Gaussian measure on Banach space Assume we have a Gaussian measure $\mu$ supported on a Banach space $X$. Can we always find a Hilbert space $H$ embedded in $X$ sch that $\mu$ is also supported on $H$?
 A: No, not necessarily.  It is shown in Examples 3.6.6 and 3.6.7 of Bogachev's Gaussian Measures that if $X = C([0,1])$ and $\mu$ is classical Wiener measure, then for any Hilbert space $H$ embedded in $X$, we have $\mu(H) =0$, so that $\mu$ is not supported on $H$.
A: For the always  existence of Hilbert space ‎$H‎$  need some conditions about Banach space  and structure of Gaussian measure to be satisfied as montioned in the below theorem where the uniqness is the strong condition for existence but in your case it's not always  Hilbert space ‎$H‎$ exist  :

Theorem:
  Suppose that ‎$E‎$ is a separable, real Banach space and that is a centered Gaussian measure which is non-degenerate.Then there exists a unique Hilbert space H such that ‎$(H;E;W)‎$ is an abstract Wiener.

For more informations look this paper about "Gaussian Measures on a Banach  Space" theorem 8.2.5 with a complet proof.
Edit: I edited my answer because I don't meant the existence of Hilbert space $H‎$ related to the given question but i meant the existence will be satisfied if and only if the conditions  cited in theorem 8.2.5 hold 
