If for any $g$ in $L^q$, $fg$ is in $L^1$, prove that $f$ is in $L^p$.
$p>1$, $p$ and $q$ are conjugate.
It is sort of an inverse version of Holder inequality.
We are considering Lebesgue measure in $\mathbb{R}^n$.
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Doris
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Any assumptions on the underlying measure space? It is wrong at least if the measure space has infinite atoms (take $f$ the be the characteristic function of one). I am not sure if σ-finiteness is required. Or is it just Lebesgue measure on $\mathbb{R}^n$? – Harald Hanche-Olsen Dec 23 '13 at 08:18
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@HaraldHanche-Olsen just Lebesgue measure in $R^n$ – Doris Dec 23 '13 at 08:20
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You can use the closed graph theorem. By assumption, the map $g\mapsto fg$ is an everywhere defined linear map $L^q\to L^1$. To see that the graph is closed, assume that $g_n\to g$ in $L^q$ and $fg_n\to h$ in $L^1$. There is a subsequence with $g_n\to g$ a.e. and $fg_n\to h$ a.e., and these imply $fg=h$ a.e.
Now the map is bounded by the closed graph theorem. Can I ask you to take it from there? It should be enough to use the consequence that $g\mapsto\int_{\mathbb{R}^n}fg\,dx$ is a bounded linear functional on $L^q$.
Harald Hanche-Olsen
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Sorry, I haven't studied the closed graph theorem before. I am just learning some real analysis. Can you use another method? – Doris Dec 23 '13 at 08:58
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I expect this can be done using more elementary methods, but I don't have any time to think of one right now, sorry. Hopefully someone else can pitch in. – Harald Hanche-Olsen Dec 23 '13 at 09:06
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As you may have noticed, this question has been asked before. There is a more elementary answer over at the duplicated. – Harald Hanche-Olsen Dec 23 '13 at 10:15
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This is not true. Pick any $f\not \in L^p$ and let $g=0$.
Maybe, you need some condition about $g\not = 0$ a.e. ?
Nicolas Bourbaki
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1“For any” in the question must be taken to mean “for all”, not “for some” as you seem to have understood it. – Harald Hanche-Olsen Dec 23 '13 at 08:05