Q:

If $f$ is an entire function with $|\,f(z)|\le|\operatorname{Re}(z)|$, then $\,f\equiv 0$.

If $f$ is an entire function so that $|\,f(z)|\le|\operatorname{Re}(z)|$ for all $\mathbb{C}$, then $f\equiv0$ on $\mathbb{C}$.
There are various ways showing the above property. For example, using Cauchy’s Inequality, Generalized Liouville’s Theorem, etc, and specific values of $f$
But, I want to show directly that $f’(0)=0$ under the conditions. If it possible, $f(0)=f’(0)=0$ implies the same conclusion.
Give some idea or adivce! Thank you!

A:

The hypothesis $|f(z)| \le |\mathrm{Re}(z)|$ implies that $f(iy) = 0$ for all $y$. This gives you $f(0) = 0$ and
$$f'(0) = \lim_{y \to 0} \frac{f(iy) - f(0)}{iy} = 0.$$

