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?

  • $\begingroup$ I doubt somebody looked at this specific condition (though I may be wrong here) but there are various observations as to when $|\varphi'|^p$ is integrable with $p>2$ (a big overkill for your purposes), examples of domains where your property fails (a long chain of blobs of diminishing sizes with very narrow bottlenecks in between), etc. So, I suggest we would go from the other end: if you tell the class of domains you are really interested in, we'll try to figure out whether the property holds for that class. If this property is the only one you need for your results, just make it a definition $\endgroup$
    – fedja
    May 16, 2019 at 18:04
  • $\begingroup$ @fedja Thank you for your comment. I would like to know the condition (A) holds for bounded uniform domains: whose boundary can be fractal-like. By the way, do you have an (interesting) example which satisfies $\int_{\mathbb{H} \cap B(z,1)}|\varphi'(w)|^{p}\,dm(w)<\infty$ with $p>2$? $\endgroup$
    – sharpe
    May 17, 2019 at 9:07
  • $\begingroup$ When you are talking about "uniform domains", do you mean that any two points $a,b$ can be joined by an arc $\Gamma$ in the domain $\Omega$ such that $\ell(\Gamma)\le C|a-b|$ and every point $x\in \Gamma$ is contained in $\Omega$ together with some ball of radius $C^{-1}\min(|x-a|,|x-b|)$ for some $C>0$ or something else? (just to make sure that we don't diverge on the terminology). $\endgroup$
    – fedja
    May 17, 2019 at 10:16
  • $\begingroup$ If so, then you have nothing to worry about: the conformal mapping is uniformly Holder, so the image of the ball centered at $z$ of radius $r<1$ has area at most $Cr^\delta$ for some $C,\delta>0$ independent of $z$, thus for all $\rho<1$ we have $\int_{H\cap B(z,\rho)}|\varphi'(w)|^2\,dm(w)\le C\rho^\delta$ and the integration by parts (or partition into dyadic rings around $z$) finishes the story. $\endgroup$
    – fedja
    May 17, 2019 at 10:43
  • $\begingroup$ @fedja Is a conformal map from upper half-plane to uniform domain Holder continuous? I did not know at all... I am very sorry but can you tell me any references? $\endgroup$
    – sharpe
    May 17, 2019 at 13:56

1 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.

  • $\begingroup$ Thank you for your very kind reply. For the proof, you use a Harnack chain with length $N$. For the construction of this chain, you use the definition of John-domain, right? I would like to ask one more question, why is $|\Phi'(w)|$ and $d/(1-|w|)$ comparable? This is a consequence of a boundary Harnack principle? Indeed, $d/(1-|w|)$ is regarded as a ratio of harmonic functions. $\endgroup$
    – sharpe
    May 18, 2019 at 6:21
  • 1
    $\begingroup$ It is just Koebe 1/4 theorem for univalent functions. $\endgroup$
    – fedja
    May 18, 2019 at 6:26
  • $\begingroup$ I see! Thank you very much for teaching me very carefully. $\endgroup$
    – sharpe
    May 18, 2019 at 6:29

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