Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
Is the measure finite? –  Dylan Moreland Sep 2 '11 at 19:22
2  
I recommend checking out this related question for inspiration. –  Jonas Meyer Sep 2 '11 at 19:29
1  
Hint: the closed graph theorem. –  Nate Eldredge Sep 2 '11 at 19:42
1  
I recommend checking out this related question for inspiration... :-) –  Did Sep 2 '11 at 21:23
3  
@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
show 4 more comments

1 Answer

up vote 15 down vote accepted

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.

share|improve this answer
    
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
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. –  robjohn May 7 '12 at 23:17
show 2 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.