Let $H$ be a Hilbert space. I am interested in isometries $f\colon H\to L^2(X,\mu)$ where $\mu$ is a probability measure on some measure space $X=(X,\mathcal F)$ where $\mathcal F$ is a $\sigma$-algebra on $X$. My question is whether there exists a canonical choice of $(X,\mu,f)$ that depends only on $H$ and not "arbitrary" choices like choosing an orthonormal basis for $H$.

For an example of what I mean, here is a construction that only works when $H$ is finite dimensional. Let $X=H$, and for $h\in H$ consider the function $f_h\colon X\to\mathbb C$ given by $f_h(x)=\langle h,x\rangle$. Let $\mu$ be the unique Borel measure on $H$ satisfying $$\int_X e^{i f_h(x)}\ d\mu(x) =e^{-\|h\|^2/2}$$ for all $h\in H$, from which it follows that $h\mapsto f_h$ is an isometry. Note that the $\mu$-expectation of the function $\|x\|^2\colon X\to\mathbb C$ equals $\dim H$.

In fact when $\dim H=\infty$, the analogue of this construction yields a measure supported on "elements of $H$ with infinite magnitude", which can be made sense of rigorously as distributions - elements of the space $\Phi^*$ in a Gelfand triple $(\Phi,H,\Phi^*)$.

Thus, a close variant of my question is to ask whether there exists a canonical construction of a Gelfand triple using only the space $H$ as input.

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    $\begingroup$ What structure does a bare Hilbert space even have, that could tell you to prefer one measure space (or one isomorphism) over another? $\endgroup$ – Nate Eldredge Dec 19 '19 at 4:35
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    $\begingroup$ Analogous to your Gaussian example, given a separable Hilbert space $H$, one may always embed it into a Banach (or Hilbert) space $W$ and equip $W$ with a Gaussian measure $\mu$, such that the embedding of $H$ into $L^2(W,\mu)$ is an isometry (not surjective). But there is no canonical way to choose $W$, and in particular no "biggest" or "smallest" such $W$. $\endgroup$ – Nate Eldredge Dec 19 '19 at 4:41
  • $\begingroup$ How can there be "a canonical construction of a Gelfand triple using only the space $H$ as input", if the Gelfand triple depends on the choice of a dense subspace $\Phi$ of $H$? Clearly, there are infinitely many such subspaces. $\endgroup$ – Iosif Pinelis Dec 19 '19 at 13:10
  • $\begingroup$ It sounds like you might want the right adjoint to the $L^2$ functor. $\endgroup$ – Oscar Cunningham Jan 1 at 13:07
  • $\begingroup$ @OscarCunningham that is a very elegant way of capturing the sort of construction I am going after. See also my question here mathoverflow.net/questions/348880/… where I attempted to formalize what I was looking for in a slightly different category theoretic manner. $\endgroup$ – pre-kidney Jan 2 at 0:33

This is impossible, at least in the case when $X=H$, as in your "finite-dimensional" example.

Indeed, suppose that $H$ is infinite dimensional. If you want the measure $\mu$ not to depend on the choice of an orthonormal basis, you have to make $\mu$ spherically invariant. But such a probability measure can only be $\delta_0$, the Dirac measure at $0$. (See details below.) And if $\mu=\delta_0$, then $L^2(H,\mu)$ is one dimensional and hence not isometric to the infinite-dimensional Hilbert space $H$.

Details: Suppose that $\mu\ne\delta_0$ is a spherically invariant probability measure on $H$. Let $\nu$ be the conditional distribution of the vector $x/|x|$ given $x\ne0$, assuming that the distribution of $x$ is $\mu$ (here, $|x|$ denotes the norm of $x$). Then $\nu$ is a spherically invariant probability measure on the unit sphere in $H$, and such a probability measure does not exist -- see e.g. Sudakov, page 24. So, indeed, the only spherically invariant probability measure on $H$ is $\delta_0$.


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