Let $B\_0(1)$ be the unit ball in $\mathbb R^n$, $n\geq2$. Let $f\in W\_0^{1,2}(B\_0(1))$, and $W^{1,2}(B\_0(1))\ni f\_i\to f$ in the sense of $L^2(B\_0(1))$-norm, as $i\to \infty$.



> 
> **Question 1**: Can we find a sequence of positive numbers $\{\delta\_i\}$ (may depends on $\{f\_i\}$), going to $0$, s.t., $$\frac{1}{\delta\_i^2}\int\_{1-\delta\_i\leq|x|\leq1}f\_i^2\,dx\to0,$$
>  as $i\to\infty$ ? 
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> 
> **Question 2**: If Question 1 is not true, we then strengthen the condition that $W^{1,2}(B\_0(1))\ni f\_i\to f$ in the sense of $W^{1,2}(B\_0(1))$-norm. Can we prove Question 1's conclusion ?
> 
> 
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Convergence of Sobolev functions near the boundary
I think Question 1 has a positive answer. Denote $B=:B\_0(1)$ and $A\_r:=\{r<\|x\|<1\} $ for $0<r<1$. For functions $f\in W^{1,2}\_0(B)$ we have a Poincaré inequality on $A\_r$ :
$$\int\_{A\_r} f^2dx\le \Big(\frac{1-r}{r}\Big)^2 \int\_{A\_r} |\nabla f|^2dx \ .$$


Now let $(f\_i)$ a sequence in $W^{1,2}(B)$ converging to $f$, so $f^2\_i$ converges to $f^2$ in $L^1(B)$. For any sequence $0<\delta\_i<1$ we have, by the above inequality:


$$\frac{1}{\delta\_i^2}\int\_{A\_{1-\delta\_i}}f\_i^2dx\le \frac{1}{\delta\_i^2}\int\_{A\_{1-\delta\_i}}f ^2dx + \frac{1}{\delta\_i^2} \|f^2-f\_i^2\|\_{1} \le$$


$$\le \frac{1}{ (1-\delta\_i)^2} \int\_{A\_{1-\delta\_i}}\|\nabla f\| ^2dx + \frac{1}{\delta\_i^2} \|f^2-f\_i^2\|\_1\ .$$


Now if $\delta\_i=o(1)$, the first term on the RHS is $o(1)$, just because $\|\nabla f\| ^2$ is integrable and $\operatorname{meas}(A\_{1-\delta\_i})=o(1)$. 


On the other hand, if we also choose the sequence $\delta\_i$ such that $ \|f^2-f\_i^2\|\_1=o(\delta\_i^2)$ as $i\to\infty$, the other term is also $o(1)$. 


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*Details on the above Poincaré inequality.* By density, one can assume $f\in C^\infty\_c(B)$. Let $0<r\le \|x\|\le 1$. So $f(x/r)=0$, and we have
$$f(x)= -\int\_1^{1/r}\partial\_t f(tx)dt=-\int\_1^{1/r}\nabla f(tx)\cdot x\ dt\ . $$ 
Thus by Cauchy-Schwarz
$$ f(x) ^2\le \Big(\frac{1}{r} -1\Big) \int\_1^{1/r}\|\nabla f(tx)\|^2 dt\ , $$ 
and 
$$ \int\_{A\_r} f(x) ^2 \ dx\le \Big(\frac{1}{r} -1\Big) \int\_{A\_r}\int\_1^{1/r}\|\nabla f(tx)\|^2 \ dt\ dx = $$
$$=\Big(\frac{1}{r} -1\Big) \int\_1^{1/r} t^{-n} \int\_{A\_{tr}} \|\nabla f( x)\|^2 \ dx \ dt $$
$$\le \Big(\frac{1}{r} -1\Big)^2 \int\_{A\_{r}} \|\nabla f( x)\|^2 \ dx \ . $$


