## Jonas Azzam

##### University of Edinburgh Lecturer (US equivalent of Assistant Professor)

JCMB 4613, Kings Buildings
j.azzam$@$ed.ac.uk ### Brief CV

2011: University of California-Los Angeles, PhD

##### PhD Students
Jean Burnazyan
Matthew Hyde
Michele Villa (PhD 2020, now at U. Helsinki)

## Research

Rectifable Sets, Harmonic Measure, General Geometric Measure Theory, Miscellaneous

### Rectifiable Sets

Given a set (or metric space) how can I tell if and how well the set resembles Rd? There are various degrees of resemblance you could demand, like being isometric. One variant that I consider is the property of being rectifiable: a set is d-rectifiable if it can be covered up to d-dimensional measure zero by Lipschitz images of Rd. One can think of these sets as measure theoretic versions of smooth manifords: a rectifiable set is one that is "locally" parametrized by Lipschitz chart maps. There are even different degrees of rectifiability (like uniform rectifiability). Much of my work concerns with classifying when a set is rectifiable and studying the properties of rectifiable sets. There is a particular emphasis on quantitative analysis, and this only became more important around the 90s in connection with boundedness of singular integrals and Painleve's probelm.

We give a characterization of $L^{p}(\sigma)$ for uniformly rectifiable measures $\sigma$ using Tolsa's $\alpha$-numbers, by showing, for $p\in(1,\infty)$ and $f\in L^{p}(\sigma)$, that $||f||_{L^{p}(\sigma)} \sim \left|\left| \left(\int_{0}^{\infty} \left(\alpha_{f\sigma}(x,r)+|f|_{x,r}\alpha_{\sigma}(x,r)\right)^2\ \frac{dr}{r} \right)^{\frac{1}{2}}\right|\right|_{L^{p}(\sigma)} .$ Arxiv.
Open Problem: The second term in the integral is unnecessary when $\sigma$ is either an Ahlfors $d$-regular measure on $\mathbb{R}^{d}$ or if its support is a chord-arc surface with small constant. We don't know whether it can be omitted in general, but if it could be omitted it would give a simpler and more natural looking estimate
We show that Ahlfors $d$-regular set $E$ in $\mathbb{R}^{n}$ is uniformly rectifiable if the set of pairs $(x,r)\in E\times (0,\infty)$ where $\mathscr{H}^{d}_{\infty}(E\cap B(x,r))<(2r)^{d}(1-\epsilon)$ is a Carleson set for every $\epsilon>0$. To prove this, we generalize a result of Schul by proving, if $X$ is a $C$-doubling metric space, $\epsilon,\rho\in (0,1)$, $A>1$, and $X_{n}$ is a sequence of maximal $2^{-n}$-separated sets in $X$, and $\mathscr{B}=\{B(x,2^{-n}):x\in X_{n},n\in \mathbb{N}\}$, then $\sum \left\{r_{B}^{s}: B\in \mathscr{B}, \frac{\mathscr{H}^{s}_{\rho r_{B}}(X\cap AB)}{(2r_{B})^{s}}>1+\epsilon\right\} \lesssim_{C,A,\epsilon,\rho} \mathscr{H}^{s}(X).$ This is a quantitative version of the classical result that for a metric space $X$ of finite $s$-dimensional Hausdorff measure, the upper $s$-dimensional densities are at most $1$ $\mathscr{H}^{s}$-almost everywhere.
We show that any $d$-Ahlfors regular subset of $\mathbb{R}^{n}$ supporting a weak $(1,d)$-Poincaré inequality with respect to surface measure is uniformly rectifiable.
Arxiv
Arxiv.