View source for Number fourteen (too lazy to come up with something else pirately to say) - Rhea

== Problem 14 ==

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

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

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

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

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

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

But recall that <math>|x e^{\imath \theta}| = |x|</math>, so:

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

Which gives precisely that <math>\widehat{f}</math> is radial.

Written by Nicholas Stull

Side note:

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