Show that $\sup_{f\in\mathcal{B}}|f'(0)| = \frac{2}{e}$, where $f$ is analytic on the unit disc.

Question

I found the following problem:

Let $\mathcal{B}$ the set of complex analytic functions from the unit disc $D$ to itself that vanish nowhere. Show that $\sup_{f\in\mathcal{B}}|f'(0)| = \frac{2}{e}$.

First of all, I tried to use the Cauchy's Integral Formula to obtain some upper bound to $|f'(0)|$.

Let $f\in \mathcal B$ and $R\in(0,1)$. Note that for all $z\in\mathbb{C}$ with $|z|=R$ we have that $$\left| \frac{f(z)}{z^2} \right| < \frac{1}{R^2}.$$ Applying the ML inequality, we get $$|f'(0)| = \frac{1}{2\pi} \left| \int_{|z|=R} \frac{f(z)}{z^2} dz \right| \leq \frac{1}{2\pi}(2\pi R)\left( \frac{1}{R^2} \right) = \frac{1}{R}.$$ Hence, $|f'(0)|\leq \frac{1}{R}$ for all $0<R<1$. However, this is far to be the desired upper bound.

I also thought it could be useful to use the function $g=\frac{1}{f}$, which is analytic because $f(z)\neq 0$ for all $z\in D$, but I have not been capable of making any progress with this idea.

Answer 1

Note that (assuming $f$ non constant) the conditions on $f$ imply that $f=e^g$ where $\Re g < 0$, so $g=\frac{h+1}{h-1}$ where $h$ is a (nonconstant) analytic map from the unit disc to itself.

Then $|f'(0)|=\frac{2|h'(0)|}{|1-h(0)|^2}e^{\Re \frac{h(0)+1}{h(0)-1}}$

By Schwarz Pick we have $|h'(0)| \le 1-|h(0)|^2$ and clearly $\Re \frac{h(0)+1}{h(0)-1}=-\Re \frac{h(0)+1}{1-h(0)} =-\frac{1-|h(0)|^2}{|1-h(0)|^2}$ since $\Re (1+a)(1-\bar a) =1-|a|^2$ so with $x=\frac{1-|h(0)|^2}{|1-h(0)|^2} >0$ we have that $|f'(0)| \le 2xe^{-x} \le \frac{2}{e}$ since by elementary calculus $a(x)=2xe^{-x}$ attains its maximum at $x=1$ on $[0, \infty)$ since $a'(x)=0$ implies $x=\pm 1$ etc

Unpacking the above to get equality, we need for example $x=1$ so $1-|h(0)|^2=|1-h(0)|^2$ as well as equality in Schwarz-Pick so $h$ is a Mobius disc automorphism, so, in particular, choosing $h(z)=z$ so $f(z)=e^{\frac{z+1}{z-1}}$ works to give such.

With more care one can show that the above conditions imply that we get equality iff $$f(z)=\alpha e^{\frac{\beta z+1}{\beta z-1}}, |\alpha|=|\beta|=1$$

The conjecture that all coefficients $a_n$ of such an $f(z)=\sum a_nz^n$ as in the OP satisfy the inequality that $a_0$ does, so $|a_n| \le \frac{2}{e}$ with equality when $f$ is above but with $z^n$ instead of $z$, so $$f(z)=\alpha e^{\frac{\beta z^n+1}{\beta z^n-1}}, |\alpha|=|\beta|=1$$ is called Krzyz conjecture and is open afaik - there is a recent book by R Peretz about it