MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
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 $t > 0$.

share|cite|improve this answer
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

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).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.