Can harmonic measure be Ahlfors regular on sets with small dimension?

2021-05-16
6 min read

In this post I describe an open problem concerning harmonic measure and its relationship with Hausdorff measure of non-integral dimension. For background on the definition, notation, and some related results, one can read through (or watch) the material for Week 10 of my geometric measure theory class.

Background

In that class, I give an overview of several results about the connection between harmonic measure $ \omega_{\Omega}$ and the geometry of its associated domain $ \Omega\subseteq \mathbb{R}^{d+1}$. One such result is that if $ E\subseteq \partial \Omega$ and $ \omega_{\Omega}|_{E}\ll\mathscr{H}^{d}|_{E}$, then $ \omega_{\Omega}|_{E}$ is $ d$-rectifiable, in that it can be covered up to harmonic measure zero by $ d$-dimensional Lipschitz graphs [2].

There are other results in this vein that find necessary or sufficient geometric conditions guaranteeing some relation between $ \omega_{\Omega}$ and $ \mathscr{H}^{d}$ (e.g. one could consider if they are $ A_{\infty}$-equivalent, or if $ \mathscr{H}^{d}\ll \omega_{\Omega}$ on the boundary instead).

The reason for focusing on $ \mathscr{H}^{d}$ is that in the classical setting (e.g. if the boundary of the domain is smooth), then not only do we know harmonic measure is mutually absolutely continuous with respect to surface measure $ \sigma$ (which itself is just a constant multiple of $ \mathscr{H}^{d}$), but we can give an explicit formula for $ \omega_{\Omega}$:

\[ d\omega_{\Omega}^{x}(\xi) = -\frac{d g_{\Omega}(x,\xi)}{d\eta} d\sigma(\xi)\]

where $ g_{\Omega}(x,y)$ is Green’s function for $ \Omega$ and $ \eta$ is the (outward) normal derivative along the boundary. So part of the motivation for [2] was understanding how non-smooth the boundary can be so that we still have absolute continuity. However, it’s natural to ask if we can have mutual absolute continuity between harmonic measure and other Hausdorff measures, i.e. with respect to $ \mathscr{H}^{\alpha}$ for some $ \alpha \neq d$.

A helpful assumption to work with when studying harmonic measure is that our domain $ \Omega$ satisfies the capacity density condition (CDC) if, for any $ \xi\in \partial \Omega$ and $ 0<r<diam \Omega$,

\[ Cap(B(\xi,r)\backslash \Omega)\geq cr^{d-1}\]

where Cap denotes Newtonian Capacity. This is basically a nondegeneracy condition, meaning that the boundary is uniformly fat in some sense (so in particular, there can’t be small parts of the boundary that are isolated from the rest of the boundary). This in particular implies that the harmonic measure is nondegenerate, in the sense that

\[ \omega_{\Omega}^{x}(B(\xi,r))\gtrsim 1 \;\; \forall \;\; x\in B(\xi,r/2)\cap\Omega,\]
\[ \xi\in \partial \Omega, 0 < r < diam \Omega. \]

In this context, we can rephrase our question as follows:

Question: Given a CDC domain $\Omega$, can $\mathscr{H}^{\alpha}$ and $\omega_{\Omega}$ be mutually absolutely continuous for some $ \alpha \neq d$?

Before you get excited, this problem is essentially solved. I’ll outline what’s known in the $\alpha>d$ and $\alpha<d$ cases separately.

A recent theorem of Xavier Tolsa [4] says that if $ \Omega\subseteq \mathbb{R}^{d+1}$, satisfies the CDC, $ \alpha>d$ and $ 0<\mathscr{H}^{\alpha}(\partial\Omega)<\infty$, then $ \omega_{\Omega}\perp \mathscr{H}^{\alpha}$. This resolves the question for $\alpha>d$.

