Difference between revisions of "Number fourteen (too lazy to come up with something else pirately to say)" - Rhea

Revision as of 19:40, 28 July 2009

Problem 14

$ \text{Let } f\in C_c^{\infty}(R^n) \text{ be radial. Show that } \widehat{f} \text{ is also radial.} $

Note that f being radial implies that for some $ f^* $, $ f(x) = f^*(|x|) $ for every $ x\in R^n $

$ \widehat{f}\left(\xi e^{\imath \theta}\right) = \int_{R^n} f^*(|x|) e^{-\imath <x,\xi e^{\imath \theta}>} dx $

$ = \int_{R^n} f^*(|x|) e^{-\imath <x e^{-\imath \theta},\xi>} dx $

Now use a change of variables, namely replace $ x $ with $ x e^{\imath \theta} $, and since this change of variables is given by a rotation (i.e. an orthogonal transformation with Jacobian 1)

$ = \int_{R^n} f^*(|x e^{\imath \theta}|) e^{-\imath <x, \xi>} dx $

But recall that $ |x e^{\imath \theta}| = |x| $, so:

$ = \int_{R^n} f^*(|x|) e^{-\imath <x,\xi>} dx = \widehat{f}(\xi) $

Which gives precisely that $ \widehat{f} $ is radial.

Written by Nicholas Stull

Side note:

I forgot to justify why the integral exists in the first place. Well, since $ f\in C_c^{\infty}(R^n) $, then there exists a compact $ K \subset R^n $ such that outside of K, $ f = 0 $, and $ |f| \leq M $ for some $ 0 < M < \infty $, and hence denoting the $ L^1 $ norm of f by $ |f|_1 $, we have $ |f|_1 \leq M m(K) < \infty $, hence $ f \in L^1(R^n) $. Note that in this assumption, we have used the definition of compact in $ R^n $ as complete and totally bounded, or closed and bounded by the Heine-Borel Theorem.