= Solution
For $n\ge0$, put
$$
p_n(w)=\|f(w)^{2^n}\|^{1/2^n},\qquad M=\sup_{w\in\partial K}r(f(w)).
$$
The bound $r(u)\le\|u\|$ 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>,
$$
q_n(w)=\max\{p_n(w),M\}.
$$
These are <continuous>, decrease pointwise to the <continuous> constant $M$, and therefore converge uniformly by the <Dini theorem>. In particular,
$$
\limsup_{n\to\infty}\sup_{w\in\partial K}p_n(w)\le M.
$$
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, gives
$$
\|f(z)^{2^n}\|^{1/2^n}\le\sup_{w\in\partial K}\|f(w)^{2^n}\|^{1/2^n}.
$$
Taking the limit on the left and the limit superior on the right proves the <spectral-radius maximum principle>:
$$
\boxed{r(f(z))\le\sup_{w\in\partial K}r(f(w)).}
$$
Back to article page
