Vol. 194, No. 2, 2000

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Changfeng Gui & Juncheng Wei

Abstract

We show that for α 1 2, the following inequality holds:

α-∫ 1 2 ′ 2 ∫ 1 1∫ 1 2g(x) 2 −1(1− x )|g (x)| dx+ −1g(x)dx − log 2 −1e dx ≥ 0,

for every function g on (1,1) satisfying g2 = 11(1 x2)|g(x)|2dx < and 11e2g(x)xdx = 0. This improves a result of Feldman et al., 1998, and answers a question of Chang and Yang in the axially symmetric case.

Authors
Changfeng Gui
Department of Mathematics
University of British Columbia
Vancouver, BC V6T 1Z2
Canada
Juncheng Wei
Department of Mathematics
The Chinese University of Hong Kong
Shatin, Hong Kong
China