Fourier inverse transform in ${C_0}$ It is well known from the Riemann-Lebesgue theorem that if $f \in {L^1}(R)$ then $\widehat f \in C(R)$ and its limit equal to $0$.
but the inverse fourier transform i defined from ${L^1}(R)$ too....Is there any function in ${C_0}$ such that its inverse doesn't exist ? thanks.
 A: Yes, $$F(\gamma) = \begin{cases} \frac{1}{\ln{\gamma}} & \gamma>e\\
\frac{\gamma}{e} & |\gamma|< e\\
-\frac{1}{\ln{-\gamma}} & \gamma<-e\end{cases}$$
is continuous and vanishes at infinity, but is not the Fourier transform of an $L^1$ function, cf., Harmonic Analysis and Applications by John J. Benedetto (https://www.crcpress.com/Harmonic-Analysis-and-Applications/Benedetto/p/book/9780849378799) examples 1.4.4 and 3.3.4a for derivation.
EDIT: For those who can't get the book, the general idea is as follows: if you have a function $f\in L^1(\mathbb{R})$ which has an odd Fourier transform $\widehat{f}(\gamma)$, then $$\widehat{f}(\gamma) = -i\int_{\mathbb{R}} f(t)\sin(2\pi\gamma t)dt.$$
This implies that \begin{align}|\int_{a}^\infty \frac{\widehat{f}(\gamma)}\gamma d\gamma| &\leq |\int_a^\infty \int_{\mathbf{R}} \frac{f(t) \sin(2\pi \gamma t)}{\gamma} d\gamma dt|\\
& = | \int_{\mathbf{R}}f(t) \int_a^\infty\frac{\sin(2\pi \gamma t)}{\gamma} dtd\gamma|\\
& \leq C ||f(t)||_{L^1},\end{align}
for $a>0$ and where the last inequality is due to the well known fact that $|\int_a^\infty \sin(x)/x dx|<\infty$. 
$F(\gamma)$ as defined above clearly violates this property as $F$ is odd and $$\int_e^\infty \frac{F(\gamma)}{\gamma} d\gamma \rightarrow \infty.$$
