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I'm trying to show that if $f\in L^1_\text{loc}(\Omega)$ and $\int_\Omega f\varphi=0$ for every $\varphi\in C_c^\infty(\Omega)$, then $f=0$ a.e. I know this to be true in the case $f\in L^2(\Omega)$. Clearly we may assume $\Omega$ is bounded for this. Intuitively, it makes sense because $f^2$ will not be "too large" on a precompact domain to integrate. I know that it's measurable. An equivalent question is: "is $L^1_\text{loc}(\Omega)$ an algebra under pointwise multiplication?"

Ryan Unger
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    Cover $\Omega$ by opens $U_i$ such that $f_{|U_i} \in L^1(U_i)$. Show that $f_{|U_i} = 0$. This is called the fundamental lemma of calculus of variations. For this you can use the density of continous functions in $L^1$. –  Nov 15 '16 at 16:46
  • @N.H. Yeah, it's obviously true for $f$ continuous as continuous functions are locally $L^2$. Will I need to use the dominated convergence theorem here? – Ryan Unger Nov 15 '16 at 16:50
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    $L^1_{\text{loc}} \not\subset L^2$. Consider $f(x) = \lvert x\rvert^{-2/3}$ on $(-1,1)$. – Daniel Fischer Nov 15 '16 at 16:52
  • I don't think you need dominated convergence (but I might be mistaken). You can approximate $f$ by $g$ continuous with small error and show that $g$ itself is small, using the hypothesis that the integral of $f$ against any smooth compactly-supported function is $0$. Then you can conclude that $f$ is arbitrary small, i.e zero. –  Nov 15 '16 at 16:56
  • @DanielFischer So the product of locally integrable functions is generally not locally integrable? – Ryan Unger Nov 15 '16 at 17:02
  • Right. You need growth conditions. If $\frac{1}{p} + \frac{1}{q} = 1$, then the product of an $L^p_{\etx{loc}}$ function and an $L^q_{\text{loc}}$ function is in $L^1_{\text{loc}}$. – Daniel Fischer Nov 15 '16 at 17:06
  • For a proof for what you're trying to show, see my answer to this question: http://math.stackexchange.com/questions/2008155/proving-that-locally-integrable-functions-are-embedded-in-the-space-of-distribut/2008467#2008467 –  Nov 15 '16 at 17:18
  • @OpenBall Nice! – Ryan Unger Nov 15 '16 at 17:28

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