However, Alexander Volberg has said he has an example of a set $ E\subseteq \mathbb{C}$ so that $ E^{c}$ satisfies the CDC and $ \omega_{E^{c}}$ and $ \mathscr{H}^{\alpha}|_{E}$ are mutually absolutely continuous where $\alpha$ is some number less than $1$. I have not seen the example, but he says it is a variant of the construction Bishop and Jones of an Ahlfors $ 1$-regular set $ E\subseteq \mathbb{C}$ that is contained in a rectifiable curve yet $ \omega_{E^{c}}$ is not absolutely continuous with respect to $ \mathscr{H}^{1}|_{E}$. In any case, I’m not aware of it being published yet, so this should be taken with a big grain of salt, and one open question could be to fill in these details, but if true it seems the question is resolved in this case as well.

One thing that I remember asking about was whether the domain constructed had Ahlfors regular boundary and this wasn’t the case. Recall a set $ E$ is Ahlfors $ \alpha$ regular if

\[ \mathscr{H}^{\alpha}(B(\xi,r)\cap E)\sim r^{\alpha}\;\; \forall \xi \in E, 0 < r < diam \; E.\]

This leads to the following question.

Question A: Given $ d\geq 2$, $ \alpha\in (d,d-1)$, is there a domain $ \Omega\subseteq \mathbb{R}^{d+1}$ so that $ \partial \Omega$ is Ahlfors $ \alpha$-regular and $ \omega_{\Omega} \ll \mathscr{H}^{\alpha}|_{\partial \Omega}\ll \omega_{\Omega}$?

Guy David also asked me this question, so I believe the problem is open.

Note that the CDC condition is implied by the Ahlfors regularity of the boundary, so this is a more restrictive assumption we are imposing.

It can be shown that if $ \partial \Omega$ is Ahlfors $ \alpha$-regular for some $ \alpha<d$, then $ \Omega$ is a one-sided NTA domain. This is the same as an NTA domain (as defined in the Week 10 notes for my GMT course) except we don’t assume exterior corkscrews. Thus, if such a domain did exist, one can use Lemma I in [3] and a bit of measure theory to blow-up harmonic measure at a point in the boundary to get a domain $ \Omega'$ whose harmonic measure satisfies

(*) $ \frac{\omega_{\Omega'}^{x}(B')}{\omega_{\Omega'}^{x}(B)}\sim \left(\frac{r_{B'}}{r_{B}}\right)^{\alpha}$ for any balls $ B'\subseteq B$ centered on $ \partial \Omega$ and $ x \in \Omega'\backslash 2B$.

Thus, Question A is equivalent to the following question.

Question B: Given $ d\geq 2$, $ \alpha\in (d,d-1)$, is there a domain $ \Omega\subseteq \mathbb{R}^{d+1}$ so that (*) holds?

If the answer to Questions A and B are false, it seems it might be easier to show B is false given that there are much more restrictive assumptions on the behavior of harmonic measure.

A negative answer to Question B has some additional consequences: in [1], I show that if $ \Omega\subseteq \mathbb{R}^{d+1}$ has Alhfors $ \alpha$-regular boundary for some $ \alpha>d$, then not only is harmonic measure singular with respect to $ \mathscr{H}^{\alpha}$, but the dimension of harmonic measure is strictly smaller than $ \alpha$, by which I mean one can find a set $ K\subseteq \partial \Omega$ with $ dim K<\alpha$ and $ \omega_{\Omega}(K^{c})=0$. In that paper, I also explain that the same techniques imply that a dimension drop occurs when $ \alpha<d$ if the answer to Question B is false.

Bibliography

  1. J. Azzam, Dimension drop for harmonic measure on Ahlfors regular boundaries, Potential Analysis (2019), 1–17.
  2. J. Azzam, S. Hofmann, J. M. Martell, S. Mayboroda, M. Mourgoglou, X. Tolsa and A. Volberg, , Rectifiability of harmonic measure, Geom. Funct. Anal. 26 (2016), 703–728. Link.
  3. J. Azzam and M. Mourgoglou, Tangent measures and absolute continuity of harmonic measure, Rev. Mat. Iberoam. 34 (2018), 305–330. Link.
  4. X. Tolsa, The Mutual Singularity of Harmonic Measure and Hausdorff Measure of Codimension Smaller than One, International Mathematics Research Notices (2019). Link.