On the growth of \(\alpha\)-potentials in \(R^ n\) and thinness of sets (Q1076199)

Let \(0<\alpha <n\) (respectively \(\alpha =n)\) and let \(\mu\) be a non- negative Radon measure in \({\mathbb{R}}^ n\) such that the \(\mu\)-integral of the function \((1+\| x\|)^{\alpha -n}\) (resp. \(\log (1+\| x\|))\) is finite. The \(\alpha\)-potential of \(\mu\) is given by \(R_{\alpha}\mu (x)=\int R_{\alpha}(x-y)d\mu (y)\), where \(R_{\alpha}(x)=\| x\|^{\alpha -n}\) (resp. \(\log (1/\| x\|)).\) Also let f: (0,\(+\infty)\to (0,+\infty)\) be non-increasing and such that f(r)\(\leq cf(2r)\) for all positive r and some constant c. The author relates (by equivalences) the boundedness in (0,1) of the function \(f^{-1}(r)A(R_{\alpha}\mu,0,r)\) \((A=volume\) mean) to the boundedness of \(f^{-1}(\| x_ k\|)R_{\alpha}\mu (x_ k)\) for a sequence \((x_ k)\) tending to 0 and also to the existence of an exceptional set E in B(0,\()\) outside which \(\limsup\{f^{-1}(\| x\|)R_{\alpha}\mu (x)\}<+\infty\) as \(x\to 0\). A similar theorem holds with ''boundedness'' and \(''<+\infty ''\) replaced by ''convergence to 0'' and \(''=0''\). In the case where \(\mu\) has finite \(\alpha\)-energy (that is, when \(\int R_{\alpha} \mu d\mu <+\infty)\) the author shows that the definition of the exceptional set has to be altered to have similar results.