There's a simple property of the Fourier transform on $\mathbb{R}^n$ that I'm having trouble establishing.
Let $\lambda > 0$. I would like to show that $(D_\lambda f)^\hat{}(\xi) = \lambda^{-n}(D_{\lambda^{-1}}\hat{f})(\xi)$, where $D_{\lambda}f(x) := f(\lambda x)$.
The substitution $u = \lambda^{-1}x$ gives me the following:
$\lambda^{-n}(D_{\lambda^{-1}}\hat{f})(\xi)$ = $\lambda^{-n} \int_{\mathbb{R}^n} f(x)e^{-2\pi i x \cdot \lambda^{-1}\xi}dx$ = $\lambda^{1-n}\int_{\mathbb{R}^n} f(\lambda u)e^{-2\pi i u \cdot \xi}du$ = $\lambda^{1-n} (D_\lambda f)^\hat{}(\xi)$, which isn't quite what I want. What am I doing wrong here?
