If $f$ is an entire function, $\lim_{z\to\infty} z^{-1} \operatorname{ Re} f(z)=0$, show $f$ is constant.
I know only the Liouville theorem: bounded entire funcion is constant. It seems that the real part of $f$ is at most linear growth around $\infty$.
If using harmonic function, it sounds easy. $f(z)=u+iv$, then using the harmonicity of $u$, we may show $u$ is constant, and thus $v$.
Is there any natural complex analysis proof? Thank you.
Let $g(z)=f(z)-f(0)$. Then $f$ is constant if and only if $g$ is constant. Besides, $g$ is an entire function too. On the other hand, $\frac{g(z)}z$ is bounded near $0$ (since $\lim_{z\to0}\frac{g(z)}z=f'(0)$). It follows from this that $\frac{\operatorname{Re}(g(z))}z$ is also bounded near $0$. But$$\lim_{z\to\infty}\frac{\operatorname{Re}(g(z))}z=0.\tag1$$Therefore, $\frac{\operatorname{Re}(g(z))}z$ is bounded. So, for some constant $M$, you have $\bigl|\operatorname{Re}(g(z))\bigr|\leqslant M|z|$. It is well-known that then $g$ is polynomial with degree at most $1$. So, $g(z)=az+b$, for some constants $a$ and $b$. But then it follows from $(1)$ that $a=0$.