5
$\begingroup$

Let $E=C[0,1]$, space of all real-valued continuous functions on $[0,1]$, $\mathcal{E}$ be its Borel $\sigma$-algebra and $\mu$ a Gaussian measure on $E$. I need help proving the following claim:

$E^*$ is the space of signed measures on $[0, 1]$ and the duality is given by: $\langle\nu,f\rangle=\nu(f)=\int_0^1f(t)\;\nu(dt),$ where of $E^*$ is a space of all continuous linear functions on $E$.

My thoughts: I know that it is an application of Riesz representation theorem. Firstly I need an inner-product on $C[0,1]$, which may as well be $\int_0^1f(t)g(t)\;dt$. Then any functional on $E$ can identified with an integral with respect to Lebesgue measure. However, the author insists on $\nu$. Thank you!

  • 3
    If you have a look at [Wikipedia](http://en.wikipedia.org/wiki/Riesz_representation_theorem), there are several theorems called Riesz representation theorem. One of them is about Hilbert spaces, another has $C(X)$ with $X$ compact as a special case.2012-01-25
  • 0
    Thank you I never knew about the $C(X)$, hence was very confused.2012-01-25
  • 1
    In fact, there was a [question](http://math.stackexchange.com/questions/36710/different-versions-of-riesz-theorems) at this site about these 3 theorems.2012-01-25

1 Answers 1

4

I don't see where the comment button is, maybe because I'm a newbie?

This is the Riesz representation theorem which states that the topological dual space of the space of continuous functions on a compact space $X$ is the space of Borel measure on $X$. You can see a proof in Real and complex analysis by Rudin.

Your thought is not true, because the topology defined by your inner product and the topology defined by the sup-norm on $C([0,1])$ are different.

  • 0
    My inner product will define a different $\mathcal{E}$? I thought it could be a problem, but I did not know the other Riesz theorem. Thank you for the reference.2012-01-25
  • 2
    The comment button says 'add comment' at the bottom of each answer or question. You need to have reputation $\geqslant 50$, see [faq](http://math.stackexchange.com/faq#reputation) for more details.2012-01-25
  • 0
    @Sasha : thank you :D2012-01-25
  • 0
    @Artiom Fiodorov : but that book is rather suitable for the ones who are interested in complex analysis. For your question, we can construct a continuous function of norm very large but with respect to your norm, it is very small. It is an elementary exercise.2012-01-25