If $f$ is an entire function with $|f(z)|\le 100\log|z|$ and $f(i)=2i$, what is $f(1)$?

Solution 1:

You can't directly use Liouville's theorem, since dividing $f$ by $\log \lvert z\rvert$ or $\log z$ doesn't produce an entire function.

But you can use the Cauchy estimates to show that that bound on $\lvert f(z)\rvert$ actually implies that $f$ is constant, for $R > 2$ and $\lvert z\rvert \leqslant \frac{R}{2}$, you can compute

$$\left\lvert f'(z) \right\rvert = \left\lvert \frac{1}{2\pi i} \int\limits_{\lvert \zeta \rvert = R} \frac{f(\zeta)}{(\zeta-z)^2}\, d\zeta\right\rvert \leqslant \frac{1}{2\pi} \int\limits_{\lvert\zeta\rvert = R} \frac{\lvert f(\zeta)\rvert}{\lvert \zeta - z\rvert^2}\, \lvert d\zeta\rvert \leqslant \frac{4R}{R^2}100\log R$$

and, since the right hand side tends to $0$ for $R \to \infty$, conclude that $f' \equiv 0$.