If $f \in L^1(\mathbb{R})$ and its Fourier transform $\hat{f} \in L^1(\mathbb{R})$, then the inverse Fourier transform of $\hat{f}(\omega)$ is defined by: $g(t) = \mathcal{F}^{-1}\{\hat{f}(\omega)\} $.
If the original $f$ is continuous, then $g(t)=f(t)$ for every $t \in \mathbb{R}$.
Question:- If, $$\hat{f}(\omega)=0$$ then$$ f(t)=\ \mathcal{F}^{-1}\{\hat{f}(\omega)\}$$
$$f(t)= \mathcal{F}^{-1}({0})$$ $$f(t)=0.$$ Is it correct?
Let $f\in L^1(\mathbb{R})\cap C(\mathbb{R})$ and $\hat{f}\in L^1(\mathbb{R})$. The Fejér kernel $$\begin{equation} F_{\lambda}(x)=\frac{1}{2\pi}\int \limits_{-\lambda}^{\lambda}\left(1-\frac{|s|}{\lambda}\right)e^{isx}dx \end{equation}$$
is an approximate identity for $L^1(\mathbb{R})$ thus $\begin{equation} \lim \limits_{\lambda \rightarrow \infty} \|F_{\lambda}*f-f\|_1=0 \end{equation}$ where $$ F_{\lambda}*f(x)=\frac{1}{2\pi}\int \limits_{-\lambda}^{\lambda}\hat{f}(s)\left(1-\frac{|s|}{\lambda}\right)e^{isx}dx.$$
Now since $\hat{f}\equiv 0$ one obtains $f\equiv 0.$ By continouity this means $f(t)=0\ \forall t\in \mathbb{R}.$
Remark: For $f\in L^1(\mathbb{R})$ the Fourier transform $\hat{f}$ is bounded and uniformly continous, so we can indeed evaluate $\hat{f}$ at any point $\omega_0\in \mathbb{R}.$