# Conformal mappings and the Phragmen-Lindelöf Theorem

I would like to go back to the Phragmen-Lindelöf theorem that I presented in a previous post. Let us recall the result. In the following, we write, for all $z\in \mathbb{C}$, $zx+iy$, with $x$ and $y$ real numbers. The open set $\Omega$ is defined as

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

$\overline{\Omega}$ denotes the closure of $\Omega$ in $\mathbb{C}$, and $\partial \Omega=\overline{\Omega}\setminus\Omega$ its boundary. I write $\mathcal{H}(\Omega)$ for the set of all holomorphic functions on $\Omega$ and $\mathcal{C}(\overline{\Omega})$ for the set of all continuous functions on $\overline{\Omega}$.

Theorem (Phragem-Lindelöf)

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.

The Phragmen-Lindelöf  method can be adapted to prove results of this type on various domains by constructing suitable families of functions $(g_{\varepsilon})_{\varepsilon>0}$ (see this same previous post for context). There is however another way to obtain a similar result for another domain. If we can find an holomorphic change of variable, that is to say a conformal mapping, that maps the domain $\Omega$ in the theorem to the domain that we are considering, we obtain the Phragmen-Lindelöf result on this last domain. Of course, the growth condition will be modified by the mapping. Let us give several examples. In the following, the original complex variable will be denoted by $z=x+iy$ as before and the new variable by $w=u+iv$.

General horizontal strip

Let $a$ and $b$ be real numbers such that $a and let $\Omega(a,b)$ be the strip

$\{z\in \mathbb{C}\,;\,a

Proposition 1

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

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

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

Proof. Let us define $\varphi:\mathbb{C}\to \mathbb{C}$ by

$\displaystyle \varphi(w)=\frac{b-a}{\pi}w+i\frac{a+b}{2}.$

It is obviously an holomorphic change of variable (it is even linear). We have $\varphi(\Omega)=\Omega(a,b).$ Let us write $g=f\circ \varphi$. The function $g$ is in $\mathcal{H}(\Omega)\cap\mathcal{C}(\overline{\Omega})$, and $|g(w)|\le 1$ for all $w \in \partial \Omega$. Let us now consider $w\in \Omega$ and $z=\varphi(w)$. We have, for the real part of $z$,

$\displaystyle x=\frac{b-a}{\pi}u.$

Since we have

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

we obtain

$\displaystyle |g(w)|\le \exp\left(A\exp\left(\frac{\alpha(b-a)}{\pi}|u|\right)\right).$

Since $\frac{\alpha(b-a)}{\pi}<1$, the Phragmen-Lindelöf Theorem yields the desired result. QED.

General vertical strip

Let $a$ and $b$ be real numbers such that $a, and let $\Pi(a,b)$ be the strip

$\{z\in \mathbb{C}\,;\,a

Proposition 2
Let $f$ be a function in $\mathcal{H}(\Pi(a,b))\cap\mathcal{C}(\overline{\Pi(a,b)})$ such that $|f(z)|\le 1$ for $z\in \partial \Pi(a,b)$, and let us assume that there exist real constants $A$ and $\alpha<\frac{\pi}{b-a}$ such that

$\displaystyle |f(z)|\le \exp\left(A\exp\left(\alpha|y|\right)\right)$

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

Proof. We define $\varphi:\mathbb{C}\to \mathbb{C}$ by $\varphi(w)=-iw$. The function $\varphi$ is an holomorphic change of variable that maps $\Omega(a,b)$ to $\Pi(a,b)$. The function $g=f\circ \varphi$ satisfies the hypotheses of Proposition 1, which yields the desired result. QED.

