I'm trying to prove the following Morrey inequality.
(Morrey's Inequality): Let $B_R(x_0) = B_R \subset \mathbb{R}^n$ denote a ball of radius $R$, and let $n < p < \infty$. For $x,y \in B_R$
\begin{equation*}
\vert u(x) – u(y) \vert \leq C|x-y|^{1-\frac{n}{p}}\Vert Du \Vert_{L^p(B_R)} \text{ for all } u \in W^{1,p}(B_R).
\end{equation*}
This following is something I have already proven
Let $B_R(x_0) = B_R \in \mathbb{R}^n$ denote the Euclidean ball of radius $R$ centered at $x_0 \in \mathbb{R}^n.$ Suppose that $u \in C^1(\overline{B_R})$ $\cap$ $W^{1,p}(B_R)$ for $p > n$. We define $\displaystyle u_{B_R} = \frac{1}{\vert B_R \vert}\int_{B_R} u(y) \, dy.$ Then for all $x \in B_R$ we have
\begin{align*}
\vert u_{B_R} – u(x) \vert \leq CR^{1-\frac{n}{p}} \Vert Du \Vert_{L_p(B_R)}.
\end{align*}
By the triangle inequality and the approximation theorem we get that
$\vert u(x) – u(y) \vert \leq CR^{1-\frac{n}{p}}\Vert Du \Vert_{L^p(B_R)} \text{ for all } u \in W^{1,p}(B_R)$
But idk how to replace $R$ with $\vert x-y\vert$ so that it holds for all $x,y \in B_R(x_0).$ Obviously it holds for $\vert x -y \vert \geq R.$ But what if $\vert x-y\vert < R.$
Best Answer
Instead of working with $B_R$, repeat the proof of the inequality $\vert u(x) - u(y) \vert \leq Cr^{1-\frac{n}{p}}\Vert Du \Vert_{L^p(B_r)}$ for any small ball $B_r(x_1,r)$ contained in $B_R$. Then given $ x$ and $y$ take $r=|x-y|$.
EDIT By a translation, you can assume that $x_0=0$. Extend $u$ to $B_{4R}$ by setting $$v(x)=u\left(\frac{R^2x}{|x|^2}\right)$$ for $x\in B_{4R}\setminus B_R$ and $v=u$ in $B_R$. You should be able to prove that $$\int_{B_{4R}}|v|^pdx\le C \int_{B_{R}}|u|^pdx$$ and $$\int_{B_{4R}}|\nabla v|^pdx\le C \int_{B_{R}}|\nabla u|^pdx.$$ So now you have that $v\in W^{1,p}(B_{4R})$. Given $x,y\in B_R$, you have that $r=|x-y|<2R$, and so $B((x+y)/2,r)\subset B_{4R}$.