,{ϵ}_k/2}$ such that $v=x+x_k+y+y_k$, and

so $v\in x+y+B_{m_k,{ϵ}_k}$. This is true for each $k$, $1\le k\le n$, 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 $\alpha \in ℂ$, $x\in X$, and $N\in ℬ$, say $N={\bigcap }_{k=1}^n{B}_{m_k,{ϵ}_k}$. Let $ϵ=\min \{{ϵ}_k:1\le k\le n\}$, let $\delta >0$ be small enough so that for each $1\le k\le n$, let , and let $U={\bigcap }_{k=1}^n{B}_{m_k,\delta }$. If $(\beta ,v)\in \mathrm{\Delta }\times (x+U)$ and $1\le k\le n$, then

$m_k(\beta v-\alpha x)$ $=m_k(\beta v-\beta x+\beta x-\alpha x)$

$\le m_k(\beta (v-x))+m_k((\beta -\alpha )x)$

$=|\beta |m_k(v-x)+|\beta -\alpha |m_k(x)$

$=\delta (\delta +|\alpha |+m_k(x))$

$\le {ϵ}_k,$

showing that $\beta v\in \alpha x+B_{m_k,{ϵ}_k}$. 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 $\mathrm{\Delta }\times (x+U)$ 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 $ℬ$ is convex. An intersection of convex sets is a convex set, so to prove that each element of $ℬ$ is convex it suffices to prove that each $B_{m,ϵ}$ is convex, $m\in ℱ$ and $ϵ>0$. If $0\le t\le 1$ and $x,y\in B_{m,ϵ}$, then

showing that $tx+(1-t)y\in B_{m,ϵ}$ and thus that $B_{m,ϵ}$ 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 ${\mu }_S:X\to [0,\mathrm{\infty })$ of an absorbing set $S$ is defined by

$${\mu }_S(x)=inf\{t\ge 0:x\in tS\},x\in X.$$

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 $|\alpha |\le 1$ 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 $0$. 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 $\mathrm{0}$, then the Minkowski functional of $U$ is a seminorm on $X$.

Proof.

Let $\alpha \in ℂ$ 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

${\mu }_U(\alpha x)$ $=inf\{t\ge 0:\alpha x\in tU\}$

$=inf\{t\ge 0:rux\in tU\}$

$=inf\{t\ge 0:x\in r^{-1}t{u}^{-1}U\}$

$=inf\{t\ge 0:x\in r^{-1}tU\}$

$=inf\{rs\ge 0:x\in sU\}$

$=rinf\{s\ge 0:x\in sU\}$

$=r{\mu }_U(x).$

Therefore, if $\alpha \in ℂ$ 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 $ϵ>0$ then $x\in (s+ϵ)U$ and $y\in (t+ϵ)U$. We have

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

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

$$x+y\in (s+t+2ϵ)U.$$

This is true for every $ϵ>0$, which means that ${\mu }_U(x+y)\le s+t$. Therefore

$${\mu }_U(x+y)\le s+t={\mu }_U(x)+{\mu }_U(y),$$

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.

Lemma 3.

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

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

$$\{{\mu }_U:U\in ℬ\}$$

is a separating family of seminorms on $X$.

Proof.

Let $U\in ℬ$. 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

Therefore, if $x\in U$ then . On the other hand, if $x\in X$ and , then there is some such that $x\in tU$. As $U$ is balanced, we have $x\in U$. Therefore, if then $x\in U$. This establishes that if $U\in ℬ$ then

If $x\ne 0$, then because singletons are closed, the set $X\setminus \{x\}$ is open and contains $0$, and thus there is some $U\in ℬ$ with $U\subseteq X\setminus \{x\}$. Hence $x\notin U$, which implies by the first claim that ${\mu }_U(x)\ge 1$. In particular, ${\mu }_U(x)\ne 0$, proving the second claim. ∎

If $X$ is a locally convex space then there is a local basis at $0$, call it $ℬ$, whose elements are balanced and convex, and we have established that $ℱ=\{{\mu }_U:U\in ℬ\}$ is a separating family of seminorms on $X$. Therefore by Theorem 1, $X$ with the seminorm topology induced by $ℱ$ is a locally convex space. The following theorem states that the seminorm topology is equal to the original topology of the space.

Theorem 4.

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

If $U\in ℬ$ and $ϵ>0$, then

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

We have shown that if $X$ is a vector space and $ℱ$ is a separating family of seminorms on $X$, then $X$ with the seminorm topology induced by $ℱ$ is a locally convex space. Furthermore, we have shown that if $X$ is a locally convex space then there is a separating family $ℱ$ of seminorms on $X$ such that the topology of $X$ is equal to the seminorm topology induced by $ℱ$. 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 $\mathrm{F}$ of seminorms on $X$, then a subset $E$ of $X$ is bounded if and only if each $m\mathrm{\in }\mathrm{F}$ is a bounded function on $E$.

