Injectivity of Fourier transform between $L^1(\mathbb{R})$ and $C_0(\mathbb{R})$

Question

The Fourier transform maps from $L^1(\mathbb{R})$ to $C_0(\mathbb{R})$ where $C_0(\mathbb{R})$ is all continuous functions that vanish as $x \rightarrow \infty$. Now given $f,g \in L^1(\mathbb{R})$, clearly we can show injectivity of the Fourier transform if $\mathcal{F}(g), \mathcal{F}(g) \in L^1(\mathbb{R})$ because the inverse Fourier transform is defined in this case. But how do we show injectivity otherwise? Every source I see seems to just say this is trivial; admittedly, I have stated this as true without (now I realize) actually understanding why. Help?

Answer 1

The Fourier transform is a linear map, you only have to check that $$\forall f\in L^1(\mathbb R), \mathcal F(f)=0 \Rightarrow f=0.$$

Let $f\in L^1(\mathbb R)$ such that $\mathcal F (f)=0$. Hence $\mathcal F(f)$ is $L^1(\mathbb R)$ since its the zero function and therefore its Fourier inverse exists. So $$f=\mathcal F^{-1}(\mathcal F(f))=\mathcal F^{-1}(0)=0.$$