# The Phragmen-Lindelöf method

A friend told me some time ago about a family of theorems that extend the maximum principle for holomorphic functions to unbounded domains. I have read about it, and I want to record some results with their proof. My main reference is the book Real and Complex Analysis, by Walter Rudin (I use the second edition).

The maximum principle for holomorphic functions

In the following, $\Omega$ denotes an open set in the complex plane $\mathbb{C}$, $\overline{\Omega}$ denotes the closure of $\Omega$ in $\mathbb{C}$, and $\partial \Omega=\overline{\Omega}\setminus\Omega$ its boundary. I denote by $\mathcal{H}(\Omega)$ the set of all holomorphic functions on $\Omega$ and by $\mathcal{C}(\overline{\Omega})$ the set of all continuous functions on $\overline{\Omega}$. I will use the the notation $z=x+iy$,  where $x$ and $y$ are real numbers, the real and imaginary parts of $z$. Let us recall the maximum principle.

Theorem 1

Let us assume that $\Omega$ is bounded and that $f$ belongs to $\mathcal{H}(\Omega)\cap\mathcal{C}(\overline{\Omega})$. Then,

$\displaystyle \sup_{z \in \Omega}|f(z)|\le \max_{z \in \partial\Omega}|f(z)|.$

Furthermore, if there exists $z_0\in \Omega$ such that $|f(z_0)|=\max_{z \in \partial\Omega}|f(z)|$, the function $f$ is a constant.

For a proof, see for instance Chapters 10 and 12 of Rudin’s book. As Rudin points out, this principle does not hold if we do not assume that $\Omega$ is bounded. He gives the following counter example:

$\displaystyle \Omega=\{z\in \mathbb{C}\,;\,-\frac{\pi}{2}

and

$\displaystyle f(z)=\exp(\exp(z)).$

The function $f$ is entire, so we obviously have $f\in \mathcal{H}(\Omega)\cap\mathcal{C}(\overline{\Omega})$. We also have

$\displaystyle f\left(x\pm i\frac{\pi}{2}\right)=\exp\left(\pm i \exp(x)\right),$

and thus

$\displaystyle \max_{z \in \partial\Omega}|f(z)|=1.$

On the other hand,

$\displaystyle f(x)=\exp(\exp(x)),$

so that

$\displaystyle \lim_{x\to +\infty}f(x)=+\infty,$

and therefore

$\displaystyle \sup_{z \in \Omega}|f(z)|=+\infty.$

The maximum principle therefore does not hold in this example.

An example of the method

In this section, as before,

$\displaystyle\Omega=\{z\in \mathbb{C}\,;\,-\frac{\pi}{2}

In the counter-example of the preceding section, the modulus of the function $f$ grows very rapidly when $x$ tends to $+\infty$. We will now show that if we prevent $|f(z)|$ from growing too rapidly when $z$ tends to $\infty$, the maximum principle holds.

To simplify, our statement, let us consider a function $f\in \mathcal{H}(\Omega)\cap\mathcal{C}(\overline{\Omega})$ such that $\left|f\left(x\pm i\frac{\pi}{2}\right)\right|\le 1$ for any $x\in \mathbb{R}$. Informally, the Phragmen-Lindelöf result states that if we know a priori that $|f(z)|$ does not grow to rapidly  when $z$ tends to $\infty$, then in fact $|f(z)|\le 1$ for all $z\in \Omega$. Furthermore, if there exists $z_0$ in $\Omega$ such that $|f(z_0)|=1$, then $f$ is a constant.

Since the precise formulation will be a bit technical, I will first try to give an outline of the method. The basic idea is to apply the standard maximum principle on a bounded domain, and to use a family of auxiliary functions $(g_{\varepsilon})_{\varepsilon>0}$. The trick is to build $(g_{\varepsilon})_{\varepsilon>0}$ such that

1. $g_{\varepsilon}\in \mathcal{H}(\Omega)\cap\mathcal{C}(\overline{\Omega})$;
2. $\left|g_{\varepsilon}\left(x\pm i\frac{\pi}{2}\right)\right|\le1$;
3. $g_{\varepsilon}(z)$ tends to $1$ when $\varepsilon$ tends to $0$ for any $z \in \Omega$;
4. $g_{\varepsilon}(z)f(z)$ tends to $0$ when $z$ tends to $\infty$, which is allowed by the a priori condition on the growth of $f(z)$.

Then, we obtain the desired result by applying the maximum principle to the function $z\mapsto f(z)g_{\varepsilon}(z)$ on the bounded rectangle

$\displaystyle \Omega_{R}=\left\{z \in \mathbb{C}\,;\, -R

and by letting $\varepsilon$ tend to $0$ and $R$ tend to $+\infty$.

Let us now give a precise statement and a proof. It will in particular show that the counter-example quoted above is in some sense optimal.

Theorem 2

Let $f$ be a function in $\mathcal{H}(\Omega)\cap\mathcal{C}(\overline{\Omega})$ such that
$\displaystyle \left|f\left(x\pm i\frac{\pi}{2}\right)\right|\le 1$
for all $x\in \mathbb{R}$, and let us assume that there exist real constants $A$ and $\alpha <1$ such that
$\displaystyle |f(z)|\le \exp(A\exp(\alpha|x|))$
for all $z \in \Omega$. Then $|f(z)|\le 1$ for all $z \in \Omega$, and, if there exists $z_0\in \Omega$ such that $|f(z_0)|=1$, $f$ is a constant.

