(New page: Back to The Pirate's Booty Suppose <math>f\in L^1({\mathbb{R}})</math> satisfies <math>f\ast f = f</math>. Show that <math>f=0</math>. Proof: We know that <math>\widehat{f\ast f}=\ha...) |
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| + | == Problem #7.6, MA598R, Summer 2009, Weigel == | ||
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but since the convolution is define everywhere and <math>f\ast f = 0\Rightarrow f = 0</math> everywhere. | but since the convolution is define everywhere and <math>f\ast f = 0\Rightarrow f = 0</math> everywhere. | ||
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Revision as of 05:37, 11 June 2013
Problem #7.6, MA598R, Summer 2009, Weigel
Back to The Pirate's Booty
Suppose $ f\in L^1({\mathbb{R}}) $ satisfies $ f\ast f = f $. Show that $ f=0 $.
Proof: We know that $ \widehat{f\ast f}=\hat{f}^2 $ so the condition on f implies that
$ \hat{f}^2=\hat{f}\Rightarrow \hat{f}(\hat{f}-1)=0 \Rightarrow \hat{f}= 0 $ or $ \hat{f}=1 $
Since we also know that $ \lim_{|x|\to\infty}\hat{f}(x)=0 \Rightarrow \hat{f}=0 $
From the Inverse Fourier Transform we clearly see that $ f\equiv 0 $ a.e.
but since the convolution is define everywhere and $ f\ast f = 0\Rightarrow f = 0 $ everywhere.
