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?
If $f$ is measurable and $fg$ is in $L^1$ for all $g \in L^q$, must $f \in L^p$?
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0Is the measure finite? – 2011-09-02
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2I recommend checking out [this related question](http://math.stackexchange.com/questions/37647/if-sum-a-n-b-n-infty-for-all-b-n-in-ell2-then-a-n-in-ell2) for inspiration. – 2011-09-02
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1Hint: the closed graph theorem. – 2011-09-02
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0If the space is $\sigma$-finite, we can use the principle of uniform boundedness. – 2011-09-02
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1I recommend checking out [this related question](http://math.stackexchange.com/questions/58565) for inspiration... :-) – 2011-09-02
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4@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). – 2011-11-24
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0@t.b. Thanks, in fact it work even if we don't assume the $\sigma$-finiteness. – 2011-11-26
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0@DavideGiraudo: Can you give an outline for a proof of the duality without assuming $\sigma$-finiteness? I found the proofs I know at this stage are overly complicated. If I only know $|g|\in L_{1}$, it is hard to show it is actually in $L_{q}$. – 2013-01-03
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0Since when $p=1$, the duality of $L^p(\mu)$ and $L^q(\mu)$ does not hold, how to prove it when $p=1$? – 2013-10-24
3 Answers
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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0Since Didier said one should let oneself be inspired by [this question](http://math.stackexchange.com/questions/58565): note that a very similar gliding hump argument is used by [Sokal's article](http://arxiv.org/abs/1005.1585) containing a proof of the uniform boundedness principle, posted in a comment there. This ties in nicely with Davide's suggestion... – 2011-09-03
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0@Theo: Thanks for the references. Reading his paper, my answer is quite similar to Alan Sokal's proof of the Uniform Boundedness Principle. – 2011-09-03
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0@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 – 2012-05-07
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0@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 $\|fg\|_{L^1}\le C\|g\|_{L^q}$. – 2012-05-07
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0sorry if I wasn't clear before. I was wondering if you could expand on the derivation of the contradiction. for example why $g=\sum\limits_{k=1}^\infty2^{-k}g_k\|g\|_{L^q}$ is less than one and why it implies that $fg\notin L^1$. – 2012-05-07
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1@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. – 2012-05-07
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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0I don't understand: are you trying to show that $f\in L^q$?. – 2014-06-13
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0You'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. – 2014-06-13
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0@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. – 2014-06-13
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0Yes, 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. – 2014-06-14
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0@robjohn, do you think my answer is totally useless and deserves to be deleted? If so, I will have no problem in deleting it. – 2014-06-14
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0I don't think it is useless, I am just trying to understand how it works. I don't have my copy of Rudin available, so I can't look up the proof of $(1)$. I would also like to understand how we can extend from specially chosen sub-sequences vanishing to the whole sequence vanishing. – 2014-06-14
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0Ok @robjohn. To prove that $h=0$ a.e., it is sufficient to prove that $\|h\|_q=0$. It can be done by means of Fatou's lemma, indeed, note that $\|h\|_q\le \|h-g_n\|_q+\|g_n\|_q$. On the other hand, to extend the result to the whole sequence, we can use the following topological lemma: Let $X$ be a metric space and $x\in X$. Assume that $(x_n)\subset X$ and that every subsequence of $x_n$ converges to $x$. Then, the whole sequence does converge to $x$. – 2014-06-14
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0Sorry @robjohn, again my notation is confused. My $h$ in the last comment is defined by $h(x)=\lim g_n(x)$, which is different of yoru $h$. Note that my $h$ is well defined, because $\sum_{n=1}^\infty |g_n(x)|\in L^q$, which implies in particular that $\sum_{n=1}^\infty |g_n(x)|<\infty$ a.e. – 2014-06-14
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0Indeed, $\|h\|_q=0\implies h=0\text{ a.e.}$ However, $\lim\limits_{n\to\infty}\|g_n\|_q=0$ does not imply that $\lim\limits_{n\to\infty}g_n(x)$ even exists anywhere, so we need more justification why we can pick a subsequence so that $g_n(x)\to0$. – 2014-06-16
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0You are right @robjohn, however, we have more than it. In fact, we have that $\sum_{n=1}^\infty |g_n(x)|\in L^q$, which implies, in particular, that this sum is finite everywhere. On the other hand, note that $$g_n(x)=\sum_{i=1}^n g_i(x)-\sum_{i=1}^{n-1}g_i(x)$$ – 2014-06-17
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0@robjohn To show that $f=\widetilde{f}$, we may nend the whole space is $\sigma$-finite. – 2015-11-12
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0@XiangYu: Since $\int\tilde{f}g\,\mathrm{d}x=Tg=\int fg\,\mathrm{d}x$ for all $g\in L^q$, we have that $\|\tilde{f}-f\|_p=0$, which means that $\tilde{f}=f$ (a.e.) – 2015-11-12
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0@robjohn How can we conclude $\|\widetilde{f}-f\|_{L^p}=0$ from $\int \widetilde{f}g=\int fg$? – 2015-11-12
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0@robjohn If $E$ is a measurable set with finite measure, then $1_E\in L^q$. Let $g=1_E$, we have $\int\widetilde{f}1_E=\int f1_E$ for all measurable set $\mu(E)<\infty$. If the whole space $X$ is $\sigma$-finite, then using proof by contradiction, we can show that $\widetilde{f}=f$ a.e. – 2015-11-12
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0Let $\overline{g}=(\tilde{f}-f)\,|\tilde{f}-f|^{p-2}$, then $0=\int(\tilde{f}-f)g\,\mathrm{d}x=\|\tilde{f}-f\|_p^p$ – 2015-11-12