# A general inequality about spherical mean of a function

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).

In this article, $u$ is a non-negative solution of $$Δu+u^{\frac{n}{n−2}}=0 \mbox{ in } B_1\setminus{0}. \tag{1}$$ So $u$ is superharmonic, $\Delta u \leq 0$. Based on Aviles' Lemma 1. Any non-negative solution of (1) satisfies $$(−\ln r)^{\frac{n−2}{2}}r^{n−2}\bar{u}(r)≤\left(\frac{n-2}{\sqrt{2}}\right)^{n−2},\mbox{ for all } 0<r<r_0$$ 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

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.
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.
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