Problem #7.5 MA598R, Summer 2009, Weigel
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Let $ f\in L^1(\mathbb{R}) $. Show that $ \hat{f}(x) $ is continuous and $ \lim_{|x|\to\infty} \hat{f}(x)=0 $.
Proof: To show continuity, we only need to show that if $ x_k\to x $ then $ \hat{f}(x_k)\to\hat{f}(x) $
$ \lim_{k\to\infty}\hat{f}(x_k)=\lim_{k\to\infty}\int e^{-ix_kt}f(t)dt = \int e^{-ixt}f(t)dt = \hat{f}(x) $
We can pass this limit through the integral since $ \hat{f} $ is dominated by $ f\in L^1 $
Now, to prove that $ \widehat{f}(\xi) \rightarrow 0 $ as $ |\xi|\rightarrow\infty $
Define the Fourier Transform to be:
$ \widehat{f}(\xi) = \int_{R} f(x) e^{-\imath 2\pi x \xi} dx $
Note that if we let $ x \rightarrow x+\frac{1}{2\xi} $, we get the following:
$ - \widehat{f}(\xi) = \int_{R} f(x+\frac{1}{2\xi}) e^{-\imath 2\pi (x+\frac{1}{2\xi}) \xi} dx $
Hence, subtracting the two, we get:
$ 2 \widehat{f}(\xi) = \int_{R} \left[f(x) - f(x+\frac{1}{2\xi})\right] e^{-\imath 2\pi x \xi} dx $
And now, we get that
$ |2 \widehat{f}(\xi)| \leq \int_{R} |\left[f(x) - f(x+\frac{1}{2\xi})\right]| dx $
Now, let $ |\xi| \rightarrow \infty $, and from a previous assignment, the right hand side goes to 0, and so too must $ \widehat{f}(\xi) $
Written by Robert and Nick
