2

Let $f\in \mathcal{S}(\mathbb{R}).$ The Fourier transform of $f$ is defined by $\hat{f}(w) := \int_{-\infty}^\infty f(x) e^{-2\pi i x w} dx$. We use the notation $f(x) \longrightarrow \hat{f}(w)$ to mean that $\hat{f}$ denotes the Fourier transform of $f$.

Show that $f(\delta x) \longrightarrow \delta^{-1} \hat{f}(\delta^{-1} w)$.

I don't see why this is true.

Fourier transform of $f(\delta x)$ is $\int_{-\infty}^\infty f(\delta x) e^{-2\pi i \delta x w} dw = \frac{1}{\delta} \int_{-\infty}^\infty f(x) e^{-2\pi ixw} dx$. On the other hand, $\delta^{-1} \hat{f}(\delta^{-1} w) = \delta^{-1} \int_{-\infty}^\infty f(x)e^{-2\pi i \delta^{-1} w}dw$. How could these two be equal? Or am I misunderstanding anything?

Community
  • 1
3x89g2
  • 7,542

1 Answers1

2

Let $\lambda > 0$, define $\delta_\lambda f(x)=f(x/\lambda)$ we show that $\widehat{\delta_\lambda f}(\xi)=\lambda^n \widehat{f}(\lambda \xi)$. Place $z=x/\lambda$, i.e. $dx=\lambda^n dz$, obviously here we are in $\mathbb{R}^n$, then

$\displaystyle \widehat{\delta_\lambda f}(\xi) := \int_{\mathbb{R}^n} e^{-2 \pi i x \cdot \xi} f(x/\lambda) dx = \int_{\mathbb{R}^n} e^{-2\pi i \lambda z \cdot \xi} \lambda^n f(z) dz = \lambda^n \int_{\mathbb{R}^n} e^{-2\pi i z \cdot \lambda \xi }f(z) dz =: \lambda^n \widehat{f}(\lambda \xi)$

Sorry for the notations, but I had written thus, surely they are similar to your case, for $n=1$.

user288972
  • 2,360
  • Pretty sure it is just $\lambda$. Not sure what kind of substitution you’re doing to get a factor of $\lambda^n$. – Ryan Mar 11 '20 at 01:26
  • He is working in R^n which generalizes the stated question, where n=1. – yberman Dec 06 '21 at 22:24