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!
Best Answer
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.