EE261 Lecture 13

课程主页：https://see.stanford.edu/Course/EE261

这次回顾第十三讲，这一讲开始引入分布的傅里叶变换。

分布的傅里叶变换

现在考虑分布的傅里叶变换

$$\langle\mathcal{F} \psi, \varphi\rangle=\int_{-\infty}^{\infty} \mathcal{F} \psi(x) \varphi(x) d x$$

假设积分都存在，由定义，我们有

$$\begin{aligned} \langle\mathcal{F} \psi, \varphi\rangle &=\int_{-\infty}^{\infty} \mathcal{F} \psi(x) \varphi(x) d x \\ &=\int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty} e^{-2 \pi i x y} \psi(y) d y\right) \varphi(x) d x \\ &=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-2 \pi i x y} \psi(y) \varphi(x) d y d x \\ &=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-2 \pi i x y} \varphi(x) \psi(y) d y d x \\ &=\int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty} e^{-2 \pi i x y} \varphi(x) d x\right) \psi(y) d y \\ &=\int_{-\infty}^{\infty} \mathcal{F} \varphi(y) \psi(y) d y \\ &=\langle\psi, \mathcal{F} \varphi\rangle \end{aligned}$$

所以

$$\langle\mathcal{F} T, \varphi\rangle=\langle T, \mathcal{F} \varphi\rangle$$

上式即为分布的傅里叶变换的定义。

类似的，定义傅里叶逆变换为

$$\left\langle\mathcal{F}^{-1} T, \varphi\right\rangle=\left\langle T, \mathcal{F}^{-1} \varphi\right\rangle$$

在上述定义下，我们有

$$\mathcal{F}^{-1} \mathcal{F} T=T \quad \text { and } \quad \mathcal{F} \mathcal{F}^{-1} T=T$$

下面验证第一个等式

$$\begin{aligned}\left\langle\mathcal{F}^{-1}(\mathcal{F} T), \varphi\right\rangle &=\left\langle\mathcal{F} T, \mathcal{F}^{-1} \varphi\right\rangle \\ &=\left\langle T, \mathcal{F}\left(\mathcal{F}^{-1} \varphi\right)\right\rangle \\ &=\langle T, \varphi\rangle \end{aligned}$$

在上述定义下，我们依然有傅里叶变换的线性性

$$\mathcal{F}\left(T_{1}+T_{2}\right)=\mathcal{F} T_{1}+\mathcal{F} T_{2} \quad \text { and } \quad \mathcal{F}(\alpha T)=\alpha \mathcal{F} T$$

证明直接由定义即可

$$\begin{aligned}\left\langle\mathcal{F}\left(T_{1}+T_{2}\right), \varphi\right\rangle &=\left\langle T_{1}+T_{2}, \mathcal{F} \varphi\right\rangle=\left\langle T_{1}, \mathcal{F} \varphi\right\rangle+\left\langle T_{2}, \mathcal{F} \varphi\right\rangle=\left\langle\mathcal{F} T_{1}, \varphi\right\rangle+\left\langle\mathcal{F} T_{2}, \varphi\right\rangle=\left\langle\mathcal{F} T_{1}+\mathcal{F} T_{2}, \varphi\right\rangle \\\langle\mathcal{F}(\alpha T), \varphi\rangle &=\langle\alpha T, \mathcal{F} \varphi\rangle=\alpha\langle T, \mathcal{F} \varphi\rangle=\alpha\langle\mathcal{F} T, \varphi\rangle=\langle\alpha \mathcal{F} T, \varphi\rangle \end{aligned}$$

$1$和$\delta $的傅里叶变换

现在考虑$\delta $的傅里叶变换，我们有

$$\mathcal{F} \delta=1$$

注意到我们有

$$\langle 1, \varphi\rangle=\int_{-\infty}^{\infty} 1 \cdot \varphi(x) d x$$

另一方面

$$\langle\mathcal{F} \delta, \varphi\rangle=\langle\delta, \mathcal{F} \varphi\rangle=\mathcal{F} \varphi(0)=\int_{-\infty}^{\infty} \varphi(x) d x$$

所以结论得证。

接着考虑$1$的傅里叶变换，我们有

$$\mathcal{F} 1=\delta$$

由定义可得

$$\langle\mathcal{F} 1, \varphi\rangle=\langle 1, \mathcal{F} \varphi\rangle=\int_{-\infty}^{\infty} \mathcal{F} \varphi(s) d s$$

