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suppose $\overline u(r)=\frac{1}{\omega_{n-1}}\int_{S^{n-1}}u(r,w)dw,0<r<1,$ is the average of $u(r,w)$ on sphere $S^{n-1}$,where $(r,w)$ are the polar coordinates in $R^n$.

My question is whether (edited) $$ u(r,w) \leq C\overline u(r),$$ where $C$ is independent of $u$. If this inequality is true, How can I prove it?

This question is from Aviles' article (see inequality(2.3)),Local Behavior of Solutions of Some Elliptic Equations, Commun.Math.Phys.108,177-192(1987).

Added information (from comments):

In this article, $u$ is a non-negative solution of \begin{equation} Δu+u^{\frac{n}{n−2}}=0 \mbox{ in } B_1\setminus{0}. \tag{1} \end{equation} So $u$ is superharmonic, $\Delta u \leq 0$. Based on Aviles' Lemma 1. Any non-negative solution of (1) satisfies \begin{equation} (−\ln r)^{\frac{n−2}{2}}r^{n−2}\bar{u}(r)≤\left(\frac{n-2}{\sqrt{2}}\right)^{n−2},\mbox{ for all } 0<r<r_0 \end{equation} for some $1>r_0>0$. In the following step, the author sets $t=−\ln|x|=−\ln r$, and $ϕ(t,w)=|x|^{n−2}u(x)$.

Obviously $r^{n−2}\bar{u}(r)≤Ct^{\frac{2−n}{2}}$, but Alives writes directly, $$ ϕ(t,w)≤Ct^{\frac{2−n}{2}}, $$ So I guess $u(r,w)≤C\bar{u}(r)$ is true...but I don't see why.

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This paper is about solutions of elliptic equations. Are you asking if it true for solutions of elliptic equations, or for any $u$? – username Oct 28 '13 at 19:57
u is solution of elliptic equation – bigheadliao Oct 29 '13 at 15:04
I can't find that article in the Internet. Could you present it in MO? – user64494 Oct 29 '13 at 16:29
@user64494 the preprint version is here – username Oct 30 '13 at 8:22
up vote 1 down vote accepted

You cannot use directly the inequality for subharmonic functions (as it is superharmonic), but if you know also that a Harnack inequality holds for this problem, that is, $$ \max_B u \leq C \min_B u $$ (possibly locally etc) then you are in business because then of course $$ u \leq \max_B u \leq C \bar{u}. $$ Looking at the preprint version, this Harnack inequality comes from Gidas and Spruck '81.

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I think you are right, thanks again. – bigheadliao Oct 30 '13 at 9:11

This inequality (as it is formulated) is not true in the general case. Here is an example in two dimensions. Your notation $\overline{u}(r,w)$ is not proper because the average of $u(r,w)$ over the sphere does not depend on $w$. Let us define $$u(r,\phi):=\begin{cases} 1,\mbox{ for }\phi \ge 0 \mbox{ and }\phi \le \epsilon,\\ 0, \mbox{ otherwise.}\\ \end{cases}$$ Then $\overline{u}(r)= \frac \epsilon {2\pi}.$ Therefore, the best possible constant $C$ is $\frac {2\pi} \epsilon$, and depends on $u$. Such inequality may be true under additional assumptions on $u$. For instance, it is true with the constant $C=1$ for subharmonic functions.

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thanks a lot, but $u$ is a solution of elliptic equation in Aviles' article – bigheadliao Oct 29 '13 at 15:07
In fact, a function is subharmonic iff $\Delta u \geq 0$, and equivalently iff everywhere within the interior of the domain, $u\leq \bar{u}$ – username Oct 29 '13 at 19:03
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 – bigheadliao Oct 30 '13 at 3:21
@bigheadliao : I have included this extra information in the question itself. – username Oct 30 '13 at 8:17

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