# Concentration of infinite-dimensional Gaussian measure

I have the question about finding the subspace of concentration of a Gaussian Measure. More precisely:

$\textbf{Question:}$ Assume we have a separable Hilbert space $\ell_2$ with Borel $\sigma$-algebra $B(\ell_2)$ and a centered Gaussian measure $\mu$, i.e. we consider the probability space $(\ell_2, B(\ell_2), \mu)$. There is also a linear continuous operator $Q:\ell_2 \rightarrow (\mathbb{R}^{\infty},\rho)$, where $\rho$ is a metric defined by the formula: $$\rho(x,y) = \sum\limits_{k=1}^{\infty}\frac{1}{2}\frac{|x_k - y_k|}{1 + |x_k - y_k|}, \, x,y\in \mathbb{R}^{\infty}.$$ Operator $Q$ pushforwards $\mu$ on $\ell_2$ to measure $Q_{*}(\mu)$ on $\mathbb{R}^{\infty}$. How to check that still measure $Q_{*}$ concentrates on $\ell_2\subset \mathbb{R}^{\infty}$, i.e. $Q_{*}(\mu)[\ell_2] = 1$?

$\textbf{Comment:}$ Thanks for the notes of Natan Eldredg from Cornell Univ. I have some ideas how to do it. At first one should say that Gaussian measure on a Hilbert space in this case defined uniquely by its covariance operator on a dual space $\Sigma^{-2} = \mathrm{Cov}(\mu)$ which is of a trace class operator. So operator $Q$ expands the norm of a Gaussian vector and geometrically the idea is to check that a new covariance operator of $Q_*(\mu)$ is also of a "trace" class.

However the propery of being a trace class operator has no rigorous meaning, since $Q_*(\mu)$ is a measure not on a Hilbert space, but on $(\mathbb{R}^{\infty},\rho)$. So one should do a bit different construction.

Having a quadratic form $\Sigma^{-2}$ on $\ell_2$ it is possible to write explicitly the new quadratic form on $(\mathbb{R}^{\infty},\rho)^*$ which is $Q\Sigma^{-2}Q^*$. It induces the Hilbert structure on dual space $(\mathbb{R}^{\infty},\rho)^*$ which is a dual space for Cameron-Martic space (ofcourse having a completion before in a new scalar product norm). Hence we can find a Cameron-Martin space of measure $Q_*(\mu)$. After that we aim to do the reconstruction of concentrtation space, via finding the measurable norm for this Cameron-Martin space -- if we want to prove that measure $Q_*$ concentrates on $\ell_2$ we check that $\|\cdot\|_2$ is a measurable norm in this CM space. If it is so, we find the completion of CM space in this norm and we are done.

$\textbf{Minor Question:}$ The last argument above is kind of obvious, but still not clear a bit. It is related to the uniqueness of reconstruction of space via CM space (obviously it is not unique), but to make the full proof one should say that the reconstructed space with reconstructed Gaussian measure coincide with the measure I began -- $Q_*(\mu)$. Any suggestions?

Also if you find any mistake in my comment, please welcome to remark. It took me some effort to get this construction (however the question sounds rather naturally for Gaussian measures) and I havent seen any links related to such question.

• From the notes of Natan Eldredge again I found a beautiful fact that one can find the support of Gaussian measure via the following way: $$supp \mu=\bigcap\limits_{q(f)=0}Ker(f),$$ where $q$ -- is the quadratic form on the dual space: $$q(f)=\int_X f^2(x)\mu(dx).$$ This formula is nice, but gives support of measure in terms of topology of $X$. But for the case when $X=R^{\infty}$ and $\mu =\prod_{k=1}^{\infty} \mathcal{N}_{\lambda_k}$, where $\lambda_k >0$ is a variance, $supp \mu=R^\infty$, which does not relfect concentration on $\ell_2$ for small $\lambda_k$ – Fedor Goncharov Jul 21 '16 at 13:32