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?

  • $\begingroup$ @runway44 I think it has no upper bound in that region, because you can consider $z=iy$ with $y$ as large as you want. $\endgroup$
    – Lord Vader
    Feb 7, 2023 at 3:24
  • $\begingroup$ Ah right I was thinking $|e^z|$ for some reason. $\endgroup$
    – anon
    Feb 7, 2023 at 3:30

2 Answers 2


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.

  • $\begingroup$ Thanks for your answer! I understand the argument that you use to prove that the coefficients of the Laurent series vanish for all $n\in\mathbb{Z}\setminus\{0\}$. However, I don't completely get the idea behind the existence of $F$. Can you tell me a little bit more about that (the theorems that you are using, and the part where the first hypothesis appears)? $\endgroup$
    – Lord Vader
    Feb 14, 2023 at 0:10
  • $\begingroup$ @LordVader The function $F$ exists for purely set-theoretic reasons: for any value $q$ the function $f$ takes the same value on all $z$ so that $q(z) = q$. It follows that if we call this value $F(q)$ then $F$ is a function; by construction we also have $F(q(z)) = f(z)$ $\endgroup$ Feb 15, 2023 at 7:10
  • $\begingroup$ The non-trivial point is that $F$ is holomorphic, and the reason for that is that while $q(z)$ is not bijective, its derivative is nowhere zero so it is locally bijective and has local inverses. In other words, for any $q_0 = q(z_0)$ there is a holomorphic function $q^{-1}$ (essentially a branch of the logarithm) defined near $q_0$ and satisfying $q^{-1}(q_0) = z_0$. Then near $q_0$ we have $F = f \circ q^{-1}$. As the composition of holomorphic functions this makes $F$ holomorphic near $q_0$, which was arbitrary. $\endgroup$ Feb 15, 2023 at 7:12
  • $\begingroup$ The only "theorem" used here is the holomorphic inverse function theorem. $\endgroup$ Feb 15, 2023 at 7:16
  • $\begingroup$ Where are you using the hypothesis that $f$ is periodic? $\endgroup$
    – Lord Vader
    Feb 15, 2023 at 21:03

Note first that $|f(x)| \le M, x \in \mathbb R$ by periodicity and assuming wlog $M \ge 1$, since $|f(z)| \le e^{|z|}$ it follows that $|f(x+iy)| \le Me^{|y|}$

(this is a simple consequence of Phragmen Lindelof applied to $f(z)e^{iz}$ in the first and second quadrants and $f(z)e^{-iz}$ in the third and fourth which for example gives $|f(z)e^{iz}| \le \max (M,1)=M$ etc), while the quadrant half-angle is $\pi/4$ so P-L applies to any function of order less than $2$, in particular to $f(z)e^{iz}$)

Let $f(0)=a$; then $g(z)=\frac{f(z)-a}{\sin \pi z}$ is entire; but now consider the disjoint discs $D(n,1/3)$ and note that outside them $$|\sin \pi z|=\frac{1}{2}|e^{2\pi iz}-1|e^{\pi|y|} \ge ce^{\pi|y|}$$ since $\min_{\theta}|e^{2\pi ie^{i\theta}/3}-1| >0$

But now $|f(x)-a| \le M_1$ on the real line and $|f(iy)-a| \le M_2e^{|y|}$ so with $M=\max (M_1, M_2)$ we have as above $$|f(x+iy)-a| \le Me^{|y|}$$ hence $$|g(z)| \le \frac{M}{c}e^{(1-\pi)|y|}$$ outside $D(n,1/3)$, but by maximum modulus that holds inside them too as it holds on the boundary, so $$|g(z)| \le \frac{M}{c} e^{(1-\pi)|y|}$$ in the full plane, hence by Liouville $g$ is constant and it must be zero by letting $y \to \infty$ so $f(x)=a$


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