The following problem comes from the Graduate Qualifying Exam in Complex analysis from Texas A&M:
Let $F$ be a function holomorphic and bounded in the upper half-plane $\mathbb{C}_+$. Suppose that $F$ has period 1, i.e., $F(z+1)=F(z)$ for all $z\in\mathbb{C}_+$. Prove that $F(z)$ has a finite limit as $\Im(z)\to\infty$.
So far, I've been able to ascertain the following:
- $F$ is bounded, which implies that its maximum occurs at its boundary, $\partial\mathbb{C}_+$. If its maximum occurs at $\infty$, then we are done, since then it will be approaching its maximum, a finite number. But if it attains its maximum on the "$x$-axis", we might not be able to say as much.
- If we have a sequence $\{z_n\}\to\infty$ in $\mathbb{C}_+$, it suffices to consider only $z_n$ in the strip $\{z\in\mathbb{C}_+\,\,:\,\,0\leq \Re(z)<1\}$, since $F$ is periodic.
Any ideas? Phragmen Lindelof vaguely came to mind.
First solution. Let me expand my comments to an answer.
Let $\mathbb{H}$ be the upper-half plane and $\mathbb{D}^{*} = \mathbb{D}\setminus\{0\} $ be the punctuated unit disc. Then we define the function $G : \mathbb{D}^* \to \mathbb{C}$ by the following relation
$$ \forall z \in \mathbb{H} \ : \qquad F(z) = G(e^{2\pi i z}) $$
We claim that $G$ is well-defined and holomorphic.
Well-definedness. For each $w \in \mathbb{D}^*$, let $z, z' \in \mathbb{H}$ satisfy $w = e^{2\pi i z} = e^{2\pi i z'}$. We know that this implies $z' = z + n$ for some integer $n$. Since $F$ has period $1$, we have $F(z') = F(z)$. Therefore the value of $G(w)$ is unambiguously defined.
Holomorphy. For each $w_0 \in \mathbb{D}^*$, fix $z_0 \in \mathbb{H}$ such that $w_0 = e^{2\pi i z_0}$. We consider the function
$$\psi(w) = z_0 + \frac{1}{2\pi i}\log(w/w_0),$$
where $\log$ is the principal logarithm. Then there is a neighborhood $U \subset \mathbb{D}^*$ of $w_0$ such that $\psi : U \to \mathbb{H}$ is holomorphic. Now we find that
$$ F(\psi(w)) = G(e^{2\pi i \psi(w)}) = G(e^{2\pi i z_0 + \log(w/w_0)}) = G(w),$$
hence it follows that $G$ is holomorphic on $U$ as well. Finally, since $G$ is holomorphic at any $w_0 \in \mathbb{D}^*$, the claim follows.
Now once we have $G$ at our hands, the proof is almost done. Since $G$ is holomorphic and bounded on $\mathbb{D}^*$, the origin $0$ is a removable singularity and hence $G$ extends to a holomorphic function on the unit disc $\mathbb{D}$. Therefore
$$ \lim_{\operatorname{Im}(z) \to \infty} F(z) \stackrel{w=e^{2\pi i z}}{=} \lim_{w \to 0} G(w) $$
converges and the claim follows.
Second solution - by brutal force computation. Fix $\epsilon > 0$ and let $M$ be a bound of $F$. If $\Gamma_n = \{ne^{i\theta} + i\epsilon : 0 \leq \theta \leq \pi\}$ denotes the upper-semicircular arc of radius $n$ at $i\epsilon$, clockwise oriented, then
$$ \left| \int_{\Gamma_n} \frac{F(\xi)}{\xi - i\epsilon} \, d\xi \right| \leq \pi M. $$
So by compactness, we can pick a subsequence $n_k$ and $\alpha \in \mathbb{C}$ such that
$$ \int_{\Gamma_{n_k}} \frac{F(\xi)}{\xi - i\epsilon} \, d\xi \xrightarrow{k\to\infty} -i\pi\alpha $$
We remark that $\alpha$ is actually the limit of $F(z)$ as $\operatorname{Im}(z) \to \infty$, though this is not required for our proof.
