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Problem #7.13, MA598R, Summer 2009, Weigel

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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)

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