Identifying a generic Hilbert space $H$ with an $L^2$ space on some measure space.

hilbert-spaceslp-spacesmalliavin-calculusprobability theorystochastic-integrals

This may be a stupid question, but I was wondering, if we are given an
infinite dimensional Hilbert space $H$, is it possible to find (or to
hypothesise that there's) a measure space $(M,\mathcal M,\mu)$ such
that $H$ coincides with $L^2(M,\mathcal M,\mu)$?

Apparently this is a basic resultm and the answer is yes.


Second question:

Let $H$ be a real Hilbert space, let $W=\{W(h):h\in H\}$ be an isonormal Gaussian process, namely a Gaussian Hilbert space indexed by $H$; then we know that $W=H^{:1:}$ (the first homogeneous chaos), and we further know that the stochastic Gaussian integral $I$ on a measure space $(M,\mathcal M,\mu)$ is an isometry from $L^2(M,\mathcal M,\mu)$ onto $H^{:1:}=W$.

Then any $W(h_i), h_i\in H$ can be represented as a stochastic Gaussian integral of some function in $L^2(M,\mathcal M,\mu)$, hence $W$ will be a Gaussian Hilbert space indexed by $L^2(M,\mathcal M,\mu)$.

My question is are $H$ and $L^2(M,\mathcal M,\mu)$ the same space?
If yes, the $W(\cdot)$ can be seen as a Gaussian stochatic integral.

Is it correct?

I may have made some mistakes on my reasoning, so if you spot something off please let me know!

Thanks in advance.

Best Answer

You get an answer to your first question in the comments so let me address the second question. First I need to clean up some notation.

$H$ isn't itself a Gaussian Hilbert space so it seems incorrect to talk about the first homogeneous chaos over $H$. The situation is that the isonormal Gaussian process $W$ is an isometry $$H \to \tilde{W} = \tilde{W}^{\mathpunct: 1 \mathpunct:} = \operatorname{Ran}(W) \subseteq L^2(\Omega)$$ for some probability space $\Omega$ (Note I change some notation to avoid giving $W$ two definitions).

Now suppose we are given a linear isometry $I: L^2(M, \mathcal{M}, \mu) \to \tilde{W}$ (i.e. a Gaussian stochastic integral on $(M, \mathcal{M}, \mu)$ into $\tilde{W}$). Note that this is an assumption and is not automatically true. What you get for free is that there is a Gaussian Hilbert space $V$ such that there is a Gaussian stochastic integral $$\tilde{I}:L^2(M, \mathcal{M}, \mu) \to V.$$ In general, it is not possible to assume $V = \tilde{W}$. A simple way to see this is to notice that $\tilde{W}$ may be finite dimensional whilst $L^2(M, \mathcal{M},\mu)$ may be infinite dimensional so that there can be no such isometry. However, for the sake of discussion, I suppose we are given such a Gaussian stochastic integral into $\tilde{W}$.

The next thing to notice is that $I$ need not be surjective. I assume you are still following "Gaussian Hilbert Spaces" by Svante Johnson. It's worth noting that when introducing Gaussian stochastic integrals he says that we can "assume without essential loss" that $I$ is surjective. He does this by noting that the range of $I$ is a Gaussian Hilbert space and we can hence just restrict the codomain. In the situation of this question however we have fixed the Gaussian Hilbert space $\tilde{W}$ and so can't restrict the codomain in this way. This means that even assuming there is a Gaussian stochastic integral into $\tilde{W}$, you don't get that $I: L^2(M, \mathcal{M}, \mu) \to \tilde{W}$ is surjective automatically. As an example of this, we may have that $H$ (and hence $\tilde{W}$) are infinite dimensional but $L^2(M, \mathcal{M}, \mu) = \mathbb{R}$ (take $M = \{0\}$ and $\mu = \delta_0$) so that $I$ cannot be surjective and also $H \not \approx L^2(M,\mathcal{M},\mu)$.

For the sake of further discussion, I will now assume that $I: L^2(M, \mathcal{M}, \mu) \to \tilde{W}$ is additionally surjective. Now the situation is simple. $W: H \to \tilde{W} \subseteq L^2(\Omega)$ is an isometric isomorphism, as is $I: L^2(M, \mathcal{M},\mu) \to \tilde{W}$. Hence $I^{-1} \circ W: H \to L^2(M,\mathcal{M},\mu)$ is an isometric isomorphism. However, the assumptions we had to make to get here mean this is far from the generic situation.

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