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Dominated convergence makes $\hat{f}$ continuous, and $\Vert \hat{f}{\Vert }_{\infty }\le \Vert f{\Vert }_1$. Smooth compactly supported functions are dense in $L^1$. Their Fourier transforms decay at infinity by integration by parts, so uniform approximation proves the Riemann-Lebesgue lemma: $\hat{f}\in C_0$.

For density, smooth compactly supported functions are also uniformly dense in $C_0$ by cutoff and mollification. Each such target $g$ is a Schwartz function; its inverse Fourier transform is another Schwartz function, hence belongs to $L^1$, and Fourier inversion recovers $g$. Therefore the image of a function of $L^1$ is dense in $C_0$.

The $L^1$ Fourier inversion theorem says that if $f,\hat{f}\in L^1$, then $f$ agrees almost everywhere with the inverse integral of $\hat{f}$. If $\hat{f}=0$, this applies and gives $f=0$ almost everywhere. Hence the Fourier transform is injective as a map of $L^1$ equivalence classes, with dense image in $C_0$.

Let $e=(1,0)$ and defineThen $\Vert g(x)-x\Vert \le 1$ everywhere, but $0$ is not in the image of a function. The map is discontinuous at $0$, giving a bounded-displacement map which is not surjective.