Fourier变换

Fourier Transform

Proof of Fourier Integral Formula

https://yannisparissis.wordpress.com/2011/03/10/dmat0101-notes-3-the-fourier-transform-on-l1/

If g(t) satisfy (I), (II) in $(-\infty, +\infty)$

(I) $\forall [a, b] \subset (-\infty, +\infty)$\
g(t) satisfy Dirichlet conditions for [a, b]:

1. g(t) is continous $\forall t \in (a, b)$;

   or finite First type of discontinuitis $\in (a, b)$

2. finite Peak values $\in (a, b)$

   • Cantor定理：f(x) 在[a, b]内定义且连续 => 在(a, b) 内一致连续
   • $||f(2ay+x)-f(x)|| e^{-ay^2} \to 0, \text{as } a\to 0$ 对连续区间内成立
   • 有界变差在连续区间内: 可以表示成两个非负单调增函数之差f(t)=f1(t) - f2(t), \
     而f1, f2均可应用Bonnet公式(微积分学教程第二卷P95)
     $\int_{a}^{b} f_1(x) g(x) d x=f_1(b) \int_{\xi}^{b} g(x) d x$\
     其中$a \leqslant \xi \leqslant b, g(x)$可积

(II) $\int_{-\infty}^{+\infty} g(t) dt < M$

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Then $\Longrightarrow$

$$\mathbb{F}[g(t)] \equiv \int_{-\infty}^{+\infty} g(t) e^{-j\omega t} dt = G(\omega)\\$$ $$\mathbb{F^{-1}}[G(\omega)] \equiv \frac{1}{2\pi} \int_{-\infty}^{+\infty} G(\omega) e^{j\omega t} d\omega =g^{*}(t)$$

where, g*(t) && g(t):

$$g^{*}(t) = \left\{ \begin{aligned} g(t) && \text{continuous at t}\\ \frac{g(t_{-}) + g(t_{+})}{2} && \text{Not continuous at t} \end{aligned} \right.$$

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Proof

preparation

$$f(x) = e^{-a x^2} \\ \begin{aligned} G(\omega) &= \int^{+\infty}_{-\infty} f(x) e^{-j\omega x} dx\\ &= \int^{+\infty}_{-\infty} e^{-a x^2} e^{-j\omega x} dx\\ &= e^{-\frac{\omega^2}{4a}} \int^{+\infty}_{-\infty} e^{-a (x+j\frac{\omega}{2a})^2} dx\\ &= e^{-\frac{\omega^2}{4a}} \int^{+\infty}_{-\infty} e^{-a x^2} dx \quad [\text{Cauchy's integral theorem}]\\ &= \pi^{\frac{1}{2}} a^{-\frac{1}{2}} e^{-\frac{\omega^2}{4a}} \end{aligned}$$

where have

$$\begin{aligned} I^2 &\equiv \int^{+\infty}_{-\infty} e^{-a x^2} dx \int^{+\infty}_{-\infty} e^{-a y^2} dy\\ &= \int^{2\pi}_{0} \int^{+\infty}_{0} e^{-a r^2} dr \cdot rd\theta\\ &= 2\pi \cdot \frac{1}{2a} \int^{+\infty}_{0} e^{-a r^2} d(ar^2)\\ &= \frac{\pi}{a} \end{aligned}\\ I = \pi^{\frac{1}{2}} a^{-\frac{1}{2}}$$