Sector

Let $\theta_1$ be a number in $]-\pi,\pi]$ and $\theta_2$ another number such that $0<\theta_2-\theta_1<2\pi$.
We define the open sector $\Sigma(\theta_1,\theta_2)$ by

$\displaystyle \Sigma(\theta_1,\theta_2)=\left\{re^{i\theta}\,;\,r>0\mbox{ and } \theta_1<\theta<\theta_2\right\},$

and we set

$\displaystyle \Sigma'(\theta_1,\theta_2)=\left\{re^{i\theta}\,;\,r>0\mbox{ and } \theta_1\le\theta\le\theta_2\right\},$

the closure of $\Sigma(\theta_1,\theta_2)$ with the origin removed.

Proposition 3

Let $f$ be a function in $\mathcal{H}(\Sigma(\theta_1,\theta_2))\cap\mathcal{C}(\Sigma'(\theta_1,\theta_2))$ such that $|f\left(re^{i\theta_1}\right)|\le 1$ and $|f\left(re^{i\theta_2}\right)|\le 1$ for all $r>0$. Let us assume that there exist real constants $A$ and $\alpha<\frac{\pi}{\theta_2-\theta_1}$ such that

$\displaystyle |f(z)|\le \exp\left(A|z|^{\alpha}\right)$

if $z \in \Sigma(\theta_1,\theta_2)$ with $|z|\ge 1$ and

$\displaystyle |f(z)|\le \exp\left(\frac{A}{|z|^{\alpha}}\right)$

if $z \in \Sigma(\theta_1,\theta_2)$ with $|z|\le 1$.
Then $|f(z)|\le 1$ for all $z\in \Sigma(\theta_1,\theta_2)$, and, if there exists $z_0\in \Sigma(\theta_1,\theta_2)$ such that $|f(z_0)|=1$, $f$ is a constant.

Let us note that this proposition implies the following weaker statement, which is often referred to as the Phragmen-Lindelöf principle.

Corollary

Let $f$ be a function in $\mathcal{H}(\Sigma(\theta_1,\theta_2))\cap\mathcal{C}(\overline{\Sigma(\theta_1,\theta_2)})$ such that $|f\left(re^{i\theta_1}\right)|\le 1$ and $|f\left(re^{i\theta_2}\right)|\le 1$ for all $r>0$. Let us assume that there exist real constants $A$ and $\alpha<\frac{\pi}{\theta_2-\theta_1}$ such that

$\displaystyle |f(z)|\le \exp\left(A|z|^{\alpha}\right)$

for all $z \in \Sigma(\theta_1,\theta_2)$.
Then $|f(z)|\le 1$ for all $z\in \Sigma(\theta_1,\theta_2)$, and, if there exists $z_0\in \Sigma(\theta_1,\theta_2)$ such that $|f(z_0)|=1$, $f$ is a constant.

Proof. Since $0<\theta_2-\theta_1<2\pi$, the exponential function is a bijection from $\Omega(\theta_1,\theta_2)$ to $\Sigma(\theta_1,\theta_2)$, and furthermore $\exp\left(\overline{\Omega(\theta_1,\theta_2)}\right)=\Sigma'(\theta_1,\theta_2)$. Let us consider $w\in \Omega(\theta_1,\theta_2)$ and $z=\exp(w)$. We have
$|z|=e^u$.

If $u\ge 0$, $|z|\ge 1$, and since

$\displaystyle |f(z)|\le \exp\left(A|z|^{\alpha}\right),$

we obtain

$\displaystyle |g(w)|\le \exp(A\exp(\alpha u)).$

If $u\le 0$, $|z|\le 1$, and since

$\displaystyle |f(z)|\le \exp\left(A|z|^{-\alpha}\right),$

we obtain

$\displaystyle |g(w)|\le \exp(A\exp(-\alpha u)).$

In both cases, we have

$\displaystyle |g(w)|\le \exp(A\exp(\alpha|u|)),$

and we can apply Proposition 2. QED.

One last remark: in my previous post, I presented several results which give a precised form of the maximum principle. This allows us to see which part of the boundary has the most weight in controlling the modulus at a given interior point. I stated the results for bounded holomorphic functions, but they are not limited to them. Indeed, we can <em<first apply one of the results in this post to show that an holomorphic function that does not grow to fast at infinity is bounded, and then apply the corresponding result in the previous post.

# Three-curves theorems

This post deals with a family of theorems that precise and extend the usual maximum modulus principle for analytic functions. Contrary to my previous post on the Phragmen-Lindelöf principle, we are less interested in the fact that the maximal modulus principle is extended to unbounded domains (although this is indeed the case in several of the following examples) than in giving a quantitative form, which will allow us to say that the parts of the boundary that are closer create a stronger constraint on the modulus of the function. In the following, $z=x+iy$ denotes a complex number, with $x$ and $y$ two real numbers, its real and imaginary part. I use as a reference Real and complex analysis, second edition, by Walter Rudin.

The three-lines theorem

Let $\Pi$ be the open vertical strip in the complex plane defined by

$\displaystyle \Pi=\left\{z \in \mathbb{C}\,;\, 0

Theorem 1

Let $f$ be a function that is holomorphic in $\Pi$, continuous on $\overline{\Pi}$ and bounded on $\overline{\Pi}.$ For $x\in [0,1]$, we set

$\displaystyle M(x)=\sup_{y\in \mathbb{R}} |f(x+iy)|.$

Then

$\displaystyle M(x)\le M(0)^{1-x}M(1)^x.$

Let us note that Theorem 1 states that $x \mapsto \log(M(x))$ is a convex function on $[0,1]$. To prove this, we will use the following Lemma.

Lemma 1

Let $f$ be a function that is holomorphic in $\Pi$, continuous on $\overline{\Pi}$ and bounded on $\overline{\Pi}.$ If $|f(z)|\le 1$ for all $z \in \partial\Pi$, then $|f(z)|\le 1$ for all $z \in \Pi$.

Proof. The above Lemma is a direct consequence of a Phragmen-Lindelöf type result on the domain $\Pi$. However, the hypothesis that $f$ is bounded allows us to give a simpler proof, taken from Rudin (Theorem 12.8).

Let us consider $z_0\in \Pi$. For any $\varepsilon>0$, we define the function

$\displaystyle g_{\varepsilon}(z)=\frac{1}{1+\varepsilon z}.$

For all $z\in \Pi$, $Re(1+\varepsilon z)=1+\varepsilon x>1$, therefore $|1+\varepsilon z|>1$ and $|g_{\varepsilon}(z)|<1$. On the other hand, for all $z \in \Pi$, $\mbox{Im}(1+\varepsilon z)=\varepsilon y$, and thus
$|g_{\varepsilon}(y)|\le \frac{1}{\varepsilon |y|}$.
We chose $R$ such that $|\mbox{Im}(z_0)| and $R\ge \frac{1}{\varepsilon }$. Let us consider the open rectangle $\Pi_{R}$ defined by

$\displaystyle \Pi_{R}=\left\{z\in \mathbb{C}\,;\,0

The function $z \to f(z)g_{\varepsilon}(z)$ is holomorphic in $\Pi_{R}$, continuous on $\Pi_{R}$, and, for all $z \in \partial \Pi_{R}$, $|f(z)g_{\varepsilon}(z)|\le 1$. Since $z_0$ is in $\Pi_{R}$, we have, according to the maximum modulus principle, $|f(z_0)g_{\varepsilon}(z_0)|\le 1$. The inequality holds for arbitrary $\varepsilon>0$, and $g_{\varepsilon}(z_0)$ tends to $1$ as $\varepsilon$ tends to $0$. Therefore, we obtain $|f(z_0)|\le 1$. QED.

Let us now prove the general case.

First, let us treat the case when $M(0)=0$, that is to say when $f$ is identically $0$ on the imaginary axis. We will show that in this case, $f$ is identically $0$ on $\Pi$, and the inequality is then trivial (by the way, this is an answer to Exercice 7, Chapter 12 in Rudin).

