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Find all entire functions such that $|f(z+z')|\leq |f(z)| + |f(z')|$, for all $z,z'\in\mathbb{C}$

In particular, let $z=z'$ yields $|f(2z)|\leq2|f(z)|$. This gives that $\frac{f(2z)}{f(z)}=c, $ for some $c\in\mathbb{C}$. By considering continuity at 0, we have $f(0)=\lim_{n\to\infty}\frac{1}{c^n} f(1)$. Then $c\in\mathbb{R}$ and $\frac{1}{c}\leq1$. If $c=1$, then $f(z)$ is the constant equating $f(0)$. But if $\frac{1}{c}<1$ we just have that $f(0)=0$. This does not seem to lead to anywhere. Is there another thing to be considered? Hints will be appreciated. Thank you.

Derewsnanu
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1 Answers1

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Let $z'=z$，then we have $$|f(2z)|\leq|2f(z)|,\quad z\in\mathbb C.$$ So $$f(2z)=2\alpha f(z),\quad z\in\mathbb C \tag{1}$$ for some $\alpha$ with $|\alpha|\leq1$(For this, we can use $\left|\frac{f(2z)}{2f(z)}\right|\leq1$, and Liouville's theorem. For detail, you can refer $|f(z)|\leq|\sin z|$).

If $\alpha=0$, there is nothing to say.

Suppose $0<|\alpha|\leq1$, by $(1)$, we can get, for $n\geq1$, $$2^nf^{(n)}(2z)=2\alpha f^{(n)}(z).$$ Take $z=0$, we have $$2^nf^{(n)}(0)=2\alpha f^{(n)}(0), \quad \forall n\geq1.$$ This implies $f^{(n)}(0)=0$ for $n\geq2$. So $$f(z)=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}z^n=f(0)+f'(0)z,\quad \forall z\in\mathbb C.$$

Back to $(1)$, we have $$f(0)+2f'(0)z=2\alpha(f(0)+f'(0)z),\quad z\in\mathbb C,$$ then we have $$ \begin{cases} f(0)=2\alpha f(0) \cr 2f'(0)=2\alpha f'(0) \end{cases}, $$ this implies $f(0)=0$ or $f'(0)=0$.

Hence we have $$f(z)\equiv f(0)\quad \text{or}\quad f(z)=f'(0)z.$$

Riemann
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