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Lemma 5.24.616E53426F2DB

Lemma 5.24.6.16E53426F2DB Let $h$ be subharmonic on $\mathbb C$, with $h(\zeta +1)=h(\zeta )$. If there are real constants $A\geq 0$ and $B$ such that

\[ h(\zeta )\leq A\log (1+|\zeta |)+B\qquad (\zeta \in \mathbb C), \]

then $h$ is constant.

Proof. By the growth bound, the real translates $h(\cdot +s)$, $s\in [0,1]$, form a family of subharmonic functions that is locally uniformly bounded above. The local upper-envelope theorem therefore shows that

\[ H(\zeta )=\bigl (\sup _{s\in [0,1]}h(\zeta +s)\bigr )^* \]

is subharmonic. By periodicity, $H$ is invariant under every real translation, so it depends only on $\operatorname {Im}\zeta $, and its subharmonicity means that it is a convex function of $\operatorname {Im}\zeta $. On the imaginary axis it satisfies $H(\mathrm i y)\leq A\log (2+|y|)+B$. Since a nonconstant convex function on $\mathbb R$ grows at least linearly at one end, this bound shows that $H$ is constant. Thus $h\leq H$ is bounded above, and lemma 5.24.5 shows that $h$ is constant. $\square$

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