 Solution

Put $Q=Q_r(x)$ and $v_Q=|Q{|}^{-1}{\int }_{Q}v$. The fundamental theorem of calculus along a line segment, followed by averaging, givesFor fixed $t$, use the change of variables formula $w=y+t(z-y)$. Its Jacobian determinant is $t^n$. The transformed domain $Q_t=y+t(Q-y)$ lies in $Q$, because a cube is a convex set. Moreover, $w\in Q_t$ implies $|w-y|\le \sqrt{n}rt$. Thus Tonelli theorem givesThe diagonal $w=y$ is a null set. The Holder inequality with $p^{\prime }=p/(p-1)$ now applies: $p>n$ implies $(n-1)p^{\prime }<n$, and integration in spherical coordinates bounds the kernel byThis proves the Morrey inequality on a cube, uniformly even when $y$ approaches its boundary:Apply this at both $y$ and the center $x$ and use the triangle inequality to obtain