| Line 11: | Line 11: | ||
<math>= \int_{R^n} f^*(|x|) e^{-\imath <x e^{-\imath \theta},\xi>} dx</math> | <math>= \int_{R^n} f^*(|x|) e^{-\imath <x e^{-\imath \theta},\xi>} dx</math> | ||
| − | Now use a change of variables, namely <math>x e^{-\imath \theta} | + | 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> | <math>= \int_{R^n} f^*(|x e^{-\imath \theta}|) e^{-\imath <x, \xi>} dx</math> | ||
Revision as of 12:19, 27 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
