0
$\begingroup$
  1. We can define a measure on $C[0,1]$ by viewing a continuous function as a path of 1-dimensional Brownian motion. This is called classical Wiener measure. I am wondering if there is a generalization to function space of higher dimension?

  2. I knew from Wikipedia that there is something called abstract Wiener space, which uses a canonical Gaussian cylinder set measure. However I do not know what is the "quotient inner product" mentioned in the construction. Could anyone explain for me what is a quotient inner product, and how does it coincide with the classical case?

$\endgroup$
  • $\begingroup$ To 1: Of course, there is a simple generalization to the $d$-dimensional case. Just take the $d$-dimensional browinan motion. The $d$-dimensional BM can be defined by setting $B_t = (B_t^{(1)}, \ldots B_t^{(d)})$, where $B_t^{(1)},\ldots B_t^{(d)}$ are continuous independent Brownian Motions. $\endgroup$ – p4sch Nov 19 '18 at 14:50
  • $\begingroup$ @p4sch But isn't that a measure on $R$ to $R^{n}$ functions instead of $R^{n}$ to $R$ functions? $\endgroup$ – 1830rbc03 Nov 19 '18 at 17:15
  • $\begingroup$ Oh, you are right! $\endgroup$ – p4sch Nov 19 '18 at 17:21
  • $\begingroup$ I find interesting the idea of ​​using a continuous function of ${C[0,1]}^n$ to $C([0,1]^n)$ to induce a distribution in $C([0,1]^n)$. $\endgroup$ – Daniel Camarena Perez Nov 20 '18 at 5:18
  • $\begingroup$ @Daniel Camarena Perez I just realized the "measure" I construct in 1 is wrong because not every function $h(x,y)$ can be represented as $f(x)+g(y)$ so please ignore it. $\endgroup$ – 1830rbc03 Nov 20 '18 at 19:08
0
$\begingroup$

Your question is ill-posed as is. There exists probability measures on $C([0,1]^n)$, for example the unit Dirac mass at the constant function equal to zero. It would help if you said which properties you would like your measure to satisfy. Note that the natural generalization of Brownian motion to higher dimensional domains is the so-called Gaussian free field but it is a measure supported on generalized functions/Schwartz distributions instead of continuous functions.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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