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 $r\to1^-$?
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.
