# Entire function constant

I came across this old exam problem. Suppose $f(z)$ is entire and $|f(z)|< e^{|z|}$. Also suppose $f(z)=f(z+1)$. Show $f(z)$ is a constant. I am able to show the singularity at infinity is not a pole. But I can't rule out it being an essential singularity.

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It seems that $e^{i 2\pi z}$ is a similar function that is not constant. So I am confused. – Digital Gal Oct 27 '12 at 6:10
Plug in z=-i and you get $|e^{2\pi}|= e^{2\pi}> e$ – blitzer Oct 27 '12 at 6:22
Btw, this statement is true if the bound satisfies the weaker condition $|f(z)| < e^{C|z|}$ where $C<2\pi$... though I don't remember the way you do the problem at the moment – blitzer Oct 27 '12 at 6:25

Since $f(z+1)=f(z)$ the function $g(z) = f\left(\frac{\log(z)}{2\pi i}\right)$ is single valued on $\mathbb{C}^{\ast}$ and

$$|g(z)| < \min_{k \in \mathbb{Z}} e^{\left|\frac{\log(z) + 2k \pi i}{2 \pi i} \right|} \leq e^{\frac{\left|\log|z|\right| + \pi}{2 \pi}} = \begin{cases} |z|^{\frac{-1}{2 \pi}} e^{\frac{1}{2}} & \textrm{for } |z| \leq 1 \\[1em] |z|^{\frac{1}{2 \pi}} e^{\frac{1}{2}} & \textrm{otherwise}. \end{cases}$$

In particular $h(z) = z g(z)$ is bounded near $z = 0$ and by Riemann's extension theorem it extends to an entire function for which $h(0)=0$. Then $g(z) = h(z)/z$ itself extends to an entire function. Likewise $g(1/z)$ also extends to an entire function and therefore $g$ is constant (it is bounded). Then $f$ is also constant.

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What is $C^{*}$ ? – Digital Gal Oct 27 '12 at 19:17
@jojo $\mathbb{C} \setminus \{0\}$. – WimC Oct 27 '12 at 20:28
Do we have to worry about log not being holomorphic on $C^{*}$? – Digital Gal Oct 28 '12 at 16:22
@jojo Thats taken care of by $f$ being periodic. The different branches of $\log$ differ by a multiple of $2 \pi i$. – WimC Oct 28 '12 at 17:00

If you don't want to use the machinery of the extension theorem, define the function

$$g(z):= \frac{f(z)-f(0)}{e^{-i2\pi z}-e^{i2\pi z}}.$$

Where $|f(z)| \leq e^{C |z|}$

(the denominator is a constant times $\sin$.)

Then it's easy to check that g(z) is entire given the periodicity conditions on $f$. Then we can show that $g(z)$ is bounded on the strips $0 \leq \Re(z) \leq 1$ (and therefore everywhere). Let $z=a + bi$ (where $a$ and $b$ are real. Then

$$|g(z)| = |g(a+ bi)| \leq \frac{e^{C|b|}+1}{e^{2 \pi b}-1}$$ when $b>0$ and $$|g(z)| \leq \frac{e^{C|n|}+1}{e^{-2\pi b}-1}$$ when $b<0$ by using the reverse triangle inequality both ways in the denominator. Then if $C<2 \pi$, the denominator will dominate for $b$ large so $g(z)$ is bounded at infinity. It is clearly bounded away from infinity.

By Liouville, $g(z) = A$ for some constant, or

$$f(z) = f(0) + A \sin(2\pi z)$$

From here it's not hard to show $A=0$

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Your own counterexample is (almost) correct. With $f(z)=e^{i2\pi z}$, you have $f(z+1)=f(z)$ and $|f(z)|=e^{-2\pi\Im z}\le e^{2\pi|z|}$, it just has the wrong exponent.

$f$ can be written as a Fourier series, i.e. a linear combination of $e^{i2k\pi z}$, which will not obey the $e^{|z|}$ bound in $i$-direction even for finite partial sums.

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