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Aug 7, 2016 at 22:15 comment added Pietro Majer Exact (more precisely, I guess I should have said "u has constant sign", so u can be assumed nonnegative w.l.o.g.)
Aug 7, 2016 at 17:15 vote accept Tomás
Aug 7, 2016 at 17:13 comment added Tomás I see. If $u$ is a maximizer, we can use the Lagrange multiplier theorem to infer that $-\Delta u=\lambda |u|^{p-2}u$. Then, an argument with the test function $v=u^-$ yields that $u\ge 0$. The rest is as you said. Thank you.
Aug 7, 2016 at 12:31 history edited Pietro Majer CC BY-SA 3.0
added 12 characters in body
Aug 7, 2016 at 8:48 history answered Pietro Majer CC BY-SA 3.0