Assume $u\_n\to u$ weakly in $H^1(\Omega)$ where $\Omega\subset \mathbb R^N$ is open bounded Lipschitz boundary. 


My goal is to find a new sequence $\bar u\_n$ and a new function $\bar u$ such that 


1. $\int\_\Omega|\nabla \bar u\_n|^2dx\leq \int\_\Omega|\nabla u\_n|^2dx$ and $\int\_\Omega|\nabla \bar u|^2dx\leq \int\_\Omega|\nabla u|^2dx$
2. $\bar{u}\_n\to \bar{u}$ weakly in $H^1$.
3. $\nabla \bar u\_n$ is $L^2$-equi-integrable, i.e., for any $\epsilon>0$ we have there exists $\delta>0$ such that for all set $T\subset \Omega$ with $\mathcal L^N(T)<\delta$ we have
$$
\sup\_{n\in\mathbb N}\int\_{T} |\nabla \bar u\_n|^2dx<\epsilon. \tag 1
$$


My idea is to define
$$
\bar u\_n:=\min\_{v\in\mathcal A(u\_n)}\left\{\int\_\Omega|\nabla v^2|\,dx\right\},\text{ and }\bar u:=\min\_{v\in\mathcal A(u)}\left\{\int\_\Omega|\nabla v^2|\,dx\right\},
$$
where 
$$ 
\mathcal A(u\_n):=\left\{v\in H^1(\Omega), T[v]=T[u\_n]\right\},
$$
and $T[\cdot]$ denotes the usual trace operator.


The property $1$ is obviously true. The prove of property $2$ I put it at the end of this post. Please help me to check whether it is correct. 


However, I can not prove property $3$. The best I can do is assuming $(1)$ does not hold, i.e., there exists a sequence of set $T\_n\subset \Omega$ such that $\lim\_{n\to 0}\mathcal L^N(T\_n)=0$ and 
$$
\lim\_{n\to\infty} \int\_{T\_n}|\nabla \bar u\_n|^2dx\geq \epsilon>0
$$
and hope to have a contradiction.


We can compute
$$
\liminf\_{n\to\infty}\int\_\Omega|\nabla \bar u\_n|^2dx\geq \liminf\_{n\to\infty}\int\_{\Omega\setminus T\_n}|\nabla \bar u\_n|^2dx+\liminf\_{n\to\infty}\int\_{T\_n}|\nabla \bar u\_n|^2dx\geq \int\_\Omega|\nabla \bar u\,|^2dx+\epsilon 
$$
but I can not get any contradiction from here. I feel I need to use the minimality of $\nabla\bar u\_n$ but I don't see how...


Any help of new idea of how to construct $\bar u\_n$ is really welcome!




---


Below is how to proof property $2$.


Now let me prove property $2$. Clearly $\bar u\_n$ is bounded in $H^1$ and hence, up to a subsequence, $\bar u\_n\to u\_0$ weakly in $H^1$. I only need to prove that $u\_0=\bar u$. To do so, I only need to prove that $u\_0$ is the weak solution of PDE
$$
\begin{cases}
-\Delta v=0, & x\in\Omega\\
v=u, & x\in\partial\Omega
\end{cases}
$$
By weak convergence in $H^1$, we have 
$$
\int\_\Omega \nabla u\_0\nabla \phi=0
$$
for all $\phi\in H\_0^1(\Omega)$. I only need to prove that $u\_0\in \mathcal A(u)$ then I would be done. To do so, I need to prove $u\_0-u\in H\_0^1(\Omega)$. I will claim
$$
\left|\int\_\Omega (u\_0-u)(x) \partial\_i\varphi(x)dx\right|\leq C\|\varphi\|\_{L^2(\Omega)}
$$
for all $\phi\in C\_c^\infty(\mathbb R^N)$.


We observe that 
\begin{multline}
\left|\int\_\Omega (u\_0-u)(x) \partial\_i\varphi(x)dx\right|=\\
\lim\_{n\to\infty}\left|\int\_\Omega (\bar u\_n-u\_n)(x) \partial\_i\varphi(x)dx\right|=\lim\_{n\to\infty}\left|\int\_\Omega \partial x\_i(\bar u\_n-u\_n)(x) \varphi(x)dx\right|\\
\leq \lim\_{n\to\infty}\|\nabla (\bar u\_n-u\_n)\|\_{L^2}\|v\|\_{L^2}\leq C\|v\|\_{L^2}
\end{multline}
as desired, where the 3rd inequality used the fact that $T[\bar u\_n-u\_n]\equiv 0$.


Hence, by the uniqueness of solution, we have $u\_0=\bar u$, and hence property $2$ is true.




---


PS: I also post this problem in Math Stack Exchange [here](https://math.stackexchange.com/questions/1336295/how-to-modify-a-h1-weak-convergence-sequence-so-that-i-have-the-l2-equi-in) because this post is just an update of my yesterday's post which exist both on math Stack Exchange as well...I will avoid this problem next time. Sorry!



How to modify a $H^1$ weak convergence sequence so that I have the $L^2$ equi-integrability of gradient?
I think that the construction that you are proposing by harmonic extension does not work. Indeed consider on the unit disk $\mathbb{B}^2 \subset \mathbb{R}^2$ the function $u\_n : \mathbb{B}^2 \to \mathbb{R}$ defined for $x = (x\_1, x\_2) \in \mathbb{B}^2$
$$
 u\_n (x\_1, x\_2) = \frac{\operatorname{Re} \bigl((x\_1 + i x\_2)^n\bigr)}{\sqrt{n}}.
$$


It can be checked that the function $u\_n$ is harmonic on the disk $\mathbb{B}^2$, therefore $\bar{u}\_n = u\_n$.


Moreover, $u\_n \to 0$ almost everywhere on $\mathbb{B}^2$ as $n \to \infty$,
$$
 \int\_{\mathbb{B}^2} \vert \nabla u\_n \vert^2 = C,
$$
where $C > 0$ does not depend on $n$, and for every compact set $K \subset \mathbb{B}^2$,
$$
 \lim\_{n \to \infty} \int\_{K} \vert \nabla u\_n \vert^2 = 0.
$$
It follows from these facts that $u\_n \rightharpoonup \bar{0} = 0$ in $H^1 (\mathbb{B}^2)$ and that the sequence $(\nabla u\_n)\_{n \in \mathbb{N}}$ is not equiintegrable.


