0
$\begingroup$

Let $(X,\mathscr{B},\mu)$ be a $\sigma$-finite measure space. Let $\gamma$ be a probability measure on $L_2(\mu)$ with $\mathrm{supp} \, \gamma = L_2(\mu)$ and existing first moment. Then

$$ f \mapsto \|f\| := \int |\langle f,g\rangle|\,\gamma(\mathrm{d}g) $$

is a norm on $L_2(\mu).$ Does this norm have a name? Is $L_2(\mu)$ with that norm complete? Is it complete with a modified version like

$$ f \mapsto \|f\|' := \sqrt{\int |\langle f,g\rangle|^2\,\gamma(\mathrm{d}g)}, $$

where we assume that $\gamma$ has a finite second moment?

Can anyone give me some references where such normed vector spaces were studied?

Why is that interesting? And what makes me think that this or related stuff was already studied?

By $\|f \|$ and $\|f\|'$ we consider (somehow) the usual $p$-norm of $f$ under the coordinate transformation $f \mapsto (\langle f,g\rangle)_{g \in L_2}.$ Actually, if $F \subset L_2(\mu)$ separates points it might be sufficient to consider a coordinate transformation $f \mapsto (\langle f,g\rangle)_{g \in F}.$ Correspondingly, consider then a probability measure $\gamma$ on $F$ with $\mathrm{supp}\, \gamma = F$ and existing second moment. After all it seems to me that considering $\|\cdot\|'$ on $L_2(\mu)$ instead of $\|\cdot\|_2$ does not change much. In fact, I think that $(L_2(\mu),\|\cdot\|')$ is a Banach space, however, it is not trivial.

$\endgroup$
  • 4
    $\begingroup$ We aren't really using the measure space structure on $L_2(\mu)$ here, so we might as well view it as a generic Hilbert space $H$. Then this norm really just corresponds to identifying $H = H^* \subset L^1(H, \gamma)$. $\endgroup$ – Nate Eldredge Dec 29 '18 at 23:00
  • $\begingroup$ Thanks for your hint! I was wondering if there are conditions such that the Hilbert space $H$ with this norm is still complete? Are there examples such that $H$ is not complete anymore? $\endgroup$ – H17 Dec 30 '18 at 9:36
4
$\begingroup$

Let's write $H$ for $L_2(\mu)$, since its Hilbert space structure is all we care about. I assume $\gamma$ is a Borel measure.

By duality, you can naturally identify $H$ with $H^*$, and $H^*$ is a set of real-valued continuous, hence $\gamma$-measurable, (linear) functions on $H$, which are integrable by your first-moment assumption. So $H^*$ naturally admits the $L_1(\gamma)$ norm, or in other words, you can identify it as a linear subspace of $L^1(H,\gamma)$. So I would just call it the $L_1(\gamma)$ norm. Your "modified" norm $\|\cdot\|'$ is just the $L_2(\gamma)$ norm.

It is typically not complete. Because of the closed graph theorem, if you take a Banach space and put another norm on it, then the second norm will either be incomplete, or equivalent to the first one, or some sort of horrible axiom of choice monster.

For example, let $H$ be separable ($H=\ell^2$ if you like) and $e_1, e_2, \dots$ be an orthonormal basis for $H$. Let $\gamma$ be a Gaussian probability measure on $H$ with diagonal covariance form $K(e_i, e_j) = 4^{-i} \delta_{ij}$ (it can be shown that such a measure exists, because the form is trace class). Let $e_i^*(x) = \langle e_i, x\rangle$ be the element of $H^*$ corresponding to $e_i$, so that the $e_i^*$ are independent $N(0, 4^{-i})$ random variables on the probability space $(H,\gamma)$.

Now your first moment assumption implies that the identity bijection $(H^*, \|\cdot\|_{H^*}) \to (H^*, \|\cdot\|_{L_1(\gamma)})$ is continuous. If the $L_1(\gamma)$ norm is complete, then by the open mapping theorem the two norms are equivalent. However, $\|e_i^*\|_{L_1(\mu)} \sim 2^{-i}$ whereas $\|e_i^*\|_{H^*} = 1$.

(I think you could also show directly that the $L_1(\gamma)$ norm is incomplete, by considering a sum like $\sum_i i e_i^*$. It is Cauchy in $L_1(\gamma)$ norm, and if it converges in $L_1(\gamma)$ norm, then it also converges pointwise $\gamma$-a.e. on $H$. But you should be able to show that its limit couldn't be $\gamma$-a.e. equal to a continuous linear functional on $H$.)

A similar argument shows that your $L_2(\gamma)$ norm is not complete either.

However, it is often of interest to consider the completion, e.g. the closure of $H^*$ in $L_2(H, \gamma)$. For a Gaussian measure, this gives you the first level of the Wiener chaos decomposition of $L_2(H,\gamma)$.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.