Take $\Omega\subset \mathbb R^N$, open bounded, smooth boundary. Take $u_n\subset L^1$ a sequence of functions so that $u_n\to u$ strongly in $L^1$ and $$ \sup_n\int_\Omega |\nabla u_n|^2dx<\infty,\,\,\int_\Omega |\nabla u|^2dx<\infty $$

My question: Can we show that $$ \liminf \int_\Omega |\nabla u_n|^2dx\geq \int_\Omega |\nabla u|^2dx $$

I got stuck on how to show that $u_n\in H^1(\Omega)$. Because $u_n$ is not in $L^2$, I don't know whether the embedding can work or not.


The $L^2$-norm is weakly sequentially lower semicontinuous, then you can prove the desired inequality with the liminf provided you show that $\nabla u_n \rightharpoonup \nabla u$ in $L^2$. The fastest way to achieve this is probably to show that $\{u_n\}$ is bounded in $H^1(\Omega)$ (and then use the sequential Banach-Alaoglu theorem together with uniqueness of the weak limit, which will imply that $\nabla u_n$ weakly converges to the right stuff, i.e. $\nabla u$).

Then let me answer the question in the title: take $v \in L^1(\Omega)$ such that $\nabla v \in L^2(\Omega)$ and let's prove that $v \in H^1(\Omega)$. I am going to apply Poincare's inequality, so we really need the smoothness assumption on the domain (Lipschitz is all we need actually).

Let $v_k$ be a truncation of $v$, defined as follows:

$$ v_k(x) = \begin{cases} -k &\ \text{if}\ v(x) < -k \\ k &\ \text{if}\ v(x) > k \\ v(x) &\ \text{otherwise}. \end{cases} $$ Then we have that $v_k \in L^2(\Omega)$ and $|\nabla v_k| \le |\nabla v|$, indeed

$$ \nabla v_k(x) = \begin{cases} 0 &\ \text{if}\ v(x) < -k \\ 0 &\ \text{if}\ v(x) > k \\ \nabla v(x) &\ \text{otherwise}. \end{cases} $$

Now let $E$ be your favorite set of positive measure contained in $\Omega$ and let $$w_E := \frac{1}{\mathcal{L}^N(E)}\int_Ew(x)\,dx.$$ Then, by Poincare's inequality we get $$\int_{\Omega}|v_k(x) - (v_k)_E|^2\,dx \le C(\Omega,E)\int_{\Omega}|\nabla v_k(x)|^2\, dx \le C(\Omega,E)\int_{\Omega}|\nabla v(x)|^2\, dx$$ By Fatou's lemma we then get $$\int_{\Omega}|v(x) - v_E|^2\,dx \le C(\Omega,E)\int_{\Omega}|\nabla v(x)|^2\, dx < \infty.$$

This shows that if $\Omega$ is regular enough $$\{u : u \in L^1(\Omega), \nabla u \in L^2(\Omega)\} = H^1(\Omega).$$

  • $\begingroup$ From your last inequality, how do you get ride of $v_e$? $\endgroup$ – spatially Nov 9 '15 at 17:55
  • $\begingroup$ $L^2$ is a vector space and the constant $v_E$ belongs to the space since the domain is bounded :) $\endgroup$ – Giovanni Nov 9 '15 at 18:42
  • $\begingroup$ I know it is bounded, but the upper bounded of $\|v\|_{L^2}$ will be depends on the average value of $v$ right? In this case we may not have a uniform bound of the $L^2$ norm of $u_n$. ($u_n$ is the sequence I mentioned in my post) $\endgroup$ – spatially Nov 9 '15 at 19:12
  • $\begingroup$ The fact that "depends" on the average is not a problem, indeed $u_n$ converges strongly in $L^1$, then the $L^1$-norm of $u_n$, which is larger than the average, can be bounded by $C\|u\|_{L^1}$. $\endgroup$ – Giovanni Nov 9 '15 at 19:16
  • $\begingroup$ Yes, but in your equation, we shall have the $L^2$ norm of $v_E$ right? Which can not be bounded by $L^1$ norm of it. $\endgroup$ – spatially Nov 9 '15 at 19:48

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