Picard's theorem and Rickman's theorem by way of Harnack's inequality

John L. Lewis
1994 Proceedings of the American Mathematical Society  
In this note we give a very elementary proof of Picard's Theorem and Rickman's Theorem which uses only Harnack's inequality.
doi:10.1090/s0002-9939-1994-1195483-3 fatcat:wh5txxsutredpnv2lli7zrgzmq