Let $f$ be an entire function. Suppose there exists integer $n>0$ such that $|f(a+b)|^{\frac{1}{n}}\le |f(a)|^{\frac{1}{n}} + |f(b)|^{\frac{1}{n}}$ for all $a,b\in\mathbb{C}$. Show that $f$ is a polynomial with degree at most $n$.
My attempt is to show that $|f^{(k)}(0)|=0$ for all $k>n$ by considering the expansion of $f$ at $0$, but I'm stuck after writing out the expansion.
By induction $|f(ka)|^{1/n} \leq k |f(a)|^{1/n}$ for any positive integer $k$. Let $M$ be a bound for $|f(z)|$ on the closed unit disk. Then, we get $|f(z)|^{1/n} \leq |z/a|M^{1/n}$ by taking $a=\frac z k$ and taking $k=[|z|]+1]$ so that $|a| \leq 1$. Note that $|a| \to 1$ as $|z| \to \infty$. Thus, $|f(z)| \leq C|z|^{n} $ for some $C <\infty$ provided $|z|$ is suffciently large. This implies that $f$ is a polynomial of degree at most $n$. [Ref.: https://math.stackexchange.com/questions/143468/entire-function-bounded-by-a-polynomial-is-a-polynomial?rq=1 ].
