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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!

SBF
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Tom Artiom Fiodorov
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1 Answers1

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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.

Martin Sleziak
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Đức Anh
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  • 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. – Tom Artiom Fiodorov Jan 25 '12 at 17:47
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    The comment button says 'add comment' at the bottom of each answer or question. You need to have reputation $\geqslant 50$, see faq for more details. – Sasha Jan 25 '12 at 17:51
  • @Sasha : thank you :D – Đức Anh Jan 25 '12 at 18:42
  • @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. – Đức Anh Jan 25 '12 at 18:47