Let us consider the function $\varphi:\mathbf{C} \to \mathbf{C}$ defined by $\varphi(w)=-iw$. We denote by $\Omega^{+}$ the horizontal strip defined by

$\displaystyle \Omega^{+}=\left\{z \in \mathbb{C}\,;\, 0

We have $\varphi(\Omega^{+})=\Pi$. We now set $g=f\circ \varphi$. The function $g$ is continuous on $\overline{\Omega^{+}}$, holomorphic in $\Omega$, and is identically $0$ on the real axis. According to the Schwarz reflexion principle (see for instance Rudin, Theorem 11.17), there exists a function $G$, holomorphic on the strip

$\Omega=\left\{z \in \mathbb{C}\,;\, -1

such that $G(z)=g(z)$ for all $z \in \Omega^{+}$. We know (by taking the limit of $G(z)$ as $\mbox{Im(z)}$ tends to $0$) that $G$ is identically $0$ on the real axis. By the isolated zeros theorem, $G$ is identically $0$ on $\Omega$, and in particular $g$ is identically $0$ on $\Omega^{+}$, and thus $f$ is identically $0$ on $\Pi$. If $M(1)=0$, we obtain that $f$ is identically $0$ on $\Pi$ by a symmetry argument: considering $z\mapsto g(1-z)$ brings us back to the previous case.

We now assume that $M(0)>0$ and $M(1)>0$. We set $g(z)=M(0)^{1-z}M(1)^{z}$. The function $g$ is entire. We have

$\displaystyle \left|M(0)^{1-z}M(1)^{z}\right|=M(0)^{1-x}M(1)^{x}$

for all $z \in \mathbb{C}$, and in particular $|g(z)|\ge \min(M(0),M(1))>0$. This implies that $g$ has no zero and that $\frac{1}{g}$ is bounded. Furthermore, for all $y \in \mathbb{R}$, $|g(iy)|=M(0)$ and $|g(1+iy)|=M(1)$. This implies that $|f(z)/g(z)|\le 1$ for all $z \in \partial \Pi$. According to Lemma 2, we have

$\displaystyle \frac{|f(z)|}{|g(z)|}\le 1$

for all $z \in \Pi$. Thus if $x \in ]0,1[$, we have, for all $y \in \mathbb{R}$,

$\displaystyle |f(x+iy)|\le M(0)^{1-x}M(1)^{x}.$

This gives us the desired result.

From Theorem 1, we obtain a more general three-line theorem simply by using a linear change of variable (actually, we already used one that to apply Schwartz reflexion principle). Let $a$ and $b$ be real numbers with $a and let $\Pi(a,b)$ be the open vertical strip defined by

$\displaystyle \Pi(a,b)=\left\{z\in \mathbb{C}\,;\, a

Theorem 2

Let $f$ be a function that is holomorphic in $\Pi(a,b)$, continuous on $\overline{\Pi(a,b)}$ and bounded on $\overline{\Pi(a,b)}$. For $x\in [a,b]$, we set

$\displaystyle M(x)=\sup_{y\in\mathbb{R}}|f(x+iy)|.$

We have

$\displaystyle M(x)\le M(a)^{\frac{b-x}{b-a }}M(b)^{\frac{x-a}{b-a}}.$

Proof. We define $\varphi(w)=(b-a)w+a$. We have $\varphi(\Pi)=\Pi(a,b)$. We set $g=f\circ \varphi$ and apply Theorem 1 to $g$. QED.

In the same way, we can formulate and prove a theorem for horizontal strips. Let $a$ and $b$ be real numbers with $a and let $\Omega(a,b)$ be the open horizontal strip defined by

$\displaystyle \Omega(a,b)=\left\{z\in \mathbb{C}\,;\, a

Theorem 3

Let $f$ be a function that is holomorphic in $\Omega(a,b)$, continuous on $\overline{\Omega(a,b)}$ and bounded on $\overline{\Omega(a,b)}$. For $y\in [a,b]$, we set

$\displaystyle M(y)=\sup_{x\in\mathbb{R}}|f(x+iy)|.$

We have

$\displaystyle M(y)\le M(a)^{\frac{b-y}{b-a }}M(b)^{\frac{y-a}{b-a}}.$

Proof. We use the change of variable $z=iw$ and apply Theorem 2. QED.

