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 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 $aU\subset V$.


