# Canonical embedding of Hilbert space in $L^2$ space

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.

• What structure does a bare Hilbert space even have, that could tell you to prefer one measure space (or one isomorphism) over another? – Nate Eldredge Dec 19 '19 at 4:35
• 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 $W$. – Nate Eldredge Dec 19 '19 at 4:41
• 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. – Iosif Pinelis Dec 19 '19 at 13:10
• It sounds like you might want the right adjoint to the $L^2$ functor. – Oscar Cunningham Jan 1 at 13:07
• @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. – 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 $$\delta_0$$.