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Fourrier Tranform

James

New member
Jan 30, 2012
15
I"m want to get a better understanding of the fourrier transform so I've started to do some problems.
^_f=fourrier transform of f.

The function f belongs to the schwartz space and k>0 f_k(x)=f(kx).
1)show that f_k also belongs to the schwartz space and ^_f(e)=(1/k)^_f(e/k)
2)the fourrier transform of exp((−x^2)/2) is sqrt(2pi)*exp((−e^2)/2) use the first part to obtain the fourrier transform for exp(−ax^2)

Attempt:
f belongs to the schwartz space then f is infinitly diff also f(kx)=kf(x) which belongs to the schwartz space.
then f_k(x)=f(kx)=kf(x) which belongs to the schwartz space.
I don't know if this is correct or how to continue...any help will be great.Thank you
 

ThePerfectHacker

Well-known member
Jan 26, 2012
236
Can you explain your notation? What does it mean f_k ?
 

James

New member
Jan 30, 2012
15
Sorry,f_k=f<sub>k
sqrt= square root
(Thinking)...is there something else?
 
Last edited:

ThePerfectHacker

Well-known member
Jan 26, 2012
236
Sorry,f_k=f<sub>k
sqrt= square root
(Thinking)...is there something else?
What does f_k mean? I understand you want to say $f_k$, but what is this the notation for?
 

James

New member
Jan 30, 2012
15
I'm sorry but that's how the problem is stated....f_k is defined to be f_k(x)=f(kx).
 

ThePerfectHacker

Well-known member
Jan 26, 2012
236
Let $S$ be the Schwartz class of functions. By definition $\varphi:\mathbb{R}\to \mathbb{R}$ belongs to $S$ means that $\varphi$ is smooth (has infinitely many derivatives) and that it satisfies the following very strong boundness condition: $|x|^n |\varphi^{(m)} (x)| \leq M_{n,m}$ for all non-negative integers $n,m$.

Let $f$ be a function belonging to $S$. Fix $k>0$ and define the function $f_k$ as follows: $f_k(x) = f(kx)$. We need to show that $f_k$ belongs to $S$ also. Thus, we need to show that $f_k$ is smooth and satisfies this boundness condition. To show this we need to use the chain rule. As $f_k(x) = f(kx)$ it means that $f_k'(x) = k f'(kx)$, and also $f_k''(x) = k^2 f''(kx)$, in general, $f_k^{(m)} (x) = k^m f^{(m)}(km)$.

Now we have,
$$ |x|^n |f_k^{(m)} (x)| = \frac{1}{k^n} \cdot | kx|^n \cdot k^m |f^{(m)}(kx)| \leq k^{m-n} M_{n,m}$$

Thus, this shows that $|x|^n |f_k^{(m)}(x)|$ is bounded for all $n,m\geq 0$. This means that $f_k$ also belongs to $S$.
 

ThePerfectHacker

Well-known member
Jan 26, 2012
236
If $\varphi$ belongs to $S$ then we can define its Fourier transform $\hat{\varphi} : \mathbb{R} \to \mathbb{R}$ as follows:
$$ \hat{\varphi} (\omega) = \int \limits_{-\infty}^{\infty} \varphi(x) e^{-i\omega x} ~ dx $$
This means that the Fourier transform of $f_k$ is:
$$ \hat{f_k}(\omega) = \int \limits_{-\infty}^{\infty} f_k(x) e^{-i\omega x} ~ dx = \int \limits_{-\infty}^{\infty} f(kx) e^{-i\omega x} ~ dx $$

Now let $y = kx$ as the substitution function, so $x = \frac{y}{k}$ and $dx = \frac{1}{k} dy$, also the limits stay the same (why?):
$$ \hat{f_k}(\omega) = \int \limits_{-\infty}^{\infty} f(y) e^{-i\omega y/k} \frac{1}{k} dy = \frac{1}{k} \int\limits_{-\infty}^{\infty} f(y) e^{-iy(\omega/k)} ~ dy = \frac{1}{k} \hat{f}\left( \frac{\omega}{k} \right) $$