Proof.

Suppose that $E$ is bounded and $m\in ℱ$. The set is an open neighborhood of $0$, so there is some $t>0$ such that $E\subseteq tU$. Hence if $x\in E$ then , so $m$ is a bounded function on $E$.

Suppose that for each $m\in ℱ$ there is some $M_m$ such that $x\in E$ implies that $m(x)\le M_m$. If $U$ is an open neighborhood of $0$, then there are $m_1,\mathrm{\dots },m_n\in ℱ$ and ${ϵ}_1,\mathrm{\dots },{ϵ}_n>0$ such that

Let $M=\max \left\{\frac{M_{m_k}}{{ϵ}_k}:1\le k\le n\right\}$. For $t>M$,

i.e.,

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

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

$$d(x,y)=\sum _{n=1}^{\mathrm{\infty }}{c}_n\frac{m_n(x-y)}{1+m_n(x-y)},x,y\in X,$$

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 $\mathrm{0}$ are balanced.

Proof.

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

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

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

so

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

Also, it is straightforward to check that the function $f:[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ defined by $f(a)=\frac{a}{1+a}$ satisfies $f(a+b)\le f(a)+f(b)$. Define $d_0(x)=d(x,0)$. If $x,y\in X$, then

$d_0(x+y)$ $={\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{c}_n{\displaystyle \frac{m_n(x+y)}{1+m_n(x+y)}}$

$\le {\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{c}_n{\displaystyle \frac{m_n(x)+m_n(y)}{1+m_n(x)+m_n(y)}}$

$\le {\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{c}_n{\displaystyle \frac{m_n(x)}{1+m_n(x)}}+c_n{\displaystyle \frac{m_n(y)}{1+m_n(y)}}$

$=d_0(x)+d_0(y).$

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

$$d(x,z)=d_0(x-y+y-z)\le d_0(x-y)+d_0(y-z)=d(x,y)+d(y,z),$$

showing that $d$ satisfies the triangle inequality.

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

$$\sum _{n=1}^{\mathrm{\infty }}{c}_n\frac{m_n(x-y)}{1+m_n(x-y)}=0.$$

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 $x-y\ne 0$ then there is some $m_n$ with $m_n(x-y)\ne 0$, 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 $|\alpha |\le 1$ and $x\in X$, then

$d_0(\alpha x)$ $={\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{c}_n{\displaystyle \frac{m_n(\alpha x)}{1+m_n(\alpha x)}}$

$={\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{c}_n{\displaystyle \frac{|\alpha |m_n(x)}{1+|\alpha |m_n(x)}}$

$={\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{c}_n{\displaystyle \frac{m_n(x)}{\frac{1}{|\alpha |}+m_n(x)}}$

$\le {\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{c}_n{\displaystyle \frac{m_n(x)}{1+m_n(x)}}$

$=d_0(x).$

Thus, if and $|\alpha |\le 1$ then , so the open ball

is balanced.

$(X,\tau )$ has a local basis at $0$ whose elements are finite intersections of sets of the form . Suppose that $ϵ>0$, let $N$ be large enough so that , and let $M$ be large enough so that . If , then

$d(x,0)$ $={\displaystyle \sum _{n=1}^N}{c}_n{\displaystyle \frac{m_n(x)}{1+m_n(x)}}+{\displaystyle \sum _{n=N+1}^{\mathrm{\infty }}}{c}_n{\displaystyle \frac{m_n(x)}{1+m_n(x)}}$

$=ϵ.$

This shows that

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

Suppose that and $N\in \mathbb{N}$. If , then of course for each $n$ we have

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

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

Therefore,

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 $\mathrm{(}X\mathrm{,}\tau \mathrm{)}$ 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 $0$, and because $V$ is bounded so is $U$. We define $\parallel x\parallel ={\mu }_U(x)$, where ${\mu }_U$ is the Minkowski functional of $U$. If $x\ne 0$, 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 $x\notin \frac{1}{t}U$, i.e., $tx\notin U$. As $U$ is balanced, by Lemma 3 we get ${\mu }_U(tx)\ge 1$. ${\mu }_U$ is a seminorm, so ${\mu }_U(x)\ge \frac{1}{t}>0$, showing that if $x\ne 0$ then ${\mu }_U(x)>0$, and hence that $\parallel \cdot \parallel$ is a norm on $X$. Also, we check that

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

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

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

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

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, 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, so $X$ is finite dimensional. ∎

3 Continuous functions on the unit disc

Let , the open unit disc. Let $C(D)$ be the set of continuous functions $D\to ℂ$. $C(D)$ is a complex vector space. If $K$ is a compact subset of $D$, define

