Is the weak convergence of the gradients sufficient for this inequality? I am trying to understand if
$$
\lim_{n \to \infty}\int_{B_2 \setminus B_1} |\nabla[\phi u + (1-\phi)u_n]|^2 \leq \lim_{n \to \infty}\int_{B_2\setminus B_1} |\nabla u_n|^2
$$
up to subsequences.
Here $B_2$ and $B_1$ denote the balls of radius $2$ and $1$ in $\mathbb{R^2}$, $\phi \in C^{\infty}_c(B_2)$, $0\leq \phi \leq 1$ in $B_2$, $\phi \equiv 1 $ in $B_1$, $\{u_n\}_n$ is a sequence of functions in $W^{1,2}(B_2)$ such that $\int_{B_2} |\nabla u_n|^2 \leq C$ for every $n$, and $u_n \rightharpoonup u$ in $W^{1,2}$.
Since $W^{1,2}(B_2)$ is compactly embedded in $L^2(B_2)$, we also have $u_n \to u$ in $L^2(B_2)$. Moreover,
$$
\nabla[\phi u + (1-\phi)u_n] = \phi \nabla u + u \nabla \phi + (1-\phi) \nabla u_n - u_n \nabla \phi \rightharpoonup \nabla u 
$$
in $L^2(B_2 \setminus B_1)$.
If we had $\nabla u_n \to \nabla u$ in $L^2(B_2 \setminus B_1)$, the result would be true, since then the above convergence would be strong and we would have
$$
\lim_{n \to \infty} \int_{B_2 \setminus B_1} |\nabla[\phi u + (1-\phi)u_n]|^2 = \lim_{n \to \infty}  \int_{B_2\setminus B_1} |\nabla u_n|^2  = \int_{B_2 \setminus B_1} |\nabla u|^2.
$$
But is the weak convergence $\nabla u_n \rightharpoonup \nabla u$ in $L^2(B_2 \setminus B_1)$ sufficient to prove the inequality at the beginning of the question?
By the weak convergence, we would have
$$
\int_{B_2 \setminus B_1} |\nabla u|^2 \leq \liminf_{n \to \infty} \int_{B_2 \setminus B_1} |\nabla u_n|^2
$$
and
$$
\int_{B_2 \setminus B_1} |\nabla u|^2 \leq \liminf_{n \to \infty} \int_{B_2 \setminus B_1} |\nabla[\phi u + (1-\phi)u_n]|^2, 
$$
but this does not seem to help.
Another approach would be to observe that, for a subsequence such that the limit exists,
\begin{align}
&\lim_{n \to \infty } \|\phi \nabla u + u \nabla \phi + (1-\phi) \nabla u_n - u_n \nabla \phi \|_{L^2(B_2 \setminus B_1)} \\
&\leq \lim_{n \to \infty} \| (u-u_n) \nabla \phi \|_{L^2(B_2 \setminus B_1)} +\lim_{n \to \infty}  \|\phi \nabla u + (1-\phi) \nabla u_n \|_{L^2(B_2 \setminus B_1)} \\
&= \lim_{n \to \infty}  \|\phi \nabla u + (1-\phi) \nabla u_n \|_{L^2(B_2 \setminus B_1)}, 
\end{align}
since $u_n \to u$ in $L^2(B_2 \setminus B_1)$.
So we are left to show that
$$
\lim_{n \to \infty} \int_{B_2 \setminus B_1} |\phi \nabla u + (1-\phi) \nabla u_n|^2 \leq \lim_{n \to \infty} \int_{B_2 \setminus B_1} |\nabla u_n|^2
$$
Any ideas? Do you think the inequality at the beginning holds or not?
Thanks for your help.
 A: I think I got the solution.
\begin{equation}
\begin{split}
&\int_{B_2 \setminus B_1} |\phi \nabla u + (1-\phi)\nabla u_n|^2 \\
&= \int_{B_2 \setminus B_1} [|\phi \nabla u|^2 + |(1-\phi)\nabla u_n|^2 + 2 \phi(1-\phi)\nabla u_n \cdot \nabla u].
\end{split}
\end{equation}
Now, since $\nabla u_n \rightharpoonup \nabla u$ in $L^2(B_2 \setminus B_1)$, we have
$$
\lim_{n \to \infty} \int_{B_2 \setminus B_1} 2\phi(1-\phi) \nabla u_n \cdot \nabla u = \int_{B_2 \setminus B_1} 2\phi(1-\phi) |\nabla u|^2.
$$
Therefore, for a subsequence such that all the limits below exist, we have
\begin{equation}
\begin{split}
&\lim_{n \to \infty} \int_{B_2 \setminus B_1} |\phi \nabla u + (1-\phi)\nabla u_n|^2 \\
& = \int_{B_2\setminus B_1} |\phi \nabla u|^2 + \lim_{n \to \infty} \int_{B_2 \setminus B_1}|(1-\phi)\nabla u_n|^2 + \int_{B_2 \setminus B_1}2\phi(1-\phi) |\nabla u|^2 \\ 
& \leq \lim_{n \to \infty} \left( \int_{B_2 \setminus B_1} |\phi \nabla u_n|^2 + \int_{B_2 \setminus B_1}|(1-\phi)\nabla u_n|^2 +\int_{B_2 \setminus B_1}2\phi(1-\phi) |\nabla u_n|^2 \right) \\
&= \lim_{n \to \infty} \int_{B_2 \setminus B_1} |\phi \nabla u_n + (1-\phi) \nabla u_n|^2 = \lim_{n \to \infty} \int_{B_2 \setminus B_1} |\nabla u_n|^2.
\end{split}
\end{equation}
