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 $aU\subset V$.
