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

Question

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!

Answer 1

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

Answer 2

Unfortunately, no, the assumption $f(0)=f′(0)=0$ does not imply the same conclusion. Take as a counterexample $$f(z)=z^2$$

Answer 3

$f$ is an entire function which VANISHES on the imaginary axis – The function $g\equiv 0$, also VANISHES on the imaginary axis, and as $f(z)=g(z)$, in a set of points with accumulation points in $\mathbb C$, then Uniqueness Theorem implies that $$ f\equiv g\equiv 0, $$ in the whole of $\mathbb C$.