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Let $(\Omega,\mathcal A,\mu)$ be a measure space, $p,q\ge1$ with $p^{-1}+q^{-1}=1$ and $f:\Omega\to\mathbb R$ be $\mathcal A$-measurable with $$\int|fg|\:{\rm d}\mu<\infty\;\;\;\text{for all }g\in L^q(\mu)\tag1.$$ By $(1)$, $$L^q(\mu)\ni g\mapsto fg\tag2$$ is a bounded linear fuctional and hence there is a unique $\tilde f\in L^p(\mu)$ with $$(f-\tilde f)g=0\;\;\;\text{for all }g\in L^q(\mu)\tag3.$$

Can we conclude that $f=\tilde f$?

EDIT: As we can see from this answer, we need to impose further assumptions; but which do we really need?

0xbadf00d
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  • Why does (1) imply the functional is bounded? (I believe that's so, but it seems to me to require some sort of argument...) – David C. Ullrich Jul 18 '19 at 02:57
  • Ah. For example, define $T:L^q\to L^1$ by $Tg=fg$. It takes only one teensy trick to use the Closed Graph Theorem to show $T$ is bounded. – David C. Ullrich Jul 18 '19 at 03:17
  • @DavidC.Ullrich It can be proved like in this answer: https://math.stackexchange.com/a/61549/47771. – 0xbadf00d Jul 18 '19 at 04:16
  • Indeed. How is this question not a duplicate of that earlier question? (And given that one of the answers to the earlier question shows that the answer is no without further assumptions, why are you asking again, dup or not?) – David C. Ullrich Jul 18 '19 at 13:12
  • @DavidC.Ullrich I don't see an answer to the other question stating that the answer is no in general. Moreover, the reason why I'm asking is that the particular point why $f=\tilde f$ is not clear to me from the accepted answer to the other question (and, as you've shown, it is wrong in general). – 0xbadf00d Jul 18 '19 at 15:22
  • Did you see the answer from Danny Pak-Keung Chan? – David C. Ullrich Jul 18 '19 at 15:25
  • @DavidC.Ullrich I did not. Edited the question. – 0xbadf00d Jul 18 '19 at 15:31
  • @DavidC.Ullrich I'm not sure if I agree this is a duplicate (assuming OP intended to put the right assumptions on the space to make it true). This question really asks "why is it enough to apply Riesz representation theorem?" to conclude the argument in the other question. – Rhys Steele Jul 18 '19 at 15:46
  • @RhysSteele I don't see how you know that that's what the question really asks. But yes, the (new) "we need to impose further assumptions; but which do we really need?" probably makes it a non-dup. – David C. Ullrich Jul 18 '19 at 17:03
  • @DavidC.Ullrich The accepted answer to the other question (which tacitly assumes semifiniteness) shows that the functional OP starts with is bounded and then says you can conclude by the Riesz representation theorem. RRT shows you that there is an $\tilde{f} \in L^p$ that induces the same functional i.e. $\int (f-\tilde{f}) g d\mu = 0$ for all $g \in L^q$. It remains to make some argument to see that $f = \tilde{f}$ and that is what is asked here and is not done in the other question. The fact that OP asks about a missing detail there makes this not a dupe (with the correct assumptions). – Rhys Steele Jul 18 '19 at 17:27

2 Answers2

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Assume that we don't have $f = \tilde{f}$ almost everywhere. Then there is an $\varepsilon > 0$ and $E$ with finite measure at least $\varepsilon$ such that $|f-\tilde{f}| \geq \varepsilon$ on $E$. Then $g = \operatorname{sign}(f- \tilde{f}) \chi_E$ is in $L^q(\mu)$ and $$\int (f-\tilde{f}) g d\mu \geq \varepsilon^2 > 0.$$

Edit: Note that this answer assumes that the measure space is semifinite to assume pathological examples. At the point it was written, it was clear from discussion with OP that this is what OP wants.

