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I have a bounded sequence $\{u_n\}$ in $H_0^1(\Omega)$. Then, a general result about Hilbert spaces and weak convergence, implies that there exists a subsequence $\{u_{\sigma(n)}\}$ in $H_0^1(\Omega)$ such that $u_{\sigma(n)} \rightharpoonup u$ where $u \in H_0^1(\Omega)$.

Then, as $H_0^1(\Omega)$ is compactly embedded in $L^2(\Omega)$, the sequence $\{u_{\sigma(n)}\}$ converges strongly to $u$ in $L^2(\Omega)$ (meaning in the $L^2$ norm).

I would now like to deduce that $\| \nabla u \|_{L^2} \leq \lim \inf \| \nabla u_{\sigma(n)} \|_{L^2} $. This would be an easy consequence of $u_{\sigma(n)} \rightharpoonup u$ in $L^2(\Omega)$, itself immediate if $u_{\sigma(n)} \to u$ in $L^2(\Omega)$, therefore this is what I would like to prove.

I proceed as follows.

(1) For convenience, relabel $v_n := u_{\sigma(n)}$ and $v:=u$. Then, since $v_n$ and $v$ are in $H_0^1(\Omega)$, they posses weak first derivatives satisfying, for each $i$:

$$\int_\Omega v_n \frac{\partial \varphi}{\partial x_i} \, \mathrm{d}x = - \int_\Omega \varphi \frac{\partial v_n}{\partial x_i} \, \mathrm{d}x \, , \qquad \forall \, \varphi \in C_0^\infty(\Omega)\, .$$

Hence we can write, for any $\varphi \in C_0^\infty(\Omega)$: \begin{align} \int_\Omega v \frac{\partial \varphi}{\partial x_i} \, \mathrm{d}x &= \lim_{n \to \infty} \int_\Omega v_n \frac{\partial \varphi}{\partial x_i}\, \mathrm{d}x \tag{2} \\ &= - \lim_{n \to \infty} \int_\Omega \varphi \frac{\partial v_n}{\partial x_i} \, \mathrm{d}x \tag{3} \\ &= - \int_\Omega \varphi \left( \lim_{n \to \infty}\frac{\partial v_n}{\partial x_i} \right) \mathrm{d}x \tag{4} \end{align} so that indeed: $$\frac{\partial v}{\partial x_i} = \lim_{n \to \infty}\frac{\partial v_n}{\partial x_i} \tag{5}$$ for all $i$ and hence (6) $\nabla v_n \to \nabla v$ in $L^2(\Omega)$.

There are a few points which do not completely convince me.

  1. I hope everything before (1) is correct.
  2. This should be true since $v_n \to v$ in $L^2(\Omega)$.
  3. Here I am using the definition of weak derivative.
  4. I am not completely convinced that is allowed. Also, I get a bit confused since the previous two limits where limits in $\mathbb{R}$ (the first one the definition of weak convergence in $L^2(\Omega)$, and since $v_n \to v$ then most certainly $v_n \rightharpoonup v$), while this one is a limit in $L^2(\Omega)$ I guess...
  5. Linked to the previous point, I am not sure if this limit is $L^2(\Omega)$ and if so, why...
  6. I presume that since each of the components of the first converge to the corresponding component of the second, the limit holds.

Any clarification on these points would be much appreciated.

EDIT: Thanks to Fnacool's answer, I understand that we don't need to establish convergence in $L^2$ to get the result that I wanted. However, I am still curious as to whether or not we can infer that $v_n$ converges (weakly or strongly) in $L^2$.

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    $\begingroup$ The whole thing with integrals and derivatives is complicated. It's easier to work in greater generality: if $x_n\rightharpoonup x$ and $T:X\to Y$ is a bounded linear operator, then $Tx_n\rightharpoonup Tx$ in $Y$. Works for any two Banach spaces. $\endgroup$
    – user147263
    Commented Mar 22, 2016 at 1:03

1 Answer 1

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Suppose $$x_n \to x $$ weakly. Then $$(x_n , x) \to (x,x)=\|x\|^2.$$

However, $|(x_n,x)|\le \|x_n\|\|x\|$. Therefore, $\liminf \|x_n\|\ge \|x\|$.

Going back to your problem, taking $x_n$ to be the weakly convergent sequence $(u_{\sigma(n)})$ in $H_0^1$, with limit $u$, we have

$$\liminf\left ( \int |\nabla u_{\sigma(n)}|^2 + \int u_{\sigma(n)} ^2 \right)\ge \int |\nabla u|^2 + \int u^2 .$$

But the second integral on LHS converges to the second integral on RHS, so we obtain

$$ \liminf \int |\nabla u_{\sigma(n)}|^2 \ge \int |\nabla u|^2.$$

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  • $\begingroup$ Thank you very much for this, I understand that we don't need to establish convergence in $L^2$ to get the result that I wanted. However, I am still curious as to whether or not we can infer that $v_n$ converges (weakly or strongly) in $L^2$. $\endgroup$
    – Dem K.
    Commented Mar 22, 2016 at 11:36

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