Conformal mappings and its singularity

Question

I have a question about singularities of conformal mappings.

Let $\mathbb{H} \subset \mathbb{C} \cong \mathbb{R}^2$ be the upper half-place and let $D$ be a Jordan domain. Let $\varphi:\mathbb{H} \to D$ denote a conformal map.

I am concerned with the following quantity: \begin{align*} I(z,r)=\int_{\mathbb{H} \cap B(z,r)}\log(|z-w|^{-1})|\varphi'(w)|^2\,dm(w),\quad z \in \bar{\mathbb{H}},\ r>0. \end{align*} Here $m$ denotes the two-dimensional Lebesgue measure and $B(z,r)$ denotes the open ball centered at $z$ with radius $r>0$. Of course, $I(z,r)$ is a variant of logarithmic potential. $I(z,r)$ roughly represents a singularity of $\varphi$ around the point $z$.

In fact, the quantity $I(z,r)$ naturally appears in the context of "random time-change" in probability theory. This controls local behaviors of the reflected Brownian motion in $D$ in some sense.

I am interested when $$\text{(A)}\quad \lim_{r \to 0}I(z,r)=0 \text{ uniformly in $z$ over each compact subset of $\bar{\mathbb{H}}$}$$ (of course, when $\partial D$ is smooth, this question is not interesting).

Has such a thing been studied in the context of conformal mappings and logarithmic potential theory?

Answer 1

I did not know at all... I am very sorry but can you tell me any references

Ah-oh. It is hard to find a decent reference now because nobody cares about writing down such trivialities any more: they are talking about domains in $\mathbb R^n$ and quasiconformal mappings, which is a huge overkill for you, so I'll simply give a proof. First of all, it will be enough to do it for the unit disk instead of half-plane since the conformal mapping of the half-plane to the disk is uniformly Lipschitz.

Let $\Omega$ be a bounded John domain (every uniform domain is John). Let $z_0$ be a John center of $\Omega$ and let $\Phi$ be the conformal mapping of $\mathbb D$ to $\Omega$ such that $\Phi(0)=z_0$. Let $w\in\mathbb D\setminus\{0\}$ and $z=\Phi(w)$. Note that $U(x)=\Re(1-\frac{\bar w}{|w|})$ is a non-negative harmonic function in $\mathbb D$ such that $U(0)=1$, $U(w)=1-|w|=dist(w,\partial\mathbb D)$. Thus $V=U\circ \Phi^{-1}$ is non-negative and harmonic in $\Omega$ with $U(z_0)=1$, $U(z)=1-|w|$. Let $d=dist(z,\partial\Omega)$. Then we can find a sequence of disks $D_j\subset\Omega$ of length $N\le C\log \frac 1d$ such that $D_0$ is centered at $z$, $D_N$ is centered at $z_0$ and the center of $D_{j+1}$ is inside $\frac 12D_j$. Going from each disk to the next and using the Harnack inequality we conclude that $1=V(z_0)\le Cd^{-C}V(z)=Cd^{-C}(1-|w|)$ for some large $C$ whence $d\le C(1-|w|)^\delta$ for some small but fixed $\delta>0$. Since $|\Phi'(w)|$ is comparable to $\frac d{1-|w|}$, we get the growth bound $|\Phi'(w)|\le C(1-|w|)^{\delta-1}$, which is equivalent to $\Phi$ being $\delta$-Holder.

As to the integration by parts, if it gives you trouble, forget about it and just split into dyadic rings.