(forgot to say why any of this integration crap made sense) |
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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> | 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>\ | + | Let <math>A\in SO(n)</math> be a square orthogonal n-dimensional matrix with the following properties: |
| − | <math>= | + | <math>A^T A = 1</math> |
| − | + | <math>det J(Ax) = 1</math> | |
| − | |||
| − | + | ||
| + | <math>\widehat{f}\left(A \xi\right) = \int_{R^n} f^*(|x|) e^{-\imath <x,A \xi>} dx</math> | ||
| + | |||
| + | <math>= \int_{R^n} f^*(|x|) e^{-\imath <A^T x ,\xi>} dx</math> | ||
| + | |||
| + | Now use a change of variables, namely replace <math>x</math> with <math>A^T x</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^*(|A^T x|) e^{-\imath <x, \xi>} det J(Ax) dx</math> | ||
| + | |||
| + | But recall that <math>|A^T x| = |x|</math>, and <math>det J(Ax) = 1</math> so: | ||
<math>= \int_{R^n} f^*(|x|) e^{-\imath <x,\xi>} dx = \widehat{f}(\xi)</math> | <math>= \int_{R^n} f^*(|x|) e^{-\imath <x,\xi>} dx = \widehat{f}(\xi)</math> | ||
Revision as of 08:00, 29 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 $
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.
