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In real analysis we know that if $f\in L_{loc}^1(\Omega)$ and $\int_\Omega f\varphi=0$ for any $\varphi\in C_0^\infty(\Omega)$, then $f=0$ a.e. I am thinking if it is possible to find an example to show that the condition $f\in L_{loc}^1(\Omega)$ is necessary.

Martin Sleziak
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    I believe the $f\in L_{loc}^1(\Omega)$ is just necessary such that $\int_\Omega f\varphi$ is well-defined. – SmileyCraft Dec 24 '18 at 23:58
  • I suppose here we mean $\int_\Omega f\ \mathsf d\varphi$ where $\varphi$ is some measure? – Math1000 Dec 25 '18 at 00:03
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    @SmileyCraft why is that? I mean $f$ is measurable (presumably) and $\varphi \in C_0^\infty(\Omega)$. So $f\varphi$ is measurable. So we just need to know whether $f\varphi \in L^1(\Omega)$. And it's not clear to me that it isn't. Consider, for example, $\Omega = {|z| < 1} \subseteq \mathbb{R}^2$ with $f$ blowing up at the boundary. Then isn't it possible that for all smooth functions $\varphi$ vanishing at the boundary of $\Omega$, that $f\varphi \in L^1(\Omega)$? – mathworker21 Dec 25 '18 at 00:11
  • @mathworker21 Interesting point! But I do believe, but correct me if I'm wrong, that the only reason the $f\in L_{loc}^1$ restriction is given is such that $\int_\Omega f\varphi$ is well-defined. You can look through the proof and see if this is indeed the only use of the given. If I am correct, then the condition $f\in L_{loc}^1$ is not necessary, and can be replaced by "$f\varphi\in L^1(\Omega)$ for all $\varphi\in C_0^\infty(\Omega)$". – SmileyCraft Dec 25 '18 at 00:27
  • @SmileyCraft Well, that would answer the OP's question then. However, I am wondering whether there does actually exist some $f \not \in L^1_{loc}$ but with $f\varphi \in L^1(\Omega)$ for all $\varphi \in C_0^\infty(\Omega)$. Like, maybe for my example above, we could find a $\varphi$ that goes to 0 "more slowly" than how quickly $f$ blows up at the boundary. – mathworker21 Dec 25 '18 at 00:30
  • @mathworker21 That is indeed a very interesting question. Maybe post it separately? And add a link here to the question. – SmileyCraft Dec 25 '18 at 00:34
  • @SmileyCraft now I'm confusing myself. If $f$ just blows up at the boundary, it is in $L^1_{loc}$, no? – mathworker21 Dec 25 '18 at 00:42
  • @mathworker21 Right. So if $f\varphi\in L^1(\Omega)$ for all $\varphi\in C_0^\infty(\Omega)$ then for any $x\in\Omega$ we can take any compact neighborhood $K$ of finite measure and then $f$ is probably integrable in $K$, since $f\chi_K\in L^1(\Omega)$. Only problem is $\chi_K$ is not smooth. – SmileyCraft Dec 25 '18 at 00:52
  • @SmileyCraft I'm not too worried about that though. There are $C^\infty$ versions of Urysohn's lemma. Ok, so I think we've shown $f\varphi \in L^1(\Omega)$ for all $\varphi \in C_0^\infty(\Omega)$ implies $f \in L^1_{loc}(\Omega)$. Feel free to post (a slightly more detailed) proof as an answer I guess. – mathworker21 Dec 25 '18 at 01:03
  • @mathworker21 I posted a proper answer summarizing our conversation rigorously. I think we should remove all our comments here except the first three. – SmileyCraft Dec 25 '18 at 01:34
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    @SmileyCraft may I ask why? I generally think more information is better, and in this case, I think seeing productive math conversations and thought processes is helpful. I'm fine deleting though, if you want. – mathworker21 Dec 25 '18 at 01:43
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    @mathworker21 This seems like something for a meta post. https://math.meta.stackexchange.com/questions/29537/what-to-do-about-long-discussions-in-comments-and-how-to-avoid – SmileyCraft Dec 25 '18 at 02:31

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The condition $f\in L_{loc}^1(\Omega)$ is necessary for the statement to even make sense. If $f\varphi$ is integrable for all $\varphi\in C_0^\infty(\Omega)$, then $f\in L_{loc}^1(\Omega)$. Since we are considering the properties of locally integrableness and smoothness, I assume we are considering open sets $\Omega\subseteq\mathbb{R}^n$.

Proof of claim: Let $\Omega\subseteq\mathbb{R}^n$ open and $f:\Omega\to\mathbb{R}$ such that $f\varphi$ is integrable for all $\varphi\in C_0^\infty(\Omega)$. Let $K\subseteq\Omega$ compact. We can find (*) $U\subseteq\Omega$ open and $L\subseteq\Omega$ compact such that $K\subseteq U\subseteq L$. By smooth Urysohn we find $\varphi\in C^\infty$ such that $\varphi|_{K}\equiv1$ and $\mbox{supp}(\varphi)\subseteq U$. Because $U\subseteq L\subseteq\Omega$ we get $\varphi\in C_0^\infty(\Omega)$. By hypothesis, $f\varphi$ is integrable on $\Omega$, and hence also on $K$. Since $(f\varphi)|_K=f|_K$ we find that $f$ is integrable on $K$. Hence $f\in L_{loc}^1(\Omega)$.

(*) This is a bit tedious. Since $K$ is compact, it is contained in some ball $B_R(0)$. Then $S=(\Omega\cap B_R(0))^c$ is closed and disjoint from $K$. Because $\mathbb{R}^2$ is normal, we find disjoint open sets $U,V\subseteq\mathbb{R}^2$ such that $K\subseteq U$ and $S\subseteq V$. Choosing $L=V^c$ we get the desired properties.

SmileyCraft
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  • +1 nice! Just one (most likely stupid) question. Is it obvious that [$\forall x$ there exists compact $K \ni x$ s.t. $\int_K |f(y)|dy < \infty$] implies [for all compact $K, \int_K |f(y)|dy < \infty$]. The latter is what I thought the definition of $L^1_{loc}$ is. – mathworker21 Dec 25 '18 at 01:40
  • Good point, and it is less obvious than you might think, so I edited the post. I think the argument even becomes a bit nicer. – SmileyCraft Dec 25 '18 at 02:04