Given an $n \times n$ matrix $M(z)$ with holomorphic entries, I want to show the map
\begin{equation*}
f: z \mapsto \|M(z)\| := \sup_{\|v\| = 1} \|M(z)v\|
\end{equation*}
is subharmonic. I have shown that the pointwise supremum of subharmonic functions is subharmonic (showing that the pointwise supremum satisfies the sub-mean value property using some approximation arguments). Naturally one defines $f_v : z \mapsto \|M(z)v\|$, I want to show that this map is subharmonic. However, I'm having problem proving this, if it's even true at all.
We can write $\|M(z)v\| = \sqrt{|g_1 (z)|^2 + \dots + |g_n (z)|^2}$ for some holomorphic maps $g_i$, and I know that $|g_i|^2$ is subharmonic, but this is as far as I got. I believe the square root of a non-negative subharmonic function is not necessarily subharmonic, so I think I need to somehow appeal to the fact that the $g_i$ are holomorphic, rather than simply they have harmonic real and imaginary parts.
Best Answer
The key here is that if $g$ is holomorphic, then $|g|$ is not only subharmonic but also logarithmically subharmonic (used to be called $PL$ subharmonic too in earlier literature) as obviously $\log |g|$ is subharmonic.
The class of these subharmonic functions (obviously if $\log f$ is subharmonic, $f$ is too because the exponential is convex increasing but the converse is false when $f \ge 0$ general subharmonic) is closed under addition, multiplication and arbitrary positive powers as opposed to the usual subharmonic class which is closed only under addition (all is obvious from the logarithm properties but addition and that follows from the characterization of the $PL$ class as those functions $u \ge 0, u$ upper continuous in some domain $D$, for which given any subdomain $D'$ and any harmonic $h$ in $D'$, $ue^h$ is subharmonic in $D'$)
This immediately implies the result you want since all $|g_k|$'s are logarithmically subharmonic, hence $\|M(z)v\|$ is too.