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the following is an assertion in the book of Haim Brezis.I want to know how to prove it.Thank you very much!

Assertion:

When $1＜p\le \infty$,it suffices to know that $u_n\to u$ in $L^p(\Omega)$ and that $(\nabla u_n)$ is bounded in $(L^p(\Omega))^N$ to conclude that $u\in W^{1,p}(\Omega)$.

Here $\Omega$ is an open set in $\mathbb{R}^N$.

A little complement:

If $(\nabla u_n)$ converges to some limit in $(L^p(\Omega))^N$,then the result is obvious.The assertion wants to tell us that we don't need this strong condition when $1＜p\le \infty$.

Eric
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1 Answers1

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Oh,I got an idea.If there's something wrong,hope any friend can point it out.

From the conditions above,we get $$\begin{aligned}\int u\frac{\partial\phi}{\partial x_i}&=\lim_{n\to \infty}\int u_n\frac{\partial \phi}{\partial x_i}\\&=\lim_{n\to \infty} \int\frac{\partial u_n}{\partial x_i}\phi \end{aligned}$$

We can regard the last one as a continuous linear functional of $\phi$,since by Holder inequality,we have $$\int\frac{\partial u_n}{\partial x_i}\phi\le \vert\vert \frac{\partial u_n}{\partial x_i}\vert\vert_p\ \vert\vert \phi\vert\vert_q$$,here $\frac 1p+\frac1q=1$.

Then by Riesz respentation theorem,there exists $g\in L^p$ ,st.$$\int g\phi=\lim_{n\to \infty}\int\frac{\partial u_n}{\partial x_i}\phi=\int u\frac{\partial\phi}{\partial x_i}$$

So we get the weak partial derivatives of $u$,which are in $L^p$,and we reach the result.

Eric
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