One-sided product bound for an entire function
= One-sided product bound for an entire function
If an <entire function> $f=u+iv$ has $uv$ bounded above, then $f$ is constant. Indeed $F=\exp(-if^2)$ is entire and $|F|=\exp(2uv)$ is bounded, so the <Liouville theorem> makes $F$ constant. Differentiating gives $ff'=0$. The derivative of $f^2$ is therefore zero, and continuity on the connected complex plane makes $f$ itself constant. No separate boundedness of $u$ or $v$ is required.