Let $\mathcal{G}$ be the set of functions $g\in L^1(\mathbb{R})$ whose Fourier transforms are continuous and have compact support (i.e. $\hat{g}\in C_c(\mathbb{R}))$. I am trying to prove a claim that $\mathcal{G}$ is dense in the set of continuous functions that vanish at infinity, $C_o(\mathbb{R})$
Here is a summary of my attempt.
Let us take $f\in C_0(\mathbb{R})$ and seek to establish that a member of $\mathcal{G}$ is arbitrarily close in $\|\cdot\|_{\infty}$. Without loss of generality let $f\in C_c^{\infty}(\mathbb{R})$, since $C_c^{\infty}(\mathbb{R})$ is dense in $C_0(\mathbb{R})$. Then $f$ belongs to the Schwartz space $\mathscr{S}(\mathbb{R})$ (the rapidly decaying functions) so $\hat{f}\in\mathscr{S}(\mathbb{R}')$ and the Fourier transformation is an 'onto' isomorphism.
I will leave out the finesse and simply identify the dual space with $\mathbb{R}$ itself. Since $\mathscr{S}(\mathbb{R})\subset C_0(\mathbb{R})$, $C_c^{\infty}(\mathbb{R})$ is also dense in $\mathscr{S}(\mathbb{R})$, so we can find a function $\tilde{g}$ in $C_c^{\infty}(\mathbb{R})$ arbitrarily close to $\hat{f}$. The isomorphism between $\mathscr{S}(\mathbb{R})$ and $\mathscr{S}(\mathbb{R'})$ then means there is a $g\in\mathscr{S}(\mathbb{R})$ of which $\tilde{g}$ is the Fourier transform, so we rename $\tilde{g}$ into $\hat{g}$. (The measure used for integrating over the function space is of course chosen to be compatible with the Lebesgue measure over the reals.)
It is now straightforward to establish that $\|g-f\|_{\infty} \le \int_{\mathbb{R}}\|\hat{g}-\hat{f}\|_{\infty} \hspace{0.1cm} d\xi \le \int_{\mathbb{R}}\delta\hspace{0.1cm} d\xi$ for any $\delta$. However, while the support of $\hat{g}$ is compact, there is nothing to say the support of $\hat{f}$ is compact. As a result, I can't restrict the region of integration to a set of finite measure, and get an infinite bound.