I found this problem in a Ph.D. Qualifing Exam:
Let $f$ be an entire function. Suppose that $f(z)=f(z+1)$ and $|f(z)|\leq e^{|z|}$ for all $z\in\mathbb{C}$. Prove that $f$ is a constant function.
I guess that we need to use the Louville's Theorem in order to prove that statement, so it is missing to show that $f$ is a bounded function. However, I am not sure how to apply the hypothesis to do that. Can you give me some advice to complete that proof?
Best Answer
Let $q(z) = e^{2\pi iz}$, defined on $\mathbb{C}$ with range $\mathbb{C}\setminus \{0\}$. For any $z_0\in \mathbb{C}$ we have $q^{-1}\left(q(z_0)\right) = z_0+\mathbb{Z}$, and it follows that there exists a function $F(q)$ such that $f(z) = F(q(z))$. This function is holomorphic on the punctured plane since $q(z)$ is a local biholomorphism (its derivative is everywhere nonzero). We would like to show that $F$ is a constant function.
Like any function on the punctured plane, $F$ has a Laurent expansion $$F(q) = \sum_{n\in\mathbb{Z}} a_n q^n\,,$$ which converges uniformly absolutely in any annulus $\{ A < |q| < B \}$. Next, writing any $q$ in the form $q = e^{2\pi i z}$ with $|\Re{z}|\leq \frac12$ we have $|q| = e^{-2\pi\Im z}$.
Taking $z$ with negative imaginary values we then get $$|q| = e^{2\pi|\Im(z)|}\geq e^{-\pi} e^{2\pi|\Im(z)|+2\pi|\Re(z)|}\geq e^{-\pi} e^{2\pi|z|}\,.$$ It follows that if $|q|>1$ with $z$ chose as above we have $$|F(q)| = |f(z)| \leq e^{|z|} \leq e^{1/2} |q|^{1/2\pi}\,.\tag{1}\label{eq:one}$$ Taking $z$ with positive imaginary value we similarly get for $|q|<1$ that $$|F(q)| \leq e^{1/2} |q|^{-1/2\pi}\,.\tag{2}\label{eq:two}$$
The key fact here is that $2\pi > 1$ (and this is the point of the exercise: the argument would work with any bound of the type $|f(z)| \leq e^{\alpha|z|}$ with $\alpha<2\pi$ but fail at $2\pi$ due to the existence of the exponential $q$ itself.
We finally recall that $a_n = \frac{1}{2\pi i} \oint_{|q|=R} q^{-(n+1)}f(q)dq$. If $n\geq 1$ we let $R\to\infty$ and use \eqref{eq:one} and the triangle inequality to get $$|a_n| \ll \frac{1}{2\pi} R^{-(n+1)} R^{1/2\pi} (2\pi R) \ll R^{-(1-1/2\pi)}\xrightarrow[R\to\infty]{}0\,.$$ Similarly if $n\leq -1$ we have $$|a_n| \ll \frac{1}{2\pi} R^{-(n+1)} R^{-1/2\pi} (2\pi R) \ll R^{(1-1/2\pi)}\xrightarrow[R\to0]{}0\,.$$
It follows that $F(q) = a_0$ so $f(z) = F(q(z)) = a_0$ and we are done.