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Jun 20, 2012 at 8:50 comment added user23078 @Bazin,I found myself not completely understand it yet,I think your definition of $\hat u$ is not unique,since it depends on the choice of $ \chi(\xi)$.so what's the problem behind this ? thanks in advance
Apr 30, 2012 at 19:17 comment added Bazin To Huang Shanlin: the function $\chi$ is smooth with compact support and is 1 near 0, in such a way that you may add (I should have done this, but I wanted to insist on the way to get rid of the singularity) on the rhs of the above equality $$ \int\vert \xi\vert^{-a-n}\phi(\xi)(1-\chi(\xi)) d\xi. $$ Best, Bazin.
Apr 27, 2012 at 7:10 vote accept user23078
Apr 27, 2012 at 2:55 comment added user23078 yes,when a is an even positive integer,the case is easy and seems degenerate. i wanna understand in your last formula,is $\chi(\xi)$ standing for any cut-off funtion which vanish at 0 ? I'm really appreciated that if you can show me some references about this question..
Apr 26, 2012 at 19:30 comment added Bazin Well if a is an even positive integer, then you get an homogeneous polynomial in $x$ of degree $a$ and its Fourier transform is a linear combination of derivatives of order $a$ of he Dirac mass at 0. When $a$ is an odd positive integer, then the Fourier transform will be indeed radial and homogeneous with degree $-a-n$: you can see that as a sort of finite part with a formula such as $$ \langle\hat u,\phi\rangle=c_{a,n}\int\vert \xi\vert^{-a-n} \bigl(\phi(\xi)-\sum_{0\le j\le N}\phi^{(j)}(0)\xi^j/j!\bigr) \chi(\xi)d\xi $$
Apr 26, 2012 at 9:00 comment added user23078 @Bzain,thank you very much.your answer is really helpful.consider again $|x|^a$,where a is any positive integer,now we can't use the extension mathod to determine its Fourier transform.indeed, as you have mentioned above,it is coincides with $c|\xi|^{-a-n}$ on $R^n \backslash 0$,but can we write the explicit expression just as the case in one dimension.
Apr 25, 2012 at 19:58 history answered Bazin CC BY-SA 3.0