Picard’s theorem and Rickman’s theorem by way of Harnack’s inequality
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- by John L. Lewis PDF
- Proc. Amer. Math. Soc. 122 (1994), 199-206 Request permission
Abstract:
In this note we give a very elementary proof of Picard’s Theorem and Rickman’s Theorem which uses only Harnack’s inequality.References
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Additional Information
- © Copyright 1994 American Mathematical Society
- Journal: Proc. Amer. Math. Soc. 122 (1994), 199-206
- MSC: Primary 30C65
- DOI: https://doi.org/10.1090/S0002-9939-1994-1195483-3
- MathSciNet review: 1195483