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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.

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No, for pretty much the reason you suggest. $\mu(x+\cdot)$ need not be absolutely continuous with respect to $\mu$. –  George Lowther Jun 2 '12 at 21:32
    
Quite right. I fixed it. –  Nate Eldredge Jun 3 '12 at 5:32
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up vote 8 down vote accepted

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 $t > 0$.

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Thanks, very nice. In fact I knew about this idea but I got caught up thinking about the effect of translation, that I forgot about the effect of dilation. One can do the same for any space $W$ and I'll add an answer to this effect. –  Nate Eldredge Jun 3 '12 at 0:46
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George Lowther's answer (thanks George!) set me on the right track. Here's a note that the same argument can be used for arbitrary infinite-dimensional $W$.

Let $\mu_t(A) = \mu(\frac{1}{\sqrt{t}} A)$, so that $P_t F(x) = \int_W F(x+y) \mu_t(dy)$. Since $\mu_t$ is a convolution semigroup, we have $\int_W P_t F(x)\mu(dx) = \int_W F(x) \mu_{1+t}(dx)$. (Or written another way, it's $P_{1+t}F(0)$.)

We can choose a sequence $f_1, f_2, \dots \in W^\*$ which, viewed as random variables on the probability space $(W,\mu)$, are iid $N(0,1)$. (Equip $W^*$ with the $L^2(\mu)$ inner product $\langle f ,g \rangle = \int_W f(x) g(x) \mu(dx)$ and use Gram-Schmidt.) Then on $(W,\mu_t)$ they are iid $N(0,t)$. Let $$A_t = \left\{x \in W : \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n |f_k(x)|^2 = t\right\}.$$ Clearly the $A_t$ are Borel, pairwise disjoint, and by the strong law of large numbers $\mu_t(A_t)=1$. If we take $F$ to be the indicator of $A_1^C$, then $F = 0$ $\mu$-a.e., but for every $t > 0$, $$\int_W P_t F(x) \mu(dx) = \mu_{t+1}(A^C) = 1.$$ Indeed, since $P_t F \le 1$, we have $P_t F = 1$ $\mu$-a.e.

This also works if $W$ is replaced by any reasonable infinite-dimensional TVS (say, Hausdorff and locally convex).

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