In special form

考虑

$$\begin{aligned} \int_{-\infty}^{+\infty} G(\omega) e^{-a\omega^2} e^{j\omega x} d\omega &= \int_{-\infty}^{+\infty} \int^{+\infty}_{-\infty} f(y) e^{-j\omega y} dy e^{-a\omega^2} e^{j\omega x} d\omega\\ &= \int_{-\infty}^{+\infty} f(y) [\int^{+\infty}_{-\infty} e^{-a\omega^2} e^{-j\omega (y-x)} d\omega] dy\\ &= \pi^{\frac{1}{2}} a^{-\frac{1}{2}} \int_{-\infty}^{+\infty} f(y) e^{-\frac{(y-x)^2}{4a}} dy\\ &= \pi^{\frac{1}{2}} a^{-\frac{1}{2}} \int_{-\infty}^{+\infty} f(y+x) e^{-\frac{y^2}{4a}} dy \quad [y = 2ay']\\ &= \pi^{\frac{1}{2}} a^{-\frac{1}{2}} \cdot 2a \int_{-\infty}^{+\infty} f(2ay+x) e^{-ay^2} dy \\ &= 2\pi^{\frac{1}{2}} a^{\frac{1}{2}} \int_{-\infty}^{+\infty} f(2ay+x) e^{-ay^2} dy \end{aligned}$$

In 1st way

$$\int_{-\infty}^{+\infty} | \int_{-\infty}^{+\infty} G(\omega) e^{-a\omega^2} e^{j\omega x} d\omega - 2\pi f(x)| dx\\ \begin{aligned} &= \int_{-\infty}^{+\infty} | 2\pi^{\frac{1}{2}} a^{\frac{1}{2}} \int_{-\infty}^{+\infty} f(2ay+x) e^{-ay^2} dy - 2\pi f(x)| dx\\ &= 2\pi^{\frac{1}{2}} a^{\frac{1}{2}} \int_{-\infty}^{+\infty} |\int_{-\infty}^{+\infty} f(2ay+x) e^{-ay^2} dy - \pi^{\frac{1}{2}} a^{-\frac{1}{2}} f(x)| dx \quad [ \int_{-\infty}^{+\infty} e^{-ay^2} dy= \pi^{\frac{1}{2}} a^{-\frac{1}{2}}] \\ &= 2\pi^{\frac{1}{2}} a^{\frac{1}{2}} \int_{-\infty}^{+\infty} |\int_{-\infty}^{+\infty} [f(2ay+x)-f(x)] e^{-ay^2} dy| dx\\ &\leq 2\pi^{\frac{1}{2}} a^{\frac{1}{2}} \int_{-\infty}^{+\infty} [\int_{-\infty}^{+\infty} |f(2ay+x)-f(x)| dx] e^{-ay^2} dy\\ &= 2\pi^{\frac{1}{2}} a^{\frac{1}{2}} \int_{-\infty}^{+\infty} ||f(2ay+x)-f(x)|| e^{-ay^2} dy \end{aligned}$$

当$\lim_{a \to 0}$ 时

$$\begin{aligned} &\lim_{a\to 0} \int_{-\infty}^{+\infty} | \int_{-\infty}^{+\infty} G(\omega) e^{-a\omega^2} e^{j\omega x} d\omega - 2\pi f(x)| dx\\ &\leq 2\pi^{\frac{1}{2}} \lim_{a\to 0}a^{\frac{1}{2}} \cdot \lim_{a\to 0} \int_{-\infty}^{+\infty} ||f(2ay+x)-f(x)|| e^{-ay^2} dy\\ &= 2\pi^{\frac{1}{2}} \cdot 0 \cdot 0 \end{aligned}$$

考虑 $||f(2ay+x)-f(x)||$, 因为

$$||f(2ay+x)-f(x)|| e^{-ay^2} \leq 2 ||f(x)||e^{-ay^2}\\ 2 ||f(x)||e^{-ay^2} \text{is integrable function}\\ ||f(2ay+x)-f(x)|| e^{-ay^2} \to 0, \text{as } a\to 0\\$$

满足 Lebesgue’s Dominated Convergence Theorem

$$\lim_{a\to 0} \int_{-\infty}^{+\infty} ||f(2ay+x)-f(x)|| e^{-ay^2} dy\\ = \int_{-\infty}^{+\infty} \lim_{a\to 0} ||f(2ay+x)-f(x)|| e^{-ay^2} dy\\ = \int_{-\infty}^{+\infty} 0 dy \quad [\text{finite discontinuties}] = 0$$