Rhys Steele
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  • Aargh. Yes, sure. For some reason, I thought there would be more on that. – 0xbadf00d Jul 17 '19 at 18:23
  • With your edit you mean, for example, $\mu\equiv0$? – 0xbadf00d Jul 17 '19 at 18:32
  • Not quite that, since then all functions are equal to each other almost everywhere. I use that $fg \in L^1$ for every $g \in L^q$ implies that $f \in L^p$ (this is really the non-trivial detail in the whole thing). This isn't true in complete generality since you can do stupid things like give every non-empty set infinite measure so that $L^p = {0}$ for every $p$. The condition you want is something like semifiniteness of the measure. – Rhys Steele Jul 17 '19 at 18:36
  • Now I'm struggling again. Actually (as you may have guessed), I need the result of the question as a last step in the proof of the implication $(\forall g\in L^q:fg\in L^1)\Rightarrow f\in L^p$. So, this seems to be circular reasoning, if you need this in your argument. – 0xbadf00d Jul 17 '19 at 19:03
  • Oh I see, that makes sense. I hadn't realised that from your question. So you have an argument that's something like in this answer and you want to know why Riesz Representation theorem is enough to conclude I guess? – Rhys Steele Jul 17 '19 at 19:09
  • I made an edit to give an argument that isn't circular for your purposes. – Rhys Steele Jul 17 '19 at 19:13
  • @DavidC.Ullrich I need only semifiniteness. Obviously OP is missing assumptions they intend to include on their measure space, as evidenced by a previous version of this answer where I point out this isn't true in generality. The point is that previous discussion here indicated OP really wanted those assumptions. The above discussion shows that they are trying to prove a result that is untrue without semifiniteness and I give an example to show this. $\sigma$-finiteness is a much stronger assumption than needed to get this result. – Rhys Steele Jul 18 '19 at 09:36
  • @DavidC.Ullrich Nothing here is an answer to deleted comments. The only thing I can imagine is confusing is that (as I note in the comments) at some point I change my answer and so earlier comments refer to a previous edit. From your last comment, I wonder if you for some reason you can't see all comments though? Something like that is definitely visible. It is contained in my first comment on this answer where I essentially describe a counterexample of the same type as in your answer to the claim that OP eventually wants to establish (for which the question is the only missing piece). – Rhys Steele Jul 18 '19 at 13:06
  • either way, I hope that the latest edit satisfies you. – Rhys Steele Jul 18 '19 at 13:15
  • @RhysSteele Where do you use the semifiniteness? I suppose for the existence of $\varepsilon$, since it is not clear to me why $\varepsilon$ exists. – 0xbadf00d Jul 18 '19 at 15:19
  • @0xbadf00d In the original not-quite-right version he states that $\chi_E\in L^q$, which is not true. If $\mu$ is semifinite then $\mu(E)>0$ shows that there exists $F\subset E$ with $0<\mu(F)<\infty$, and now $g = \operatorname{sign}(f- \tilde{f}) \chi_F$ works. – David C. Ullrich Jul 18 '19 at 15:49
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    @0xbadf00d No, that $\varepsilon$ always exists since ${|f-\tilde{f}| \neq 0} = \bigcup_n {|f-\tilde{f}| \geq \frac 1n}$. Indeed, this shows that if we don't have $f = \tilde{f}$ a.e. then there is an $n$ such that ${|f-\tilde{f}| \geq \frac 1n}$ has positive measure. Now existence of $\varepsilon$ is trivial. I use semifiniteness to take $E$ of finite measure. – Rhys Steele Jul 18 '19 at 15:50
  • @0xbadf00d (Too late to edit the comment.) Note that in that comment I was referring to the notation in a previous version of the answer, where it was possible that $\mu(E)=\infty$. – David C. Ullrich Jul 18 '19 at 15:57
  • @RhysSteele Ah, sure. Thank you, once again. – 0xbadf00d Jul 18 '19 at 17:32
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Actually the answer is yes if you assume something about the measure, for example $\sigma$-finiteness is enough. But it's no to the question as stated, even for $p=q=2$.

Say $X=\{0,1\}$, $\mu(\{0\})=\infty$, $\mu(\{1\})=1$. Let $f=1$. Then $fg\in L^1$ for every $g\in L^2$ but $f\not\in L^2$.

Edit: The question has changed; now the OP acknowledges that some additional assumption is needed and asks what assumption. It's easy to see, as mentioned above, that $\sigma$-finiteness is enough. In fact it's sufficient too assume $\mu$ is semi-finite, and that's exactly right: If there exists $E$ such that $\mu(E)>0$ and $\mu(F)$ is $0$ or $\infty$ for every $F\subset E$ then $f=\chi_E$ gives a counterexample.

David C. Ullrich
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