Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 7 / 4

Fix $z\in K$. If $z\in \partial K$, the conclusion is immediate. Otherwise $z$ is in the interior of $K$. If $f(z)=0$, nonnegativity of the boundary supremum suffices. For $f(z)\ne 0$, the complex Hahn-Banach theorem gives a bounded linear functional $\ell :A\to \mathbb{C}$ with $\Vert \ell \Vert =1$ and $\ell (f(z))=\Vert f(z)\Vert$. The scalar function $g=\ell \circ f$ is holomorphic.

Let $V$ be the connected component of the interior of $K$ containing $z$. It is bounded, its closure is contained in $K$, and $\partial V\subseteq \partial K$. To see the last inclusion, a boundary point of $V$ lying in the interior of $K$ would have a small connected open ball in that interior meeting $V$; the ball would belong to the same component, contradicting that it is a boundary point. The scalar maximum modulus principle on a bounded domain therefore givesFor completeness, this boundary form of the maximum modulus principle follows by taking the maximum on the compact closure of $V$. An interior maximum makes $g$ constant on $V$, in which case continuity gives the same value on its nonempty boundary. Thus the norm maximum principle for a Banach-space-valued holomorphic function applies to arbitrary compact $K$, even when it is disconnected or has nonsmooth boundary:

For $n\ge 0$, putThe bound $r(u)\le \Vert u\Vert$ makes $M$ finite. Each $p_n$ is continuous; submultiplicativity gives $p_{n+1}\le p_n$, and the spectral radius formula gives $p_n(w)\to r(f(w))$. Spectral radius need not be continuous, so applying the Dini theorem directly to $p_n$ would not be justified. Instead define, on the compact boundary,These are continuous, decrease pointwise to the continuous constant $M$, and therefore converge uniformly by the Dini theorem. In particular,Multiplication is a continuous bilinear operation in a Banach algebra, so the product rule shows that $w\mapsto f(w{)}^{2^n}$ is a Banach-space-valued holomorphic function. Part (i), applied to this power, givesTaking the limit on the left and the limit superior on the right proves the spectral-radius maximum principle:

Apply the spectral-radius maximum principle to the entire Banach-space-valued holomorphic function $f(z)=a+zb$ on the closed disk $|z|\le \epsilon$. On its boundary the assumed estimate gives $r(f(z))\le C\epsilon$. Thereforefor every $\epsilon >0$. Letting $\epsilon \downarrow 0$ provesThe puncture at $z=0$ in the hypothesis causes no difficulty: the bound is used only on the nonzero boundary circle.