Let $D$ denote the open unit disc centered at $0\in \mathbb{C}$ and suppose $f:D\to D-\lbrace 0 \rbrace$ is analytic and $f(0)=1/2$. Show $\vert f(1/2)\vert \ge 1/8$.

The only techniques that come to mind when dealing with inequalities of $\vert f \vert$, when  is defined on $D$, are Schwarz's Lemma and Schwarz-Pick Lemma.

By the Schwarz-­Pick Lemma $$\Bigg \vert \frac{f(1/2)-f(0)}{1-\overline{f(1/2)}\cdot f(0)}\Bigg\vert\le 1/2$$ so $\vert f(1/2)-1/2\vert \le \frac{\vert 1-f(1/2)/2\vert}{2}=1/2-f(1/2)/4$.

We don't know whether the leftmost side is (a) $\vert f(1/2)\vert-1/2$ or (b) $1/2-\vert f(1/2)\vert.$

In case (a) we have $-1/4\le 5\vert f(1/2)\vert/4$, which doesn't necessarily imply the desired inequality. In case (b) we have $0\le 3\vert f(1/2)\vert/4$ which also doesn't imply the desired inequality. So neither case is of any use.

Another idea is to consider the function $\frac{1}{f(z)}-2$ since it maps $0$ to $0$ (in an attempt to apply Schwarz) but I can't show that this function maps into $D$.

How can I proceed?


1 Answer 1


We must use Harnack's inequality rather than Schwarz's lemma.
We can write $f(z)=e^{g(z)}$ for some $g(z)$ analytic in $D$ since $f(z)\ne 0$ in $D.$ Let $u(z)=-\operatorname{Re}\,g(z),$ then $u(z)$ has the properties that $u(z)$ is harmonic in $D$, $u(z)\ge 0$ and $u(0)=\log 2$ because $|f(z)|=e^{\operatorname{Re}g(z)}=e^{-u(z)}\le 1$ and $|f(0)|=1/2$.
By Harnack's inequality $u(z)\le \frac{1+|z|}{1-|z|}u(0)$ we have$$u(1/2)\le 3u(0)=\log 8.$$Therefore $|f(1/2)|=e^{-u(1/2)}\ge e^{-\log 8}=\frac{1}{8}.$

  • $\begingroup$ No everyone is familiar with this inequality, do you have any other solutions maybe? $\endgroup$
    – UserB95
    Commented Mar 12, 2015 at 13:44
  • $\begingroup$ @UserB95 what other inequalities do you know in complex? $\endgroup$ Commented Mar 13, 2015 at 0:34

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .