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).
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. 
 A: 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.  
A: 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.
