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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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    Oh my mistake. Fixed.2012-05-28

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.2012-05-28
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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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    Are you saying that my equality should be $\hat f(\lambda t) = \hat f_{1/\lambda}(t) = (1/\lambda)^df(t/\lambda)$?2012-05-28