An inner product?

Question

Let $S$ be any complete, separable metric space. Let
\begin{align*} \mathcal{M}(S)&=\{\mu: \mu \text{ is a measure on } S\}\\ \mathcal{C}(S)&=\{f:S\to \mathbb R: f\text{ is bdd and continuous with sup norm}\} \end{align*}

Now my question is, if someone defines as follows, what does it mean? Is it inner product between two vector space? what this quantity specify?

$\langle f,\mu\rangle=\int_{S}f(x)\mu(dx); f\in \mathcal{C}(S), \mu\in \mathcal{M}(S)$

Thank you.

Answer 1

An inner product in a vector space $X$ is a map $X\times X\rightarrow\mathbb{F}$ with $\mathbb{F}=\mathbb{R}\vee\mathbb{C}$. This is not the case here.

In this case you are looking at the vector space $C(S)$ of continuous bounded functions on $S$. This has as dual space the space of all Radon measures with bounded variations. So your case is not so much as looking at an inner product, but at linear functionals.

As it happens $\langle \mu,f\rangle$ is notation used to denote the application of linear functional $\mu$ to $f$.

You can find more at: https://mathoverflow.net/questions/89274/the-dual-space-of-cx-x-is-noncompact-metric-space.