Prove that $\ f*g = f$

Question

Good morning,

We have : $g \in$ $L^{p}(\mathbb{R}^n)$, $\forall p \in [0;\infty)$ and $f\in$ $L^{1}(\mathbb{R}^n)$, $\widehat{f}(x) =0$ if $|x| \geq R $, $\varphi = \widehat{g}$ such as $\varphi(x) = 1, \left | x \right | \leqslant R $ and $ x \in \mathbb{R}^n$, $\varphi\in {C_{c}^{\infty }(\mathbb {R} ^{n})}$

So I did a Fourier Transform of $\ f*g$ (Convolution product) and got $\widehat{f*g}(x)=\widehat{f}$ How can I say that $\ f*g = f$ with what I found ?

Thank you in advance !

What do you mean by "$\widehat{f(x)} =0$ if $|x|\ge R$"? Do you mean that "$\widehat{f}\color{red}{(x)} =0$ if $|x|\ge R $"? — Sam Wong, Dec 16 '20 at 09:08

score 1 · Accepted Answer · answered Dec 16 '20 at 08:56

1

The Fourier transform is injective (in the relevant function spaces). If $\hat h=\hat f$, then $h=f$. Now you have $h=f*g$.

answered Dec 16 '20 at 08:56

Joonas Ilmavirta

25,809

Thank you very much ! – AFrenchstudent Dec 16 '20 at 08:59
@AFrenchstudent I'm glad it helped! Please accept an answer when it settles your question. You also have enough reputation to vote up any questions and answers you like. – Joonas Ilmavirta Dec 16 '20 at 09:01
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Just a supplement: A proof that Fourier Transform is injective on $L^1$ can be found here. – Sam Wong Dec 16 '20 at 09:05

Prove that $\ f*g = f$

1 Answers1