Does it make sense to talk about $L^2$ inner product of two functions not necessarily in $L^2$?

Question

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!

Answer 1

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 $p'$.

Answer 2

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.