# 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 $\mathcal{B}$ of open neighborhoods of $p$ such that if $U$ is an open neighborhood of $p$ then there is some ${U}_{0}\in \mathcal{B}$ 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 $\u2102$ is a vector space $X$ over $\u2102$ that is a topological space such that singletons are closed sets and such that
vector addition $X\times X\to X$ and scalar multiplication $\u2102\times X\to X$ 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.^{1}^{1}
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 $\mathcal{B}$ 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 $\mathcal{B}$ 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 $\mathcal{F}$ is a set of seminorms on $X$,
we say that $\mathcal{F}$ is a separating family if $x\ne 0$ implies that
there is some $m\in \mathcal{F}$ 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 $0$.^{2}^{2}
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 $\mathrm{F}$.

###### 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 $N\in \mathcal{B}$ satisfying $x+N\subseteq U$. If $\mathcal{U}$ is a subset of $\tau $ and $x\in {\bigcup}_{U\in \mathcal{U}}U$, then there is some ${U}_{0}\in \mathcal{U}$ with $x\in {U}_{0}$, and there is some ${N}_{0}\in \mathcal{B}$ satisfying $x+{N}_{0}\subseteq {U}_{0}$. We have

$$x+{N}_{0}\subseteq {U}_{0}\subseteq \bigcup _{U\in \mathcal{U}}U,$$ |

which tells us that ${\bigcup}_{U\in \mathcal{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 ${N}_{1},\mathrm{\dots},{N}_{n}\in \mathcal{B}$ satisfying $x+{N}_{k}\in {U}_{k}$ for $1\le k\le n$. But the intersection of finitely many elements of $\mathcal{B}$ is itself an element of $\mathcal{B}$, so $N={\bigcap}_{k=1}^{n}{N}_{k}\in \mathcal{B}$, 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 ${m}_{y}\in \mathcal{F}$ with ${\u03f5}_{y}={m}_{y}(x-y)\ne 0$; there is such a seminorm because $\mathcal{F}$ is a separating family. Then ${U}_{y}=y+{B}_{{m}_{y},{\u03f5}_{y}}$ 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 $N\in \mathcal{B}$. There are ${m}_{k}\in \mathcal{F}$ and ${\u03f5}_{k}>0$, $1\le k\le n$, such that $N={\bigcap}_{k=1}^{n}{B}_{{m}_{k},{\u03f5}_{k}}$. Let $U={\bigcap}_{k=1}^{n}{B}_{{m}_{k},{\u03f5}_{k}/2}$. If $v\in (x+U)+(y+U)$ and $1\le k\le n$, then there are ${x}_{k}\in {B}_{{m}_{k},{\u03f5}_{k}/2}$ and ${y}_{k}\in {B}_{{m}_{k},{\u03f5}_{k}/2}$ such that $v=x+{x}_{k}+y+{y}_{k}$, and

$$ |

so $v\in x+y+{B}_{{m}_{k},{\u03f5}_{k}}$. 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 $\alpha \in \u2102$, $x\in X$, and $N\in \mathcal{B}$, say $N={\bigcap}_{k=1}^{n}{B}_{{m}_{k},{\u03f5}_{k}}$. Let $\u03f5=\mathrm{min}\{{\u03f5}_{k}:1\le k\le n\}$, 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

${m}_{k}(\beta v-\alpha x)$ | $={m}_{k}(\beta v-\beta x+\beta x-\alpha x)$ | ||

$\le {m}_{k}(\beta (v-x))+{m}_{k}((\beta -\alpha )x)$ | |||

$=|\beta |{m}_{k}(v-x)+|\beta -\alpha |{m}_{k}(x)$ | |||

$$ | |||

$=\delta (\delta +|\alpha |+{m}_{k}(x))$ | |||

$$ | |||

$\le {\u03f5}_{k},$ |

showing that $\beta v\in \alpha x+{B}_{{m}_{k},{\u03f5}_{k}}$. 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 $\mathcal{B}$ is convex. An intersection of convex sets is a convex set, so to prove that each element of $\mathcal{B}$ is convex it suffices to prove that each ${B}_{m,\u03f5}$ is convex, $m\in \mathcal{F}$ and $\u03f5>0$. If $0\le t\le 1$ and $x,y\in {B}_{m,\u03f5}$, then

$$ |

showing that $tx+(1-t)y\in {B}_{m,\u03f5}$ and thus that ${B}_{m,\u03f5}$ 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 $0$.^{3}^{3}
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 $\mathrm{0}$, then the Minkowski functional of $U$ is a seminorm on $X$.

###### Proof.

Let $\alpha \in \u2102$ 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 $\alpha \in \u2102$ 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 $\u03f5>0$ then $x\in (s+\u03f5)U$ and $y\in (t+\u03f5)U$. We have

$$x+y\in (s+\u03f5)U+(t+\u03f5)U=\{(s+\u03f5)u+(t+\u03f5)v:u,v\in U\},$$ |

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 ${s}^{\prime}=s+\u03f5$ and ${t}^{\prime}=t+\u03f5$, so

$$x+y\in (s+t+2\u03f5)U.$$ |

This is true for every $\u03f5>0$, 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.^{4}^{4}
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 $\mathrm{0}$, then

$$ |

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

$$\{{\mu}_{U}:U\in \mathcal{B}\}$$ |

is a separating family of seminorms on $X$.