In 2nd way

$$\begin{aligned} &| \int_{-\infty}^{+\infty} G(\omega) e^{-a\omega^2} e^{j\omega x} d\omega - \int_{-\infty}^{+\infty} G(\omega) e^{j\omega x} d\omega|\\ &\leq \int_{-\infty}^{+\infty} |G(\omega) [1-e^{-a\omega^2}] e^{j\omega x}| d\omega \end{aligned}$$

考虑

$$|G(\omega) [1-e^{-a\omega^2}] e^{j\omega x}| \leq 2 |G(\omega)|\\ 2 |G(\omega)| \text{ is integrable function}\\ |G(\omega) [1-e^{-a\omega^2}] e^{j\omega x}| \to 0, \text{as } a\to 0\\$$

满足 Lebesgue’s Dominated Convergence Theorem
https://en.wikipedia.org/wiki/Dominated_convergence_theorem

$$\lim_{a\to 0} | \int_{-\infty}^{+\infty} G(\omega) e^{-a\omega^2} e^{j\omega x} d\omega - \int_{-\infty}^{+\infty} G(\omega) e^{j\omega x} d\omega|\\ \leq \lim_{a\to 0} \int_{-\infty}^{+\infty} |G(\omega) [1-e^{-a\omega^2}] e^{j\omega x}| d\omega\\ = \int_{-\infty}^{+\infty} \lim_{a\to 0} |G(\omega) [1-e^{-a\omega^2}] e^{j\omega x}| d\omega\\ = \int_{-\infty}^{+\infty} 0 d\omega = 0$$

所以有

$$\lim_{a\to 0} \int_{-\infty}^{+\infty} G(\omega) e^{-a\omega^2} e^{j\omega x} d\omega = \int_{-\infty}^{+\infty} G(\omega) e^{j\omega x} d\omega$$

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Sum Up

我们知道

$$\lim_{a\to 0} \int_{-\infty}^{+\infty}|g(x, a) - 2\pi f(x)| dx = 0\\ \lim_{a\to 0} g(x, a) = g(x)\\$$

此处 $a\to 0$, 一致收敛 $g(x, a) \to g(x)$\
<=> $\forall \epsilon$, 可以找到 $\delta$\
只要 $0<a<\delta$, 对 $\forall x, |g(x, a) - g(x)|<\epsilon$

一致收敛short proof\
因为 $\int_{-\infty}^{+\infty} |G(\omega)|d\omega < M$, 所以可以找到A\
$\int_{-\infty}^{-A} |G(\omega)|d\omega < \frac{\epsilon}{6}, \int_{A}^{+\infty} |G(\omega)|d\omega < \frac{\epsilon}{6}$\
把$|g(x, a) - g(x)|$分拆

$$\begin{aligned} |g(x, a) - g(x)|&\leq \int_{-\infty}^{-A}|G(\omega) [1-e^{-a\omega^2}] e^{j\omega x}| d\omega\\ &+ \int_{A}^{+\infty} |G(\omega) [1-e^{-a\omega^2}] e^{j\omega x}| d\omega\\ &+ \int_{-A}^{+A}|G(\omega) [1-e^{-a\omega^2}] e^{j\omega x}| d\omega\\ &\leq 2 \int_{-\infty}^{-A} |G(\omega)|d\omega + 2 \int_{A}^{+\infty} |G(\omega)|d\omega \\ &+ M \int_{-A}^{+A} [1-e^{-aA^2}] d\omega\\ &\lt \frac{2\epsilon}{3} + 2MA[1-e^{-aA^2}] \lt \epsilon\\ \end{aligned}$$

