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

Latest revision as of 05:56, 11 June 2013

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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 $

Let $ A\in SO(n) $ be a square orthogonal n-dimensional matrix with the following properties:

$ A^T A = 1 $

$ det J(Ax) = 1 $

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

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

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

$ = \int_{R^n} f^*(|A^T x|) e^{-\imath <x, \xi>} det J(Ax) dx $

But recall that $ |A^T x| = |x| $, and $ det J(Ax) = 1 $ 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.

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