Proof. Let us first pick $z_0\in \Omega$. We want to prove that $|f(z_0)|\le 1$. In accordance with the method outlined above, let us choose some $\beta$ such that $\alpha <\beta$ and $\beta <1$. For all $\varepsilon>0$, let us set

$\displaystyle g_{\varepsilon}(z)=\exp(-\varepsilon\cosh(\beta z)).$

For the moment, the parameter $\varepsilon$ is fixed. The function $g_{\varepsilon}$ is entire and therefore belongs to $\mathcal{H}(\Omega)\cap\mathcal{C}(\overline{\Omega})$. Furthermore, we have

$\displaystyle|g_{\varepsilon}(z)|=\exp(-\varepsilon\mbox{Re}(\cosh(\beta z)))=\exp(-\varepsilon\cosh(\beta x)\cos(\beta y)).$

This implies

$\displaystyle|g_{\varepsilon}(z)|\le \exp(-\delta \exp(\beta|x|))$

with

$\displaystyle\delta=\frac{\varepsilon}{2}\cos\left(\frac{\beta\pi}{2}\right)>0.$

This implies that

$\displaystyle\left|f\left(x\pm i\frac{\pi}{2}\right)g_{\varepsilon}\left(x\pm i\frac{\pi}{2}\right)\right|\le 1$

for all $x \in \mathbb{R}$, and that

$\displaystyle|f(z)g_{\varepsilon}(z)|\le \exp\left(A\exp(\alpha|x|)-\delta\exp(\beta|x|)\right)$

for all $z \in \Omega$.
Let us now chose $R>0$ large enough so that $R> |z_0|$ and

$\displaystyle \exp\left(A\exp(\alpha R)-\delta\exp(\beta R)\right)\le 1.$

Such an $R$ exists since $\beta>\alpha$. Of course, it will depend on $\varepsilon$. The bounded rectangle $\Omega_{R}$ is defined as above. The function $z \mapsto f(z)g_{\varepsilon}(z)$ is in $\mathcal{H}(\Omega_{R})\cap\mathcal{C}(\overline{\Omega_{R}})$ and $|f(z)g_{\varepsilon}(z)|\le 1$ for all $z$ in $\partial \Omega_{R}$, the boundary of $\Omega_{R}$. According to the maximum principle (Theorem 1), since $z_0\in \Omega_{R}$, $|f(z_0)g_{\varepsilon}(z_0)|\le 1$. This last inequality has been proved for an arbitrary $\varepsilon$, and $f_{\varepsilon}(z_0)\to 1$ when $\varepsilon \to 0$, therefore $|f(z_0)|\le 1$. This allows us to conclude that $|f(z)|\le 1$ for all $z \in \Omega$.

Let us now assume that there exists $z_0\in \Omega$ such that $|f(z_0)|=1$. Let us pick some $R>0$ such that $R>|x_0|$. Using our previous notation we have $z_0\in \Omega_{R}$, and, since $|f(z)|\le 1$ for all $z\in \overline{\Omega}$, we have in particular $|f(z)|\le 1$ for all $z\in \partial \Omega_R$. According to the maximum principle, this implies that $f$ is a constant on $\Omega_{R}$. Since $\Omega_{R}$ is an open set in $\Omega$, the principle of analytic continuation tells us that $f$ is a constant on $\Omega$. This concludes the proof.

Let us note if $\alpha\ge 1$, the theorem is false, as shown by the counter-example of the first section. The method is obviously very flexible. With little change in the proof, we can obtain results dealing with a domain that is bounded on one side, or with a function that satisfies the inequality only for some values of the modulus. To be more explicit, I will state two results, but we can imagine much more. In the following, we consider $\Omega'$, the open half-strip defined by

$\displaystyle \Omega'=\{z\in \mathbb{C}\,;\,0

Proposition 3

Let $f$ be a function in $\mathcal{H}(\Omega')\cap\mathcal{C}(\overline{\Omega'})$ such that
$|f(z)|\le 1$ for all $z \in \partial \Omega'$,
and let us assume that there exist real constants $A$ and $\alpha<1$ such that

$\displaystyle |f(z)|\le \exp(A\exp(\alpha x))$

for all $z \in \Omega'$. Then $|f(z)|\le 1$ for all $z \in \Omega$, and, if there exists $z_0\in \Omega'$ such that $|f(z_0)|=1$, $f$ is a constant.

Proposition 4

Let $f$ be a function in $\mathcal{H}(\Omega')\cap\mathcal{C}(\overline{\Omega'})$ such that $|f(z)|\le 1$ for all $z \in \partial \Omega'$, and let us assume that there exist real constants $A$ and $\alpha<1$, and a sequence $(R_n)$ of positive numbers tending to $+\infty$ such that, for all $n \in \mathbb{N}$,

$\displaystyle |f(z)|\le \exp(A\exp(\alpha R_n))$

for all $z \in \Omega'$ such that $x=R_n$. Then $|f(z)|\le 1$ for all $z \in \Omega$, and, if there exists $z_0\in \Omega'$ such that $|f(z_0)|=1$, $f$ is a constant.