Local ellipticity, $L^2$ convergence

Question

Let $U \subset \mathbb{R}^{d}$ be open and $a_{ij} \in L^{1}_{loc}(U; dx)$, $a_{ij}=a_{ji}$, $1\leq i,j \leq d$ ( $dx$ is Lebesgue measure on $U$ ).

Suppose that the following condition hold:

For every $K \subset U$, $K$ compact, there exists a constant $C_{K}$ such that

\begin{eqnarray*} {\rm for}\,\mu-a.e. x \in K &&\\ \sum_{i,j=1}^{d}a_{ij}(x)\xi_{i}\xi_{j} &\geq& C_{K} \| \xi \|^{2}_{\mathbb{R}^{d}}\,{\rm for \,all\,}\,\xi=(\xi_{1},...,\xi_{d}) \in \mathbb{R}^{d} \end{eqnarray*}

Define \begin{eqnarray*} \mathcal{A}(u,v)=\sum_{i,j=1}^{d} \int_{U} \frac{\partial u}{\partial x_{i}}\frac{\partial v}{\partial x_{j}} a_{ij} dx \quad(u,v \in C_{0}^{\infty}(U)) \end{eqnarray*}

I want to prove the following　assertion:

If for all $u_{n} \in C_{0}^{\infty}(U),\,n\in \mathbb{N}$, such that $\lim_{n,m \to \infty}\mathcal{A}(u_{n}-u_{m},u_{n}-u_{m})= 0$ and $u_{n} \to 0$ in $L^{2}(U,dx)$, it follows that $\mathcal{A}(u_{n},u_{n})\to 0$

My unfinished proof is as follows.

Write $U$ as a countable union of compact sets $K_{l}$ with $K_{1} \subset K_{2} \subset ... $

Then \begin{eqnarray*} C_{1} \int_{K_{1}} \sum_{i,j=1}^{d} \left |\frac{\partial }{\partial x_{i}}(u_{n}-u_{m}) \right|^{2}dx &\leq& \int_{K_{1}} \sum_{i,j=1}^{d} \frac{\partial }{\partial x_{i}}(u_{n}-u_{m})\frac{\partial }{\partial x_{j}}(u_{n}-u_{m})a_{ij} dx \\ &\leq&\int_{K_{2}} \sum_{i,j=1}^{d} \frac{\partial }{\partial x_{i}}(u_{n}-u_{m})\frac{\partial }{\partial x_{j}}(u_{n}-u_{m})a_{ij} dx \\ &...&\\ &\leq&\int_{U} \sum_{i,j=1}^{d} \frac{\partial }{\partial x_{i}}(u_{n}-u_{m})\frac{\partial }{\partial x_{j}}(u_{n}-u_{m})a_{ij} dx\\ &=&\mathcal{A}(u_{n}-u_{m},u_{n}-u_{m}) \to 0 \end{eqnarray*}

Therefore $(\frac{\partial u_{n}}{\partial x_{i}})_{n=1}^{\infty}$ is Cauchy seq. in $L^{2}(K_{1}; dx)$. By the completeness of $L^{2}(K_{1};dx)$, we can find $f_{i}$ such that $\frac{\partial u_{n}}{\partial x_{i}}\to f_{i}$ in $L^{2}(K_{1};dx)$.

If "$\frac{\partial u_{n}}{\partial x_{i}}\to f_{i}$ in $L^{2}(U;dx)$ and $f_{i}=0\,dx-a.e.$" then we can find subseq. $\frac{\partial u_{n_{k}}}{\partial x_{i}} \to 0 \,dx-a.e. $ and get

\begin{eqnarray*} \mathcal{A}(u_{n},u_{n})&=&\int_{U} \lim_{k \to \infty} \sum_{i,j=1}^{d} \frac{\partial }{\partial x_{i}}(u_{n}-u_{n_{k}})\frac{\partial }{\partial x_{j}}(u_{n}-u_{n_{k}})a_{ij} dx \\ &\leq&\liminf_{k \to \infty} \int_{U} \sum_{i,j=1}^{d} \frac{\partial }{\partial x_{i}}(u_{n}-u_{n_{k}})\frac{\partial }{\partial x_{j}}(u_{n}-u_{n_{k}})a_{ij} dx \\ &=&\liminf_{k\to \infty}\mathcal{A}(u_{n}-u_{n_{k}},u_{n}-u_{n_{k}})\to 0\quad(n \to \infty) \end{eqnarray*}

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How do I get to " $\frac{\partial u_{n}}{\partial x_{i}}\to f_{i}$ in $L^{2}(U;dx)$ and $f_{i}=0\,dx-a.e.$" ? Thanks.

Answer 1

Since $\partial u_n/\partial x_i$ is $L^2$ convergent on $K_N$ for every $N$, you can use a diagonal argument to find a subsequence $(u_{n_k})$ so that $\partial u_{n_k}/\partial x_i \to f_i$ almost everywhere on $U$. If you've chosen your compact sets $K_1\subset K_2\subset \cdots$ wisely, every $v\in C^\infty_0(U)$ has its support in some $K_N$, and so
$$\int f_i\, v\,dx=\lim_k\int {\partial u_{n_k}\over \partial x_i}\, v\,dx =-\lim_k\int u_{n_k}\,{\partial v\over \partial x_i}\,dx=0.$$ This shows that $f_i=0$ almost everywhere on $U$, and now you can draw your conclusion.