# Pointwise lower bound for holomorphic functions on the unit disk

I am looking at Problem 8C in August 2012 complex analysis qualifying exam from University of Wisconsin-Madison (direct link to the file).

Let $f$ be a holomorphic function on the unit disc $\mathbb{D}$. Fix $z_0\in\mathbb{D}$. Suppose that $f(0) = \frac{1}{2}$, $f$ does not vanish on $\mathbb{D}$ and $|f(z)|\leqslant 1$. Show that $|f(z_0)| > c$ for some positive constant $c$ independent of $f$.

Here $\mathbb{D}$ denotes the open unit disk centred at the origin, as usual. The part "independent" of $f$ is the tricky part for me. So if I understand the problem correctly, suppose $z_0=i/2\in\mathbb{D}$. Then there is no way to find a sequence of holomorphic functions $f_{n}$ such that $f_{n}(0)=\frac{1}{2}$, $0<|f(z)|\leqslant 1$ for all $z\in\mathbb{D}$, and that $|f_{n}(i/2)|\to 0$ as $n\to\infty$.

I tried using Schwarz Lemma, but to no avail. I would appreciate any hints. :)

## 1 Answer

Hints:

1. Consider $\log f$, normalised to $\log f(0) = -\log 2 \in \mathbb{R}$.
2. Compose with a Möbius transformation.
3. Schwarz lemma.
4. Transform back to get a bound on $\lvert \log f(z_0)\rvert$.

Assuming the OP has meanwhile completed the proof, an elaboration won't harm:

Since $\mathbb{D}$ is simply connected and $f$ has no zero, holomorphic logarithms of $f$ exist. Let $g$ be such that $g(0) = \log\frac12$ is real. Since $\lvert f(z)\rvert < 1$ for $z\in\mathbb{D}$, we have $\operatorname{Re} g(z) < 0$ for $z\in\mathbb{D}$. The Möbius transformation

$$T(w) = \frac{w + \log 2}{w-\log 2}$$

maps the left half plane biholomorphically to the unit disk, and $T(\log\frac12) = 0$, so $h = T\circ g$ is a holomorphic function $\mathbb{D} \to\mathbb{D}$ with $h(0) = 0$. By the Schwarz lemma, we have $\lvert h(z_0)\rvert \leqslant \lvert z_0\rvert$. Then

$$g(z_0) = T^{-1}(h(z_0)) = \log 2\cdot \frac{h(z_0)+1}{h(z_0)-1}$$

can be bounded

$$\lvert g(z_0)\rvert \leqslant \log 2 \cdot \frac{1+\lvert z_0\rvert}{1-\lvert z_0\rvert}$$

and

$$\left\lvert f(z_0)\right\rvert = \left\lvert e^{g(z_0)}\right\rvert = e^{\operatorname{Re} g(z_0)} \geqslant e^{-\lvert g(z_0)\rvert} \geqslant \exp \left(-\log 2 \cdot \frac{1+\lvert z_0\rvert}{1-\rvert z_0\rvert}\right).$$

The bound is sharp, it is attained for $f = \exp \circ T^{-1} \circ \rho$, where $\rho$ is the rotation mapping $z_0$ to $\lvert z_0\rvert$.

• Thanks for impressively fast answer! (+1) – Prism Oct 27 '13 at 22:02