ThmDex – An index of mathematical definitions, results, and conjectures.
Fourier transform under scaling
Formulation 0
Let $M = (\mathbb{R}^N, \mathcal{L}, \ell)$ be a D1744: Lebesgue measure space such that
(i) $f : \mathbb{R}^N \to \mathbb{C}$ is an D1921: Absolutely integrable function on $M$
(ii) \begin{equation} \mathfrak{F}_f(\xi) : = \int_{\mathbb{R}^N} f(x) e^{- 2 \pi i x \cdot \xi} \, \ell(d x) \end{equation}
(iii) \begin{equation} \mathfrak{G}_f(\xi) : = \int_{\mathbb{R}^N} f(x) e^{- i x \cdot \xi} \, \ell(d x) \end{equation}
Let $a \in \mathbb{R} \setminus \{ 0 \}$ be a D993: Real number.
Then
(1) \begin{equation} \mathfrak{F}_{x \mapsto f(x / a)} (\xi) = |a|^N \mathfrak{F}_f (a \xi) \end{equation}
(2) \begin{equation} \mathfrak{F}_{x \mapsto f(a x)} (\xi) = |a|^{- N} \mathfrak{F}_f \left( \frac{\xi}{a} \right) \end{equation}
(3) \begin{equation} \mathfrak{G}_{x \mapsto f(x / a)} (\xi) = |a|^N \mathfrak{G}_f (a \xi) \end{equation}
(4) \begin{equation} \mathfrak{G}_{x \mapsto f(a x)} (\xi) = |a|^{- N} \mathfrak{G}_f \left( \frac{\xi}{a} \right) \end{equation}
Proofs
Proof 0
Let $M = (\mathbb{R}^N, \mathcal{L}, \ell)$ be a D1744: Lebesgue measure space such that
(i) $f : \mathbb{R}^N \to \mathbb{C}$ is an D1921: Absolutely integrable function on $M$
(ii) \begin{equation} \mathfrak{F}_f(\xi) : = \int_{\mathbb{R}^N} f(x) e^{- 2 \pi i x \cdot \xi} \, \ell(d x) \end{equation}
(iii) \begin{equation} \mathfrak{G}_f(\xi) : = \int_{\mathbb{R}^N} f(x) e^{- i x \cdot \xi} \, \ell(d x) \end{equation}
Let $a \in \mathbb{R} \setminus \{ 0 \}$ be a D993: Real number.
Fix $\xi \in \mathbb{R}^N$. As a consequence of R3020: Complex Lebesgue integral under scaling, we have \begin{equation} \begin{split} \mathfrak{F}_{x \mapsto f(x / a)} (\xi) & = \int_{\mathbb{R}^N} f(x / a) e^{- 2 \pi i x \cdot \xi} \, \ell(d x) \\ & = \int_{\mathbb{R}^N} f(x / a) e^{- 2 \pi i (x / a) \cdot (a \xi)} \, \ell(d x) \\ & = |a|^N \int_{\mathbb{R}^N} f(x) e^{- 2 \pi i x \cdot (a \xi)} \, \ell(d x) \\ & = |a|^N \mathfrak{F}_f (a \xi) \end{split} \end{equation} This establishes the second claim. The second claim is a corollary to this which one obtains by applying the above procedure for a constant $b : = 1 / a$. The cases for $\mathfrak{G}$ are proved in exactly the same manner. $\square$