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Let $(\mathcal{H},\langle \cdot, \cdot \rangle)$ be an arbitrary Hilbert space. Can one construct two independent and identically distributed random elements $X,Y:(\Omega,\mathbb{F},P)\to (\mathcal{H},\langle \cdot, \cdot \rangle)$ with $\text{supp}(X)\not = \{0\}$, such that $$ \langle X(\omega),Y(\omega) \rangle =0 $$ for almost all $\omega\in\Omega$, i.e. $X$ and $Y$ are almost surely orthogonal.

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I have shown that this can not be done for separable Hilbert spaces $\mathcal{H}$, but is it possible to make such a construction in non-separable Hilbert spaces?

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  • $\begingroup$ I'm not sure what you mean by a random function taking values in a Hilbert space. Usually people mean the Bochner integrable functions on a probability space. And their range is, by definition, inside a separable subspace almost surely. $\endgroup$
    – Stephen Montgomery-Smith
    Nov 8, 2016 at 17:54
  • $\begingroup$ @StephenMontgomery-Smith - $X$ and $Y$ are random Borel elements in $\mathcal{H}$ (not random functions, but random variables if you will, such that each $\omega$ maps to $X(\omega)\in\mathcal{H}$). We don't assume that $X$ and $Y$ are bochner integrable. This would simply lead the the arguments for separable Hilbert spaces, in which case I have already shown non-existence. I would like to make an argument that almost surely orthogonal random i.i.d. elements are not in general degenerate, but they are if $\mathcal{H}$ is separable (or the random elements are Bochner-measurable) they are. $\endgroup$
    – John
    Nov 8, 2016 at 19:23
  • $\begingroup$ What precisely do you mean by "random element?" I assume the map $\omega \mapsto X(\omega)$ is measurable, but with respect to which topology on $\mathcal H$? The norm topology or the weak topology? (I think the Borel sets generated can be different for non-separable spaces.) $\endgroup$
    – Stephen Montgomery-Smith
    Nov 8, 2016 at 21:08
  • $\begingroup$ Measurable with respect to the sigma algebra generated by the norm topology. Sorry if this was not explicitly written in the question. $\endgroup$
    – John
    Nov 8, 2016 at 21:18

2 Answers 2

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Counterexample. This counterexample is inspired by class notes by D.J.H. Garling from the 1980s.

Let $\kappa$ be a measurable cardinal https://en.wikipedia.org/wiki/Measurable_cardinal which is also a limit ordinal. Thus there is a $\{0,1\}$ valued measure on $\kappa$ in which every subset is measurable.

Let $\mathcal H$ be the Hilbert space whose basis is of cardinality $\kappa$, and let $(e_{\alpha})_{\alpha \in \kappa}$ be a basis. Let $X,Y:\kappa \to \mathcal H$ be the functions $X(\alpha) = e_{\alpha}$ and $Y(\alpha) = e_{\alpha'}$, where $\alpha'$ denotes the successor ordinal of $\alpha$.

Clearly $\langle X,Y\rangle = 0$ everywhere. It remains to show that $X$ and $Y$ are independent. But this follows because any two subsets of $\kappa$ are independent (since their measures can only take the values $0$ or $1$).

It does use measurable cardinals, whose existence cannot be proved. But most likely, if their existence can be disproved, then probably the same proof will show ZF is inconsistent.

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    $\begingroup$ As an aside, if $X$ and $X'$ are independent copies of $X$ defined on the first and second coordinates of the product space $\kappa \times \kappa$, then $\langle X,Y\rangle$ is non-measurable! This is essentially the point D.J.H. Garling was trying to make in his notes - separability is no mere technicality. $\endgroup$
    – Stephen Montgomery-Smith
    Nov 10, 2016 at 20:01
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Let $\mathcal{A} = \mathbb{R}^{[0,1]}$ be the set of all functions from $[0,1]$ to $\mathbb R$, not necessarily continuous, equipped with the product Gaussian measure. Let $\mathcal H$ be $L^2(\mathcal{A})$. $\mathcal H$ is not separable.

Let $x$ be uniformly distributed on $[0,1]$ and let $X_x$ be the function that takes $a \in \mathcal A$ to $a(x)$. The marginal distribution of each $X$ is a Gaussian, so $X_x \in \mathcal H$. Furthermore, $X_x$ is independent of $X_y$ unless $x=y$, which happens with probability zero.

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