I am studing for exams and am stuck on this problem.
Suppose $f$ is an entire function s.t. $f(z) =f(z+1)$ and $|f(z)| < e^{|z|}$. Show $f$ is constant.
I've deduced so far that: a) $f$ is bounded on every horizontal strip b) for every bounded horizontal strip of length greater than 1 a maximum modulus must occur on a horizontal boundary.
I'm a little wary of Liouville Theorem approaches... if you choose $f(z) = {1 \over 2}\sin(2\pi z)$ then it satisfies the conditions of the problem except $|f(z)| < e^{2\pi |z|}$ instead of $|f(z)| < e^{|z|}$.
A suggestion: try showing $f(z) = g(e^{2\pi iz})$ where $g(z)$ is analytic except at $z = 0$. Then translate the condition $|f(z)| < e^{|z|}$ into growth conditions of $|g(z)|$ as $z \rightarrow \infty$ and $z \rightarrow 0$ and show that if they occur $g(z)$ must be constant.
This is a highly non-trivial theorem in complex analysis. It is called Carlson's Theorem. Roughly it states that if an entire function vanishes at integer points and have an exponential growth, then the function is zero.
Consider $$ g(z) = \frac{f(z) - f(0)}{\sin(\pi z)} $$ This is an entire function, since $\sin(\pi z)$ has poles at the integers which are cancelled by the zeros of $f(z) - f(0)$ which also occur at every integer. We have $g(z + 2\ell) = g(z)$ and also $g(x + iy) \rightarrow 0$ when $|y| \rightarrow \infty$ and $|x| \leq B$ for any fixed $B$. Therefore $g$ is bounded. Hence by Liouville $g = C$ with $C$ constant. We must have $C = 0$ because otherwise $f$ is of order greater than $e^{|z|}$ (i.e it would be of order at least $e^{|\pi z|}$). Therefore $C = 0$ and $f(z) = f(0)$ as desired.
The following lemma may help.
"If $f$ is an entire function and and $Re(f)$ is bounded, then $f$ is constant."
Write $f(z)=g(z)+ih(z)$ where $g$ and $h$ are real valued functions. Now use the information in the question to prove that $g$ is bounded which completes the answer.
To prove the lemma consider $e^{f(z)}$ and use Liouville's theorem.
Let me try: Let $0<\epsilon<1$. Then
Claim: If $f$ is entire with $f(z) =f(z+1)$ such that
$$|f(z)|\leq e^{\epsilon|z|},$$
then $f$ is constant.
$(\because)$ Since $f$ is $1$-periodc, $g(z) = f\left({\log z\over 2\pi i}\right)$ is a well-defined holomorphic function except for the origin regardless of the branch cut. Check that $g(e^{2\pi iz}) = f(z)$. Then,
$$|g(z)| = \left|f\left({\log z\over 2\pi i}\right)\right|\leq \exp\left({\epsilon}\left|{\log |z|+i\arg(z)\over 2\pi i}\right|\right) = e^{\epsilon{\arg(z)\over 2\pi}}e^{\epsilon{|\log|z||\over 2\pi}}.$$
Hence, if $|z|<1$ then
$$|g(z)|\leq C |z|^{-{\epsilon\over 2\pi}}.$$
Hence,
$$|zg(z)|\leq C|z|^{1-{\epsilon\over 2\pi}}\to 0,\quad |z|\to 0.$$
Hence, $g$ can be extended analytically through the origin, i.e., $g$ is entire.
Now, if $|z|>1$ then as $|g(z)|\leq C|z|^{-{\epsilon\over 2\pi}}$ and $g$ is entire, by Cauchy inequality, $g$ is constant. Hence, $f$ is constant.
Now, letting $\epsilon\to 1$, this completes the proof.
