An entire function with periodic bounds

This is a bit of a repost from an old question, but it doesn't seem like it was fully answered before and this is a bit of an abstraction from that post. I'm trying to show the following:

Let $f(z)$ be an entire function such that $f(z)=f(z+1)$ for all $z$. If there exists $c\in\mathbb{R}$ with $c$ not a multiple of $2\pi$ such that $|f(z)|\le e^{c|z|}$ for all $z$, then $f(z)=0$.

The previous post was here: Entire "periodic" function

I've tried using Zarrax's method in that post:

The periodicity of $f(z)$ allows us to write $f(z)=g(e^{2\pi i z})$, since $e^{2\pi i (x+iy)}=e^{2\pi i (x+1+iy)}$. Now we have: $$\left|g(e^{2\pi i z})\right|\le e^{c|z|}$$

Now make a change of variables, re-writing $z=(2\pi i)^{-1}\log z$: $$\left|g(z)\right|\le e^{c\left|\frac{\log z}{2\pi i}\right|}$$

Now assume $|z|\ge 1$: $$\left|g(z)\right|\le e^{\frac{c}{2\pi}(\log |z|+2\pi)}=|z|^\frac{c}{2\pi}e^c$$

So now I would say that this implies $g(z)$ has to be a polynomial, but I don't really see how I can derive a contradiction from that.

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Well, I might as well finish it up now. The result is only true if $c < 2\pi$, as $\sin(2\pi z)$ has period 1 and satisfies $|\sin(2\pi z)| \leq e^{c |z|}$ for any $c \geq 2\pi$.

You're on the right track; as you said up there $g(z)$ has to be a polynomial. Note that for $y > 0$ that $f(-iy) = g(e^{2\pi y})$. If $g(z)$ is of degree $n$, then $|f(-iy)|$ grows as $Ce^{2\pi ny}$ as $y \rightarrow \infty$. So if $n > 0$, $f(z)$ will violate the growth condition. Thus $n =0$, so that $g(z)$ (and therefore $f(z)$) is constant. Note that $f(z)$ doesn't have to be zero.

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Got it! Thank you; this has been racking my brain on and off for weeks now. – Connor May 22 '12 at 4:06
Sorry for commenting on an old post, but I believe this proof is subtly flawed. The problem is with "$g(z)$ has to be a polynomial." I don't see any obvious way to derive this simply from the growth condition $|g(z)| \leq e^c|z|^\frac{c}{2\pi}$ when $|z| \geq 1$. In fact, if $f(z) = \sin(2\pi z)$ then $g(z)$ is not a polynomial. – 4ae1e1 Dec 11 '13 at 21:07
Also, if we already have $g$ is a polynomial, then we do not even need to go back to $f$. $c/2\pi<1$ and $|g(z)| \leq e^c |z|^{c/2\pi}$ already shows that $g$ is a constant. – 4ae1e1 Dec 11 '13 at 23:24
It follows from the fact that $\left|g(z)\right|\le e^{c\left|\frac{\log z}{2\pi i}\right|}$. Replace $z$ by ${1 \over z}$ and you see that the Laurent expansion of $g(z)$ has no negative powers either. – Zarrax Dec 17 '13 at 4:44

It's not true. Note that $e^{c|z|} \ge e^{2\pi |z|}$ if $c > 2 \pi$.

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How is that a contradiction though? Maybe I'm missing your implication, but that's not an analytic function. – Connor May 22 '12 at 3:28
If $f(z)$ is a periodic function and there is $c$ such that $|f(z)| \le e^{c|z|}$ for all $z$, then this bound also works for all greater $c$. It can't just be true for multiples of $2 \pi$. – Robert Israel May 22 '12 at 3:52