# An entire function that must be a polynomial

An entire function $f$ is of exponential type zero if, for every $\epsilon$ > $0$ there exists a positive constant $C_{\epsilon}$ such that $|f(z)|$ < $C_{\epsilon}$$e^{\epsilon |z|}$, $z$ $\in$ $\mathbb{C}$.

Let $f$ be an entire function of exponential type zero such that its restriction to the real line belongs to $L^{p}$ for some $p$ > $1$. Well, by the Pólya-Plancherel theorem $f$ is bounded on the real axis, so by the Bernstein's inequality we have $f'(x)$ = $0$ for x real, so $f'(z)$ = $0$ taking the analytic extension, i.e., $f$ is constant.

The problem I'm trying to solve is a little more general: if $f$ has exponential type zero and some derivative $f^{(k)}$ belongs to $L^{p}$ for some $p$ > $1$, then f is a polynomial of degree at most $k$. In this case, is it true that $f^{(k)}$ also has exponential type zero? If so, I can apply the simpler case to solve this one.

• Doesn't exponential type $0$ just mean the function is bounded, in which case the statement follows from Liouville? – Nishant May 7 '15 at 0:20
• But every function of exponential type zero would have to be constant, which is not true because any polynomial is of exponential type zero. – Br09 May 7 '15 at 0:24
• Wikipedia says that "exponential type $C$" means "bounded by $e^{C|z|}$." If $C=0$, this means it's bounded by $1$. Are you using a different definition? – Nishant May 7 '15 at 1:08
• I edited the question with the definition that I'm using :) – Br09 May 7 '15 at 1:18