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We put, $M(\mathbb R)= $The set of bounded complex Borel measure $\mu$ on $\mathbb R$ and for $\mu \in M(\mathbb R)$, we define $||\mu||:= |\mu| (\mathbb R) = \text {total variation of } \ \mu $; and $$B(\mathbb R) := \{f:\mathbb R \to \mathbb C : \exists \ \mu \in M(\mathbb R) \ \text {such that} \ f(y) = \hat{\mu}(y) \};$$ where, $\hat{\mu}(y)= \int_{\mathbb R} e^{-iyx} d\mu(y),$ and for $f\in B(\mathbb R)$, we define $$||f||=||\hat{\mu}||:= ||\mu||.$$

My Question : Put, $\mathbb T:= \{z\in \mathbb C : |z|=1 \}$ and suppose $f$ is function on the circle $\mathbb T$ and its Fourier series is absolutely convergent, that is, $\sum_{n\in \mathbb Z}|\hat{f}(n)|< \infty$; where, $\hat{f}(n)=\frac{1}{2\pi}\int_{-\pi}^{\pi} f(t) e^{-int} dt, \ (n\in \mathbb Z)$; , and it satisfies the property $f(e^{i\theta})=0$ for $\pi-\delta \leq \theta \leq \pi + \delta $; ($0<\delta < \pi $).

Define $F:\mathbb R \to \mathbb C$ such that $$F(x):= f(e^{ix})$$ for every $x\in \mathbb R$.

How to show that $F\in B(\mathbb R)$; and $||F||= \sum_{n\in \mathbb Z}|\hat{f}(n)|$.

Vague Attempt : To show $F\in B(\mathbb R)$, that is , by definition of $B(\mathbb R)$, I must show, there exists complex Borel $\mu$ on $\mathbb R$ such that, $$F(y)=\hat{\mu} (y)= \int_{\mathbb R} e^{-iyx} d\mu(x);$$ keeping this goal in mind, inversion formula suggests me that, I should take, $\mu $, such that, $ \int_{\mathbb R} e^{-iyx} d\mu(x) = \int_{\mathbb R} e^{-ixy} \hat{F}(y) dy $ ( the inversion formula valid provided $F, \hat{F} \in L^{1} (\mathbb R)$, and but here in our case this may not be the case and this may be the real difficulties); If this is the case, then I guess, we can define, $\mu (E)=\mu_{F} (E)= \int_{E} \hat{F} (y) dy $, for Borel set $E$ and also I guess, $||\mu||= ||\hat{F}||_{L^{1}(\mathbb R)}$; but I really don't see be any connection between $||F||$ and $\sum_{n\in \mathbb Z}|\hat{f}(n)|$;

Thanks,

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put on hold as off-topic by Yemon Choi, Chris Godsil, Stefan Kohl, S. Carnahan 2 days ago

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This question appears to be off-topic because it is rather basic; that is, the solution is straightforward given the original definitions –  Yemon Choi Aug 16 at 14:38

1 Answer 1

I suspect that you meant

\begin{equation*} \widehat{f}(n) = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(e^{i\theta}) e^{-in\theta} d\theta \end{equation*} in which case \begin{equation*} F(x) = f(e^{ix}) = \sum_{n\in\mathbb{Z}} \widehat{f}(n) e^{inx}. \end{equation*}

The measure $\mu$ is then a series of shifted Diracs

\begin{equation*} \mu = \sum_{n\in\mathbb{Z}} \widehat{f}(n) \delta(\cdot+n) \end{equation*}

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