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

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 $f(z)\ne 0$, 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 $K_n=\{z\in ℂ:|z|\le 1-\frac{1}{n}\}$, $n\ge 1$. If $K$ is a compact subset of $D$, then there is some $n$ with $K\subseteq K_n$, so ${\nu }_K(f)\le {\nu }_{K_n}(f)$, and hence

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 $\{{\nu }_K:K\text{ is compact}\}$, 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

$$d(f,g)=\sum _{n=1}^{\mathrm{\infty }}{c}_n\frac{{\nu }_{K_n}(f-g)}{1+{\nu }_{K_n}(f-g)},f,g\in C(D).$$ (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 $i,j\to \mathrm{\infty }$. $C(K_n)$ 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 $i\to \mathrm{\infty }$. We define $f:D\to ℂ$ 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 $n\ge m$. $f$ 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 $C\mathit{}\mathrm{(}D\mathrm{)}$ 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 ${\nu }_{K_n}(f)\le M_n$.

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

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

and let ${\varphi }_1,{\varphi }_2$ be a partition of unity subordinate to this open cover of $D$. For any constant $M>0$, the restriction of $M{\varphi }_2$ to $K_N$ is $0$ and hence belongs to $U$. But ${\nu }_{K_{N+1}}(M{\varphi }_2)=M$, 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 $n\ge m$ and $f\in C(K_n)$, let $r_{n,m}(f)$ be the restriction of $f$ to $K_m$. For $n\ge m$, the function $r_{n,m}$ is a continuous linear map $C(K_n)\to C(K_m)$, and if $n\ge m\ge l$ 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.

$C(D)=\underleftarrow{\lim }C(K_n)$.

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 $n\ge m$ then $r_m=r_{n,m}\circ r_n$. Suppose the $Y$ is a locally convex space, that ${\varphi }_n:Y\to C(K_n)$ are continuous linear maps, and that if $n\ge m$ then

$${\varphi }_m=r_{n,m}\circ {\varphi }_n.$$ (2)

If $z\in K_m$ and $n\ge m$, then by (2) we have ${\varphi }_n(y)(z)={\varphi }_m(y)(z)$. For $z\in D$, eventually $z\in K_n$, and define $\varphi (y)(z)$ to be ${\varphi }_n(y)(z)$ 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 $\varphi (y)$ to $K_n$ is equal to ${\varphi }_n(y)$, hence $\varphi (y)$ is continuous at $z$. Therefore $\varphi (y)\in C(D)$, so $\varphi :Y\to C(D)$.

Suppose that $y_1,y_2\in Y$ and $\alpha \in ℂ$. If $z\in D$, then there is some $n$ with $z\in K_n$, and because ${\varphi }_n$ is linear,

$$\varphi (\alpha y_1+y_2)(z)={\varphi }_n(\alpha y_1+y_2)(z)=\alpha {\varphi }_n(y_1)(z)+{\varphi }_n(y_2)(z)=\alpha \varphi (y_1)(z)+\varphi (y_2)(z).$$

Therefore $\varphi$ is linear.

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

$${\nu }_{K_n}(\varphi (y_{\alpha })-\varphi (y))={\nu }_{K_n}({\varphi }_n(y_{\alpha })-{\varphi }_n(y)),$$

and ${\varphi }_n(y_{\alpha })\to {\varphi }_n(y)$ because ${\varphi }_n$ is continuous. Therefore, for each $n\in \mathbb{N}$ we have ${\nu }_{K_n}(\varphi (y_{\alpha })-\varphi (y))\to 0$, so $\varphi$ 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.

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. Therefore, ${\prod }_{n=1}^{\mathrm{\infty }}C(K_n)$ is separable. Because each $C(K_n)$ is a metric space, this countable product ${\prod }_{n=1}^{\mathrm{\infty }}C(K_n)$ 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 ${\prod }_{n=1}^{\mathrm{\infty }}C(K_n)$ 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 $D\to ℂ$. $H(D)$ 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 $C(D)$. 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 $C\mathit{}\mathrm{(}D\mathrm{)}$.

