Does every continous function in L1 is a fourier transform?

Question

Let $f: \mathbb{R}\rightarrow\mathbb{C}$ denote a continous function in $L_1(\mathbb{R})$. Does this imply that there exists a function $G: \mathbb{R}\rightarrow\mathbb{C}$ in $L_1(\mathbb{R})$ such that $f$ is $G$'s fourier transform?

Answer 1

No. For one thing, the Riemann-Lebesgue lemma says the Fourier transform of an $L^1$ function goes to $0$ at $\pm \infty$.

But even if you make that assumption, not every continuous function that goes to $0$ at $\pm \infty$ is the Fourier transform of an $L^1$ function. One way to prove this is to show that there are $L^1$ functions of norm $1$ whose Fourier transforms have arbitrarily small supremum norm.