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The $L^2$-inner product of two real functions $f$ and $g$ on a measure space $X$ with respect to the measure $\mu$ is given by $$ \langle f,g\rangle_{L^2} := \int_X fg d\mu, $$

When $f$ and $g$ are both in $L^2(X)$, $|\langle f,g\rangle_{L^2}|\leqslant \|f\|_{L^2} \|g\|_{L^2} < \infty$.

I was wondering if it makes sense to talk about $\langle f,g\rangle_{L^2}$ when $f$ and/or $g$ may not be in $L^2(X)$? What are cases more general than $f$ and $g$ both in $L^2(X)$, when talking about $L^2(X)$ makes sense?

Thanks and regards!

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Tim, there are obvious examples. Just take $g=0$ then take $f$ to be whatever you want. Or, take $f=\frac{1}{\sqrt{x}}$ and $g=e^{-x}$ on $X=[1,\infty)$ with the usual Lebesgue measure. That said, if you want $\langle f,g\rangle$ to be finite for every $g\in L^2$ that seems like a bit more of an interesting question. –  Alex Youcis Mar 12 '13 at 16:52
    
@AlexYoucis: Thanks! I'd like to hear what can be said about the more interesting question if I want ⟨f,g⟩ to be finite for every $g∈L^2$? –  Tim Mar 12 '13 at 16:54
    
I would suggest editing your question, or perhaps asking a new question since it appears people have already answered your question. –  Alex Youcis Mar 12 '13 at 16:55
    
If you were going to ask such a new question, I would suggest asking the even more interesting question about, given $f$ measurable on $X$, what is the minimal subset $S$ of $L^2(X)$ for which $\langle f,g\rangle<\infty$ for all $g\in S$ implies that $f\in L^2(X)$. –  Alex Youcis Mar 12 '13 at 16:57
    
@AlexYoucis: The question you suggested has been posted here math.stackexchange.com/questions/328842/… –  Tim Mar 12 '13 at 22:33
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2 Answers

up vote 3 down vote accepted

Yes it does make sense. For example, if you take $f\in L^p(X)$ and $g\in L^{p'}(X)$ with $\frac{1}{p}+\frac{1}{p'}=1$, then $\langle f,g\rangle_{L^2} $ is well defined and $$\langle f,g\rangle_{L^2} <\|f\|_p\|g\|_{p'}$$

With this notation it is said the the inner product on $L^2$ induces the duality $p$ and $p'$.

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When you take $f\notin L^{2}(X)$ the problem is that you can not ensure $fg$ is integrable, but if $f\in L^{p}(X)$ then you can choose $g\in L^{q}(X)$ with $\frac{1}{q}+\frac{1}{p}=1$, then you can ensure $fg$ is integrable.

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