只需合适的选取A, 满足

$$[a \lt \frac{-\ln(1-\frac{\epsilon}{6MA})}{A^2} \Rightarrow 2MA[1-e^{-aA^2}] \lt \frac{\epsilon}{3}]$$

END一致收敛short proof

因为此处 $a\to 0$, 一致收敛 $g(x, a) \to g(x)$\
所以成立

$$\lim_{a\to 0} \int_{-\infty}^{+\infty}|g(x, a) - g(x)| dx \\ = \int_{-\infty}^{+\infty}|\lim_{a\to 0} g(x, a) - g(x)| dx \\ = \int_{-\infty}^{+\infty}0dx = 0\\$$

进而有

$$\int_{-\infty}^{+\infty}|2\pi f(x) - g(x)| dx \\ = \lim_{a\to 0}\int_{-\infty}^{+\infty}|2\pi f(x) - g(x)| dx \\ \leq \lim_{a\to 0} \int_{-\infty}^{+\infty}|g(x, a) - 2\pi f(x)| dx \\+ \lim_{a\to 0} \int_{-\infty}^{+\infty}|g(x, a) - g(x)| dx\\ = 0$$

而$2\pi f(x)$ 和 $g(x)$都是分段连续函数,\
$|2\pi f(x) - g(x)|$也是分段连续函数

So, 每一段x连续区间都有

$$2\pi f(x) = g(x)$$

定义g(x, a), g(x)

$$g(x, a) \equiv \int_{-\infty}^{+\infty} G(\omega) e^{-a\omega^2} e^{j\omega x} d\omega\\ g(x) \equiv \int_{-\infty}^{+\infty} G(\omega) e^{j\omega x} d\omega$$

也就是，每一段x连续区间都有

$$f(x) = \frac{1}{2\pi} \int_{-\infty}^{+\infty} G(\omega) e^{j\omega x} d\omega$$

END Proof

Appendix: G(w) 的性质

• $G(\omega)$ is uniformly continuous
  

• $\lim_{\omega\to \infty} G(\omega) = 0$

  (Riemann-Lebesgue lemma)

  use Riemann–Lebesgue lemma($||f(x)||_{L1}< M$)
  https://en.wikipedia.org/wiki/Riemann%E2%80%93Lebesgue_lemma

  https://yannisparissis.wordpress.com/2011/03/10/dmat0101-notes-3-the-fourier-transform-on-l1/

补充说明：间断点处

必须满足Dirichlet 条件：

$f(x+),f(x-)$存在

此时有
$\frac{f(x+) + f(x-)}{2} = \frac{1}{2\pi} \int_{-\infty}^{+\infty} G(\omega) e^{j\omega x} d\omega$

参见微积分学教程第三卷P353, 354, 358, 359

Proof

一方面

$$\frac{1}{2\pi} \int_{-\infty}^{+\infty} G(k) e^{jk x} d\omega = \lim_{K\to \infty} \frac{1}{2\pi} \int_{-K}^{+K} G(k) e^{jk x} d\omega = \lim_{K\to \infty} S_{K}(x)$$

另一方面

$$\begin{aligned} S_{K}(x) &=\int_{-\infty}^{\infty} d t f(t) \int_{-K}^{K} \frac{e^{i k(x-t)}}{2 \pi} d k \\ &=\left.\int_{-\infty}^{\infty} d t f(t) \frac{e^{i k(x-t)}}{2 \pi i(x-t)}\right|_{-K} ^{K} \\ &=\int_{-\infty}^{\infty} d t f(t) \frac{\sin K(x-t)}{\pi(x-t)}\\ &=\underbrace{\int_{-\infty}^{x} f(t) \frac{\sin K(x-t)}{\pi(x-t)} d t}_{J_{K}(x)}+\underbrace{\int_{x}^{\infty} f(t) \frac{\sin K(x-t)}{\pi(x-t)} d t}_{I_{K}(x)} \end{aligned}$$

