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, $\hat{f}(\xi_{0})=0.$