Show that $H(U)$ the space of holomorphic functions on $U$ is not normable. I'm trying to show the following result:
Let $U \subset \mathbb{C}$ be an open set. Show that the vector space $H(U)=\{f:U \to \mathbb{C} \; \text{analytic}\}$ is not normable.
I know that $H(U)$ is a Frechet space. If we take $A_{k}:=\{z \in U\ : |z| \leq k \; , \; \text{dist}(\partial U,z) \geq \frac{1}{k} \}$ and $p_{k}(f)=\sup_{z \in A_{k}}|f(z)|$ for each $k \in \mathbb{N}$, we have
$\cup_{k}A_{k}=U$ and metric
$$d(f,g)=\sum_{k=1}^{\infty}\frac{1}{2^{k}} \frac{p_{k}(f-g)}{1+p_{k}(f-g)}.$$
I'm not sure how to work with this. Any help would be greatly appreciated.
 A: An easy way to understand this (if the required topology is uniform convergence on compacts) is to consider the "derivative" operator $d:H(U) \rightarrow H(U)$. Suppose $H(U)$ is normable. Then $d$ is sequentially continuous so has a bounded operator norm.
In particular, its eigenvalues are bounded.
But, for each $\lambda \in \mathbb{C}$, $\lambda$ is an eigenvalue of $d$ with eigenvector $z \in U \longmapsto e^{\lambda z}$, so we get a contradiction.
A: Supposing  by contradiction that $H(U)$ is normable via some norm $\|{\cdot}\|$, observe that  the set
$$
  B:= \{f\in H(U): \|f\|<1\}
  $$
is open, so there exists some $\varepsilon >0$, such that
$$
  \Omega :=\{f\in H(U): d(f, 0)≤\varepsilon \}\subseteq  B.
  $$
Notice moreover  that, for every $z$ in $U$, the point evaluation
$$
  \delta _z: f\in  H(U) \mapsto  f(z)\in  \mathbb C
  $$
is a continuous linear functional.  So,
given any $f$ in $\Omega $, and any $z\in U$, we  have that
$$
  |f(z)|  \leq  \|\delta _z\| \|f\| \leq   \|\delta _z\|.
  \tag 1
  $$
This is the first indication that something is wrong, since the functions in $H(U)$ are not necessarily bounded.
Choose $n$ such that $\sum_{k=n+1}^\infty 2^{-k}≤\varepsilon /2$ and notice that, since
$$
  A_1\subseteq A_2\subseteq \cdots \subseteq A_n
  $$
we have that
$p_1\leq p_2\leq \cdots \leq p_n$, whence for every $f$ in $H(U)$ we have
$$
  d(f,0) =
  \sum_{k=1}^\infty \frac1{2^k} \frac{p_k(f)}{1+p_k(f)} \leq  $$$$ \leq
  \sum_{k=1}^n\frac1{2^k} \frac{p_k(f)}{1+p_k(f)} +   \sum_{k=n+1}^n\frac1{2^k} \leq  $$$$ \leq
  \Big(\sum_{k=1}^n\frac1{2^k}\Big) \frac{p_n(f)}{1+p_n(f)} +  \frac\varepsilon 2 \leq
  \frac{p_n(f)}{1+p_n(f)} +  \frac\varepsilon 2.
  $$
Chosing $\delta >0$ such that
$$
  0\leq x\leq \delta \Rightarrow \frac{x}{1+x}\leq \varepsilon /2,
  $$
we then deduce that
$$
  p_n(f)\leq \delta  \Rightarrow d(f,0)\leq \varepsilon.
  $$
Summing up  we have by (1) that
$$
  \sup_{z\in A_n}|f(z)| \leq  \delta  \Rightarrow  f\in \Omega   \Rightarrow  |f(z)| \leq  \|\delta _z\|,\quad \forall z\in  U.
  \tag 2
  $$
Picking any point $z_0$ in $U\setminus A_n$ it is now easy to choose some holomorphic function $f$ on $U$ that is
bounded by $\delta $ on $A_n$, and takes on an arbitrarily large value on $z_0$, e.g. bigger that $ \|\delta _{z_0}\|$, thus
contradicting (2).
A: It is easy to prove this by using Montel theorem. It states that if $f_n \in H(U)$ is a bounded sequence (i.e. it is bounded by means of seminorms $p_k$) then there is a convergent subsequence $f_{n_m}$. Thus, if $H(U)$ is normable then its unit ball is a bounded subset and, therefore, by Montel theorem it is compact. Now we observe that a normable space with compact unit ball is finite dimensional. Therefore, $H(U)$ is not normable.
There is an abstract notion of Montel space: a barrelled locally convex space in which all closed bounded subsets are compact. $H(U)$ is a Montel space (a simple corollary of Montel theorem). The foregoing argument shows that a normable space is Montel iff it is finite dimensional. By the same reason the space $C^\infty(U)$ is not normable (with the topology of uniform convergence of all derivatives on compact subsets of $U$).
