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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?

QuantumDots
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1 Answers1

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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 $$f=\mathcal F^{-1}(\mathcal F(f))=\mathcal F^{-1}(0)=0.$$

Bebop
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  • Hi! It is not direct from linearity, since $\mathcal{F}(f)=0$ implies $f\in\text{Ker}\mathcal{F}$. The problem again reduces to proving that: $\text{Ker}\mathcal{F}={0}$. How can the latter be done? – p-adic-manimanito Oct 03 '22 at 04:10
  • @P-Adic-Gatito No, the problem does not "again reduce to" that. Bebop's argument adds some useful information about the elements $f\in\ker\mathcal F:$ since their Fourier transform ($\mathcal Ff=g=0$) is integrable, $\mathcal F^{-1}$ applies to it, and $f=\mathcal F^{-1}g=\mathcal F^{-1}0=0.$ P.S. Up to normalization choices, $\mathcal F^{-1}$ is defined on $\L^1$ by $(\mathcal F^{-1}g)(x):=(\mathcal F g)(-x).$ – Anne Bauval Mar 13 '23 at 13:19