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We fix the point $\xi_{0}\in \mathbb R.$

Choose sequence $\{f_{n}\}_{n\in \mathbb N}\subset L^{1}(\mathbb R)$ with the following property :

(1) $\|f_{n}\|_{L^{1}(\mathbb R)} \leq 1, $ for $n\in \mathbb N;$ and (2) $\hat{f_{n}}(\xi_{0})= 0,$ for every $n\in \mathbb N.$

Define $f:\mathbb R \to \mathbb C$ as follows: $$f(x)=\sum_{n=1}^{\infty}\frac{2^{n-1}}{3^{n}} f_{n}(x), (x\in \mathbb R).$$

Since $\|f_{n}\|_{L^{1}}\leq 1$, and $\sum_{n=1}^{\infty} \frac{2^{n-1}}{3^{n}}<\infty,$ the series, $ \sum_{n=1}^{\infty}\frac{2^{n-1}}{3^{n}} f_{n}(x) $ is absolutely summable in $L^{1}(\mathbb R);$ and hence, $f\in L^{1}(\mathbb R).$

My Question is: Can we expect $\hat{f}(\xi_{0})= 0$ ?

Ruff attempt: By definition of Fourier transform, we have,

$\hat{f}(\xi_{0})= \int_{\mathbb R} f(x) e^{-2\pi i \xi_{0} \cdot x} dx = \int_{\mathbb R}(\sum_{n=1}^{\infty}\frac{2^{n-1}}{3^{n}} f_{n}(x)) e^{2\pi i \xi_{0}\cdot x} dx; $ now from here, if I guess, we can interchange sum and integral; (but I don't know how to justify it); then by the hypothesis (2), $\hat{f_{n}}(\xi_{0})=0;$ it will follows that, $\hat{f}(\xi_{0})=0.$

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  • $\begingroup$ The series converges in $L^1$. The map $g \mapsto \int g(x)e^{2\pi i \xi\cdot x}\,dx$ is continuous for every $\xi$. That allows the interchange. $\endgroup$ – Daniel Fischer Jul 25 '14 at 15:46
  • $\begingroup$ @DanielFischer; thanks, sorry I could not follow you, would you please tell me how does it help ? $\endgroup$ – Inquisitive Jul 25 '14 at 15:54
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    $\begingroup$ Since the series converges in $L^1$, we have $$\int \left(\lim_{N\to\infty} \sum_{n=1}^N f_n(x)\right)e^{2\pi i\xi\cdot x}\,dx = \lim_{N\to\infty} \sum_{n=1}^N \int f_n(x)e^{2\pi i\xi \cdot x}\,dx$$ by continuity of the integral as a function of $g\in L^1$. $\endgroup$ – Daniel Fischer Jul 25 '14 at 16:02
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Answered in comments:

The series converges in $L^1$. The map $g \mapsto \int g(x)e^{2\pi i \xi\cdot x}\,dx$ is continuous for every $\xi$. That allows the interchange. ...we have $$\int \left(\lim_{N\to\infty} \sum_{n=1}^N f_n(x)\right)e^{2\pi i\xi\cdot x}\,dx = \lim_{N\to\infty} \sum_{n=1}^N \int f_n(x)e^{2\pi i\xi \cdot x}\,dx$$ by continuity of the integral as a function of $g\in L^1$. -Daniel Fischer

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