The three-circles theorem

The theorem is due to Jacques Hadamard. There exist several proofs, here we will deduce it from Theorem 2 (see Rudin, Chapter 12, Exercise 8).

We consider $0 and we denote by $A(r_1,r_2)$ the open annulus defined by

$\displaystyle A(r_1,r_2)=\{z\in \mathbb{C}\,;\,r_1<|z|

Theorem 4

Let $f$ be a function that is holomorphic in $A(r_1,r_2)$ and continuous on $\overline{A(r_1,r_2)}$. For $r \in [r_1,r_2]$, we set

$\displaystyle M(r)=\max_{\theta\in]-\pi,\pi]}|f(re^{i\theta})|.$

Then, we have

$\displaystyle M(r)=M(r_1)^{\frac{\log(r_2)-\log(r)}{\log(r_2)-\log(r_1)}}M(r_2)^{\frac{\log(r)-\log(r_1)}{\log(r_2)-\log(r_1)}}.$

Let us note that in this case, $f$ is clearly bounded on $\overline{A(r_1,r_2)}$, since it is continuous and $\overline{A(r_1,r_2)}$ is compact. Furthermore, the maximum modulus principle for an holomorphic function on a bounded domain tells us that the maximum of $f$ is reached on one of the boundary circles $\{z\,;\,|z|=r_1\}$ and $\{z\,;\,|z|=r_2\}$. The improvement resides in the fact that Theorem 4 is quantitative: it tells us for instance that if we are closer to the inner circle, the maximum of the modulus of $f$ on this circle has more weight in controlling the modulus of $f$.

Let us also note that Theorem 4 can be expressed as the fact that $M(r)$ is a convex function of $\log(r)$. Indeed, it was first stated in this form by Hadamard.

Proof. The exponential function maps the open vertical strip $\Pi(\log(r_1),\log(r_2))$ to $A(r_1,r_2)$ and its closure $\overline{\Pi(\log(r_1),\log(r_2))}$ to $\overline{A(r_1,r_2)}$ (the mapping is not one-to-one, but it doesn’t matter here). We set $g=f\circ \exp$ and apply Theorem 2 to $g$. QED.

The three-rays theorem

I haven’t found this last result stated anywhere in quite this way, but it a straightforward transposition of Theorem 3. Let $\theta_1$ be a number in $]-\pi,\pi]$ and $\theta_2$ another number such that $0<\theta_2-\theta_1<2\pi$.
We define the open sector $\Sigma(\theta_1,\theta_2)$ by

$\displaystyle \Sigma(\theta_1,\theta_2)=\left\{re^{i\theta}\,;\,r>0\mbox{ and } \theta_1<\theta<\theta_2\right\}.$

Theorem 5

Let $f$ be a function that is holomorphic in $(\Sigma(\theta_1,\theta_2)$, continuous and bounded on $\overline{\Sigma(\theta_1,\theta_2)}$. For $\theta\in[\theta_1,\theta_2]$, we set

$\displaystyle M(\theta)=\sup_{r>0}|f(re^{i\theta})|.$

Then, we have

$\displaystyle M(\theta)\le M(\theta_1)^{\frac{\theta_2-\theta}{\theta_2-\theta_1}}M(\theta_2)^{\frac{\theta-\theta_1}{\theta_2-\theta_1}}.$

Proof. The exponential function sends the horizontal strip $\Omega(\theta_1,\theta_2)$ to the open sector $\Sigma(\theta_1,\theta_2)$. The closure $\overline{\Omega(\theta_1,\theta_2)}$ of this strip is sent to the closure of the sector minus the origin. We set $g=f\circ \exp$ and apply Theorem 3 to $g$. QED.

