Sobolev functions in the critical case are uniformly continuous in \(s\)-Ahlfors regular metric spaces when \(s\leq 1\) (Q2832826)

When it comes to metric spaces, there are different concepts of Sobolev spaces. One concept that gives a nontrivial Sobolev space if the underlying metric space is not connected is the one of Hajłasz spaces. In a suitable setting, if the exponent of the Sobolev space is strictly larger than the dimension, then each Sobolev function has a continuous representative. Questions about existence of continuous representatives have mainly been studied in spaces of dimension larger or equal than \(1\). In the article under review, the author looks at metric spaces with dimension less or equal than \(1\).NEWLINENEWLINEMore precisely, the result is as follows:NEWLINENEWLINELet \((X,d,\mathcal{H}^s)\) be an \(s\)-Ahlfors regular metric space and \(0<s\leq 1\). If \(u\in M^{1,s}(X,d,\mathcal{H}^s)\), then \(u\) is uniformly continuous. Moreover, there exists a constant \(C>0\) such that for any ball \(B\subset X\), NEWLINE\[NEWLINE\sup_{x,\,y\in B}|u(x)-u(y)|\leq C{\bigg(\int_{2B} g^s\, d\mathcal{H}^s\bigg)}^{\frac{1}{s}},NEWLINE\]NEWLINE where \(g\) is a Hajłasz upper gradient in \(L^s(X)\).NEWLINENEWLINETwo important ingredients in the proof are a chaining argument and a reverse Minkowski inequality.NEWLINENEWLINEContrasting the result, the author conjectures that on \(s\)-Ahlfors regular spaces there are always discontinuous Sobolev functions in \(M^{1,s}(X)\) provided \(s>1\).