> [!definition] > > Let $f \in L^1(\real^d)$ be a [[Integrable Function|integrable function]]. Define the **Fourier transform**[^2] of $f$ by > $ > \hat f(\xi) = \int_{\real^d} f(x)e^{-2\pi i\angles{\xi, x}}dx > $ > then $\normn{\wh f}_u \le \norm{f}_1$ and is [[Continuity|continuous]][^1]. # Basic Properties > [!theorem] > > Let $f \in L^1(\real^d)$, $y \in \real^d$, and $\xi, \eta \in \real^d$, then > $ > (\wh{\tau_yf})(\xi) = e^{-2\pi i \angles{\xi , y}} \wh f(\xi) \quad \tau_\eta \wh f(\xi) = \braks{e^{2\pi i \angles{\eta, \cdot}}f(\cdot)}^{\wedge} > $ > *Proof*. By the translation invariance of the Lebesgue measure, > $ > \begin{align*} > (\wh{\tau_y f})(\xi) &= \int_{\real^d}f(x - y)e^{-2\pi i \angles{\xi, x}}dx = \int_{\real^d}f(x)e^{-2\pi i \angles{\xi, x + y}} \\ > &= e^{-2\pi i \angles{\xi, y}}\int_{\real^d}f(x)e^{-2\pi i \angles{\xi, x}} = e^{-2\pi i \angles{\xi, y}} \wh f(\xi) > \end{align*} > $ > On the other hand, > $ > \begin{align*} > \tau_\eta \wh f(\xi) &= \wh f(\xi - \eta) = \int_{\real^d} f(x)e^{-2\pi i\angles{\xi - \eta, x}}dx \\ > &= \int_{\real^d} f(x)e^{2\pi i \angles{\eta, x}} \cdot e^{-2\pi i \angles{\xi, x}}dx= \braks{e^{2\pi i\angles{\eta, \cdot}}f(\cdot)}^\wedge > \end{align*} > $ > [!theorem] > > Let $T \in \laut{\real^d}$ be a [[Space of Toplinear Isomorphisms|linear isomorphism]], and $f \in L^1(\real^d)$. Denote $S = (T^*)^{-1}$, then > 1. $(\wh{f \circ T}) = \abs{\det T}^{-1} \cdot \wh f \circ S$. > 2. If $T$ is unitary, then $(\wh{f \circ T}) = \wh f \circ T$. > 3. If $Tx = t^{-1}x$ ($t > 0$), then $(\wh{f \circ T})(\xi) = t^d\wh f(t\xi)$. > > *Proof*. > $ > \begin{align*} > (\wh f \circ T)(\xi) &= \int_{\real^d} f(Tx)e^{-2\pi i \angles{\xi, x}}dx \\ > &= \abs{\det T}^{-1}\int_{\real^d}f(x)e^{-2\pi i \angles{\xi, T^{-1}x}}dx\\ > &= \abs{\det T}^{-1}\int_{\real^d}f(x)e^{-2\pi i \angles{(T^{-1})^*\xi, x}}dx \\ > &= \abs{\det T}^{-1} \cdot \wh f \circ S > \end{align*} > $ > [!theorem] > > Let $f, g \in L^1(\real^d)$, then $\wh f \cdot \wh g = \wh{f * g}$. > > *Proof*. Using the [[Fubini-Tonelli Theorem]], and the translation invariance of the Lebesgue integral, > $ > \begin{align*} > \wh{f * g}(\xi) &= \int_{\real^d} \int_{\real^d}f(x - y)g(y)dy \cdot e^{-2\pi i \angles{\xi, x}}dx \\ > &= \int_{\real^d}\int_{\real^d} f(x - y)e^{-2\pi i \angles{\xi, x - y}}dx \cdot g(y)e^{-2\pi i \angles{\xi, y}} dy \\ > &= \wh f(\xi) \cdot \int_{\real^d}g(y)e^{-2\pi i \angles{\xi, y}}dy = \wh f(\xi) \cdot \wh g (\xi) > \end{align*} > $ > [!theorem] > > Let $f, g \in L^1$, then $\anglesn{\wh f, g} = \anglesn{f, \wh g}$. > > *Proof*. Using the [[Fubini-Tonelli Theorem]], > $ > \begin{align*} > \int_{\real^d} f(x) \wh g(x)dx &= \int_{\real^d} f(x) \int_{\real^d} g(y)e^{-2\pi i \angles{x, y}}dydx \\ > &= \int_{\real^d} g(y) \int_{\real^d} f(x)e^{-2\pi i \angles{x, y}}dxdy \\ > &= \int_{\real^d} g(y)\wh f(y)dy > \end{align*} > $ # Smoothness and Decay The smoothness of the Fourier transform depends on the rate of decay of the function, and the rate of decay of the Fourier transform depends on the smoothness of the function. > [!theorem] > > Let $f \in L^1(\real^d)$ such that $x^\alpha f(x) \in L^1$ for all $\abs{\alpha} \le p$[^3], then $\wh f \in C^p(\real^d)$ with $\partial^\alpha \wh f = \braks{(-2\pi i x)^\alpha f}^\wedge$. > > *Proof, by induction on $\abs \alpha$*. Suppose that the proposition holds for all $\abs \alpha \le p - 1$, and let $1 \le j \le d$, then > $ > \begin{align*} > \braks{\partial^{\alpha + e_j}\wh f}(\xi) &= \partial^{e_j} \braks{(-2\pi i x)^\alpha f(x)}^\wedge \\ > &= \lim_{h \to 0}\frac{1}{h}\int_{\real^d}(-2\pi i x)^\alpha f(x)\braks{e^{-2\pi i \angles{\xi + e_jh, x}} - e^{-2\pi i \angles{\xi, x}}}dx > \end{align*} > $ > where > $ > \abs{e^{-2\pi i \angles{\xi + e_jh, x}} - e^{-2\pi i \angles{\xi, x}}} \le 4\pi \abs{\angles{e_jh, x}} \le 4\pi \abs{h} \cdot \abs{x} > $ > If $x^{\alpha + e_j}f(x) \in L^1$, then by the [[Dominated Convergence Theorem]], > $ > \begin{align*} > \braks{\partial^{\alpha + e_j}\wh f}(\xi) &= \int_{\real^d} (-2\pi i x)^\alpha f(x)\braks{\partial^{e_j}_\xi e^{-2\pi i \angles{\xi, x}}}dx \\ > &= \int_{\real^d} (-2\pi i x)^{\alpha + e_j}f(x)e^{-2\pi i \angles{\xi, x}}dx > \end{align*} > $ > [!theorem] > > Let $f \in L^1(\real^d) \cap C^p(\real^d)$ such that $\partial^\alpha f \in L^1(\real^d)$ for all $\abs \alpha \le p$, and $\partial^\alpha \in C_0(\real^d)$ for all $\abs \alpha < p$, then $(\wh{\partial^\alpha f})(\xi) = (2\pi i \xi)^\alpha \wh f(\xi)$. > > *Proof, by induction on $\abs \alpha$*. Suppose that the proposition holds for all $\abs \alpha \le p - 1$, and let $1 \le j \le d$. Let $x' = (x_1, \cdots, x_{j - 1}, x_{j+1}, \cdots, x_n)$, then splitting the integral with $x = (x', x_j)$ gives > $ > \begin{align*} > (\wh{\partial^\alpha f})(\xi) &= \int_{\real^d} \braks{\partial^{\alpha + e_j}_xf(x)} e^{-2\pi i \angles{\xi, x}}dx = \int_{\real^d} \braks{\partial^{e_j}_x\partial^{\alpha}_xf(x)} e^{-2\pi i \angles{\xi, x}}dx \\ > &= \int_{\real^{d - 1}} \int_\real \braks{\partial_{x_j} \partial^{\alpha}_{(x', x_j)}f(x', x_j)}e^{-2\pi i \angles{\xi, (x', x_j)}}dx_jdx' > \end{align*} > $ > Using integration by parts on the inner integral gives > $ > \begin{align*} > &\int_\real \braks{\partial_{x_j} \partial^{\alpha}_{(x', x_j)}f(x', x_j)}e^{-2\pi i \angles{\xi, (x', x_j)}}dx_j \\ > &= \braks{\partial^{\alpha}_{(x', x_j)}f(x', x_j)}e^{-2\pi i \angles{\xi, (x', x_j)}}\bigg|_{-\infty}^\infty - \int_{\real}\braks{\partial^{\alpha}_{(x', x_j)}f(x', x_j)}\braks{\partial_{x_j}e^{-2\pi i \angles{\xi, (x', x_j)}}}dx_j \\ > &= (2\pi i \xi)^{e_j}\int_\real \braks{\partial^{\alpha}_{(x', x_j)}f(x', x_j)}e^{-2\pi i \angles{\xi, (x', x_j)}}dx_j > \end{align*} > $ > where $\partial^\alpha f$ vanishes at infinity and $e^{-2\pi i \angles{\xi, x}}$ is bounded. Plugging the result back into the iterated integral yields > $ > (\wh{\partial^\alpha f})(\xi) = (2\pi i \xi)^{e_j}\int_{\real^d}\braks{\partial^\alpha_x f(x)} e^{-2\pi i \angles{\xi,x}}dx = (2\pi i \xi)^{\alpha + e_j} \wh f(\xi) > $ # On the [[Schwartz Space]] > [!theorem] > > The Fourier transform is a toplinear isomorphism on $\cs$. > > *Proof*. Let $f \in \cs$, then for any multi-indices $\alpha$ and $\beta$, $x^\alpha \partial^\beta f \in L^1 \cap C_0$. Therefore $\wh f \in C^\infty$ with > $ > (x^\alpha \partial^\beta f)^\wedge(\xi) = (-1)^{\abs \alpha} \cdot (2\pi i)^{\abs \beta - \abs \alpha} \cdot \partial^\alpha(\xi^\beta \wh f(\xi)) > $ > Since $x^\alpha \partial^\beta f \in L^1$, $(x^\alpha \partial^\beta f)^\wedge$ and $\partial^\alpha (\xi^\beta \wh f (\xi))$ are bounded, so $\wh f \in \cs$. Moreover, there exists $N \ge 0$ such that $\int (1 + \abs{x})^{-n} dx < \infty$, so > $ > \begin{align*} > \normn{\paren{x^\alpha \partial^\beta f}^\wedge}_u &\le \normn{x^\alpha \partial^\beta f}_1 \le C\normn{ (1 + \abs x)^N x^\alpha \partial^\beta f(x)}_u \\ > \normn{\wh f}_{(n, \beta)} &\le C' \sum_{\abs \gamma \le \abs \beta}\normn{f}_{(N + n, \gamma)} > \end{align*} > $ > hence the Fourier transform is a continuous linear map. > > Since the [[Fourier Inversion Theorem|inverse Fourier transform]] is the composition of a reflection and the original Fourier transform, it is continuous as well. Therefore the Fourier transform is a toplinear isomorphism. [^1]: [[Dominated Convergence Theorem]] [^2]: Imposing a convention: Latin letters for elements in the real space, Greek letters for elements in the frequency space. In the inner product, put the vector in the frequency space before the vector in the real space. [^3]: This corresponds to $f$ being the density of a $L^p$ random variable.