If $f$ is an entire function such that $|Re(f(z))| \leq |z|^n, \forall z,$ then $f$ is a polynomial of degree at most $n$

Solution 1:

Better than differentiating the usual Poisson formula is using the representation of holomorphic functions

$$f(z) = i\operatorname{Im} f(0) + \frac{1}{2\pi} \int_0^{2\pi} \frac{Re^{it}+z}{Re^{it}-z} \operatorname{Re} f(Re^{it})\,dt,\quad \lvert z\rvert < R\tag{1}$$

which for $R = 1$ was the subject of this question. The generalisation to arbitrary $R > 0$ is straightforward.

This yields

$$\lvert f(z)\rvert \leqslant \lvert f(0)\rvert + C\cdot \lvert z\rvert^n\tag{2}$$

for some constant $C$ by setting $R = 2\lvert z\rvert$ in $(1)$ for $z\neq 0$. That $(2)$ implies that $f$ is a polynomial of degree $\leqslant n$ should be familiar, it is a direct consequence of the Cauchy estimates.