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I have a simple question about integration by parts in multi differentiation case.
Suppose $f\in S$ (Schwartz space, i.e. space of rapidly decreasing functions).
When proving the Fourier Transform:
$$\widehat{D^\alpha f}=\xi ^\alpha \hat f(\xi)$$ I use the notation $D_j=\frac{1}{i}\partial_j \text{ and } D^\alpha=D^\alpha_1 D^\alpha_2... $ and $\alpha$ is a multi-index
So we have
$$\widehat{D^\alpha f}(\xi)=\int e^{-ix\xi}D_x^\alpha f(x)dx=(-1)^{|\alpha|}\int D_x^\alpha e^{-ix\xi}f(x)dx=(-1)^{|\alpha|}\int (-1)^{|\alpha|}\xi^\alpha e^{-ix\xi}f(x)dx=\xi^\alpha \hat f(\xi)$$ I guess that the second equal sign comes from integration by parts, but could someone explain more in details? I mean there should be a constant part (without the integral sign) in the integration by part formula?

And I'm also confused about the third equal sign, could someone help me understand?

I greatly appreciate your time and help!

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  • $\begingroup$ Didn't you mean $D_j=\frac{1}{j}\partial_j$? Anyway, are you sure you want to divide by the index? It doesn't seem to make sense. $\endgroup$ – tomasz May 2 '17 at 2:34
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    $\begingroup$ Also, the non-integral part vanishes because it is an expression of the form $g(x)|_{x=-\infty}^{\infty}$ for $g$ in the Schwartz class, which is just $0-0=0$. $\endgroup$ – tomasz May 2 '17 at 2:37
  • $\begingroup$ Start with $S(\mathbb{R})$, then extend to $S(\mathbb{R}^2)$ and $S(\mathbb{R}^n)$ $\endgroup$ – reuns May 2 '17 at 3:02
  • $\begingroup$ @tomasz Thanks for your answer! But in my notes, it is 1/i where i is the imaginary unit. I am also a bit confused about why dividing this i, is it because this is in the complex field? And why the non-integral part vanishes? Do all rapidly decreasing functions have compact support? But my definition for S space is the functions in $C^\infty$ and $sup |x^\alpha \partial ^\beta f|< \infty$. And if all functions in S have compact support, what would be the difference between S space and the test function space (D)? $\endgroup$ – J.Y May 2 '17 at 17:39
  • $\begingroup$ @J.Y remember that these are improper integrals, so you're not evaluating $g$ at infinity when looking at $g(x)|_{-\infty}^\infty$ you're looking at it's value when you take the limit as $x$ goes to $\pm$ infinity. For $g\in \mathcal{S}$, this limit is always zero. $\endgroup$ – mathisfun May 5 '17 at 20:33

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