Let $B\_0(1)$ be the unit ball in $\mathbb R^n$, $n\geq2$. $h\in W\_0^{1,2}(B\_0(1))$. For $r\in (0,1)$, define a function $f\_r(x):[0,1]\rightarrow \mathbb R$ by 
\begin{equation}
f\_r(x):=
\begin{cases}
1,&\text{when} \ x\in(0,r],\\
\frac{1-x}{1-r},& \text{when} \ x\in(r,1].
\end{cases}
\end{equation}
Let $g\_r(x):B\_0(1)\rightarrow \mathbb R$ to be $g\_r(x):=f\_r(|x|)\cdot h(x)$. 



> 
> Question: Can we prove that $\int\_{B\_0(1)\backslash B\_0(r)}|\nabla g\_r|^2dx\to0$, as $r\to1^-$?
> 
> 
> 



Convergence of energy of Sobolev functions near the boundary
Yes, this works. By computing the gradient with the product rule, this boils down to showing that
$$
\frac{1}{\delta^2} \int\_{1-\delta\le |x|\le 1} |h(x)|^2 \to 0
$$
as $\delta\to 0$ for $h\in W^{1,2}\_0$. If $h$ is also smooth, then, since $h(y)=0$ on $|y|=1$,
$$
h(x) = -\frac{x}{|x|}\cdot\int\_{|x|}^1 \nabla h(tx/|x|)\, dt .
$$
So $|h(x)|^2\le (1-|x|)\int\_{|x|}^1 |\nabla h|^2\, dt$, and if we now integrate this over $1-\delta\le |x|\le 1$ (in spherical coordinates), we find that
$$
\int\_{1-\delta\le |x|\le 1} |h(x)|^2\, dx \lesssim \int\_{1-\delta}^1 (1-r)\, dr \int\_{1-\delta\le |x|\le 1} |\nabla h|^2\, dx ,
$$
and this gives the claim because the first integral on the RHS is $O(\delta^2)$ and the second one is $o(1)$ as $\delta\to 0$. The general case now follows by approximation.


