Let $\mu$ be a finite complex valued measure on $\mathbb{R}$ and let $\hat{\mu}$ be it's Fourier–Stieltjes transform $$ \hat{\mu}(\omega)= \int_{\mathbb{R}} e^{it\omega} d \mu(t) $$ Question: Does $\hat{\mu}$ uniquely determine $\mu$? I am fairly sure that it does. However, I was not able to locate my standard references. Is there a good place where I can find proof of this fact?
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1$\begingroup$ As a variant on Vincius Novelli's answer: instead of using the Schwarz class, note that VN's argument actually proves that if $\mu$ is a finite measure on ${\mathbb R}$ then it annihilates every $f\in C_0({\mathbb R})$ which is the (inverse) Fourier transform of an $L^1$-function on $\widehat{\mathbb R}$. The class of all such $f$ is denoted by $A({\mathbb R})$, this is the Fourier algebra of ${\mathbb R}$, which is known to be dense in $C_0({\mathbb R})$ using e.g. Stone-Weierstrass. One advantage of this approach is that it works on any locally compact abelian group. $\endgroup$– Yemon ChoiCommented Jan 9, 2023 at 0:26
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I assume that $\mu$ is a regular complex Borel measure. Assume that $\widehat{\mu}=0$. Let $f \in \mathcal{S}(\mathbb{R})$ be a Schwartz class function. Then, writing $f=\widehat{g}$ for $g\in \mathcal{S}(\mathbb{R})$, we have from Fubini's theorem that $$ \int_{\mathbb{R}}f(t)d\mu(t) = \int_{\mathbb{R}}\int_{\mathbb{R}}g(s)e^{-its}dsd\mu(t) = \int_{\mathbb{R}}g(s)\widehat{\mu}(-s)ds = 0. $$ This implies $\mu$ is orthogonal to the Schwartz class, and by density, it's orthogonal to the space $C_0(\mathbb{R})$ of continuous functions vanishing at infinity, which implies by Riesz's theorem that $\mu=0$.
answered Jan 7, 2023 at 16:09
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$\begingroup$ What do you mean 'by the density'. I couldn't follow that sentence fully. Also, do you have a good reference where I can find this proof. I want to cite something. $\endgroup$– BobyCommented Jan 7, 2023 at 16:15
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1$\begingroup$ @Boby I mean that $\mathcal{S}(\mathbb{R}) \subset C_0(\mathbb{R})$ is dense when $C_0(\mathbb{R})$ is endowed with the Banach norm $\|f\|_{C_0(\mathbb{R})}=\sup_{\mathbb{R}}|f|$. This has a standard proof using mollifiers, see here for example. For Riesz's theorem you can check Rudin's Real and Complex Analysis, chapter 6. For this proof in particular I don't know a reference, but it's so simple that I doubt you need one. $\endgroup$ Commented Jan 7, 2023 at 16:20