Extensions of convex and semiconvex functions and intervally thin sets
A set \(A\subset \mathbb R^N\) is calledintervally thin if for all \(x,y\in\mathbb R^N\) and \(\varepsilon >0\) there exist \(x'\in B(x,\varepsilon)\) and \(y'\in B(y,\varepsilon)\) such that the segment \([x',y']\) does not intersect \(A\). Let \(\omega:\mathbb R_+\to\mathbb R_+\) be a nondecreasing upper semicontinuous function such that \(\omega(0)=0.\) A function \(f:S\subset\mathbb R^N\to\mathbb R\) is said to be \(\omega \)-semiconvex if \( f(\alpha x+(1-\alpha )y)\leq \alpha f(x)+(1-\alpha )f(y)+ \alpha (1-\alpha )\|x-y\| \omega (\|x-y\|)\) for \(\alpha\in [ 0,1]\) and \(x,y\in S\) such that \([x,y]\subset S\). The main result in this paper states that every \(\omega \)-semiconvex function \(f:U\setminus A\to\mathbb R\) defined on the complement in an open set \( U\subset \mathbb R^N\) of a closed intervally thin set \(A\subset U\) has a unique \(\omega \)-semiconvex extension onto \(U\). The authors also prove that \( U\setminus A\) is connected if \(U\) so is. Another result states that sets with \((N-1)\)-dimensional Hausdorff measure zero are intervally thin. As a consequence, the authors deduce that if \(U\subset\mathbb R^N\) is an open convex set, \(A\) is a closed subset of \(U\) with \((N-1)\)-dimensional Hausdorff measure zero, and \( f:U\setminus A\to\mathbb R\) is convex on every segment contained in \(U\setminus A\), then \(f\) has a unique convex extension onto \(U\).
scientific article; zbMATH DE number 5673252