Ergodicity of Convoluted White Noise I have a question regarding ergodicity in infinite dimensional spaces.
Let $\mathcal{D}$ be the space of distributions on a Schwartz space, and let $\mu$ be the white noise process which exists by the Bochner-Minlos theorem.
We can define the translation $\tau_x \phi$ of a distribution $\phi$  by:
$$ \langle \tau_x \phi, \varphi \rangle = \langle \phi, \tau_x \varphi \rangle, $$
for all $\varphi \in C^\infty_c(\mathbb{R}^n)$, where $\tau_x \varphi$ is the usual translation of a function.  The set of translations form a group acting on $\mathcal{D}$.  
My first question is:
Is the group $\lbrace \tau_x : x\in \mathbb{R}^n \rbrace$ ergodic with respect to the white noise measure?
Now let $\varphi \in C^\infty_c(\mathbb{R}^n)$, then we can define the convolution $\varphi * \zeta(\omega)$, where $\zeta$ is a white-noise distributed variable.  This is a $C^\infty$ random variable. My second question is:
Is  $\varphi * \zeta(\omega)$ ergodic with respect to translations? 
My intuition is that the first answer is true, and the second answer is "true, assuming the support is sufficiently small", but I don't even know what tools to use to tackle this problem.   
 A: The answer is yes for both situations because mixing implies ergodic.
To make things precise, let $S=S(\mathbb{R}^n)$ be the Schwartz space of
smooth real valued functions on $\mathbb{R}^n$ which decay faster than any negative power of the
Euclidean norm. Let $S'$ be the dual space of tempered distributions.
For $f\in S$ and $\phi\in S'$ denote by $\phi(f)$ the duality pairing. 
Let the cylindrical $\sigma$-algebra $\mathcal{C}$ on $S'$ be
the smallest such that
$\phi\mapsto \phi(f)$ from $(S',\mathcal{C})$ to $(\mathbb{R}, {\rm Borel\ sets})$
is measurable, for any $f\in S$. 
I assume the measure $\mu$ you are talking about is the centered Gaussian probability
measure on $(S',\mathcal{C})$ with covariance
$$
\int_{S'} \phi(f)\phi(g)\ {\rm d}\mu(\phi)=
\int_{\mathbb{R}^n} f(x) g(x) \ {\rm d}^n x\ .
$$
Let $x\in\mathbb{R}^n$ we define translation by $x$ by:
$(\tau_x f)(y)=f(x-y)$ for test functions, $(\tau_x \phi)(f)=\phi(\tau_{-x} f)$
for distributions, and by $(\tau_x F)(\phi)=F(\tau_{-x}\phi)$
for complex valued $\mathcal{C}$-measurable functionals on $S'$.
To show the mixing property, you need to prove that for any such functionals
$F$, $G$ in $L^2(S',{\rm d}\mu)$ one has
$$
\int_{S'} F(\tau_x \phi) G(\phi)\ {\rm d}\mu(\phi)\rightarrow \left(\int_{S'} F(\phi)\ {\rm d}\mu(\phi)\right) \times\left(\int_{S'} G(\phi)\ {\rm d}\mu(\phi)\right)\ \ \ (\ast)
$$
when $x$ goes to infinity.
First note that the span $V$ of monomials $\phi(f_1)\cdots \phi(f_n)$
is dense in $L^2(S', {\rm d}\mu)$. That's basically the Wiener-Bargmann-Ito-Segal
isomorphism with Fock space. 
Also note that $||\tau_x F||_2=||F||_2$. So if you approach $F$ with $F_n$
in $V$ the norm of the difference is the same for all translates.
So a simple three epsilon argument allows to reduce the proof of $(\ast)$ to the case
of monomials.
If you evaluate
$$
\int_{S'}
\phi(\tau_x f_1)\cdots\phi(\tau_x f_n)
\phi(g_1)\cdots\phi(g_m)
\ {\rm d}\mu(\phi)
$$
using the so-called Wick theorem (due in fact to Isserlis)
you will see the factors
$$
\int_{S'} \phi(\tau_x f_i)\phi(g_j)=
\int_{\mathbb{R}^n} f_i(y-x)g_j(y)\ {\rm d}^n y
$$
which couple the $f$'s and the $g$'s go to zero.
The same proof goes in the mollified case too.
The essential ingredient here is that the kernel
of the covariance decays fast enough.
Also remark that this generalizes to measures which are not Gaussian.
One just needs to decompose moments into cumulants aka connected correlation
functions. The key input which generalizes the single factor decay 
between $f_i$ and $g_j$ above is called the clustering property. It is one of the Osterwalder-Schrader
axioms in constructive quantum field theory which physically corresponds to
the uniqueness of the vacuum.