# 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.

# The fondamental theorem of algebra

Ok, I’ve not quite given up yet. I’ll keep doing this blog a little bit more and see if something comes out of it. Since I don’t have anything special to say at this point, I’ll post about something quite simple and well known. It is a proof of the fundamental theorem of algebra that uses  only freshman calculus. Actually I learned about it during my second year of prep school, but it is really quite simple, an doesn’t use any complex analysis. Let me recall the result.

Fondamental theorem of algebra

Let $P(X)=X^d+a_{d-1}X^{d-1}+\dots+a_1 X+a_0$ be a polynomial with $d$ a non negative integer, $a_i$ complex numbers.
Then there exist $z \in \mathbb{C}$ such that $P(z)=0$.

The proof goes like this. You consider the function $z \mapsto |P(z)|$ on the complex plane. As the fact that $\mathbb{C}$ is a field does not play any role at this point, you can identify $\mathbb{C}$ to $\mathbb{R}^2$. You prove that this fuction admits a global minimum. You then prove that for any complex number $z_0$ such that $P(z_0)\neq 0$, there exists a neighbouring number $z_1$ such that $|P(z_1)|<|P(z_0)|$. But this means that the global minimum that we found must be a zero of $P$, and that is the expected result.

Step one

The fact that $P$ has a global minimum is a consequence of the following lemma

Lemma 1

Let $f: \mathbb{R}^n \mapsto \mathbb{R}$ be a continuous function such that $\lim_{\|x\|\to +\infty}f(x)=+\infty$. Then $f$ has a global minimum.

Proof: There is an $R \ge 0$ such that for all $\|x\| > R$, $f(x) \ge f(0)$. Now $f$ has a minimun on the compact set $\{x : \|x\|\le R\}$ (the closed ball of radius $R$). Let $x_0$ be a point where the minimal value is reached. For any point $x \in \mathbb{R}^n$, either $\|x\|>R$, in which case $f(x_0)\le f(0)\le f(x)$, or $\|x\|\le R$, and then $f(x_0)\le f(x)$. Either way, $f(x_0) \le f(x)$, and therefore $x_0$ is a point of global minimum.

Now to apply this lemma, we need only to check that $\lim_{|z|\to +\infty}|P(z)|=+\infty$, and this is made obvious by the reversed triangle inequality :

$\displaystyle |P(z)| \ge |z|^d-|a_{d-1}| |z|^{d-1}- \dots -|a_1| |z| - |a_0|.$

Step 2

This part is the closest in spirit to complex analysis, but since we deal with polynomials we won’t need the whole theory. We will prove a version of the minimum principle.

Lemma 2

Let $P$ be a polynomial that satisfies the hypothesis of the fondamental theorem (with complex coefficients and not a constant), and $z_0$ a complex number such that $P(z_0)\neq 0$. Then there exists a complex number $z_1$ such that $|P(z_1)|<|P(z_0)|$.

Proof: If we divide by $P(z_0)$, we only need to study the case where $P(z_0)=1$. By the usual factorisation argument, there is an integer $m \ge 1$ and a polynomial $Q$ with $Q(z_0)\neq 0$ such that for any $z$,

$\displaystyle P(z)=1+(z-z_0)^m Q(z).$

We will now use the polar representation to find $z_1$ such that $|P(z_1)|<1$. By continuity of $Q$, for any given $\varepsilon >0$, there is an $r>0$ such that for any $|z-z_0|\le r$, $|Q(z)-Q(z_0)|\le \varepsilon$. We now set $z=z_0+=re^{\i \theta}$, and we choose $\theta$ to get

$\displaystyle (z-z_0)^mQ(z)=r^me^{\i m \theta}Q(z_0)=-r^m|R(z_0)|$

(which of course can be done). Then by the triangle inequality,

$\displaystyle |P(z)| \le 1-r^m|Q(z_0)|+\varepsilon .$

If, at the beginning of this argument , we choose $\varepsilon$ to be $1/2r^m|Q(z_0)|$, we get

$\displaystyle |P(z)| \le 1-1/2 r^m |Q(z_0)|.$

This ends the proof of the second lemma and of the theorem.