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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 $\sum_{n\in \mathbb Z}|\hat{f}(n)|$;


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closed as off-topic by Yemon Choi, Chris Godsil, Stefan Kohl, S. Carnahan Aug 18 '14 at 11:45

  • This question does not appear to be about research level mathematics within the scope defined in the help center.
If this question can be reworded to fit the rules in the help center, please edit the question.

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 '14 at 14:38

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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