# 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.

# 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.