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Easy question, but can't seem to find it online.

Informally, let $\Omega$ denote a probability space and $H$ a Hilbert space. Then what exactly does $L^2(\Omega ; H)$ mean? I presume it is a Hilbert space under the inner product $\langle f,g \rangle_{L^2(\Omega;H)} = \mathbb E(\langle f, g\rangle_H)$. (This is an inner product, right?)

This question arose from reading Neulart's "Lectures on Malliavin Calculus and Application to Finance" page 20 - in the definition of the divergence operator $\delta$ as the adjoint of the Malliavin derivative $D$ it was stated $\mathbb E[F\delta(u)] = \mathbb E[\langle D(F), u\rangle]$. http://www.math.wisc.edu/~kurtz/NualartLectureNotes.pdf

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  • $\begingroup$ Do you have an example in which this notation is used? $\endgroup$ – Marcus M Oct 23 '17 at 15:57
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You may consider an H-valued random variable as a measurable function from $\Omega$ to $H$, where $H$ is equipped with its Borel $\sigma$-algebra. Then as you say, $L^2(\Omega; H)$ is the set of (equivalence classes of) random variables $f$ such that $E[\|f\|^2] < \infty$.

The inner product is $E[\langle f, g\rangle]$ as you guessed; this is indeed an inner product (easy exercise), and moreover it is complete, so $L^2(\Omega; H)$ is a Hilbert space itself.

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