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I was reading a paper and I think that they used the following theorem:

Let $G$ compact group and $\mu$ a probability measure on $G$. If $$\hat{\mu}(\xi)= \int_G \overline{\xi(x)} d\mu(x) = \begin{cases} 0 & \text{ if }\xi = 0 \\ 1 & \text{ if } \xi\neq 0\end{cases} \quad \xi \in \hat{G}$$ then $\mu$ is the Haar measure.

Is this theorem true?

I looked on the book An Introduction to Harmonic Analysis of Katznelson that my teacher recommended and I only found the converse of the theorem on a exercise.

Any help will be appreciated.

user111216
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  • From the context I concluded that $G$ is abelian, even though you did not say so. You may want to edit to clarify this. –  Oct 12 '14 at 22:31

1 Answers1

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The linear span of characters is dense in $C(G)$ (the space of complex-valued continuous functions on $G$), by the special case of Peter-Weyl theorem (for abelian groups, all irreducible representations are one-dimensional, i.e., they are the characters). This is specifically addressed in the lecture notes by Dikran Dikranjan.

By the exercise in Katznelson you know that the Haar measure indeed has the property stated in your question. If two Radon measures agree on a dense subset of $C(G)$, they agree on all of $C(G)$ and therefore are the same measure (this is one part of the Riesz representation theorem).

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