Smoothness of the convolution of a singular measure with itself - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T05:48:25Z http://mathoverflow.net/feeds/question/100739 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/100739/smoothness-of-the-convolution-of-a-singular-measure-with-itself Smoothness of the convolution of a singular measure with itself user17240 2012-06-27T03:25:57Z 2012-07-04T23:59:06Z <p>Let $\gamma:I\subset\mathbb{R}\rightarrow\mathbb{R}^2, s\mapsto \gamma(s)$, denote the arclength parametrization of a smooth, convex curve $\Gamma:=\gamma(I)\subset\mathbb{R}^2$. Equip $\Gamma$ with arclength measure $\sigma$.</p> <p>Here is my question: what extra conditions on $\Gamma$ will ensure that the triple convolution $\sigma\ast\sigma\ast\sigma$ is a <em>smooth</em> function of $x\in\mathbb{R}^2$ inside its support (i.e. the set $\Gamma+\Gamma+\Gamma\subset\mathbb{R}^2$), up to the boundary? I am also interested in partial regularity results, and counterexamples.</p> <p>Here is a possible approach: for any $x\in\mathbb{R}^2$ and $\epsilon>0$, <code>$$\int_{|y-x|&lt;\epsilon} \sigma\ast\sigma\ast\sigma(y)dy=|\{(s,s',s'')\in I^3: |\gamma(s)+\gamma(s')+\gamma(s'')-x|&lt;\epsilon\}|.$$</code></p> <p>Understanding how these sets look like will probably help. I am familiar with results in this spirit for <em>double</em> convolutions (e.g. Fefferman 1970), but not for triple ones. </p> <p>References would be much appreciated.</p> <p>Thank you.</p>