Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $u\in L^{1}_{loc}(\Omega):=\{f:\Omega \to \mathbb R\;| \int_{K}|f(x)|dx<\infty,\;K\subset\Omega\; \mathtt{compact}\} $, where $\Omega\subset\mathbb{R}^{n}$, and let $\phi$ be a test function on $\Omega$. Show that $$ \left(\forall\phi\in \mathbb{D}(\Omega):\int\limits_{\Omega}u(x)\phi(x)dx=0\right)\implies \left(u =0\; \mathtt{a.e.}\right) $$ Here's how I started. Take some $\phi$ in $\mathbb{D}(\Omega)$. Then there is a compact set $K$ such that $\{x: \phi(x)\neq 0\}\subset K$. Since $\phi$ is continous, it is bounded on $K$. Let $M=\mathtt{sup}\{|\phi(x)|:x\in K\}$ $$ 0=|\int_{\Omega}u(x)\phi(x)dx|\leq\int_{\Omega}|u(x)||\phi(x)|dx\leq M\int_{K}|u(x)|dx$$ But the last expression is greater or equal to $0$, hence $u$ must vanish almost everywhere in $K$. Now, my questions are: 1) Is it enough to show that $u$ disappears a.e. on any compact subset of $\Omega$? 2) If so, is there a way to show this using what I've done above?

share|improve this question
    
See here. –  Giuseppe Negro Jan 19 '13 at 21:16
    
Thank you for the link :) –  czachur Jan 20 '13 at 21:06

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.