Bounds of $\,f'(a)$, where $f$ is a conformal mapping

Question

Let $D\subsetneq\mathbb C$ be a simply connected region and $a\in D$. The Riemann mapping theorem guarantees the existence of a biholomorphic $f:D\to B(0,1)$, such that $f(a)=0, f'(a)>0$.

Show that $$ \min_{z \in \partial D}|z-a|\le \frac{1}{f'(a)}\le \max_{z\in \partial D} |z-a|. $$

I tried to use the Taylor expansion of $f$ near $a$ and estimate the error terms but seems not work a lot. Thanks for any help.

Answer 1

If $r=\min_{z\in\partial D}$, then $\overline{B}(a,\varrho)\subset D$, for all $\varrho<r$ and by virtue of Cauchy Integral Formula $$ f'(a)=\frac{1}{2\pi i}\int_{|z-a|=\varrho}\frac{f(z)\,dz}{(z-a)^2} $$ which implies that $$ f'(a)=|\,f'(a)|\le \frac{\max_{|z-a|=\varrho} |\,f(z)|}{\varrho}\le \frac{1}{\varrho}, $$ and since this is true for all $\varrho<r$, then $$ f'(a)\le \frac{1}{r}. $$

If $g=f^{-1}: B(0,1)\to D$, then $g(0)=a$ and $g'(0)=1/f'(a)$ and, if $h(z)=g(z)-a$, then for any $R\in (0,1)$ we have $$ g'(0)=h'(0)=\frac{1}{2\pi i}\int_{|z|=R}\frac{h(z)\,dz}{|z|^2}=\frac{1}{2\pi i}\int_{|z|=R}\frac{(g(z)-a)\,dz}{|z|^2}. $$ Hence $$ \frac{1}{f'(a)}=g'(0)=|g'(0)|\le\frac{\max_{|z|=R}|g(z)-a|}{R}\le \frac{\max_{w\in\partial D}|w-a|}{R}, $$ since $|g(z)-a|\le\max_{w\in\partial D}|w-a|$ (Maximum Principle.) Since the inequality above holds for all $R\in (0,1)$, then $$ \frac{1}{f'(a)}\le \max_{w\in\partial D}|w-a|. $$

Answer 2

We can do this with the Schwarz Lemma (SL): Let $r,R$ be the minimum and maximum distances of $a$ to $\partial D.$ Define $g : \mathbb D \to D$ by $g(z) = a + rz.$ Then $f\circ g$ is a holomorphic map of $\mathbb D$ into $\mathbb D$ with $f\circ g (0) = 0.$ By SL, $|(f\circ g)'(0)| \le 1.$ I.e.,

$$|f'(g(0))g'(0)| = f'(a)r \le 1.$$

This gives the inequality on the left.

For the other inequality, note that $f^{-1}$ maps $\mathbb D$ into $D$ with $f^{-1}(0)=a.$ Because $|f^{-1}(z) -a|<R$ for $z\in \mathbb D,$ we have $g(z) =(f^{-1}(z) -a)/R$ mapping $\mathbb D$ into $\mathbb D,$ with $g(0)=0.$ Again using SL, we have $|g'(0)| \le 1.$ I.e,

$$|(f^{-1})'(0)/R| = \frac{1}{f'(a)R}\le 1.$$

This gives the inequality on the right.