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$f$ is an entire function such that $|f(z+w)| \leq |f(z)| + |f(w)|$ for any $z, w $ in $\mathbb{C}$. I need to show that $f(z) = az +b$ for some complex numbers $a$ and $b$.

So, it suffices to show that the entire function $f'$ is bounded, since then the rest follows from Liouville's Theorem. We can assume that $f(0)=0$, without loss of generality. I thought I could get a bound using the inequality on the definition of $f'(z)$ or on the Cauchy Integral formula for $f^{(n)}(z)$. But now I can not see how.

Can you please help me with a way to use the inequality to get a bound on $f'$ ?

Thanks in advance !

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Notice that if $f$ satisfies your condition and $f(z)=0$ with $z\neq 0$ then $f(nz)=0$ for all $n\in \mathbb{Z}^+$, in particular we must have $b=0$. – Jose27 Jan 29 at 6:46
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Fix an upper bound $B$ of $f$ on the unit circle. Use the given condition to get a bound of $f$ on a circle of radius $r$ (something like $Br$). Now use Cauchy's integral formula for $f'(z)$, and try to show that $f'(z)$ is bounded. – Sanchez Jan 29 at 6:59
@Jose27 Yes, we will have that $f(nz_0) = 0$ for any $ n \in \mathbb{Z}^+$ whenever $z_0$ is a root of $f$. But if I understand correctly, if $z_0 \neq 0$, it then implies that $f$ is the zero function. Is it true ? – CB_Student Jan 29 at 7:06
@Sanchez Thanks a lot ! Yes, that works. It turns out that $f'(z)$ will be bounded by $B$ itself. – CB_Student Jan 29 at 7:23

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