# 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.