# Quick criterion to decide whether a limit of functions in $W^{1,p}(\Omega)$ is in that space

Let $\Omega\subset \mathbb R^N$ be an open set. Suppose we are given a sequence $u_n$ in $W^{1,p}(\Omega)$, $1<p\leq+\infty$, such that $u_n\to u$ in $L^p(\Omega)$ and such that $(\nabla u_n)$ is bounded in $\left(L^p(\Omega)\right)^N$. Then how can we conclude that $u\in W^{1,p}(\Omega)$?

I've found this question as q remark on the Brezis, however I've thought for a while and I didn't come up with an answer which should be pretty easy I think. In particular then I would like to know what goes wrong for $p=1$.

Thanks

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Suppose $p<\infty$: If you consider $L^p(\Omega)$ as the dual of $L^q(\Omega)$, then boundedness of $g_n\in L^p(\Omega)$ implies that a subsequence $g_{n_k}$ converges weakly to some $g\in L^p(\Omega)$.

Applying this to the sequence $\nabla u_n$ you obtain a subsequence which converges weakly in $L^p(\Omega)^N$ to some $v$ and $$\int v \varphi = \lim_{k} \int \nabla u_{n_k} \varphi = \lim_k \; - \int u_{n_k} \nabla \varphi =- \int u \nabla \varphi \quad \forall \varphi \in C_c^\infty(\Omega)$$ proves $\nabla u = v$ and $u\in W^{1,p}(\Omega)$.

For $p=\infty$, you could boot-strap the result for $p<\infty$: First pick any $q<\infty$, then it follows by the above and the inclusion $L^\infty(\Omega')\subset L^q(\Omega')$, that $u\in W^{1,q}(\Omega')$ for any $\Omega'\subset \subset \Omega$. And furthermore $$\Vert \nabla u \Vert_{L^q(\Omega')} \le \sup_n \Vert \nabla u_n\Vert_{L^q(\Omega')} \le \sup_n \Vert u_n\Vert_{L^\infty(\Omega)} \; |\Omega'|^{1/q}$$ Letting $q\to \infty$ gives $\Vert \nabla u\Vert_{L^\infty(\Omega')} \le \sup_n \Vert u_n\Vert_{L^\infty(\Omega)}$. Since this is true for any $\Omega'\subset\subset\Omega$, we conclude that $\Vert \nabla u\Vert_{L^\infty(\Omega)} \le \sup_n \Vert u_n\Vert_{L^\infty(\Omega)}$ and hence $u\in W^{1,\infty}(\Omega)$.

I'm not sure if this is the easiest way to see it, though.

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For $p=\infty$ we can use the same argument Sam L. used in the case $1<p<\infty$, using the fact that a bounded sequence in $L^\infty$ is weakly-$\ast$ relatively compact. The problem in the case $p=1$ is the lack of compactness, i.e. we can not extract a convergent subsequence from a bounded sequence in $L^1$.

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