I'm preparing for a PhD prelim in Complex Analysis, and I encountered this question from an old PhD prelim:
Suppose $f(z)$ is an entire function such that $|f(z)| \leq \log(1+|z|) \forall z$. Show that $f \equiv 0$.
Well, for $z=0$, $|f(0)| \leq 0$. On the other hand, for $z \neq 0$, $\log(1+|z|) > 0$, a positive constant. I'm guessing this would mean that $f$ turns out to be a bounded entire function, so then by Liouville's theorem, $f$ is constant, but this doesn't necessarily mean that $f \equiv 0$, does it? Am I wrong somewhere? Some guidance would be much appreciated!
Best Answer
Since $f$ is entire it can be expressed as a power series that converges everywhere: $f(z) = \sum_{n=0}^\infty a_n z^n$. From $|f(0)| \leq 0$ we know that $f(0)=0$, hence $a_0 = 0$. So $g(z) := f(z)/z$ can be continued to an entire function that satisfies $$|g(z)| \leq \frac{\log(1+|z|)}{|z|} \quad \text{for all } z\neq 0.$$ The right side converges to zero for $|z| \to \infty$, in particular it is bounded. By Liouville's theorem, $g$ is constantly zero and so is $f$.