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Let $f$ be a measurable function on a measure space $X$ and suppose that $fg \in L^1$ for all $g\in L^q$. Must $f$ be in $L^p$, for $p$ the conjugate of $q$? If we assume that $\|fg\|_1 \leq C\|g\|_q$ for some constant $C$, this follows from the Riesz Representation theorem. But what if we aren't given that such a $C$ exists?

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Is the measure finite? –  Dylan Moreland Sep 2 '11 at 19:22
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I recommend checking out this related question for inspiration. –  Jonas Meyer Sep 2 '11 at 19:29
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Hint: the closed graph theorem. –  Nate Eldredge Sep 2 '11 at 19:42
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I recommend checking out this related question for inspiration... :-) –  Did Sep 2 '11 at 21:23
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@Davide: in case you haven't seen robjohn's answer here: for the duality between $L^p$ and $L^q$ with $1 \lt p \lt \infty$ you don't need any hypotheses on the measure space. For the duality between $L^1$ and $L^\infty$ you need something, $\sigma$-finiteness is enough. In fact you can show that the natural map $L^\infty \to (L^1)^\ast$ is an isomorphism if and only if the measure space is "localizable" (this condition allows you to patch together an arbitrary family of measurable functions to a measurable function defined on the entire space). –  t.b. Nov 24 '11 at 16:30

2 Answers 2

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Suppose that $fg\in L^1$, but there is no $C$ so that $\|fg\|_{L^1}\le C\|g\|_{L^q}$. Without loss of generality, we can assume all functions are positive. Suppose we have a sequence of $L^q$ functions $\{g_k:\|g_k\|_{L^q}=1\}$ where $\int|fg_k|\;\mathrm{d}x>3^k$. Set $g=\sum\limits_{k=1}^\infty2^{-k}g_k$. $\|g\|_{L^q}\le1$ yet $fg\not\in L^1$. Thus, there must be a $C$ so that $\|fg\|_{L^1}\le C\|g\|_{L^q}$. Then, as you say, apply the Riesz Representation Theorem.

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Since Didier said one should let oneself be inspired by this question: note that a very similar gliding hump argument is used by Sokal's article containing a proof of the uniform boundedness principle, posted in a comment there. This ties in nicely with Davide's suggestion... –  t.b. Sep 3 '11 at 9:23
    
@Theo: Thanks for the references. Reading his paper, my answer is quite similar to Alan Sokal's proof of the Uniform Boundedness Principle. –  robjohn Sep 3 '11 at 13:10
    
@robjohn: do you mind elaborating on your proof a bit. the part where you showed that $\|fg\|_{L^1} \leq C \|g\|_{L^q}$? Thanks –  Kuku May 7 '12 at 22:08
    
@kuku: This is an indirect argument. I started by assuming there was no such $C$ so that $\|fg\|_{L^1}\le C\|g\|_{L^q}$ for all $g\in L^q$. Using that assumption, I derived a contradiction, which shows that the assumption was false. Thus, there must be a $C$ so that $\|fg\|_{L^1}\le C\|g\|_{L^q}$. –  robjohn May 7 '12 at 22:37
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@kuku: Since $\|g_k\|_{L^q}=1$, $$ \begin{align} \|g\|_{L^q} &\le\sum_{k=1}^\infty 2^{-k}\|g_k\|_{L^q}\\ &=1 \end{align} $$ Since $\int fg_k\;\mathrm{d}x>3^k$ (we were assuming all functions were non-negative) $$ \begin{align} \int fg\;\mathrm{d}x &=\sum_{k=1}^\infty2^{-k}\int fg_k\;\mathrm{d}x\\ &>\sum_{k=1}^\infty2^{-k}3^k \end{align} $$ which diverges. –  robjohn May 7 '12 at 23:17

I would like to add another answer to this question. Consider the linear functional $T:L^q\to\mathbb{C}$ defined by $$Tg=\int gf.$$

It is sufficient to prove that $T$ is bounded. To this end, assume that $g_n\in L^q$ is such that $\|g_n\|_q\to 0$.

We can extract a subsequence of $g_n$ (not relabeled) such that $$g_n(x)\to 0,\ |g_n(x)|\le h(x),\ \mbox{a.e.},\tag{1}$$

where $h\in L^q$ (this partial converse of Lebesgue theorem, can be found, for example, in Rudin's book "Real and Complex Analysis", Theorem 3.12, or in Brezis book "Functional Analysis", Theorem 4.9). It follows from $(1)$ that $$g_n(x)f(x)\to 0,\ |g_n(x) f(x)|\leq h(x)|f(x)|,\ \mbox{a.e}.\tag{2}$$ Note that by hypothesis, $h|f|\in L^1$, therefore, we can apply Lebesgue theorem to conclude that $$Tg_n\to 0.\tag{3}$$

As every subsequence of $g_n$, has a subsequence which satisfies $(1)$, we conclude that $(3)$ is true for the whole sequence.

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I don't understand: are you trying to show that $f\in L^q$?. –  robjohn Jun 13 at 20:05
    
You've shown that $T$ is continuous. Therefore, it is bounded. Thus, you can use Riesz to get an $\tilde{f}\in L^p$ so that $\int\tilde{f}g\,\mathrm{d}x=Tg=\int fg\,\mathrm{d}x$ for all $g\in L^q$. Then you can show that $f=\tilde{f}$ a.e. so that $f\in L^p$. However, the proof of $(1)$ seems quite non-trivial. –  robjohn Jun 13 at 23:36
    
@robjohn, I would say that non-trivial is too much. If you take a look in the proof of $(1)$, you will see that all amounts to prove that, any subsequence $g_{n_i}$ of $g_n$, which satisfies $$\|g_{n_i+1}-g_{n_i}\|_q<2^{-i},$$ satisfies $(1)$ and it can be proved only by using the basics of measure theory. –  Tomás Jun 13 at 23:52
    
Yes, we can choose a sub-sequence $\|g_n\|_q\le2^{-n}$ and let $h(x)=\sum\limits_{n=1}^\infty|g_n(x)|$. Then $h\in L^q$ and $|g_n(x)|\le h(x)$. How do we arrange it so that $\lim\limits_{n\to\infty}g_n(x)=0$ for all (or even almost all) $x$? That is the part that seems less than trivial. –  robjohn Jun 14 at 0:52
    
@robjohn, do you think my answer is totally useless and deserves to be deleted? If so, I will have no problem in deleting it. –  Tomás Jun 14 at 1:00

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