For $\xi \geq 0$, define $ g(\xi)=\left\{\begin{array}{ll}\frac{1}{\ln \xi}, & \xi>e \\ \frac{\xi}{e}, & 0 \leq \xi \leq e\end{array} \right.$, when $ \xi<0 $ , define $g(\xi)=-g(-\xi)$. Prove that There is no $f(x) \in L^{1}(\mathbb{R})$, s.t. $\hat{f}(\xi)=g(\xi)$, where $\hat{f}(\xi)=\int_{\mathbb{R}} f(x) e^{-i x \xi} \mathrm{d} x, \xi \in \mathbb{R}, i=\sqrt{-1}$.
I know that $g(\xi) $ is not in $L^1(\mathbb{R})$ , and $\|\hat{f}(\xi)\|_{L^\infty}\leq C\|f\|_{L^1}$. But these seem to be useless for this problem , can anyone help me?