= Solution
Dominated convergence makes $\widehat f$ <continuous>, and $\|\widehat f\|_\infty\leq\|f\|_1$. 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>: $\widehat f\in C_0$.
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 $f,\widehat f\in L^1$, then $f$ agrees almost everywhere with the inverse <integral> of $\widehat f$. If $\widehat f=0$, this applies and gives $f=0$ almost everywhere. Hence \b[the <Fourier transform> is <injective> as a map of $L^1$ equivalence classes, with dense image in $C_0$].
Back to article page
