Can the topological algebra of analytic functions be endowed with a norm that defines the natural topology?

For those who don't have the book (or have the wrong version), here is the proof that the topological vector space of holomorphic functions on the unit disk is not normable (i.e. whose topology is not defined by a norm).

Definition: A topological vector space (over $\mathbb R$ or $\mathbb C$) is locally bounded iff there is a neighborhood $U$ of 0 such that for any neighborhood $V$ of 0, there is an $\epsilon>0$ such that for any $a$ with $|a|<\epsilon$, $aU\subset V$. (We say a set $U$ is bounded iff it satisfies the aforementioned condition, so a topological vector space is locally bounded iff it has a bounded neighborhood of 0.)

It follows from first principles that a normed space is locally bounded, so to show that the space of holomorphic functions on the unit disk is not normable, it suffices to show that this space is not locally bounded.

Proof. Suppose $U$ is a bounded neighborhood of 0. Since by definition any subset of a bounded set is bounded, we can shrink $U$ so that

$$ U=U(K,\epsilon)=\{f: |f(z)|<\epsilon, \forall z\in K\}, $$

where $\epsilon>0$ and $K$ is a compact subset of the unit disk. Clearly the larger $K$ is, the smaller $U(K,\epsilon)$ is, and any such $K$ is contained in the (closed) disk $D(r)$ of radius $r<1$. Thus we can again shrink $U$ to $U(D(r),\epsilon)$ for some $r<1$. Now

$$aU=U(D(r),a).$$

Let

$$V=U(D(\sqrt r),1/2).$$

No matter how small $a>0$ is, we can choose $n$ such that $(\sqrt r)^n<a$, so that

$$ f:=\frac12(z/\sqrt r)^n\in aU\backslash V, $$

so we cannot have $aU\subset V$.


By Montel's theorem, every bounded set (w.r.t. the topology of uniform convergence on compact sets) in the space of holomorphic functions is relatively compact. If the space were normed its closed unit ball would be compact which implies that the space is finite dimensional.


There is an elementary answer. Let $D$ be any domain of $\mathbb{C}$. The usual derivation operator $\partial : \mathcal{O}(D)\to \mathcal{O}(D)$ is continuous for the topology of uniform convergence (use Cauchy integral formula), but not for any norm. Consider indeed the sequence of functions $f_n:=\frac{e_n}{||e_n||}$ where $e_n~:~x\mapsto \exp(nx)$.