(New page: Let <math>f\in L^1(\mathbb{R}^n)</math>. Prove: a) If <math>f\geq 0</math> then <math>\|\hat{f}\|_{\infty}=\hat{f}(0)=\|f\|_1</math> b) If <math>f</math> is continuous at <math>0</math> ...) |
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b) From the reverse Fourier Transform we know that <math>f(t)=\int\hat{f}(x)e^{2\pi ixt}dx</math> | b) From the reverse Fourier Transform we know that <math>f(t)=\int\hat{f}(x)e^{2\pi ixt}dx</math> | ||
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| + | <math>f(0)=\int \hat{f}(x)dx = \|\hat{f}\|_1</math> (since <math>\hat{f}\geq 0</math>) | ||
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| + | (I'm not sure how much of this is correct since I didn't use the continuity of f at 0, just that it's defined there) | ||
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| + | --[[User:Rlalvare|Rlalvare]] 22:42, 28 July 2009 (UTC) | ||
Revision as of 18:42, 28 July 2009
Let $ f\in L^1(\mathbb{R}^n) $. Prove:
a) If $ f\geq 0 $ then $ \|\hat{f}\|_{\infty}=\hat{f}(0)=\|f\|_1 $
b) If $ f $ is continuous at $ 0 $ and $ \hat{f}\geq 0 $ then $ \|\hat{f}\|_1 = f(0) $
Proof: a)
$ \|\hat{f}\|_{\infty} = |\int e^{-ixt}f(t)dt|\leq \int |f(t)|dt = \|f\|_1 = \hat{f}(0) $ (since $ f\geq 0 $)
$ \|\hat{f}\|_{\infty}\geq \hat{f}(0) = \int f(t)dt = \|f\|_1 $ (since $ f\geq 0 $)
b) From the reverse Fourier Transform we know that $ f(t)=\int\hat{f}(x)e^{2\pi ixt}dx $
$ f(0)=\int \hat{f}(x)dx = \|\hat{f}\|_1 $ (since $ \hat{f}\geq 0 $)
(I'm not sure how much of this is correct since I didn't use the continuity of f at 0, just that it's defined there)
--Rlalvare 22:42, 28 July 2009 (UTC)
