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

We define $\mathcal{S}:C_{0}^{\infty}(D) \times C_{0}^{\infty}(D)\to \mathbb{R}$ by

\begin{eqnarray*} \mathcal{S}(u,v)=\sum_{i,j=1}^{d} \int_{D} \frac{\partial u}{\partial x_{i}} \frac{\partial v}{\partial x_{j}}a_{i,j} \,d \mu\quad(u,v \in C_{0}^{\infty}(D)) \end{eqnarray*}

We assume the following :

$\cdot$ There exists $(u_{n})_{n=1}^{\infty} \subset C_{0}^{\infty}(D)$ such that for all $K \subset D$, compact

\begin{eqnarray*} \sum_{i,j=1}^{d} \int_{K} \frac{\partial u_{n}}{\partial x_{i}} \frac{\partial u_{n}}{\partial x_{j}}a_{i,j} \,d \mu\ \to 0 \quad(n\to \infty) \end{eqnarray*}

For $(u_{n})_{n=1}^{\infty}\subset C_{0}^{\infty}(D)$ above, I want to prove $\mathcal{S}(u_{n},u_{n})\to 0 \quad(n \to \infty)$

Proof (unfinished)

Take compact sets $(K_{l})_{l=1}^{\infty}$ such that $K_{1} \subset K_{2} \subset \cdots,\, \cup_{l=1}^{\infty}K_{l}=D$ then

\begin{eqnarray*} \sum_{i,j=1}^{d} \int_{K_{l}} \frac{\partial u_{n}}{\partial x_{i}} \frac{\partial u_{n}}{\partial x_{j}}a_{i,j} \,d \mu\ \to \mathcal{S}(u_{n},u_{n}) \quad(l\to \infty) \end{eqnarray*} and for each $l$

\begin{eqnarray*} \sum_{i,j=1}^{d} \int_{K_{l}} \frac{\partial u_{n}}{\partial x_{i}} \frac{\partial u_{n}}{\partial x_{j}}a_{i,j} \,d \mu\ \to 0 \quad(n\to \infty) \end{eqnarray*}

I think these are useful but It has gone wrong. Is there any good method ?　Thanks.

ko4
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1 Answers1

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You cannot prove that, because it is false. For a counterexample, let us consider the case $d=1$ and $a_{1,1}=1$ and $D=\left(0,1\right)$, i.e. $$ \mathcal{S}\left(u,v\right)=\int_{0}^{1}u'\left(x\right)\cdot v'\left(x\right)\, dx. $$

Now choose a sequence of functions $\left(u_{n}\right)_{n\in\mathbb{N}}$ in $C_{c}^{\infty}\left(D\right)$ with ${\rm supp}\left(u_{n}\right)\subset\left(0,\frac{1}{n}\right)$ and $$ \mathcal{S}\left(u_{n},u_{n}\right)=\int_{0}^{1}\left|u_{n}'\left(x\right)\right|^{2}\, dx=1 $$ for all $n\in\mathbb{N}$ (this can surely be done by rescaling).

Then for $K\subset D$ compact, we have $K\subset\left(\varepsilon,1\right)$ for suitable $\varepsilon>0$ and thus ${\rm supp}\left(u_{n}\right)\cap K=\emptyset$ for large enough $n\in\mathbb{N}$ (depending on $\varepsilon$ and hence on $K$) and thus $$ \int_{K}u_{n}'\left(x\right)\cdot u_{n}'\left(x\right)\, dx=0 $$ for large enough $n\in\mathbb{N}$ (depending on $K$).

PhoemueX
  • 35,087