... implies that the Alexander operator is convex.References1 F. Ronning, “Uniformly convex functions and a corresponding class of starlike functions,” Proceedingsof the American Mathematical Society, ... 1993.2 D. Breaz and N. Breaz, Two integral operators, Studia Universitatis Babesá-Bolyai, Mathematica, vol. 47,no. 3, pp. 13–19, 2002. Daniel Breaz 3This relation is equivalent tozFnzFnz ... Spβ and consider the integral operator of Alexander, Fzz0ft/tdt.In this condition, F is convex by the order β.Proof. We havezFzFzzfzfz− 1. 2.13 2 Journal...