# Qualifying exam problem (complex analysis)

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?

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

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.

• 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)? Feb 14, 2023 at 0:10
• @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 $F(q(z)) = f(z)$ Feb 15, 2023 at 7:10
• 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. Feb 15, 2023 at 7:12
• The only "theorem" used here is the holomorphic inverse function theorem. Feb 15, 2023 at 7:16
• Where are you using the hypothesis that $f$ is periodic? 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 $$f(x)=a$$