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Here is a part that I am having hard time to understand. Actually, it is basically the general trick in Schwartz space that is just integration by parts.

However, I would like to see how we get rid of the second part of the solution we got after integration by parts

the following is what I have understood so far, could you clarify the points that I am missing throughout the expansion of the proof that is given in our text.thx in advance.

The main objective below is proving the fact that Fourier Transform maps Schwartz Function to Schwartz Functions:

What we have to prove is that $\big|\xi^{\alpha}\partial_{\xi}^{\beta}\hat{\phi}(\xi)\big|\le C_{\alpha\beta},$ for any $\alpha,\beta.$ The following is written in the text:

$F:=$ Fourier Transform

1)$\big|\xi^{\alpha}\partial_{\xi}^{\beta}\hat{\phi}(\xi)\big|\\=\big|\xi^{\alpha}F((-x)^{\beta}\phi)\big|\\=\big|F(D_{x}^{\alpha}(-x)^{\beta}\phi)\big|\\=\big|\int_{R^{n}}{e^{-i<x,\xi>}\frac{1}{<x>^{n+1}}{<x>^{n+1}}(D_{x}^{\alpha}(-x)^{\beta}\phi)}\big|=*$

2) Now, after the last step we have to apply Integration by parts in order to be able to prove that the very first statement is bounded.

3) In order to apply the integration by parts we have to choose our $u$ such that it is supposed to be smooth then we wil be able to differentiate it, and the rest of the integrand will be our $dv,$

4) I picked $u={e^{-i<x,\xi>}{<x>^{n+1}}(D_{x}^{\alpha}(-x)^{\beta}\phi)}$ and $\frac{1}{<x>^{n+1}}dx=dv,$

5) In our text $<x>=(1+|x|^{2})^{\frac{1}{2}}.$

6)$ *= \\\big|{e^{-i<x,\xi>}{<x>^{n+1}}(D_{x}^{\alpha}(-x)^{\beta}\phi)}\int_{R^{n}}\frac{1}{<x>^{n+1}}dx-\\ \int_{R^{n}}\int_{R^{n}}{\frac{1}{<x>^{n+1}}D({{e^{-i<x,\xi>}{<x>^{n+1}}(D_{x}^{\alpha}(-x)^{\beta}\phi)}}})dx_{1}^{2}...d{x_{n}}^{2}\big|$

7) What I really want to see that the part that I subtracted in the above statement is $0$, I guess it is because of the compact support argument of $u$ but I could not see it very clearly,

8) Since the professor repeated it over and over again just by saying 'apply IBP then check that the second part tends to zero,' I would like to understand it.

9) Any kinds of help is greatly appreciated, thx!

share|improve this question
    
This is a very common argument which relies on Stokes' Theorem. Basically, the integral over $\mathbb{R}^n$ of the derivative of a compactly supported function (or form), is equal to the integral of the function (or form) over the boundary, which is empty, and hence the integral is zero. –  Michael Albanese Sep 2 '13 at 23:19
    
Why integrate by parts? $D_x^\alpha\left((-x)^\beta\phi\right)$ is a Schwartz function, in particular $L^1$, so its Fourier transform is bounded. Oh, and you don't have anything you can assume has compact support, unless I'm missing something. –  Daniel Fischer Sep 2 '13 at 23:20
    
However, even if I know that $\phi$ is a Schwartz function it does just mean that it is smooth so I don't understand how I will be able to use the fact that out of a compact set the function has a measure zero which will lead me to say it is zero. –  user92604 Sep 3 '13 at 0:09
    
I don't understand what you're trying to do. Two basic facts: a) the product of a Schwartz function and a polynomial is Schwartz, b) any derivative of a Schwartz function is Schwartz. The Fourier transform of an $L^1$ function is bounded. Thus $\xi^\alpha \partial_\xi^\beta \mathcal{F}[\phi] = \mathcal{F}[D_x^\alpha((-x)^\beta\phi)]$ is bounded. Case closed. –  Daniel Fischer Sep 3 '13 at 10:21
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