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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 Mathoverflow here because this post is just an update of my yesterday's post which exist on both set... Sorry for double posting here! I will avoid this situation for my next post.

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  • $\begingroup$ What is $T[-]$? $\endgroup$ – paul garrett Jun 23 '15 at 14:48
  • $\begingroup$ @paulgarrett It is trace operator. I also update it in my question. Thanks for pointing out! $\endgroup$ – spatially Jun 23 '15 at 14:55
  • $\begingroup$ Is it supposed to be $|\nabla v^2|$ or $|\nabla v|^2$? $\endgroup$ – Jose27 Jun 24 '15 at 14:17
  • $\begingroup$ @Jose27 I just completely updated my post. Please have a look. Thank you! $\endgroup$ – spatially Jun 24 '15 at 14:43
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    $\begingroup$ As stated, your goal could be reached by defining $\bar u_n=0$ and $\bar u=0$. Perhaps there is more that you want? Also, in your definition of $\bar u_n$ you probably want $\operatorname{argmin}$, not $\min$. Finally, I don't think this construction works. And if you insist on the "bar" functions having the same trace (apparently this is what you want), I don't think there is any way to achieve uniform integrability, Boundary values exhibit strong control on where the energy is large. Consider $u_n$ being an approximation to the Poisson kernel in the disk, suitably normalized. $\endgroup$ – user147263 Jun 25 '15 at 5:02
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This question has been asked and answered on MathOverflow. I have replicated the accepted answer by Jean Van Schaftingen below.

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.

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