TITLE: Wiener measure and Bochner Minlos
QUESTION [10 upvotes]: I am reading probability theory and I have a question.  The Bochner-Minlos theorem roughly says that if we have $E \subset H \subset E^*$ where $H$ is a Hilbert space, then there is a unique measure on $E^*$ satisfying some nice properties.  
On the other hand, the Wiener measure is supported on the space of continuous functions.  Does that mean Wiener measure is not Bochner-Minlos type of measure?  In other words, is the Wiener measure the result of the Bochner-Minlos construction?
I apologize if this question is too simple or doesn't even make sense.  I am a beginner in this.

REPLY [5 votes]: Wiener measure can most definitely be characterised as the only probability measure $\mathbf{P}$ on the space $C_0$ of continuous functions starting at the origin and such that the identity
$$
\int_{C_0} \exp\Big(i \int f(t)\,\mu(dt)\Big)\,\mathbf{P}(df) = \exp \Big(-{1\over 2} \iint (s\wedge t)\,\mu(ds)\,\mu(dt)\Big)\;,
$$
holds for any compactly supported finite measure $\mu$. This is a consequence of some variant of Bochner's theorem, although it may well be that the one stated in Hida does not exactly fit the bill. (I do not have that book at hand right now.) A good reference with all you ever wanted to know about Gaussian measures (and probably much more) is the book by Bogachev.