考虑$ \mathcal{F} \varphi$的傅里叶逆变换在$0$的值，

$$\mathcal{F}^{-1} \mathcal{F} \varphi(t)=\int_{-\infty}^{\infty} e^{2 \pi i s t} \mathcal{F} \varphi(s) d s \quad \text { and at } t=0 \quad \mathcal{F}^{-1} \mathcal{F} \varphi(0)=\int_{-\infty}^{\infty} \mathcal{F} \varphi(s) d s$$

而

$$\mathcal{F}^{-1} \mathcal{F} \varphi(0)=\varphi(0)$$

所以

$$\langle\mathcal{F} 1, \varphi\rangle=\varphi(0)=\langle\delta, \varphi\rangle$$

$e^{2 \pi i x a}$和$\delta_a$的傅里叶变换

考虑$\delta_a$定义的傅里叶变换，其中$\delta_a $的定义为

$$\left\langle\delta_{a}, \varphi\right\rangle=\varphi(a)$$

我们有

$$\left\langle\mathcal{F} \delta_{a}, \varphi\right\rangle=\left\langle\delta_{a}, \mathcal{F} \varphi\right\rangle=\mathcal{F} \varphi(a)=\int_{-\infty}^{\infty} e^{-2 \pi i a x} \varphi(x) d x=\left\langle e^{-2 \pi i a x}, \varphi\right\rangle$$

即

$$\mathcal{F} \delta_{a}=e^{-2 \pi i s a}$$

接着考虑$e^{2 \pi i x a}$的傅里叶变换，结论为

$$\mathcal{F} e^{2 \pi i x a}=\delta_{a}$$

这是因为

$$\begin{aligned}\left\langle\mathcal{F} e^{2 \pi i x a}, \varphi\right\rangle &=\left\langle e^{2 \pi i x a}, \mathcal{F} \varphi\right\rangle \\ &=\int_{-\infty}^{\infty} e^{2 \pi i x a} \mathcal{F} \varphi(x) d x \end{aligned}$$

最后一项可以看成$\mathcal F \varphi$的傅里叶逆变换在$a$处的值，所以

$$\left\langle\mathcal{F} e^{2 \pi i x a}, \varphi\right\rangle = \mathcal F^{-1}\mathcal F \varphi(a) =\varphi(a)=\left\langle\delta_{a}, \varphi\right\rangle$$

从而结论成立。

$\sin$和$\cos$的傅里叶变换

利用$\delta_a$的傅里叶变换，我们可得

$$\begin{aligned} \mathcal{F}\left(\frac{1}{2}\left(\delta_{a}+\delta_{-a}\right)\right) &=\frac{1}{2}\left(e^{-2 \pi i s a}+e^{2 \pi i s a}\right)=\cos 2 \pi s a \\ \mathcal{F}\left(\frac{1}{2 i}(\delta(x+a)-\delta(x-a))\right) &=\frac{1}{2 i}\left(e^{2 \pi i s a}-e^{-2 \pi i s a}\right)=\sin 2 \pi s a \end{aligned}$$

反之，

$$\begin{aligned} \mathcal{F} \cos 2 \pi a x &=\mathcal{F}\left(\frac{1}{2}\left(e^{2 \pi i x a}+e^{-2 \pi i x a}\right)\right)=\frac{1}{2}\left(\delta_{a}+\delta_{-a}\right)\\ \mathcal{F} \sin 2 \pi a x &=\mathcal{F}\left(\frac{1}{2 i}\left(e^{2 \pi i x a}-e^{-2 \pi i x a}\right)\right)=\frac{1}{2 i}(\delta_a-\delta_{-a}) \end{aligned}$$

微分运算

假设

$$\lim_{x \rightarrow \pm \infty} f(x) \varphi(x) \rightarrow 0$$

那么

$$\begin{aligned} \int_{-\infty}^{\infty} f^{\prime}(x) \varphi(x) d x &=[f(x) \varphi(x)]_{-\infty}^{\infty}-\int_{-\infty}^{\infty} f(x) \varphi^{\prime}(x) d x \\ &=-\int_{-\infty}^{\infty} f(x) \varphi^{\prime}(x) d x \end{aligned}$$

结合速降函数性质，我们由此给出分布函数导数的定义

$$\left\langle T^{\prime}, \varphi\right\rangle=-\left\langle T, \varphi^{\prime}\right\rangle$$

