Holomorphic functions

Let $\mathbb{D}= \{z \in \mathbb{C}: |z|<1 \}$ be the complex unit disk and let $0<a<1$ be a real number. Suppose that $f:\mathbb{D} \rightarrow \mathbb{C}$ is a holomorphic function such that $f(a)=1$ and $f(-a)=-1$.

a) Prove that $\sup \limits_{z \in \mathbb{D}} \left| f(z) \right| \geq \frac{1}{a}$.
b) Prove that if $f$ has no root , then:

$$\sup_{z \in \mathbb{D}} \left|f(z)\right|\geq \exp \left( \frac{1-a^2}{4a} \pi \right)$$

(Schweitzer Competition, 2012)

Solution

a) Let $g(z)= \frac{f(z)-f(-z)}{2}, \; z \neq 0$ and $g(0)=f'(0)$.  This is also a holomorphic function satisfying $g(a)=\frac{1}{a}$. For $a<r<1$ by the triangle inequality and the maximal principal we have:

\begin{align*}
\sup_{z \in \mathbb{D}}\left | f(z) \right | &\geq \max_{\left | z \right |=r} \left | f(z) \right | \\
 &\geq  r \max_{\left | z \right |=r}\frac{\left | f(z) \right |+\left | f(-z) \right | }{2r} \\
 &\geq r \max_{\left | z \right |=r} \left | g(z) \right | \\
 &\geq r g(a) = \frac{r}{a}
\end{align*}

As $ r \rightarrow 1-0$ the result follows.

b) Let $M= \sup \limits_{z \in \mathbb{D}} \left|f(z)\right|$ . Since $f$ is non constant , $\left|f \right| < M$ everywhere in $\mathbb{D}$. In particular , from $f(a)=1$ we see that $M>1$. Since $f$ is non zero on the simply connected disk $\mathbb{D}$ it has a logarithm, that is there exists a holomorphic function $g$ such that $f(z)=\exp (g(z))$. Without loss of generality we can assume that $g(a)=0$. From $f(-a)=-1$ we conclude that $g(-a)=\kappa \pi i $ for some odd integer $\kappa$. Also from the fact that $\left| f \right| < M$ we get that $\mathfrak{Re}(g)< \log M$. Let $\mathbb{H}$ denote the half plane $\mathfrak{Re}(g)< \log M$ . Therefore $g$ is a $\mathbb{D} \rightarrow \mathbb{H}$ function.

Now, define the linear fractional transformations $\varphi:\mathbb{D} \rightarrow \mathbb{D}$ and $\varphi^{-1}:\mathbb{D} \rightarrow \mathbb{D}$ as:

$$\varphi(z)= \frac{z+a}{1+az} \quad \text{and} \quad \varphi^{-1}(z)= \frac{z-a}{1-az} $$

Also define the linear fractional transformation $\psi:\mathbb{H} \rightarrow \mathbb{D}$ as

$$\psi(z)= \frac{z}{2\log M - z}$$

Now we consider the $\mathbb{D} \rightarrow \mathbb{D}$ function $h=\psi \circ g \circ \varphi$. Since $\varphi(0)=a$, $g(a)=0$ and $\psi(0)=0$ we have that $h(0)=0$. We apply Schwartz's lemma on $h$ and the point $\varphi^{-1}(-a)=-\frac{2a}{1+a^2}$ thus:

$$ \left | h\left ( -\frac{2a}{1+a^2} \right ) \right |\leq \frac{2a}{a^2+1}$$

Hence:

\begin{align*}
\frac{2a}{a^2+1} &\geq \left | h\left ( \varphi^{-1}\left ( -a \right ) \right ) \right | \\
 &= \left | \psi \left ( g(-a) \right ) \right |\\
 &= \left | \frac{\kappa \pi i }{2\log M - \kappa \pi i } \right |\\
 &=\frac{1}{\sqrt{\left ( \frac{2 \log M}{|\kappa | \pi} \right )^2 +1}}
\end{align*}

So, finally,

$$ \log M \geq \frac{\left | \kappa  \right | \pi}{2}\sqrt{\left ( \frac{1+a^2}{2a} \right )^2-1}= \frac{|\kappa| \pi}{2}\cdot \frac{1-a^2}{2a} \geq \frac{1-a^2}{4a} \pi $$

Remark: The estimates in the problem are very sharp. For example we have equality for $f(z)=\frac{z}{a}$ in part a) and for $\displaystyle f(z)= -i \exp \left ( \frac{iz-a^2}{iz+1}\cdot \frac{\pi}{2a} \right )$ in part b).