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This is a small part of a larger problem I am trying to solve. This is stated as a basic property of the fourier transform. First we define for $f \in L^1(\mathbb{R}^d)$ and $\lambda \neq 0$, $$ \hat f(x) := \int_{\mathbb{R}^d} f(t) e^{-2\pi i t x} dt \quad\text{and}\quad \hat f_\lambda(x) := \lambda f(\lambda x). $$

The property is that, $\hat f_\lambda(x) = \hat f(x/\lambda)$

I don't see how to prove this, not sure where to start really.

EDIT: The property I am trying to prove is the scaling property of this wikipedia article. http://en.wikipedia.org/wiki/Fourier_transform#Basic_properties

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Check your variable of integration and edit. –  ncmathsadist May 28 '12 at 1:45
    
Oh my mistake. Fixed. –  Danikar May 28 '12 at 1:48
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2 Answers 2

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Following the notation of the wiki article. If $h(x)=f(ax)$, then $\hat{h}(\xi)=|a|^{-1}\hat{f}\left(\frac{\xi}{a}\right)$.

Proof:

$\hat{h}(\xi)=\int h(x)e^{-2\pi i\xi x}\, dx=\int f(ax)e^{-2\pi i\xi x}\, dx=|a|^{-1}\int f(y)e^{-2\pi i\xi \left(\frac{y}{a}\right)}\, dy$.

This becomes: $|a|^{-1}\int f(y)e^{-2\pi i\left(\frac{\xi}{a}\right)y}\, dy=|a|^{-1}\hat{f}\left(\frac{\xi}{a}\right)$.

Note this was for functions taking values in $\mathbf{R}$, in higher dimensions, say $n$, the $|a|$ gets an exponent of $-n$. I think that is the only change.

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Thanks this clarifies a lot. –  Danikar May 28 '12 at 3:36
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In this case $$\hat f(\lambda t) = \int_{\mathbb{R}^d} f(x)e^{2\pi i \lambda x t}\,dx = \int_{\mathbb{R}^d} f(x/\lambda)e^{2\pi i xt}{dx\over \lambda^d} $$ Note the $d$ power owing to the Jacobian of the transformation on $\mathbb{R}^d$.

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Note $\lambda \not= 0$. –  ncmathsadist May 28 '12 at 1:56
    
I think I am lost with the notation. You have $x$ outside the integral and your are integrating over $dx$. That is why I used $t$, but it seems like something is lost there because I have two variables on my power of $e$ but you only have 1. My book used defined it this way $\hat f(\xi) := \int_{\mathbb{R}^d} f(x) e^{-2\pi i x \cdot \xi} dx $. I changed it to $x$ and $t$ to match it up with where I saw this basic property, and so I didn't have to use $\xi$. –  Danikar May 28 '12 at 2:05
    
check edit. It gets late and my brain fogs. –  ncmathsadist May 28 '12 at 2:10
    
After the second equality should the $\lambda$ in the power of $e$ be canceled? So $e^{2\pi i x t}$ instead? After than we have that $\hat f(\lambda t) = \hat f(x/\lambda)/\lambda^d$ right? That seems like it is close to where I want to be, do I need to take a few more steps? EDIT: Actually nvm, I dont know how I got that equality. I'm still just confused how this shows the property that I am trying to show. –  Danikar May 28 '12 at 2:15
    
you can turn $\lambda$ into $1/\lambda$ if need be. –  ncmathsadist May 28 '12 at 2:16
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