Fourier Transform on Schwarz Space

Fourier Transform

If $f\in \mathcal{S}(\mathbb{R})$, we define its Fourier Transform for $\xi \in \mathbb{R}$, by $\hat{f}(\xi )={\int }_{-\infty }^{\infty }f(x)e^{-2\pi ix\xi }\text{d}x.$

Proposition

If $f\in \mathcal{S}(\mathbb{R})$, then: (i) $\hat{f}{e}^{2\pi ih\xi }$ is the Fourier transform of $f(x+h)$ wherever $h\in \mathbb{R}$ (ii) $\hat{f}(x+h)$ is the Fourier transform of $f(x)e^{-2\pi ixh}$ wherever $h\in \mathbb{R}$ (iii) ${\delta }^{-1}\hat{f}({\delta }^{-1}\xi )$ is the Fourier transform of $f(\delta x)$wherever $\delta \in \mathbb{R}$ (iv) $2\pi i\xi \hat{f}(\xi )$ is the Fourier transform of $f^{\prime }(x)$ (v) $\frac{d}{d\xi }\hat{f}(\xi )$ is the Fourier transform of $-2\pi ixf(x)$.

Proof (i) and (ii) directly follow from definitions. For (iii) similarly $\int f(\delta x)e^{-2\pi ix\xi }dx={\delta }^{-1}\int f(\delta x)e^{-2\pi ix\xi {\delta }^{-1}}d\delta x={\delta }^{-1}\hat{f}({\delta }^{-1}\xi ).$ For (iv), through integration by parts, ${\int }_{-N}^N{f}^{\prime }(x)e^{-2\pi ix\xi }dx={\left[f(x)e^{-2\pi ix\xi }\right]}_{-N}^N+2\pi i\xi {\int }_{-N}^{N}f(x)e^{-2\pi ix\xi }dx,$ so letting $N$ goes to infinity, we get desired relation.

For (v), we want to show that the derivative of $\hat{f}(\xi )$ is $-2\pi ixf(x)$, let $ϵ>0$ and consider $\begin{array}{rl}\frac{\hat{f}(\xi +h)-\hat{f}(\xi )}{h}-(\widehat{-2\pi ixf})(\xi )& =\\ & {\int }_{-\infty }^{\infty }f(x)e^{-2\pi ix\xi }\left[\frac{e^{-2\pi ixh}-1}{h}+2\pi ix\right]dx.\end{array}$

Since $f(x)$ and $xf(x)$ are of rapid decrease, there exists an integer $N$ so that ${\int }_{|x|\ge N}|f(x)|dx\le ϵ$ and ${\int }_{|x|\ge N}|x||f(x)|dx\le ϵ$. Moreover, for $|x|\le N$, $(e^{-2\pi ixh}{)}^{\prime }(0)=-2\pi i$, so ${\lim }_{h\to 0}\left|\frac{e^{-2\pi ixh}-1}{h}-(-2\pi i)\right|=0$, then there exists $h_0$ so that $|h|<h_0$ implies

$\left|\frac{e^{-2\pi ixh}-1}{h}+2\pi ix\right|\le \frac{ϵ}{N}.$

Hence for $|h|<h_0$ we have

$\begin{array}{rl}& \left|\frac{\hat{f}(\xi +h)-\hat{f}(\xi )}{h}-(\widehat{-2\pi ixf})(\xi )\right|\\ & \le {\int }_{-N}^N\left|f(x)e^{-2\pi ix\xi }\left[\frac{e^{-2\pi ixh}-1}{h}+2\pi ix\right]\right|dx+Cϵ\\ & \le C^{\prime }ϵ.\end{array}$

Theorem

If $f\in \mathcal{S}(\mathbb{R})$, then $\hat{f}\in \mathcal{S}(\mathbb{R})$.

The proof is an easy application of the fact that the Fourier transform interchanges differentiation and multiplication. In fact, note that if $f\in \mathcal{S}(\mathbb{R})$, its Fourier transform $\hat{f}$ is bounded; $|\hat{f}|\le \int |f{e}^{-2\pi i\xi }|=\int |f|<\infty$ then also, for each pair of non-negative integers $\ell$ and $k$, the expression ${\xi }^k{\left(\frac{d}{d\xi }\right)}^{\ell }\hat{f}(\xi )$ is bounded, since by the last proposition, ${\xi }^k{\left(\frac{d}{d\xi }\right)}^{\ell }\hat{f}(\xi )={\xi }^k(\widehat{(-2\pi ix{)}^{l}f(x)})$ so it is the Fourier transform of $\frac{1}{(2\pi i{)}^k}{\left(\frac{d}{dx}\right)}^k[(-2\pi ix{)}^{\ell }f(x)].$ With $f\in \mathcal{S}(\mathbb{R})$, we get $\frac{1}{(2\pi i{)}^k}{\left(\frac{d}{dx}\right)}^k[(-2\pi ix{)}^{\ell }f(x)]\in \mathcal{S}(\mathbb{R})$, so the Fourier transform is bounded, thus we get $\hat{f}\in \mathcal{S}(\mathbb{R})$.