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 $h\to 0$, $h\in ℂ$. 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

$$f_j(w)=\frac{1}{2\pi i}{\int }_{\gamma }\frac{f_j(\zeta )}{\zeta -w}𝑑\zeta ,w\in B_r(z).$$

Therefore

$f(w)-{\displaystyle \frac{1}{2\pi i}}{\displaystyle {\int }_{\gamma }}{\displaystyle \frac{f(\zeta )}{\zeta -w}}𝑑\zeta$ $=$ $f(w)-f_j(w)+{\displaystyle \frac{1}{2\pi i}}{\displaystyle {\int }_{\gamma }}{\displaystyle \frac{f_j(\zeta )}{\zeta -w}}𝑑\zeta -{\displaystyle \frac{1}{2\pi i}}{\displaystyle {\int }_{\gamma }}{\displaystyle \frac{f(\zeta )}{\zeta -w}}𝑑\zeta$

$=$ $f(w)-f_j(w)+{\displaystyle \frac{1}{2\pi i}}{\displaystyle {\int }_{\gamma }}{\displaystyle \frac{f_j(w)-f(w)}{\zeta -w}}𝑑\zeta .$

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

$$\left|f(w)-\frac{1}{2\pi i}{\int }_{\gamma }\frac{f(\zeta )}{\zeta -w}𝑑\zeta \right|\le |f(w)-f_j(w)|+\frac{1}{2\pi }\cdot 2\pi r\cdot \frac{{\nu }_{\gamma }(f_j-f)}{r-|w-z|}.$$

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

$$f(w)=\frac{1}{2\pi i}{\int }_{\gamma }\frac{f(\zeta )}{\zeta -w}𝑑\zeta ,w\in B_r(z).$$ (3)

Applying (3), we have for ,

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

hence

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

$=$ ${\displaystyle \frac{1}{2\pi i}}{\displaystyle {\int }_{\gamma }}f(\zeta )\left({\displaystyle \frac{1}{\zeta -(z+h)}}-{\displaystyle \frac{1}{\zeta -z}}\right)𝑑\zeta$

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

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 $\left|\frac{f(\zeta )}{(\zeta -(z+h))(\zeta -z)}\right|\le \frac{{\nu }_{\gamma }(f)}{(r-|h|)r}$, and so by the dominated convergence theorem we get

$$\underset{h\to 0}{lim}\frac{f(z+h)-f(z)}{h}=\frac{1}{2\pi i}{\int }_{\gamma }\frac{f(\zeta )}{{(\zeta -z)}^2}𝑑\zeta .$$

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. An equivalent statement is called Montel’s theorem.

Theorem 13 (Heine-Borel property).

The Fréchet space $H\mathit{}\mathrm{(}D\mathrm{)}$ 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.

Theorem 15.

Differentiation $H\mathit{}\mathrm{(}D\mathrm{)}\mathrm{\to }H\mathit{}\mathrm{(}D\mathrm{)}$ 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 $n\ge m$ 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 $n\ge m\ge l$ 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 $\underleftarrow{\lim }\overline{H}(K_n)$. The following theorem states that this projective limit is equal to the Fréchet space $H(D)$.

Theorem 16.

$H(D)=\underleftarrow{\lim }\overline{H}(K_n)$.

5 Dual spaces

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

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

$$\{p_E:E\text{ is a bounded subset of }X\}.$$

(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 prove that $X_{\beta }^{*}$ is normable with the operator norm

$$\parallel \lambda \parallel =sup\{|\lambda x|:\parallel x\parallel \le 1\}.$$

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 theorem yields that $X^{*}$ separates $X$: if $x\ne 0$ then there is some $\lambda \in X^{*}$ with $\lambda x\ne 0$. 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. 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. 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 $\varphi :X\to Y$ is a morphism, the dual of $\varphi$ is the morphism

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

defined by

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

One verifies that ${\varphi }^{*}$ is in fact a morphism. If the spaces $X_j$ and the morphisms ${\varphi }_{i,j}:X_i\to X_j$, $i\ge j$, are a projective system in the category of locally convex spaces, then the dual spaces ${(X_j)}_{\beta }^{*}$ and the morphisms ${\varphi }_{i,j}^{*}:{(X_j)}_{\beta }^{*}\to {(X_i)}_{\beta }^{*}$, $i\ge j$, 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. 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:

$$H^{*}(D)=\underrightarrow{\lim }{(\overline{H}(K_n))}_{\beta }^{*}.$$

Cooper 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 $𝔄$ be those sequences $a\in {ℂ}^{\mathbb{N}}$ satisfying

$$lim\; sup{|a_n|}^{1/n}\le 1.$$

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 $\ge 1$, and $𝔄$ is a complex vector space. The map

$$a\mapsto \sum _{n=0}^{\mathrm{\infty }}{a}_n{z}^n$$

is linear and has the linear inverse

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

so $H(D)$ and $𝔄$ are linearly isomorphic. For , define

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

Each $q_r$ is a norm, yet we do not give $𝔄$ the norm topology. Rather, we give $𝔄$ the seminorm topology induced by the family , and with this topology $𝔄$ 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 $𝔄$. Then, one proves that the dual space of $𝔄$ are those sequences $b\in {ℂ}^{\mathbb{N}}$ such that

and $b$ corresponds to

$$\sum _{n=0}^{\mathrm{\infty }}{b}_n{\left(\frac{1}{z}\right)}^{n+1}.$$