Logarithm of the norm of a Banach space-valued holomorphic function is subharmonic?

Question

Given a holomorphic function $f$ in some open subset $G\subset\mathbb{C}$, it is well-known that the real-valued function $z\in G\mapsto\log|f(z)|\in\mathbb{R}$ is a \emph{subharmonic} function, i.e., for every closed disk $D(z,r)\subset G$ of radius $r$ centered in $z$, \begin{equation} \log|f(z)|\leq\frac{1}{2\pi}\int_0^{2\pi}|\log f(z+re^{i\theta})|\,\mathrm{d}\theta. \end{equation} I am wondering if a similar property holds if we consider an holomorphic function $F:G\subset\mathbb{C}\mapsto X$, with $X$ being a Banach space with norm $\|\cdot\|$. Naively, I would expect the function $z\in G\mapsto\log\|F(z)\|\in\mathbb{R}$ to be subharmonic as well, i.e. \begin{equation} \log\|F(z)\|\leq\frac{1}{2\pi}\int_0^{2\pi}\|\log F(z+re^{i\theta})\|\,\mathrm{d}\theta. \end{equation} Is this true in general? If not, are there sufficient conditions to ensure that?

Answer 1

$\log ||F||$ is indeed subharmonic (and moreover it is plurisubharmonic if we have open $G \subset Y$ some other Banach space rather than $\mathbb C$ and the proof is a slight elaboration of the one below using some other equivalent definitions of plurisubharmonicity which is the right generalization of subharmonicity to many complex variables).

The proof here is easy as we just take $\psi \circ F$ for any $\psi$ in the dual of $X$ and that is usual holomorphic so $\log |\psi \circ F|$ is subharmonic hence $\log |\psi \circ F(z)| \le \frac{1}{2\pi}\int_0^{2\pi} \log |\psi \circ F(z+re^{i\theta})|\mathrm{d}\theta$

By Hahn-Banach $\log ||F(w)||=\sup \log |\psi \circ F(w)|, \psi \in X', ||\psi||=1$, hence for any $\epsilon >0$ there is $\psi_{\epsilon,z} \in X', ||\psi_{\epsilon,z}||=1$ s.t

$\log ||F(z)|| \le \log |\psi_{\epsilon,z} \circ F(z)| + \epsilon \le \frac{1}{2\pi}\int_0^{2\pi} \log |\psi_{\epsilon,z} \circ F(z+re^{i\theta})|\mathrm{d}\theta+ \epsilon \le$

$ \le \frac{1}{2\pi}\int_0^{2\pi} \log||F(z+re^{i\theta})||\mathrm{d}\theta+ \epsilon$

and we let $\epsilon \to 0$