What is known about the Gaussian measure of the unit ball in a Hilbert Space? Let $X$ be an infinite dimensional separable Hilbert Space with norm $||\cdot||$ and let $\mu$ be a Gaussian measure on $X$ such that $\mu(X) = 1$.  What do we know about $\mu(B(0,1))$, where $B(0,1)$ is the unit ball w.r.t the norm?
This seems to me like a fundamental question but I cannot seem to find anything.  Any information/references would be most appreciated.
EDIT: A related question which is of interest to me:  Do there exist asymptotically tight bounds to 
$\int_{||u||> K}||u||^2 \mu(du)$?
 A: In the book Kazhdan’s Property (T) (Appendix A7) by Bekka, de la Harpe and Valette the symmetric Fock space on a Hilbert space is $H$ studied as the analogue of a space of measurable functions on a Hilbert space $H$. This is called the Gaussian construction and quite important if one wants to pass from unitary representations of a group $G$ to actions of $G$ on a probability measure space. This is probably not quite what you want, but serves as a suitable replacement of the Gaussian measure (on a finite-dimensional Hilbert space) for many purposes.
In case $H$ is finite-dimensional, it precisely corresponds to the study of the Gaussian measure on $H$. Here, the correspondence is clear: If $G$ acts by unitary operators on $H$, then it preserves the Gaussian measure $\mu$ on $H$ and hence, there is an associated action on the probability space $(H,\mu)$.
A: There is no Gaussian measure on an infinite dimensional Hilbert space, or rather the Gaussian measure is identically zero. (Proof:  If the Gaussian measure of a ball of radius r on a 1-dimensional Hilbert space is c<1, then that of a ball in n dimensional is less than cn, so in infinite dimensions any ball has measure 0, so the measure of the whole space is 0.)  You can put a non-zero Gaussian measure on a larger space (see http://en.wikipedia.org/wiki/Rigged_Hilbert_space) and the unit ball of Hilbert space is a subset of this, but has measure 0 by the above argument.
A: I think you can see the articles
entitled "concentration of measure phenomenon ". The idea is as
follows: Let $(X,d,\mu)$ be a metric measure space, such as
$\mu(X)=1$. Let
$$\alpha(\epsilon) = \sup {\mu(X \backslash A_\epsilon) \  | \ 
\mu(A) = 1/2 }$$ where
$$A_\epsilon = { x \  | \  d(x, A) < \epsilon }$$ is
the $\epsilon$-extension of a set $A$. The function $\alpha(.)$ is
called the concentration rate of the space $E$. The following
equivalent definition has many applications:$$\alpha(\epsilon) =
\sup { \mu( { F >= M + \epsilon }) },$$ where the supremum is
over all $1$-Lipschitz functions $F: X \to \mathbb{R}.$ For example
the median (or Levy mean) $M = \mathop{Med}(F) $ is defined by the
inequalities $$\mu ( F \geq M ) >= 1/2, \  \mu ( F <= M ) \geq
1/2.$$ More precisely, the space $X$ exhibits a concentration
phenomenon if $\alpha(\epsilon)$ decays very fast as $\epsilon$
grows. More formally, a family of metric measure spaces
$(X_n,d_n,\mu_n)$ is called a Levy family if the corresponding
concentration rates $\alpha(\epsilon)$ satisfy
$$\forall \epsilon > 0 \ \  \alpha_n(\epsilon) \to 0,$$ and a normal
Levy family if $$ \forall \epsilon \to 0 \ \  \alpha_n(\epsilon) =
O(\exp(-C n \epsilon^2))$$ for $C$ some positive constant. the last
inequality is obtained bay applying the "Hoeffding inequality" and in
the  case of  Hilbert space with concentration in small balls we do :
$$\forall (x_1,x_2)\in X^2, \ \  d(x_1,x_2)=\|x_1,x_2\|< r.$$
A: You can't talk about "the" Gaussian measure on an infinite-dimensional Hilbert space, for the same reason that you can't talk about a uniform probability distribution over all integers. It doesn't exist; see Richard's answer. However, there are a lot of non-uniform Gaussian measures on infinite dimensional Hilbert spaces. 
Consider the measure on $\mathbb{R}^\infty$ where the $j$th coordinate is a Gaussian with mean 0 and variance $\sigma_j^2$, where $\sum_{j=1}^{\infty} \sigma_j^2 < \infty$ (and different coordinates are independent). This is almost surely bounded in the $\ell_2$ metric, and any projection onto a finite-dimensional space has a Gaussian distribution. The squared length of a vector drawn from this measure is a sum of squares of Gaussians, and so follows some kind of generalized $\chi$-square distribution. If I knew more about generalized $\chi$-square distributions, I might be able to tell you what the measure of the unit ball was.
This kind of Gaussian distribution is very important in quantum optics. In fact, in quantum optics, a thermal state is Gaussian, so "the" Gaussian measure actually makes some sense.
