Do two functions have to be an element of a space so that you can take their inner-product on that space?

Question

For example: say we have two functions $f, g \in (\Omega,\Sigma,\mu)$ where $\Omega\subset\mathbb{R}^n$ and $\mu$ is the Lebesgue measure. We would like to take the $L^2$ inner product of $f$ and $g$, namely $$\langle f,g\rangle = \int_{\Omega} f\cdot g \ d\mu.$$ Is it necessary that $f,g\in L^2$ or can one endow $\mathbb{R}^n$ with the $L^2$ inner product arbitrarily? Can we not just compute the above integral for two arbitrary functions in euclidian space?

Answer 1

If I get your question right, the answer is "no" in general. Consider $\Omega := [1, \infty)$ with the standard Borel $\sigma$-algebra $\Sigma$ and the Lebesgue measure $\lambda$, and let $$ f(z) := g(z) := \ln(z) $$ for $z \in \Omega$. Then clearly, $f$ and $g$ are well-defined on $\Omega$, but their inner product (in the $L^2$-sense) is $$ \langle f, g \rangle = \int_\Omega f \cdot g \ \mathrm{d}\lambda = \int \limits_1^\infty \ln^2(z) \ \mathrm{d} z = + \infty $$ Hence, the inner product is not defined - in other words, $\ln \not \in L^2(\Omega)$.