Taking the expectation out of the inner product

Question

Suppose that $X$ is a zero mean random element with values in a separable Hilbert space $\mathbb H$ such that $\operatorname E\|X\|^2<\infty$. The covariance operator $C:\mathbb H\to\mathbb H$ of $X$ is defined as $C(f)=\operatorname E[\langle X,f\rangle X]$ for each $f\in\mathbb H$. I would like to show that $$ \langle C(f),f\rangle=\operatorname E|\langle X,f\rangle|^2 $$ for each $f\in\mathbb H$. I would like to proceed in the following way $$ \langle C(f),f\rangle=\langle \operatorname E[\langle X,f\rangle X],f\rangle=\operatorname E\langle \langle X,f\rangle X,f\rangle=\operatorname E[\langle X,f\rangle\langle X,f\rangle]=\operatorname E|\langle X,f\rangle|^2. $$ However, I do not know how I can justify taking the expectation out of the inner product. If I investigate a particular Hilbert space, for example, $L^2([0,1],\mathbb R)$, then I can use Fubini's theorem to justify the interchange of the order of integration, but how can I do that in the general case? Is it possible to justify taking the expectation out of the inner product in the general case?

Thank you very much for your help!

Answer 1

Too long for a comment, but not a complete answer by any means.

Additionally, I don't know the answer to this question, but I have some suggestions that you could perhaps consider.

1. The covariance is well-defined on the Hilbert space, so this may help with applications of Fubini/Tonelli.
2. You have a separable Hilbert space, so taking some (countable) basis $f_1, f_2, ...$ might not be a bad shout. Maybe point 1. can be used to take this infinite sum out in some way, cancelling cross-terms. This may help as you need only consider certain $f$.
3. Moreover, you may consider $X = X^+ - X^-$ in the usual way, and consider just non-negative $f$ similarly. Of course, while $\Bbb{E}(X) = 0$, you don't necessarily have $\Bbb{E}(X^{\pm}) = 0$.

Anyway, I hope that these points may be of some help to you. Sorry that I can't be of more help!