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.

  • 1
    $\begingroup$ From this point of view the Wiener measure is a measure on $E^*$ supported on a much smaller subspace of $E^*$. A good place to look is volume 4 of Gelfand's classic Generalized functions. This volume is written with Vilenkin. In Chapter 3 has a very nice discussion of these issues. $\endgroup$ Dec 27, 2013 at 16:12
  • $\begingroup$ Could you give a reference for the version of Bochner-Minlos that you have in mind? $\endgroup$ Dec 28, 2013 at 6:29
  • $\begingroup$ I am attempting to read "Brownian Motion" by Hida. Wiener measure is covered in Chapter 2 and Bockner-Minlos is in Chapter 3. Thanks. $\endgroup$ Dec 29, 2013 at 0:17

2 Answers 2


Yes. Wiener measure can be arrived at using the Bochner-Minlos Theorem in at least two ways.

  1. One can consider the bilinear form on $S(\mathbb{R})$ $$ C(f,g)=\int_{\mathbb{R}}\ \overline{\widehat{f}(\xi)} \widehat{g}(\xi)\ \frac{d\xi}{2\pi} $$ then the BM Theorem applied to the characteristic function $\exp\left(-\frac{1}{2}C(f,f)\right)$ builds white noise. Namely, one gets a random distribution $W$ in $S'(\mathbb{R})$. Brownian motion is obtained as $W(f)$ where the "test-function" $f$ is the charcteristic function of the interval $[0,t]$. Of course one has to work a bit to show one can indeed feed $W$ a singular object like ${\bf 1}_{[0,t]}$.
  2. One can consider the subspace $$ S_0(\mathbb{R})=\left\{f\in S(\mathbb{R})| \int_{\mathbb{R}} f(x)\ dx=0\right\} $$ and the bilinear form $$ B(f,g)=\int_{\mathbb{R}}\ \frac{\overline{\widehat{f}(\xi)} \widehat{g}(\xi)}{|\xi|^2}\ \frac{d\xi}{2\pi} $$ on this subspace. The BM Theorem then gives Brownian motion directly as a random element $\phi$ in the dual of $S_0(\mathbb{R})$. For $\phi(f)$ to make sense as a random variable one needs the test function $f$ to have a zero integral. An extreme case (which can be made sense of) is when $f$ is a delta function at $t$ minus a delta function at $s$, say with $s<t$. Then $\phi(f)=\phi(t)-\phi(s)$. The increments of the process $\phi$ make sense but not the process itself. Finally Brownian motion can be defined as $B(t)=\phi(t)-\phi(0)$. In other words one can lift the ambiguity of being able to add a constant by imposing the initial condition $B(0)=0$.

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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