显然导数的线性性满足，这里不再复述。

单位跃迁函数的导数

考虑单位跃迁函数

$$H(x)=\left\{\begin{array}{ll}{0} & {x \leq 0} \\ {1} & {x>0}\end{array}\right.$$

那么

$$\langle H, \varphi\rangle=\int_{-\infty}^{\infty} H(x) \varphi(x) d x=\int_{0}^{\infty} \varphi(x) d x$$

考虑其导数，我们得到

$$\left\langle H^{\prime}, \varphi\right\rangle=-\left\langle H, \varphi^{\prime}\right\rangle=-\int_{-\infty}^{\infty} H(x) \varphi^{\prime}(x) d x=-\int_{0}^{\infty} 1 \cdot \varphi^{\prime}(x) d x=-(\varphi(\infty)-\varphi(0))=\varphi(0)$$

从而

$$\left\langle H^{\prime}, \varphi\right\rangle=\varphi(0)=\langle\delta, \varphi\rangle$$

即

$$H^{\prime}=\delta$$

单位坡道(ramp)函数的导数

考虑单位坡道函数

$$u(x)=\left\{\begin{array}{ll}{0} & {x \leq 0} \\ {x} & {x>0}\end{array}\right.$$

计算其导数可得

$$\begin{aligned} \left\langle u^{\prime}(x), \varphi(x)\right\rangle &=-\left\langle u(x), \varphi^{\prime}(x)\right\rangle\\ &=-\int_{-\infty}^{\infty} u(x) \varphi^{\prime}(x) d x\\ &=-\int_{0}^{\infty} x \varphi^{\prime}(x) d x \\ &=-\left([x \varphi(x)]_{0}^{\infty}-\int_{0}^{\infty} \varphi(x) d x\right)\\ &=\int_{0}^{\infty} \varphi(x) d x \\ &=\langle H, \varphi\rangle \end{aligned}$$

所以

$$u'=H$$

$\text{sign}$函数的导数

考虑$\text{sign}$函数

$$\operatorname{sgn}(x)=\left\{\begin{array}{ll}{+1} & {x>0} \\ {-1} & {x<0}\end{array}\right.$$

计算其导数得到

$$\begin{aligned} \left\langle\mathrm{sgn}^{\prime}, \varphi\right\rangle &=-\left\langle\operatorname{sgn}, \varphi^{\prime}\right\rangle\\ &=-\int_{-\infty}^{\infty} \operatorname{sgn}(x) \varphi^{\prime}(x) d x \\ &=-\left(\int_{-\infty}^{0}(-1) \varphi^{\prime}(x) d x+\int_{0}^{\infty}(+1) \varphi^{\prime}(x) d x\right) \\ &=(\varphi(0)-\varphi(-\infty))-(\varphi(\infty)-\varphi(0))\\ &=2 \varphi(0)\\ &=\langle 2 \delta, \varphi\rangle \end{aligned}$$

即

$$\operatorname{sgn}^{\prime}=2 \delta$$

$\delta$函数的导数

$$\left\langle\delta^{\prime}, \varphi\right\rangle=-\left\langle\delta, \varphi^{\prime}\right\rangle=-\varphi^{\prime}(0)$$

包含经典傅里叶变换的广义傅里叶变换

考虑$\mathcal F_f$定义的增缓分布，

$$\begin{aligned} \left\langle T_{\mathcal{F} f}, \varphi\right\rangle &=\int_{-\infty}^{\infty} \mathcal{F} f(s) \varphi(s) d s \\ &=\int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty} e^{-2 \pi i s x} f(x) d x\right) \varphi(s) d s\\ &=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-2 \pi i s x} f(x) \varphi(s) d s d x \\ &=\int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty} e^{-2 \pi i s x} \varphi(s) d s\right) f(x) d x\\ &=\int_{-\infty}^{\infty} \mathcal{F} \varphi(x) f(x) d x\\ &=\left\langle T_{f}, \mathcal{F} \varphi\right\rangle \end{aligned}$$

根据傅里叶变换的定义，我们有

$$\left\langle T_{f}, \mathcal{F} \varphi\right\rangle=\left\langle\mathcal{F} T_{f}, \varphi\right\rangle$$

