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\title {MENGER CURVATURE ON RECTIFIABLE AND FRACTAL SUBSETS OF THE PLANE}
\author{Pertti Mattila, Mathematics Department, University of
Jyv\"askyl\"a, P.O. Box 35, FIN-40351 Jyv\"askyl\"a, Finland, email:
{\tt pmattila@jylk.jyu.fi} }
\markboth{Santa Barbara Symposium -- P.~Mattila}{Santa
Barbara Symposium -- P.~Mattila}
\begin{document}\label{zf-conf-pm}
\maketitle
%------
The Menger curvature $c(x,y,z)$ of the triple $x, y, z\in\R^2$ is
the reciprocal of the radius of the circle passing through $x$, $y$
and $z$. See \cite{K} for background and history. Note that $c(x,y,z)=0$
if and only if $x$, $y$ and $z$ are collinear. Melnikov found in
\cite{M} the following remarkable identity relating Menger
curvature to the Cauchy kernel $1/z$, $z\in\C$:
$$
c(z_1,z_2,z_3)=\sum_\sigma\frac1{(z_{\sigma(1)}-z_{\sigma(3)})\,
\overline{(z_{\sigma(2)}-z_{\sigma(3)})}}
$$
where $z_1, z_2, z_3\in\C$ and $\sigma$ runs through all six
permutations of $\{1,2,3\}$. This has had applications to
Cauchy singular integral operator and analytic
capacity. We first give some definitions.
Let $E\subset\R^2$ be compact. We say that $E$ is {\it
rectifiable\/} if there are rectifiable curves $\Gamma_1,\Gamma_2,\dots$
such that
$$
\ch^1\bigg(E\setminus\bigcup^\infty_{i=1}\Gamma_i\bigg)=0.
$$
$\ch^s$ stands for the $s$-dimensional Hausdorff measure. We say
that $E$ is {\it uniformly rectifiable\/} if there is one
rectifiable curve $\Gamma$ such that $E\subset\Gamma$ and there is
$C$ such that
$$
\ch^1\big(\Gamma\cap B(x,r)\big)\le Cr
$$
for all $x\in\R^2$ and $r>0$. We say that $E$ is {\it regular\/}
if there is $C$ such that
$$
r/C\le\ch^1\big(E\cap B(x,r)\big)\le Cr
$$
for all $x\in E$, $00$.
\end{itemize}
\end{theorem}
The last condition means that the Cauchy singular integral
operator is bounded in $L^2(E)$.
We say that $E$ is removable for bounded analytic function if
every bounded analytic function $f\:U\setminus E\to\C$, where $U$
is open with $E\subset U$, has an analytic extension to $U$. These
are the null-sets of analytic capacity.
\begin{theorem}
{\rm ([D])} Suppose $\ch^1(E)<\infty$. Then $E$
is removable for bounded analytic functions if and only if $\ch^1(E\cap\Gamma)=0$
for every rectifiable curve $\Gamma$.
\end{theorem}
The last condition says that $E$ is a so-called purely
unrectifiable set, or irregular in the sense of Besicovitch. The
proof of Theorem~2 uses the following rectifiability result of
David and L\'eger.
\begin{theorem}
{\rm (\cite{L})} Suppose $\ch^1(E)<\infty$.
Then $E$ is rectifiable if and only if $E$ can be written as
$E=\bigcup^\infty_{i=1}E_i$ where for each $i$
$$
\int_{E_i}\int_{E_i}\int_{E_i}c(x,y,z)^2\,d\ch^1x\,d\ch^1y\,d\ch^1z<\infty.
$$
\end{theorem}
In \cite{Li} Lin studied the behavior of
$\int_E\int_E\int_E c^{2s}\,d\ch^s\,d\ch^s\,d\ch^s$ on the
$s$-dimensional subsets $E$. He showed, with some extra
assumptions, that this integral is often $\infty$ if $s>1$, and if
$s<1$ and if it is finite then $E$ lies close to lines in small
scales.
\begin{thebibliography}{MMV}
\bibitem[D]{D} G.~David, \textit{Unrectifiable $1$-sets have vanishing
analytic capacity}, to appear in Rev. Mat. Iberoamericana
\bibitem[K]{K} S.~Kass, \textit{Karl Menger}, Notices Amer. Math. Soc.
\textbf{43} (1996), 558--561.
\bibitem[L]{L} J.-C. L\'eger, \textit{Menger curvature and
rectifiability}, preprint
\bibitem[Li]{Li} Y.~Lin, \textit{Menger curvature, singular integrals
and analytic capacity}, Ann. Acad. Sci. Fenn. Ser. A I Math.
Dissertationes \textbf{111}, 1997.
\bibitem[MMV]{MMV} P. Mattila, M. S. Melnikov and J. Verdera,
\textit{The Cauchy integral, analytic capacity, and uniform
rectifiability}, Ann. of Math. \textbf{144} (1996), 127--136.
\bibitem[M]{M} M. S. Melnikov, \textit{Analytic capacity: discrete
approach and curvature of measure}, Sbornik Mathematics \textbf{186}
(1995), 827--846.
\end{thebibliography}
\end{document}