In my MSE answer here, I discussed the example of compactly supported continuous function
$$g(x)=
\begin{cases}
\dfrac{\frac12 -x}{\log(x)},&0<x\leq1/2\\
0,&\text{otherwise}
\end{cases}$$
which is not the Fourier transform of any $f\in L^1(\mathbb R)$, i.e. $g\not\in \mathcal F(L^1(\mathbb R))$. Looking through the proof, I think that one can smooth out $g$ at $x=\frac 12$, and get a function $\tilde g \in C_c(\mathbb R)$, supported on $[0,\frac 12]$, infinitely differentiable at EVERY point besides $x=0$ (though not even once differentiable at $x=0$), which is not in $ \mathcal F(L^1(\mathbb R))$.
I am now wondering if there is an example of compactly supported function $G\in C_c(\mathbb R)$ which is differentiable at every point, which is is still not in $ \mathcal F(L^1(\mathbb R))$. Equivalently, is there a everywhere differentiable $G\in C_c(\mathbb R)$ whose Fourier transform is not in $L^1(\mathbb R)$.
Bernstein's theorem tells us that if $G'$ is bounded, then $G$ is Lipschitz so $\hat G \in L^1(\mathbb R)$. So any potential $G$ must have unbounded derivative.
Best Answer
Yes, such differentiable compactly supported functions $G$ with $\widehat{G}\notin L^1$ exist. This follows from the fact the Fourier transform is not bounded as a map from $\{f\in C[0,1]: f(0)=f(1)=0\}$ to $L^1$ (in fact, does not even map into this space).
So we can find $f_n\in C_0^{\infty}(0,2^{-n^2})$ with $|f_n|\le 2^{-2n}$, $\|\widehat{f_n}\|_1\ge 2^{n^2}$. We can then set $$ G(x) = \sum e^{i\alpha_n x} f_n(x-2^{-n}) , $$ with $\alpha_n$'s to be chosen later.
Differentiability of $G$ is only an issue at $x=0$, but we made sure that $|G(x)|\lesssim x^2$, so this is also guaranteed.
The series converges in $L^2$ and thus the same is true for the Fourier transforms. Hence also $\sum_{n=1}^N e^{i2^{-n}\xi}\widehat{f_n}(\xi-\alpha_n) \to \widehat{G}(\xi)$ pointwise almost everywhere along a suitable subsequence. Even though $\widehat{f_n}$ is not compactly supported, it is of course still true that we can catch almost all of its $L^1$ norm by integrating over a sufficiently long but bounded interval $I_n$. Moreover, the $\widehat{f_n}$ are Schwartz functions and thus decay pointwise. We can now make use of the $\alpha_n$ to shift these quasi-supports $I_n$ to well separated regions. It follows that $\widehat{G}\notin L^1$, as desired.