# schwatz space and fourier transform: problem with induction

• October 27th 2008, 07:55 AM
marianne
schwatz space and fourier transform: problem with induction
I'll write everything down from the start.
First, the notation:

$\alpha \in \mathbb{N}^n, \alpha=(\alpha_1, \ldots, \alpha_n)$
$|\alpha|=\alpha_1 + \ldots \alpha_n$
$x \in \mathbb{R}^n$

$x^{\alpha}=x_1^{\alpha_1} \ldots x_n^{\alpha_n}$

$\partial^{\alpha}=\partial_1^{\alpha_1} \ldots \partial_n^{\alpha_n}$, where
$\partial_i^{\alpha_i}=\frac{\partial^{\alpha_i}}{\ partial x_i ^{\alpha_i}}$

Then the theorem:
If $x^{\alpha}f(x)$ is summable for every $\alpha, |\alpha|\leq k$, then
$\partial^{\alpha} \hat{f}={[(-2 \pi i x)^{\alpha})f]}^{\wedge}$, where $\hat{f}$ denotes Fourier transform of f.

Now, the problem. :(

We're supposed to prove this using induction on $\alpha$, and I just can't do it. The definition of $\partial^{\alpha}f$ is somewhat confusing, and I would b really grateful if someone could help me out, just for the basis, and I will try to work from there.

Thank you!
• October 27th 2008, 12:24 PM
Opalg
Quote:

Originally Posted by marianne
If $x^{\alpha}f(x)$ is summable for every $\alpha, |\alpha|\leq k$, then $\partial^{\alpha} \hat{f}={[(-2 \pi i x)^{\alpha})f]}^{\wedge}$, where $\hat{f}$ denotes Fourier transform of f.

[The equation in the statement of the theorem should read $\partial^{\alpha} \hat{f}={[(-2 \pi i x)^{|\alpha|})f]}^{\wedge}$, with $|\alpha|$ rather than $\alpha$ as the exponent on the right-hand side.]

I'm assuming you know how to do this in the case n=1. (For a function f(x) of one variable, differentiating the Fourier transform of f corresponds to multiplying f by -2πix, provided that xf(x) is integrable. This is proved by integration by parts.)

To prove the multi-variable case by induction, the inductive hypothesis will be that the result is true when $|\alpha|=m-1$ for some integer m>0. The inductive step goes like this. Let $\alpha$ be a multi-index with $|\alpha|=m$. Then at least one component of $\alpha$, say $\alpha_j$, must have $\alpha_j>0$. Let $\beta = (\alpha_1,\ldots,\alpha_{j-1},\alpha_j-1,\alpha_{j+1}\ldots,\alpha_n)$. Then $|\beta|=m-1$, so by the inductive hypothesis $\partial^{\beta}\hat{f} = {[(-2 \pi i x)^{|\beta|})f]}^{\wedge}$. To complete the inductive step, all you have to do now is to apply the single-variable result to the function ${[(-2 \pi i x)^{|\beta|})f]}^{\wedge}$ considered as a function of its j'th variable.
• October 27th 2008, 12:45 PM
Laurent
(there's a typo in the title: it should be Schwartz, from the name of Laurent Schwartz (Wink))
• October 27th 2008, 08:30 PM
ThePerfectHacker
Quote:

Originally Posted by Laurent
(there's a typo in the title: it should be Schwartz, from the name of Laurent Schwartz (Wink))

It would have been more interesting if the poster mispelled it as Schwarz.
Because that is a kinda well-known mathematician.
Mispelling these two names is common.
• October 28th 2008, 03:55 AM
marianne
Opalg, thank you very much, I get it now!

And sorry about the typo, latexing the question took me so much time I didin't pay much attention to the title. :)