Let $\ell^2(\mathbb{R})$ denote the space of square-summable real-valued sequences equipped with the inner product

$$ \langle x, y \rangle = \sum_{n = 1}^{\infty} x_n y_n $$

Let $\nu$ denote its canonical cylinder measure i.e. a cylinder measure defined on the weak sigma algebra whose Fourier transform has the form

$$ \exp( - \frac{1}{2} \langle x, x \rangle_{\ell^2}) ~~.$$

Does there exist a measurable norm associated to $\ell^2$ and $\nu$?

If so, is there a nice description of its associated abstract Wiener space?

If not, is there a proof that there is not?

I do know that $\ell^2$ is the Cameron-Martin space of $\mathbb{R}^{\infty}$ with the product topology and the distribution of an iid sequence of random variables $(X_n)_{n \in \mathbb{N}}$ with $X_1 \sim \mathcal{N}(0,1)$, and also that $\ell^2$ lies dense in that space w.r.t. the topology of point-wise convergence.

So it "should be" $\mathbb{R}^{\infty}$, but it is not since $\mathbb{R}^{\infty}$ is only separable Frechet and not Banach.


1 Answer 1


Sure, there are many such norms (they are highly non-unique). Indeed, take any other infinite-dimensional abstract Wiener space $(W, H, \mu)$. Since $H$ and $\ell^2$ are both infinite-dimensional separable Hilbert spaces, there is an isometric isomorphism $T : \ell^2 \to H$. Now since $\|\cdot\|_W$ is a measurable norm on $H$, it follows that $\|T \cdot\|_W$ is a measurable norm on $\ell^2$.

If you want the norm to be an inner product, let $A$ be any injective Hilbert-Schmidt operator on $\ell^2$. It is shown in Proposition 4.59 of these notes that $\langle Ax, Ay\rangle$ defines an inner product on $\ell^2$ which induces a measurable norm. (The assumption that $A$ should be injective is missing from the notes, but is necessary; otherwise we only get a measurable seminorm.)

For a specific example, fix a sequence of real numbers $a_n$ with $\sum a_n^2 < \infty$ (such as $a_n = 1/n$) and apply the proposition to the operator defined by $(Ax)_n = a_n x_n$. The resulting inner product is $\langle x,y \rangle_a = \sum a_n^2 x_n y_n$, and the completion of $\ell^2$ under this inner product is simply the weighted $\ell^2$ space $\ell^2_a$ of sequences $x$ satisfying $\sum a_n^2 x_n^2 < \infty$. It's easy to construct the associated Gaussian measure on $\ell^2_a$; let $\xi_n$ be an iid sequence of standard normal random variables on some probability space $\Omega$. If $e_n$ is the standard basis in $\ell^2$, you can easily show that $\sum \xi_n e_n$ converges in $L^2(\Omega; \ell^2_a)$, and $\mu$ is the law of its limit.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .