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| + | == Problem #7.6, MA598R, Summer 2009, Weigel == | ||
Sneak through the inky black night back to [[The_Ninja%27s_Solutions]] | Sneak through the inky black night back to [[The_Ninja%27s_Solutions]] | ||
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~Ben Bartle | ~Ben Bartle | ||
| + | ---- | ||
| + | [[The_Ninja%27s_Solutions|Back to Ninja Solutions]] | ||
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| + | [[MA_598R_pweigel_Summer_2009_Lecture_7|Back to Assignment 7]] | ||
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| + | [[MA598R_%28WeigelSummer2009%29|Back to MA598R Summer 2009]] | ||
Revision as of 04:30, 11 June 2013
Problem #7.6, MA598R, Summer 2009, Weigel
Sneak through the inky black night back to The_Ninja's_Solutions
Suppose $ f \in L^{1}(\mathbb{R}) $ satisfies $ f*f=f $. Show that $ f=0 $.
$ \hat{f} = \widehat{f*f} = \hat{f}^2 $ by problem 2.
Now $ (\hat{f}(\xi))(\hat{f}(\xi)-1)=0 $ so $ \hat{f} = \chi_{A} $ for some set $ A $
But problem 5 gives $ \hat{f} $ is continuous and the limit is zero, hence $ \hat{f}\equiv 0 $
Applying an inverse fourier transfom gives $ f = 0 $ a.e.
$ f = f*f = \int_{\mathbb{R}} f(x-y)f(y)dy = 0 $ because the integral of something that is zero a.e. is zero.
~Ben Bartle