计算 $I_{K}(x)$

$$\begin{aligned} I_{K}(x) &=\int_{0}^{\infty} f(x+u) \frac{\sin K u}{\pi u} d u \\ &=\int_{0}^{\infty} \frac{f(x+u)-f\left(x^{+}\right)}{\pi u} \sin K u d u+\frac{f\left(x^{+}\right)}{\pi} \int_{0}^{\infty} \frac{\sin K u}{u} d u \end{aligned}$$

后一部分是

$$\int_{0}^{\infty} \frac{\sin K u}{u} d u=\frac{\pi}{2}\\ \frac{f\left(x^{+}\right)}{\pi} \int_{0}^{\infty} \frac{\sin K u}{u} d u = \frac{f\left(x^{+}\right)}{2}$$

前一部分，考虑$\frac{f(x+u)-f\left(x^{+}\right)}{\pi u} \equiv G(u)$

将$\int_0^{\infty} G(u)\sin(Ku) du$拆成三部分
$\int_0^{\infty} = \int_0^{\delta} + \int_\delta^{A} + \int_A^{\infty}$

• 第一部分, 因为f(t)有界变差，对f(t)=f1(t) - f2 (t), \
  而f1, f2均可应用Bonnet公式(微积分学教程第二卷P95)
  $\int_{a}^{b} f_1(x) g(x) d x=f_1(b) \int_{\xi}^{b} g(x) d x$\
  其中$a \leqslant \xi \leqslant b, g(x)$可积\
  此处$a=0, b=\delta, g(x) = \frac{\sin(Ku)}{u}$
  可以有\
  $\int_0^{\delta} =[f(x+\delta)-f\left(x^{+}\right)]\int_{\xi}^{\delta}\frac{\sin(Ku)}{u} du < \epsilon L$

• 第二部分$\int_\delta^{A}$, 对$G(u)$在$[\delta, A]$绝对可积\
  $\int_\delta^{A}|G(u)|du \leq \int_\delta^{A}|\frac{f(x+u)}{\pi u}|du + \int_\delta^{A}|\frac{f\left(x^{+}\right)}{\pi u}|du
  \leq
  \int_{-\infty}^{+\infty}\frac{|f(x+u)|}{\pi \delta}du + \int_\delta^{A}|\frac{f\left(x^{+}\right)}{\pi u}|du < M$

  应用Riemann–Lebesgue lemma
  $\lim_{K\to\infty} \int_\delta^{A} G(u) \sin(Ku) du = 0$

• 第三部分与第二部分之和\
  $\lim_{K\to\infty}\int_\delta^{\infty} = \lim_{A\to \infty}\lim_{K\to\infty}\int_\delta^{A} = 0$

• 综上，前一部分有\
  $\lim_{K\to\infty} \int_{0}^{\infty} \frac{f(x+u)-f\left(x^{+}\right)}{\pi u} \sin K u d u = 0$

所以,

$$\begin{aligned} \lim_{K\to\infty} I_{K}(x) &= \lim_{K\to\infty} \int_{0}^{\infty} \frac{f(x+u)-f\left(x^{+}\right)}{\pi u} \sin K u d u+\frac{f\left(x^{+}\right)}{2}\\ &= \frac{f\left(x^{+}\right)}{2} \end{aligned}$$

同理

$$\lim_{K\to\infty} J_{K}(x) = \frac{f\left(x^{-}\right)}{2}$$

总之有

$$\lim_{K\to \infty} S_{K}(x) = \lim_{K\to \infty} [I_{K}(x) + J_{K}(x)] = \frac{f\left(x^{+}\right) + f\left(x^{-}\right)}{2}$$

Conclusion

$$\frac{1}{2\pi} \int_{-\infty}^{+\infty} G(k) e^{jk x} d\omega = \lim_{K\to \infty} S_{K}(x) = \frac{f\left(x^{+}\right) + f\left(x^{-}\right)}{2}$$

END Proof