Timeline for A general inequality about spherical mean of a function

8 events

when toggle format	what		by	license	comment
Oct 30, 2013 at 9:43	vote	accept	bigheadliao
Oct 30, 2013 at 9:43
Oct 30, 2013 at 8:17	comment	added	username		@bigheadliao : I have included this extra information in the question itself.
Oct 30, 2013 at 3:21	comment	added	bigheadliao		In fact, $u$ is non-negative solution of $\Delta u+u^{\frac{n}{n-2}}=0$(1.1) so we have $\Delta u \leq 0$. Based on Aviles' Lemma 1. Any non-negative solution of (1.1) satisfies $(−lnr)^{n−2/2}r^{n−2}\overline u(r)≤(\frac{n−2}{\sqrt 2})^{n−2},0<r<r0$ for some $1>r0>0$.Set $t=−ln\|x\|=−lnr,\phi(t,w)=\|x\|^{n-2}u(x)$. obviously $r^{n-2}\overline u(r) \leq Ct^{(2−n)/2}$,but Alives get$\phi (t, w) \leq C t^{(2-n)/2}$,So I guess$u(r,w) \leq C\overline u(r)$ is true
Oct 29, 2013 at 19:03	comment	added	username		In fact, a function is subharmonic iff $\Delta u \geq 0$, and equivalently iff everywhere within the interior of the domain, $u\leq \bar{u}$
Oct 29, 2013 at 15:07	comment	added	bigheadliao		thanks a lot, but $u$ is a solution of elliptic equation in Aviles' article
S Oct 28, 2013 at 21:35	history	suggested	username	CC BY-SA 3.0	fixed grammar
Oct 28, 2013 at 21:27	review	Suggested edits
S Oct 28, 2013 at 21:35
Oct 28, 2013 at 15:45	history	answered	user64494	CC BY-SA 3.0