Inequality for Hardy-Littlewood type function

Question

Consider $r\gt0,x\in R^n$ and the average of a 1-Lipschitz function $u:R^n\longrightarrow R$ over the $B(x,r) $ $$u_{r}(x)=\frac{1}{Vol(B(x,r))}\int_{B(x,r)}u(z)dz$$

Then there is an absolute constant such that $$\Vert\nabla{u_{r}}(x)-\nabla{u_{r}(y)}\Vert_{2}\leq{\frac{c\sqrt{n}}{r}}\Vert{x-y}\Vert_{2}$$

we can transform and use the unit ball instead along with $u(x+rz)$ and Leibniz rule to differentiate under the integral.Then the $\sqrt{n} $ comes from bounding the norm of the gradient of $u$...but i can't get to the ratio $1/r$ which seems to come from the ratio of surface/volume...

Answer 1

Let $V_n(r)$ denote the volume of $n$-dimensional ball of radius $r$. The gradient of $u_r$ is the integral of $\nabla u$ over $B(x,r)$, divided by $V_n(r)$. (The gradient exists almost everywhere, and is bounded by $1$.) How much do the integrals $\int_{B(x,r)}\nabla u$ and $\int_{B(y,r)}\nabla u$ differ? The volume of symmetric difference $B(x,r)\triangle B(y,r)$ is at most $2\|x-y\| V_{n-1}(r)$, because if this set is stacked onto $(n-1)$-dimensional cross section, its thickness is at most $2\|x-y\|$. Thus, $$ \left\| \int_{B(x,r)}\nabla u - \int_{B(y,r)}\nabla u \right\| \le 2\|x-y\| V_{n-1}(r) $ $$ Dividing by $V_n(r)$ yields $$ \Vert\nabla{u_{r}}(x)-\nabla{u_{r}(y)}\Vert \le \frac{2V_{n-1}(r)}{V_{n}(r)} \|x-y\| $$ There's a relevant recursion formula: $$ \frac{V_{n-1}(r)}{V_{n}(r)} = \frac{1}{r}\frac{\Gamma(n/2+1)}{\sqrt{\pi} \Gamma((n+1)/2)} $$ Stirling's formula shows that the latter ratio is asymptotic to $\sqrt{n}$.