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 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!

  • $\begingroup$ Since $u_n\to u$ in $H^1$, you have $\int|\nabla u_n|^2\leq1+\int|\nabla u|^2$ for $n$ large enough. Then by the energy minimizing property $\int|\nabla\bar u|^2$ is bounded uniformly in $n$. Is this what you want or did I miss something? $\endgroup$ Commented Jun 24, 2015 at 21:20
  • $\begingroup$ @JoonasIlmavirta ah not really. Please see equation $(1)$ for details explanation. $\endgroup$
    – JumpJump
    Commented Jun 24, 2015 at 23:30
  • $\begingroup$ @Denoising: How about $\bar u = \bar u_n \equiv 0$? This satisfies 1-3. $\endgroup$
    – gerw
    Commented Jul 14, 2015 at 10:39
  • $\begingroup$ @gerw of course you are right. but I wish to preserve the boundary condition as well. Actually I will close this problem since I need to add/cancel some assumptions. Thank you anyway! $\endgroup$
    – JumpJump
    Commented Jul 14, 2015 at 13:50
  • $\begingroup$ If you like the answer you should upvote it. $\endgroup$ Commented Apr 10, 2018 at 19:47

1 Answer 1


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.

  • $\begingroup$ I see. Yea your example works... by any chance, do you know some known ways to turn/truncate a sequence to a qui-integrable sequence and somehow has the properties that I wish to have? Thank you! $\endgroup$
    – JumpJump
    Commented Jun 25, 2015 at 12:12

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