The dimension of the kernel in planar sets which are starshaped via \(\alpha\)-paths
Let \(S\) be a plane set and \(\lambda:[p,q]\to S\) a continuous mapping from an interval \([p,q]\subset R\) to \(S:\lambda\) is the path of a curve in \(S\) from \(\lambda(p)\) to \(\lambda(q)\). If \(x,y\in S\), \(x\) sees \(y\) via \(\lambda\) if and only if the curve's range contains \(x\) and \(y\). When this occurs \(\lambda(x,y)\) will denote the subpath of \(\lambda\) from \(x\) to \(y\) and \(\|\lambda(x,y)\|\) will denote its length. A path is said to be an \(\alpha\)-path if, given a positive real number \(\alpha\), \(\|\lambda(x,y)\|\leq\alpha\). A set \(S\) is said to be \(\alpha\)-starshaped if a point \(p\in S\) sees every point \(s\in S\) via \(\alpha\)-paths; then \(\text{Ker}_\alpha S\) will denote the set of all the points of \(S\) for which \(S\) is \(\alpha\)-starshaped. The author proves that if \(S\) is a compact planar set whose boundary is a simple closed locally connected curve and \(\alpha\) is positive real number, the \(\alpha\)-kernel of \(S\) has a non-empty interior if and only if for some positive real number \(\varepsilon\) every three points of \(S\) see via \(\alpha\)-paths in \(S\) some pair of points \(a\) and \(b\) such that \(\text{dist}(a,b)\geq\varepsilon\). Furthermore, an example is exhibited which shows that three is the best possible number. A plane set \(S\) is said to be \(\alpha\)-convex if every two of its points are joined by an \(\alpha\)-path in \(S\). Then if \(S\) is a closed simply connected plane set, \(\text{Ker}_\alpha S\) is the intersection of all the maximal \(\alpha\)-convex subsets of \(S\) and \(\text{Ker}_\alpha S\) is itself \(\alpha\)-convex.
scientific article; zbMATH DE number 1214017