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

Right, so in my research in complex analysis I was puzzled by this question which may have a simple approachable answer that eludes me, but I am truly itching to find out and in need of it so I am requesting help here. The question I face is:

If we have a domain (like the unit disk) D in the complex plane and we have the topological algebra of analytic functions in one variable on D, is there a suitable norm on this algebra such that it induces the "natural topology" i.e. the topology of convergence on compact sets?

I would truly appreciate a help in the form of a yes/no response with some convincing to settle this for me. I thank all kind helpers.

• Can you define the "natural topology" on this algebra? – Liviu Nicolaescu Apr 28 '16 at 21:30
• @LiviuNicolaescu : I of course mean the topology of convergence on compact sets, or equivalently the compact open topology – Don John Prep Apr 28 '16 at 21:32
• For the negative answer, you show that the "natural topology" is not "locally bounded" and therefore not normable. [Kelley-Namioka, Problem 2M] – Gerald Edgar Apr 28 '16 at 21:44
• I believe his last answer contained the reference you're asking for. – Anthony Quas Apr 28 '16 at 22:11
• While I'm sure this is a perfectly real, valid question, all I could think when I read that title was "now I know what technobabble sounds like to other people". – Nic Hartley Apr 29 '16 at 8:17

## 3 Answers

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 $e_n~:~x\mapsto \exp(nx)$.