###### Proof.

Let $U\in \mathcal{B}$. 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 $U\in \mathcal{B}$ 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 $U\in \mathcal{B}$ 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 $\mathcal{B}$, whose elements are balanced and convex, and we have established that $\mathcal{F}=\{{\mu}_{U}:U\in \mathcal{B}\}$ is a separating family of seminorms on $X$. Therefore by Theorem 1, $X$ with the seminorm topology induced by $\mathcal{F}$
is a locally convex space. The following theorem states that the seminorm topology is equal to the original topology of the space.^{5}^{5}
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 $\mathrm{(}X\mathrm{,}\tau \mathrm{)}$ 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 $\mathcal{B}$ be a local basis at $0$ whose elements are balanced and convex and let $\mathcal{F}=\{{\mu}_{U}:U\in \mathcal{B}\}$. If $U\in \mathcal{B}$, then $$, which is an open neighborhood of $0$ in the seminorm topology induced by $\mathcal{F}$, and this implies that the seminorm topology is at least as fine as $\tau $.

If $U\in \mathcal{B}$ and $\u03f5>0$, then

$$ |

$\u03f5U\in \tau $ and $0\in \u03f5U$, and it follows that $\tau $ is at least as fine as the seminorm topology. Therefore $\tau $ is equal to the seminorm topology induced by $\mathcal{F}$. ∎

We have shown that if $X$ is a vector space and $\mathcal{F}$ is a separating family of seminorms on $X$, then $X$ with the seminorm topology induced by $\mathcal{F}$ is a locally convex space. Furthermore, we have shown that if $X$ is a locally convex space then there is a separating family $\mathcal{F}$ of seminorms on $X$ such that the topology of $X$ is equal to the seminorm topology induced by $\mathcal{F}$. 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 $m\in \mathcal{F}$. 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 $m\in \mathcal{F}$ 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 ${m}_{1},\mathrm{\dots},{m}_{n}\in \mathcal{F}$ and ${\u03f5}_{1},\mathrm{\dots},{\u03f5}_{n}>0$ such that

$$ |

Let $M=\mathrm{max}\left\{\frac{{M}_{{m}_{k}}}{{\u03f5}_{k}}:1\le k\le n\right\}$. 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 $$. Suppose that $\u03f5>0$, let $N$ be large enough so that $$, and let $M$ be large enough so that $$. If $$, then

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

$$ | |||

$$ | |||

$$ | |||

$=\u03f5.$ |

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 $0$,^{6}^{6}
6
Walter Rudin,
Functional Analysis, second ed., p. 12, Theorem 1.14. 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,^{7}^{7}
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,^{8}^{8}
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 $$, the open unit disc. Let $C(D)$ be the set of continuous functions $D\to \u2102$. $C(D)$ 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 ${K}_{n}=\{z\in \u2102:|z|\le 1-\frac{1}{n}\}$, $n\ge 1$. 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 $f:D\to \u2102$ 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 $\u03f5>0$ 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.

$C(D)=\underleftarrow{\mathrm{lim}}C({K}_{n})$.

###### 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 $\alpha \in \u2102$. 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.^{9}^{9}
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.^{10}^{10}
10
Stephen Willard, General Topology, p. 109, Theorem 16.4. 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 $D\to \u2102$. $H(D)$ 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 $C(D)$.^{11}^{11}
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 $C\mathit{}\mathrm{(}D\mathrm{)}$.

###### 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 $h\to 0$, $h\in \u2102$. 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

$${f}_{j}(w)=\frac{1}{2\pi i}{\int}_{\gamma}\frac{{f}_{j}(\zeta )}{\zeta -w}\mathit{d}\zeta ,w\in {B}_{r}(z).$$ |

Therefore

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

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

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

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

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

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

Applying (3), we have for $$,

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

hence

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

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

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

thus

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

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

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

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.^{12}^{12}
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 $H\mathit{}\mathrm{(}D\mathrm{)}$ 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.^{13}^{13}
13
Henri
Cartan, Elementary Theory of Analytic Functions of One or Several Complex Variables, p. 143, chapter V, §1.

###### 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 $\underleftarrow{\mathrm{lim}}\overline{H}({K}_{n})$.
The following theorem states that this projective limit is equal to the Fréchet space $H(D)$.^{14}^{14}
14
J. L .Taylor, Notes on locally convex topological vector spaces,
http://www.math.utah.edu/~taylor/LCS.pdf, p. 8

###### Theorem 16.

$H(D)=\underleftarrow{\mathrm{lim}}\overline{H}({K}_{n})$.

## 5 Dual spaces

The dual of a topological vector space $X$ is the set ${X}^{*}$ of continuous linear maps $X\to \u2102$. If $E$ is a bounded subset of $X$ and $\lambda \in {X}^{*}$, then $\lambda (E)$ is a bounded subset of $\u2102$ (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^{15}^{15}
15
K. Yosida,
Functional Analysis, sixth ed., p. 111, Theorem 1. 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^{16}^{16}
16
Walter Rudin, Functional Analysis, second ed., p. 59, Theorem 3.4.
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.^{17}^{17}
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.^{18}^{18}
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 $\varphi :X\to Y$ 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.^{19}^{19}
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:

$${H}^{*}(D)=\underrightarrow{\mathrm{lim}}{(\overline{H}({K}_{n}))}_{\beta}^{*}.$$ |

Cooper^{20}^{20}
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 $\U0001d504$ be those sequences $a\in {\u2102}^{\mathbb{N}}$ 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 $\U0001d504$ 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 $\U0001d504$ are linearly isomorphic. For $$, define

$${q}_{r}(a)=\mathrm{max}\{|{a}_{n}|{r}^{n}:n\in \mathbb{N}\}.$$ |

Each ${q}_{r}$ is a norm, yet we do not give $\U0001d504$ the norm topology. Rather, we give $\U0001d504$ the seminorm topology induced by the family $$, and with this topology $\U0001d504$ 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 $\U0001d504$. Then, one proves that the dual space of $\U0001d504$ are those sequences $b\in {\u2102}^{\mathbb{N}}$ such that

$$ |

and $b$ corresponds to

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