所以

$$T_{\mathcal{F} f}=\mathcal{F} T_{f}$$

对偶性

回顾对偶性

$$\begin{aligned} \mathcal{F} f(-s) &=\mathcal{F}^{-1} f(s) \\ \mathcal{F}^{-1} f(-t) &=\mathcal{F} f(t)\\ \mathcal{F}(f(-t)) &=\mathcal{F}^{-1} f(s) \\ \mathcal{F}^{-1}(f(-s)) &=\mathcal{F} f(s) \end{aligned}$$

回顾之前的记号

$$f^{-}(x)=f(-x)$$

利用上述记号，之前的结论变成

$$(\mathcal{F} f)^{-}=\mathcal{F}^{-1} f, \quad\left(\mathcal{F}^{-1} f\right)^{-}=\mathcal{F} f, \quad \mathcal{F} f^{-}=\mathcal{F}^{-1} f, \quad \mathcal{F}^{-1} f^{-}=\mathcal{F} f$$

利用后两个结论可得

$$\mathcal{F} \mathcal{F} f=f^{-}, \quad \mathcal{F}^{-1} \mathcal{F}^{-1} f=f^{-}$$

推广

现在考虑增缓分布的情形

$$\begin{aligned}\left\langle T_{f^{-}}, \varphi\right\rangle &=\int_{-\infty}^{\infty} f(-x) \varphi(x) d x \\ &=\int_{\infty}^{-\infty} f(u) \varphi(-u)(-d u) \\ &=\int_{-\infty}^{\infty} f(u) \varphi(-u) d u \end{aligned}$$

即

$$\left\langle T_{f^{-}}, \varphi\right\rangle=\left\langle T_{f}, \varphi^{-}\right\rangle$$

接着定义$(T_f)^-$

$$\left\langle\left(T_{f}\right)^{-}, \varphi\right\rangle=\left\langle T_{f}, \varphi^{-}\right\rangle$$

所以

$$\left(T_{f}\right)^{-}=T_{f^{-}}$$

一般的，假设$T$是分布，根据下式定义reversed distribution

$$\langle T^{-}, \varphi\rangle=\langle T, \varphi^{-}\rangle$$

性质

$$\begin{aligned} \langle (\mathcal{F} T)^{-}, \varphi\rangle &=\langle \mathcal{F} T, \varphi^{-}\rangle\\ &=\langle T, \mathcal{F}\left(\varphi^{-}\right)\rangle\\ &=\langle T, \mathcal{F}^{-1} \varphi\rangle\\ &=\langle\mathcal{F}^{-1} T, \varphi\rangle\end{aligned}$$

所以

$$(\mathcal{F} T)^{-}=\mathcal{F}^{-1} T$$

同理可得

$$\mathcal{F} T=\left(\mathcal{F}^{-1} T\right)^{-}$$

再看几个例子

$$\left\langle\mathcal{F}\left(T^{-}\right), \varphi\right\rangle=\left\langle T^{-}, \mathcal{F} \varphi\right\rangle=\left\langle T,(\mathcal{F} \varphi)^{-}\right\rangle=\left\langle T, \mathcal{F}^{-1} \varphi\right\rangle=\left\langle\mathcal{F}^{-1} T, \varphi\right\rangle$$

所以

$$\mathcal{F}\left(T^{-}\right)=\mathcal{F}^{-1} T$$

最后，我们有

$$\mathcal{F}^{-1}\left(T^{-}\right)=\mathcal{F} T$$

对上述两式分别作用$\mathcal F, \mathcal F^{-}$可得

$$\mathcal{F} \mathcal{F} T=T^{-}, \quad \mathcal{F}^{-1} \mathcal{F}^{-1} T=T^{-}$$

奇偶分布

我们称分布$T$是偶分布，如果$T^-=T$；奇分布如果$T^-=-T$

注意到$f(x)$决定了分布$T_f$，分布的奇偶性和$f$相同：

$$\left(T_{f}\right)^{-}=T_{f^{-}}=T_{ \pm f}=\pm T_{f}$$

偶分布的例子

$\delta $是偶分布

$$\left\langle\delta^{-}, \varphi\right\rangle=\left\langle\delta, \varphi^{-}\right\rangle=\varphi^{-}(0)=\varphi(-0)=\varphi(0)=\langle\delta, \varphi\rangle$$

