If $f$ is a smooth function and $f(0)=0$, it is clear that $\|f\|_{L^\infty(B)}$ can be estimated (for any Ball $B$) if $\|\nabla f\|_\infty$ is known.
My question is whether something similar can be done for Sobolev function and the $L^2$-norm. More precisely: Suppose $f\in H^1_0(\mathbb{R}^n\setminus B_\epsilon)$ for a ball $B_\epsilon$ of small radius. Is it possible to get an estimate like $$\|f\|_{L^2(B_r)} \leq C(r-\epsilon)\|\nabla f\|_{L^2(B_r)} $$ ?
The reason I'm asking is that the naive approach to this fails:
$\int_{B_r} |f(x)|^2 dx = \int_{B_r}\big|\int_0^1 \frac{d}{dt}f(tx)dt\,\big|^2dx$
$\leq r^2\int_{B_r}\int_0^1|\nabla f(tx)|^2dt\,dx$
Now, if one exchanges the integrals and changes the variable to $tx=y$, it fails to be integrable in $t$:
$"=" r^2\int_0^1\int_{B_r}\frac 1 t |\nabla f(y)|^2dy\,dt$