# Trick in a inequality of a paper of free boundary problem that involves the p-laplacian with 1<p<2

I tried to ask this in mathstack, but no one answered me.

Let $B = B(x_0,R) \subset \subset \Omega$ a ball in $R^n$ with $\Omega$ a domain in $R^n$ with smooth boundary and consider two functions $u,v \in W^{1,p}(\Omega)$, $(1<p \leq 2)$. From general theory we have the inequality

$$C_1\int_B (|\nabla u| + |\nabla v|)^{p-2} |\nabla u- \nabla v|^2 \leq \int_B |\nabla u|^p - |\nabla v|^p, \ \ \ C_1 = C_1 (n,p) \ \ \ (1).$$

Suppose that

$$\int_B |\nabla u|^p - |\nabla v|^p \leq C_2(n,p)R^n \ \ \ (2)$$

The author of the paper that I am reading says that it is possible to conclude that

$$\int_B |V(\nabla u) - V(\nabla v)|^2 \leq C_3(n,p)R^n, (*)$$

where $V(\xi) = |\xi|^{\frac{p-2}{2}}\xi, \xi \in R^n,$ from the inequality

$K^{-1} (|\xi|^2 + |\eta|^2)^{\frac{p-2}{2}} |\xi - \eta|^2 \leq |V(\xi) - V(\eta)|^2 \leq K(|\xi|^2 + |\eta|^2)^{\frac{p-2}{2}} |\xi - \eta|^2 , \text{where } K=K(n,p) \ \ (3),$

for $\xi , \eta \in R^n \setminus \{ 0\}$

I am not seeing how to obtain $(*)$ from (1) , (2) and (3).

The paper is this : http://arxiv.org/abs/1508.07447 and the problem is in page 6

I agree with the doubt: since $p<2$ you have that $g(t)= |t|^{p-2}$ is decreasing, hence from $|\nabla u| + |\nabla v| \geq \sqrt{|\nabla u|^2 + |\nabla v|^2}$ you deduce $(|\nabla u| + |\nabla v|)^{p-2} \leq (|\nabla u|^2 + |\nabla v|^2)^{\frac{p-2}{2}}$, which goes the opposite with respect to the wanted chain of inequalities, so I suspect there might be a problem in the argument.
This is really a comment on the response posted above: for $\xi,\eta\neq 0$, using $|\xi|\leq (|\xi|^2+|\eta|^2)^{1/2}$ and $|\eta|\leq (|\xi|^2+|\eta|^2)^{1/2}$ one obtains $$|\xi|+|\eta|\leq 2(|\xi|^2+|\eta|^2)^{1/2},$$ which, for $1<p<2$ leads to $$(|\xi|^2+|\eta|^2)^{(p-2)/2}\leq 2^{2-p}(|\xi|+|\eta|)^{p-2}.$$
From (1) and (2) it suffices to show that $$|V(\xi)-V(\eta)|^2 \lesssim (|\xi|+|\eta|)^{p-2}|\xi-\eta|^2.$$
By (3) $$|V(\xi)-V(\eta)|^2 \lesssim (|\xi|^2+|\eta|^2)^{\frac{p-2}{2}}|\xi-\eta|^2.$$
Hence, you only need to show that $$(|\xi|^2+|\eta|^2)^{\frac{p-2}{2}} \lesssim (|\xi|+|\eta|)^{p-2}.$$
Since $(p-2)/2$ is negative, the previous is true if you have $$(|\xi|^2+|\eta|^2) \geq \frac{1}{2}(|\xi|+|\eta|)^2,$$ which is true.