Let $W$ be a separable Banach space and $\mu$ a Gaussian Borel measure on $W$ which is centered and non-degenerate. For $F : W \to \mathbb{R}$ bounded Borel and $t \ge 0$, let
$$P\_t F(x) = \int\_W F(x+\sqrt{t}y)\,\mu(dy)$$
be the transition semigroup of Brownian motion on $W$.



> 
> Is $P\_t$ well defined as an operator from $L^\infty(W,\mu)$ to itself? That is, if $F=0$ $\mu$-almost everywhere, must the same be true of $P\_t F$?
> 
> 
> 


Of course, if $W$ is finite dimensional the answer is yes, because the measures $\mu(x + \cdot)$ are mutually absolutely continuous as $x$ ranges over $W$. This suggests that the answer may be no in infinite dimensions, but I can't seem to find a counterexample.



$L^\infty$ properties of an infinite-dimensional Gaussian semigroup
As suggested in the question, $P\_t$ need *not* be a well defined operator on $L^\infty(W,\mu)$. That is, $F$ can be zero $\mu$-almost everywhere, but $P\_tF$ is nonzero on a set of positive $\mu$-measure.


For example, take the Banach space $W$ to be $\ell^2$. Let $X\_0,X\_1,\cdots$ be an IID sequence of normal random variables with zero mean and unit variance, defined on some probability space $(\Omega,\mathcal{F},\mathbb{P})$. Then, set $X=(X\_0,2^{-1}X\_1,\ldots,2^{-n}X\_n,\ldots)\in\ell^2$ (a.s.), and let $\mu$ be the measure of $X$.
Define the measurable function $G\colon\ell^2\to\mathbb{R}$ by
$$
G(c)=\begin{cases}
\lim\_{n\to\infty}\frac1n\sum\_{k=0}^{n-1}4^kc\_k^2,&\textrm{if the limit exists},\cr
0,&\textrm{otherwise}.
\end{cases}
$$
By the strong law of large numbers,
$$
G(tX)=t^2\lim\_{n\to\infty}\frac1n\sum\_{k=0}^{n-1}X\_k^2=t^2\textrm{ (a.s.)}
$$
You can now define $F\colon\ell^2\to\mathbb{R}$ by $F(c)=0$ when $G(c)=1$ and $F(c)=1$ otherwise. Then, $F(X)=0$ (a.s.) and $F(tX)=1$ for $t > 1$ (a.s.), so $F=0$ $\mu$-almost everywhere. However, the measure $\int P\_t(x,\cdot)d\mu(x)$ is the same as the distribution of $(1+t)^{1/2}X$. Therefore, $P\_tF=1$ $\mu$-almost everywhere for each $t > 0$.


