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.

  • $\begingroup$ So we need to show that $f$ is bounded and entire. We are given that it is entire. It seems that $|f(z)| < e^{|z|}$ and $|f(z+1)|< e^{|z|}$. Then $|f(z+2)| < e^{|z+1|}$. $\endgroup$ – PrimeNumber Dec 15 '10 at 16:53
  • $\begingroup$ Yes. So as you said $f$ is bounded on horizontal strips. I don't know how to show boundedness for the vertical strips $\endgroup$ – Digital Gal Dec 15 '10 at 17:03
  • $\begingroup$ I think a bit of heavier machinery may be in order. You may want to estimate $\Delta\log|f|$ in the ball or radius $R$ and let $R\to\infty$. (Some familiarity with Jensen's formula may help.) $\endgroup$ – Andrés E. Caicedo Dec 15 '10 at 22:33

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.

  • $\begingroup$ Welcome to MSE! Maybe you can improve this answer by adding more details. Just a friendly suggestion. Regards $\endgroup$ – Amzoti Jul 17 '13 at 22:58
  • 1
    $\begingroup$ Carleson's theorem is an over-kill since Carlson's theorem would allow us to replace the condition $f(z + \ell) = f(z)$ for all $z \in \mathbb{C}$ and all $\ell \in \mathbb{Z}$, by the weaker condition $f(\ell) = f(0)$ for all $\ell \in \mathbb{Z}$ and obtain the same conclusion. $\endgroup$ – blabler Jul 17 '13 at 23:43

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.

  • $\begingroup$ I don't see why $Re(f)$ or $Im(f)$ is bounded. $\endgroup$ – Digital Gal Dec 15 '10 at 17:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.