Here is my question:

Let $f(z)$ be an entire function such that $f(z+1)=f(z)$ and $|f(z)| \leq A e^{c|z|}$ for all $z \in \mathbb{C}$, where $c < 2\pi$ is some real constant. Then $f$ is constant.

In fact, this question appeared twice on Math.SE.

One is this question: Entire “periodic” function, which asks for the special case $c=1$. This one is easy, because $c < \pi$ can be readily handled by Carlson's theorem. Anyway, the answer for this question that received the most upvote is due to Zarrax, who suggested considering a holomorphic function $g$ on $\mathbb{C} \setminus \{0\}$ such that

$$f(z) = g(e^{2\pi i z}),$$

and trying to show that $g$ is a constant. Zarrax did not work out the rest of the proof here.

Then there is another question: An entire function with periodic bounds, where the OP generalized to $c < 2\pi$, and followed Zarrax's suggestion to find

$$g(\zeta) \leq Ae^c |\zeta|^{c/2\pi}$$

for all $\zeta$ such that $|\zeta| \geq 1$. Then a "proof" was found following the observation that $g$ must be a polynomial due to the above bound. However, this proof is flawed due to the unwarranted observation that $g$ is a polynomial. This observation only works if $g$ is entire; but $g$ could have a pole, or even an essential singularity at 0. In fact, if we take $f(z) = \sin(2\pi z)$ (which is the reason why we need $c < 2\pi$ in the first place), then $g$ satisfies the above bound but it is clearly not a polynomial.

So, could anyone come up with a correct proof? Thanks.

  • $\begingroup$ Doesn't @blabler's answer to the first quoted question work fine? $\endgroup$
    – Igor Rivin
    Dec 12, 2013 at 0:50
  • $\begingroup$ @IgorRivin No, that approach only works for $c < \pi$ if I understood it correctly. This is because $f$ can grow as fast as $e^{2\pi z}$ here while $\sin(\pi z)$ grows only as fast as $e^{\pi z}$; so $g$, as defined by blabler, might not be bounded at infinity. $\endgroup$
    – 4ae1e1
    Dec 12, 2013 at 0:53
  • $\begingroup$ OK, what about these answers: math.stackexchange.com/questions/221926/… $\endgroup$
    – Igor Rivin
    Dec 12, 2013 at 1:01
  • $\begingroup$ @IgorRivin Wow, so there is another occurrence of the problem I missed. Thanks! By the way, in retrospect, the trick that filled the gap is rather standard. I should have looked at $g$ more seriously before asking. $\endgroup$
    – 4ae1e1
    Dec 12, 2013 at 1:11

1 Answer 1


To fill the gap described above, i.e., to show that $g$ is entire, we need to use the idea outlined in this answer. In particular, we have to not only consider the $|\zeta| \geq 1$ behavior, but also the behavior in the annulus $0 < |\zeta| < 1$, which turns out to be

$$|g(\zeta)| \leq Ae^c |\zeta|^{- c/2\pi}.$$

This might be divergent, but fortunately, since $1 - c/2\pi > 0$, we have

$$|h(\zeta)| \overset{def}{=} |\zeta g(\zeta)| \leq Ae^c |\zeta|^{1 - c/2\pi}$$

not only bounded, but converges to $0$ as $\zeta$ goes to $0$. Therefore, not only the singularity is removable, but we also have $h(0) = 0$, which implies that $g$ also has a removable singularity at $0$. Hence $g$ is entire, and the gap is filled.

  • $\begingroup$ It just occurred to me that even the argument $g$ entire implies $g$ is polynomial isn't easy. The best way to show $g$ is constant seems to be arguing that both $g(\zeta)$ and $g(1/\zeta)$ are entire (the latter could be derived with the same argument as above). $\endgroup$
    – 4ae1e1
    Dec 12, 2013 at 6:55

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