上述事实可以推导

$$\mathcal{F} 1=\delta$$

推导如下

$$\mathcal{F} 1=\left(\mathcal{F}^{-1} 1\right)^{-}=\delta^{-}=\delta$$

傅里叶变换的奇偶性

傅里叶变换产生的分布的奇偶性和原分布相同，这是因为

$$(\mathcal{F} T)^{-}=\mathcal{F} T^{-}=\mathcal{F} (\pm T) =\pm \mathcal F T$$

$\sin c$的傅里叶变换

$$\begin{aligned} \mathcal{F} \text { sinc } &=\mathcal{F}(\mathcal{F} \Pi) \\ &=\Pi^{-} \\ &=\Pi \end{aligned}$$

函数和分布相乘

考虑函数和分布相乘决定的分布

$$\left\langle g T_{f}, \varphi\right\rangle=\int_{-\infty}^{\infty} g(x) f(x) \varphi(x) d x=\int_{-\infty}^{\infty} f(x)(g(x) \varphi(x)) d x = \left\langle T_{f}, g \varphi\right\rangle$$

由此引入定义：

假设$T$是分布，$g$是速降函数，那么分布$gT$定义为

$$\langle g T, \varphi\rangle=\langle T, g \varphi\rangle$$

函数乘以$\delta $

$$\langle g \delta, \varphi\rangle=\langle\delta, g \varphi\rangle= g(0) \varphi(0)$$

因此

$$g(x) \delta=g(0) \delta$$

特别的，我们有

$$x^{n} \delta=0$$

类似的，我们证明

$$g(x) \delta_{a}=g(a) \delta_{a}$$

这是因为

$$\left\langle g \delta_{a}, \varphi\right\rangle=\left\langle\delta_{a}, g \varphi\right\rangle= g(a) \varphi(a)=g(a)\left\langle\delta_{a}, \varphi\right\rangle=\left\langle g(a) \delta_{a}, \varphi\right\rangle$$

对于函数乘以$\delta$，给出更强的结论：

• 如果$T$是分布，$xT=0$，那么$T=c\delta$

首先，我们有

$$\begin{aligned} \psi(x) &=\psi(0)+\int_{0}^{x} \psi^{\prime}(t) d t \\ &=\psi(0)+\int_{0}^{1} x \psi^{\prime}(x u) d u \\ &=\psi(0)+x \int_{0}^{1} \psi^{\prime}(x u) d u \end{aligned}$$

令

$$\Psi(x)=\int_{0}^{1} \psi^{\prime}(x u) d u$$

即

$$\psi(x)=\psi(0)+x \Psi(x)$$

特别的，如果$\psi(0)=1$，那么

$$\psi(x)=x \Psi(x)$$

现在上述事实证明之前的结论。

假设$xT=0$，那么

$$\langle x T, \varphi\rangle= 0$$

固定函数$\varphi_0$，其满足$\varphi_0(0)=1$，那么

$$\left\langle T, \varphi_{0}\right\rangle= c$$

利用恒等式，我们可得

$$\varphi(x)=\varphi(0) \varphi_{0}(x)+\left(\varphi(x)-\varphi(0) \varphi_{0}(x)\right)=\varphi(0) \varphi_{0}(x)+\psi(x)$$

而

$$\psi(0)=\varphi(0)-\varphi(0) \varphi_{0}(0)=\varphi(0)-\varphi(0)=0$$

这意味着

$$\psi(x)=x \Psi(x)$$

即

$$\varphi(x)=\varphi(0) \varphi_{0}(x)+x \Psi(x)$$

那么

$$\begin{aligned} \langle T, \varphi(x)\rangle &=\left\langle T, \varphi(0) \varphi_{0}+x \Psi\right\rangle \\ &=\left\langle T, \varphi(0) \varphi_{0}\right\rangle+\langle T, x \Psi\rangle \\ &=\varphi(0)\left\langle T, \varphi_{0}\right\rangle+\langle T, x \Psi\rangle \text { (linearity) } \\ &=\varphi(0)\left\langle T, \varphi_{0}\right\rangle+\langle x T, \Psi\rangle \text { (that's how mutiplying } T \text { by the smooth function } x \text { works) } \\ &=\varphi(0)\left\langle T, \varphi_{0}\right\rangle+0 \text { (that's how mutiplying } T \text { by the smooth function } x \text { works) } \\ &=c \varphi(0) \\ &=\langle c \delta, \varphi\rangle \end{aligned}$$

因此

$$T= c\delta$$