A question on Koebe theorem

Question

Assume that $f$ is a conformal mapping of a bounded Jordan domain $\Omega$ onto the unit disk U such that $f(0)=0$. How to prove the following inequality $(1-|f(z)|)\le K \sqrt{dist(z,\partial \Omega)}$, where $K$ does not depend on $f$ but depends on distance of 0 from boundary of $\Omega$.

Answer 1

Let $g$ be the inverse function to $f$, and $w=f(z)$, so that $g(w)=z$. Let $h$ be the automorphism of the disc sending $0$ to $w$. Then $g\circ h$ sends $0$ to $z$. Then by Koebe $1/4$ theorem $$dist(g(w),\partial\Omega)\geq (1/4)|(g\circ h)'(0)|=(1/4)|g'(w)||h'(0)|.$$ We estimate $|g'(w)|$ by the Koebe distortion theorem which gives $|g'(w)|\geq |g'(0)|(1-|w|)/8.$ Then we compute $|h'(0)|=1-|w|^2\geq 1-|w|$. Finally we estimate $|g'(0)|$ from below using Schwarz's Lemma: $|g'(0)|=1/|f'(0)|\geq R$, where $R$ is the distance from $0$ to $\partial \Omega$. Combining all these estimates, we obtain $$dist(z,\partial\Omega)\geq (R/32)(1-|w|)^2.$$

Answer 2

I am also curious to know why you claim the inequality is true without knowing how to prove it. However, when I saw Fedja's challenge I wanted to try it myself :-) so here it goes :

I can prove the inequality $$(1-|f(w)|)^2 \leq 4|f'(0)|(1+|f(w)|^2)dist(w,\partial \Omega).$$

First, recall the classical inequality for univalent functions on the unit disk $\mathbb{D}$ :

$$\frac{1}{4}(1-|z|^2)|f'(z)| \leq dist(f(z),\partial \Omega) \leq (1-|z|^2) |f'(z)|,$$ where $f$ is univalent on $\mathbb{D}$, $z \in \mathbb{D}$ and $\Omega = f(\mathbb{D})$.

The above inequality follows from Koebe's one-quarter theorem and Schwarz's lemma.

Applying the above with $f$ replaced by $f^{-1}$ and $z=f(w)$ yields $$(1-|f(w)|^2) \leq \frac{4}{|(f^{-1})'(f(w))|} dist(w,\partial \Omega)$$ i.e. $$(1-|f(w)|^2) \leq 4 |f'(w)| dist(w,\partial \Omega),$$ so that $$(1-|f(w)|) \leq \frac{4|f'(w)|}{1+|f(w)|} dist(w,\partial \Omega).$$

Now, let us combine this with Koebe's distortion inequality :

$$\frac{|(f^{-1})'(z)|}{|(f^{-1})'(0)|} \geq \frac{1-|z|}{(1+|z|)^3}$$ for $z \in \mathbb{D}$. Writing $z=f(w)$, we get $$|f'(w)| \leq \frac{(1+|f(w)|)^3}{1-|f(w)|} |f'(0)|.$$ Combining this with the other inequality proves the claim.

You can bound the first factor by a constant depending only on $\Omega$ in the sense that you want. As in Eremenko's answer, you can bound $|f'(0)| \leq 1/R$ where $R$ is the distance from $0$ to $\partial \Omega$, so that we get $$(1-|f(w)|)^2 \leq (8/R)dist(w,\partial \Omega).$$