Convergence in $L^p$ plus bounded gradient implies that the limit belongs to $W^{1,p}$?

Question

I have a question with this problem I have found in the latest edition of the book Functional analysis, Sobolev Spaces author Haim Brezis pag 264 Remark 4

Let $(u_n) \subset W^{1,p} $ such that $u_n \longrightarrow u$ in $L^p$ and $(\nabla u_n)$ is bounded in $(L^p)^N$ to conclude that $u \in W^{1,p} $.

I appreciate any help beforehand.

Answer 1

First note that the claim does not hold for $p=1$, even in the $1$-dimensional case. As a counterexample, consider

$$ u_{n}\left(x\right):=\begin{cases} nx, & x\in\left[0,\frac{1}{n}\right]\\ 1, & x\in\left[\frac{1}{n},1-\frac{1}{n}\right]\\ \left(1-x\right)n, & x\in\left[1-\frac{1}{n},1\right]\\ 0, & \text{otherwise}. \end{cases} $$

(Show that this is indeed a counterexample!)

So there has to be something special about the case $p=1$. The answer here is (the characterization of Sobolev spaces by) duality.

For $1 < p \leq \infty$ and $f \in L^p(\Omega)$, we have $f \in W^{1,p}(\Omega)$ iff

$$ \left| \int_\Omega f(x) \partial_i g(x) \, dx \right| \leq C \cdot \Vert g \Vert_{p'} $$

holds for all $i=1,\dots,N$ and all $g \in C_c^\infty(\Omega)$, where $p'$ is the conjugate exponent of $p$.

The direction "$\Rightarrow$" of the claim even holds for $p=1$ and only uses the Hölder inequality as well as the definition of the weak derivative.

For the direction "$\Leftarrow$", note that the above implies that the functional

$$ \varphi : C_c^\infty(\Omega) \rightarrow \mathbb{K}, g \mapsto \int_\Omega f(x) \partial_i g(x) \, dx $$

is bounded, when $C_c^\infty(\Omega)$ is considered as a subspaces of $L^{p'}(\Omega)$. By Hahn-Banach, there exists a continuous, linear extension $\tilde{\varphi}$ of $\varphi$ on the whole space $L^{p'}(\Omega)$.

Now we use that $1 < p \leq \infty$ and thus $1 \leq p' < \infty$, so that we now the dual of $L^{p'}$, namely $L^p$. Thus, there is some $f_i \in L^p(\Omega)$ with

$$ \tilde{\varphi}(g) = \int_\Omega f_i (x) g(x) \,dx $$

holds for all $g \in L^{p'}(\Omega)$, in particular for $g\in C_c^\infty(\Omega)$. Use the definition of $\tilde{\varphi}, \varphi$ to conclude $\partial_i f = f_i \in L^p$ and thus $f \in W^{1,p}(\Omega)$.

Finally, in your situation, you have

$$ \left| \int_\Omega u(x) \cdot \partial_i g(x) \,dx \right| = \lim_n \left|\int_\Omega u_n (x) \cdot \partial_i g(x) \,dx \right| = \lim_n \left|\int_\Omega \partial_i u_n(x) \cdot g(x) \,dx \right| \leq C \cdot \Vert g \Vert_{p'}, $$

where in the last step the uniform boundedness of $\nabla u_n$ in $L^p$ (by $C$) was used.