Now let $z \in \mathbb{C}$ with $\operatorname{Im}(z) > \epsilon$ and assume that $k$ is sufficiently large that $|z| < |n_k|$. If $L_{n_k}$ denotes the line segment from $-n_k+i\epsilon$ to $n_k+i\epsilon$, then by the residue theorem we have
We want to take limit as $k\to\infty$ and simplify both sides. To this end, we make some observations:
It is easy to check that
$$ \int_{\Gamma_{n}} \frac{F(\xi)}{\xi - z} \, d\xi = \int_{\Gamma_{n}} \frac{F(\xi)}{\xi - i\epsilon} \, d\xi + \mathcal{O}(n^{-1}) \quad \text{as} \quad n \to \infty. $$
So the RHS of $\text{(1)}$ converges to $2\pi i F(x) - i\pi \alpha$ as $k\to\infty$.
Utilizing the periodicity of $F$, we find that the LHS of $\text{(1)}$ is written as
$$ \text{[LHS of (1)]} = \int_{L_{n_k}} \frac{F(\xi)}{\xi - z} \, d\xi = \int_{i\epsilon}^{1+i\epsilon} \left( \sum_{n = -n_k}^{n_k - 1} \frac{1}{\xi + n - z} \right) F(\xi) \, d\xi. $$
By symmetrizing the summation, we can write
\begin{align*} \sum_{n = -n_k}^{n_k - 1} \frac{1}{\xi + n - z} &= -\frac{1}{\xi + n_k - z} + \frac{1}{2}\sum_{n = -n_k}^{n_k} \left( \frac{1}{\xi + n - z} + \frac{1}{\xi - n - z} \right) \\ &\hspace{5em} \xrightarrow{k\to\infty} \frac{1}{2}\sum_{n = -\infty}^{\infty} \left( \frac{1}{\xi + n - z} + \frac{1}{\xi - n - z} \right) =: K(\xi - z), \end{align*}
where the convergence is uniform in $\xi \in [0, 1]$.
Combining altogether, taking limit as $k\to\infty$ to $\text{(1)}$ yields the following identity
Now we claim that $K(\xi - z) \to i\pi$ as $\operatorname{Im}(z) \to \infty$ uniformly in $\xi$. Since $K$ has period $1$, it suffices to show that $K(x-iy) \to i\pi$ as $y\to\infty$ uniformly in $x \in [0, 1]$. We begin by writing
\begin{align*} K(x-iy) &= \frac{1}{2}\sum_{n = -\infty}^{\infty} \left( \frac{1}{x - iy + n} + \frac{1}{x - iy - n} \right) \\ &= i \sum_{n = -\infty}^{\infty} \frac{y+ix}{y^2 + n^2 - x^2 + 2ixy} \\ &= i \sum_{n = -\infty}^{\infty} \frac{1+\frac{ix}{y}}{1 + (\frac{n}{y})^2 - \frac{x^2}{y^2} + \frac{2ix}{y}} \cdot \frac{1}{y} \end{align*}
Now it is a tedious computation to check that
$$ \frac{1+\frac{ix}{y}}{1 + (\frac{n}{y})^2 - \frac{x^2}{y^2} + \frac{2ix}{y}} \cdot \frac{1}{y} = (1 + \mathcal{O}(y^{-1})) \int_{\frac{n}{y}}^{\frac{n+1}{y}} \frac{dt}{1+t^2} $$
uniformly in $n$ and $x$. So it follows that
$$ K(x - iy) \xrightarrow{y\to\infty} i \int_{-\infty}^{\infty} \frac{dt}{1+t^2} = i\pi $$
uniformly in $x$ as desired. (Alternatively, you may utilize the identity $K(z) = \pi \cot (\pi z)$ and prove this claim by pure computation.) Applying this to $\text{(2)}$, we finally obtain the desired conclusion:
(Here, $\epsilon > 0$ was a fixed but otherwise arbitrarily chosen number. So we can even take limit as $\epsilon \to \infty$ to find that the limit above is in fact $\alpha$.)