> [!definition] > > Let $F \in \cs'$ be a [[Tempered Distribution|tempered distribution]]. Define the **Fourier transform** of $F$ by > $ > \anglesn{\wh F, \phi} = \anglesn{F, \wh \phi} \quad \forall \phi \in \cs > $ > and the **inverse Fourier transform** of $F$ by > $ > \anglesn{\check F, \phi} = \anglesn{F, \check \phi} > $ > > If $F = T_f$ with $f \in L^1 + L^2$, then $\wh F = T_{\wh f}$ agrees with the ordinary [[Fourier Transform|Fourier transform]]. > [!theorem] > > Let $F \in \cs'$, $y, \eta \in \real^d$, $\alpha$ be a [[Multi-Index|multi-index]], $\psi \in \cs$, then > 1. $\wh{(\tau_y F)}(\xi) = e^{-2\pi i \angles{\xi, y}}\wh F$, $\tau_\eta \wh F = \braks{e^{2\pi i \angles{\eta, x}}F}^\wedge$. > 2. $\partial^\alpha \wh F = \braks{(-2\pi i x)^\alpha F}^\wedge$, $\wh{(\partial^\alpha F)}(\xi) = (2\pi i \xi)^\alpha \wh F$. > 3. $(F * \psi)^\wedge = \wh \psi \wh F$. > > *Proof*. Let $\phi \in \cs$, then > $ > \begin{align*} > \anglesn{\wh{\tau_yF}, \phi} &= \anglesn{\tau_y F, \wh \phi} = \anglesn{F, \tau_{-y}\wh \phi} = \anglesn{F, [e^{-2\pi i \angles{\xi, y}}\phi(\xi)]^\wedge} \\ > &= \anglesn{\wh F, e^{-2\pi i \angles{\xi, y}}\phi(\xi)} = \anglesn{e^{-2\pi i \angles{\xi, y}}\wh F, \phi(\xi)} \\ > \anglesn{\tau_\eta \wh F, \phi} &= \anglesn{\wh F, \tau_{-\eta}\phi} = \anglesn{F, \wh{\tau_{-\eta}\phi}} = \anglesn{F, e^{2\pi i \angles{\xi, \eta}}\wh \phi} \\ > &= \anglesn{e^{2\pi i \angles{\xi, \eta}} F, \wh \phi} = \anglesn{[e^{2\pi i \angles{\eta, x}} F]^{\wedge}, \phi} \\ > \anglesn{\partial^\alpha \wh F, \phi} &= (-1)^{\abs \alpha}\anglesn{\wh F, \partial^\alpha \phi} = (-1)^{\abs \alpha}\anglesn{F, \wh{\partial^\alpha \phi}} \\ > &= (-1)^{\abs \alpha}\anglesn{F, (2\pi i \xi)^\alpha \wh\phi} = (-1)^{\abs \alpha}\anglesn{(2\pi i \xi)^\alpha F, \wh\phi} \\ > &= \angles{[(-2\pi i x)^\alpha F]^\wedge, \phi} \\ > \anglesn{\wh{\partial^\alpha F}, \phi} &= \anglesn{\partial^\alpha F, \wh \phi} = (-1)^\alpha\anglesn{F, \partial^\alpha \wh \phi} \\ > &= (-1)^\alpha\angles{F, [(-2\pi i x)^\alpha\phi]^\wedge} \\ > &= (-1)^\alpha \anglesn{\wh F, (-2\pi i x)^\alpha \phi} =\anglesn{(2\pi i x)^\alpha \wh F, \phi} \\ > \anglesn{\wh{F * \psi}, \phi} &= \anglesn{F * \psi, \wh \phi} = \anglesn{F, \td \psi * \wh \phi} = \anglesn{\wh F, \check{\td{\psi}} \phi } = \anglesn{\wh \psi \wh F, \phi} > \end{align*} > $ > [!theorem] > > Let $F \in \ce'$ be [[Compactly Supported|compactly supported]], then $\wh F$ is [[Slowly Increasing Function|slowly increasing]] with $\wh F(\xi) = \angles{F, E_{-\xi}}$ with $E_{\xi}(x) = e^{2\pi i \angles{\xi, x}}$. > > *Proof*. Since the difference quotients converge to the derivatives uniformly on compact sets, $g(\xi) = \angles{F, E_{-\xi}}$ is smooth with > $ > \partial^\alpha_\xi g(\xi) = \anglesn{F, \partial^\alpha_\xi E_{-\xi}} = (-2\pi i)^{\abs{\alpha}}\angles{F, x^\alpha E_{-\xi}} > $ > By continuity, compart support, and the multi-index product rule, there exists $R, C \ge 0$ and $N \in \nat$ such that > $ > \begin{align*} > \abs{\partial^\alpha_\xi g(\xi)} &= (-2\pi i)^{\abs \alpha}\angles{F, x^\alpha E_{-\xi}} \\ > &\le C\sum_{\abs \beta \le N}\norm{\partial^\beta_x [x^\alpha E_{-\xi}(x)]}_{u, B(0, R)} \\ > &\le C'(1 + R)^{\abs \alpha}(1 + \abs \xi)^N > \end{align*} > $ > so $g$ is slowly increasing. To show that $g = \wh F$, let $\phi \in C_c^\infty$, then $g\phi \in C_c^\infty$. Using Riemann sums, > $ > \begin{align*} > \angles{g, \phi} &= \limv{n}\Delta_n \sum_{j = 1}^n\angles{F, E_{-\xi_j}}\phi(\xi_j) \\ > &= \limv{n}\Delta_n \angles{F, \sum_{j = 1}^n \phi(\xi_j)E_{-\xi_j}} = \anglesn{F, \wh \phi} > \end{align*} > $ > [!theorem] > > Let $\delta$ be the delta mass at $0$, $y \in \real^d$, then > $ > \braks{\partial^\alpha \tau_y\delta}^\wedge(\xi) = (2\pi i \xi^\alpha)e^{-2\pi i \angles{\xi, y}} > $ > *Proof*. Since $\delta \in \ce'$, > $ > \begin{align*} > \braks{\partial^\alpha \tau_y\delta}^\wedge(\xi) &= \angles{\partial^\alpha \tau_y\delta, E_{-\xi}} = (-1)^{\abs \alpha}\angles{\tau_{y}\delta,\partial^\alpha E_{-\xi}} \\ > &= (-1)^{\alpha}(-2\pi i \xi)^{\abs{\alpha}} e^{-2\pi i\angles{\xi, y}}=(2\pi i \xi)^{\abs{\alpha}} e^{-2\pi i\angles{\xi, y}} > \end{align*} > $