Comparing the deviation of a function from its mean on concentric balls Let $\Omega \subset \mathbb{R}^n$ ($n\geq 2$) a domain with smooth boundary.  Consider a ball $B(x_0,R) \Subset  \Omega$ . Is true that 
$$ \int_{B(x_0,r)} | u - u_{x_0 , r}|^p \leq \int_{B(x_0,R)} | u - u_{x_0 , R}|^p,$$
for $u \in W^{1,p}(\Omega)(p >1)$ and for all $0 < r \leq R$, where $u_{x_0 , r} = \frac{1}{|B(x_0,r)|} \int_{B(x_0,r)} u $ and  $u_{x_0 , R} = \frac{1}{|B(x_0,R)|} \int_{B(x_0,R)} u $. ?
I know that the inequality is true for $p=2$. If the inequality is true for all $p>1$, I will understand an inequation of a paper that I am reading. There is a good time that I am searching for a reference ( I think that probably is true). But I did not find anything. 
 A: The case $p=2$ is special, because the mean minimizes $\int |u-c|^2$ among all $c\in\mathbb{R}$. It does not have such a property for $p\ne 2$, and there is no reason for the stated inequality to hold. 
As PhoemueX noted, the Sobolev space is not really relevant: any $L^p$ function can be approximated by smooth functions in $L^p$ norm, and the quantities in your inequality are continuous in $L^p$ norm. So, I'm going to consider $u\in L^p$ to simplify things.
Here is a counterexample for large $p$. Let
$$u = \chi_{B(0,\rho)}+\chi_{B(0,R)\setminus B(0,r)}$$
where the radii are chosen so that the measures of balls are related as
$$|B(0,\rho)|=\frac1{10} |B(0,r)|\quad \text{ and } \quad |B(0,R)| = 2(|B(0,r)|+|B(0,\rho)|)$$
On the ball $B(0,r)$, the mean of $u$ is $1/10$, hence 
$$\int_{B(0,r)} | u - u_{0 , r}|^p \ge 
\int_{B(0,\rho)} | u - u_{0 , r}|^p = \left(\frac{9}{10}\right)^p |B(0,\rho)|\tag1$$  
On the ball $B(0,R)$, the mean of $u$ is $\frac12 $, hence 
$$\int_{B(0,R)} | u - u_{0 , R}|^p = \left(\frac12\right)^p  |B(0,R)| \tag2$$ 
When $p$ is sufficiently large, $(1)$ is greater than $(2)$.
