Inequality and a trick - Sobolev spaces with p smaller than 2

Question

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

$K^{-1} (|\xi|^2 + |\eta|^2)^{\frac{p-2}{2}} |\xi - \eta|^2 \leq |V(\xi) - V(\eta)|^2 \leq K(|\xi|^2 + |\eta|^2)^{\frac{p-2}{2}} |\xi - \eta|^2 , K=K(n,p) \ \ (3),$ where $\xi , \eta \in R^n - \{ 0 \}$

I am not seeing how to obtain $(*)$ from (1) , (2) and (3). With my computations I am getting anywhere. Someone could help me ?

Thanks in advance

Answer 1

Using (3) in the form $$ |V(\xi) - V(\eta)|^2 \leq K(|\xi|^2 + |\eta|^2)^{\frac{p-2}{2}} |\xi - \eta|^2 \le K (|\xi| + |\eta|)^{p-2} |\xi - \eta|^2 $$ we get from (1) $$ \int_B |V(\nabla u) - V(\nabla v)|^2 \le K \int_B (|\nabla u| + |\nabla v|)^{p-2} |\nabla u- \nabla v|^2 \leq \int_B |\nabla u|^p - |\nabla v|^p$$ and an application of (2) yields the claim.

Answer 2

I read $V(\nabla u)-V(\nabla u)$ as the function defined by $V((\nabla u)(\xi))$. So if $1<p<2$, then we have from the second inequality in $$c^{-1}(\left|\xi\right|^{2}+\left|\eta\right|^{2})^{\frac{p-2}{2}}\left|\xi-\eta\right|^{2}\leq\left|V(\xi)-V(\eta)\right|^{2}\leq c(\left|\xi\right|^{2}+\left|\eta\right|^{2})^{\frac{p-2}{2}}\left|\xi-\eta\right|^{2}\tag{1}$$ that \begin{align*} \left|V((\nabla u)(\xi))-V((\nabla v)(\xi))\right|^{2}&\leq c(\left|\nabla u(\xi)\right|^{2}+\left|\nabla v(\xi)\right|^{2})^{\frac{p-2}{2}}\left|(\nabla u)(\xi)-(\nabla \eta)(\xi)\right|^{2}\tag{2}\\ \end{align*} The integral over $B_{2R}(x_{0})$ of the RHS is bounded by $$\dfrac{c}{C_{1}}\int_{B_{2R}(x_{0})}\left(\left|(\nabla u)(\xi)\right|^{p}-\left|(\nabla v)(\xi)\right|^{p}\right)d\xi\leq\dfrac{c}{C_{1}}CR^{n}\tag{3}$$ by the estimate on the bottom of pg. 5.

When $p>2$, we have by Holder's inequality that \begin{align*} \left(\int_{B_{2R}(x_{0})}\left|\nabla u-\nabla v\right|^{2}\right)^{1/2}&=\left(\int_{B_{2R}(x_{0})}\chi_{B_{2R}(x_{0})}\left|\nabla u-\nabla v\right|^{2}\right)^{1/2}\\ &\leq\left(\int_{B_{2R}(x_{0})}1\right)^{\frac{1}{2(p/2)'}}\left(\int_{B_{2R}(x_{0})}\left|\nabla u-\nabla v\right|^{2\frac{p}{2}}\right)^{\frac{1}{2(p/2)}}\\ &=\left|B_{2R}(x_{0})\right|^{\frac{1}{2(p/2)'}}\left(\int_{B_{2R}(x_{0})}\left|\nabla u-\nabla v\right|^{p}\right)^{1/p}\\ &=\left|B_{2R}(x_{0})\right|^{\frac{1}{2}}\left(\dfrac{1}{\left|B_{2R}(x_{0})\right|}\int_{B_{2R}(x_{0})}\left|\nabla u-\nabla v\right|^{p}\right)^{1/p}\tag{4}\\ \end{align*} From the $p>2$ estimate $$\int_{B_{2R}(x_{0})}\left|\nabla(u-v)\right|^{p}\leq\dfrac{1}{C_{1}}\int_{B_{2R}(x_{0})}\left(\left|\nabla u\right|^{p}-\left|\nabla v\right|^{p}\right)\leq\dfrac{CR^{n}}{C_{1}} \tag{5}$$ we obtain that \begin{align*} \int_{B_{2R}(x_{0})}\left|V(\nabla u)-V(\nabla v)\right|^{2}=\int_{B_{2R}(x_{0})}\left|\nabla(u-v)\right|^{2}&\leq\dfrac{\left|B_{2R}(x_{0})\right|}{\left|B_{2R}(x_{0})\right|^{2/p}}\left(\dfrac{CR^{n}}{C_{1}}\right)^{2/p}\\ &=\left(\dfrac{C}{2^{n}C_{1}\left|B_{1}(x_{0})\right|}\right)^{2/p}(2R)^{n}\left|B_{1}(x_{0})\right|\tag{6} \end{align*}