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请教傅立叶正反变换后还原怎么证明

如题,假设傅立叶正变换定义为
\[F(\omega)=\mathcal{F}(f(t))=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(t)e^{-i\omega t}dt\]
反变换定义为
\[f(t)=\mathcal{F}^{-1}(F(\omega))=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}F(\omega)e^{i\omega t}d\omega\]

现在想证明$f(t)=\mathcal{F}^{-1}[\mathcal{F}[f(t)]]$,我是这样做的:
\begin{align*}
\mathcal{F}^{-1}[\mathcal{F}[f(t)]]&=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\left[\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(t)e^{-i\omega t}dt\right]e^{i\omega t'}d\omega\\
&=\frac{1}{2\pi}\int_{-\infty}^{\infty}f(t)dt\int_{-\infty}^{+\infty}e^{-i\omega(t-t')}d\omega\\
&=\frac{1}{2\pi}\int_{-\infty}^{\infty}f(t)dt\sqrt{2\pi}\delta(t-t')=\frac{1}{\sqrt{2\pi}}f(t')
\end{align*}
请问是哪里出了问题?
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回复 1# abababa


\[\int_{-\infty}^{\infty}e^{-i\omega(t-t')}d\omega=2\pi\delta(t-t')\]

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回复 2# 战巡

谢谢,但我还是不懂,这个积分不是$1$的傅立叶变换吗,按1楼的定义,它的结果不应该是$\sqrt{2\pi}\delta(t-t')$吗?

我用Mathematica算的时候是这样的:
FourierTransform[1, w, t-t']

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本帖最后由 战巡 于 2020-4-5 12:02 编辑

回复 3# abababa


对啊,但按你1楼的定义
\[F(\omega)=\sqrt{2\pi}\delta(t-t')=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}1\cdot e^{-i\omega(t-t')}dt\]
不就有
\[\int_{-\infty}^{+\infty}1\cdot e^{-i\omega(t-t')}dt=2\pi\delta(t-t')\]

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回复 4# 战巡

谢谢,我终于明白了,我总是差那个系数忘了乘上去。

我在网上查了很多证明,最终都是依赖于这个$\delta$函数的,只是前面那个系数有区别。如果不通过$\delta$函数,要如何证明傅立叶正反变换后还原回函数本身这个问题呢?

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这是Brad Osgood教授的一本书上给出的“证明”, 不知道具不具备你要求的严格性,供你参考。

InverseFT

a.png
2020-4-13 11:48

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回复 6# 业余的业余

谢谢,这种方式以前看过,应该是一种定义傅立叶正反变换的方式,就是把那个级数按周期无穷的形式写出来,然后说里面的积分是正变换,外面的积分是反变换,或者交换正反变换。
但如果不这样从级数定义,而是直接给出一楼那样的积分定义,就是要证明$f(t)$经过两个积分后能还原回本身,也就是求证(避免符号冲突,当$f(t)$和$f(x)$是同一映射时):
\[f(t) = \frac{1}{2\pi}\int_{-\infty}^{+\infty}\left[\int_{-\infty}^{+\infty}f(x)e^{-i\omega x}dx\right]e^{i\omega x}d\omega\]
这样如果不依赖$\delta$函数,还能证明吗?

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我相信是走不通的,否则那么多作者为什么不走这显明的路呢?

手里有Walter Rudin 的泛函分析 (Functional Analysis), 我的水平离看这书还远。看了一下目录,有讲到Tempered Distribution 和 傅里叶变换, 我想它的证明一定达到了分析的严谨性。转过来供你参考。我的意思是真正要严谨的证明(那个用Dirac-$\delta$的我个人估计应该达不到分析的严谨性), 很可能需要勒贝格积分和复分析的基础。

a.png
2020-4-13 22:07

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前段时间偶然一搜,找到了一个傅立叶逆变换定理的比较基础的证明。觉得值得分享以下。链接 https://statweb.stanford.edu/~ca ... tures/Lecture02.pdf

proof of fourier inversion thereom

finversion.png
2021-9-11 09:19

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那个从 $\delta$ 借力的招数好理解,难想到。这个证明可算是对主楼证明的严谨化。

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回复 9# 业余的业余
(我抄一下)
Theorem 1 (Fourier inversion formula). Suppose $f, \hat{f} \in L^{1}(\mathbb{R}) .$ Then
$$
f(t)=\frac{1}{2 \pi} \int \hat{f}(\omega) e^{i \omega t} d \omega .
$$
Proof. We make the additional assumption that $f$ and $\hat{f}$ are continuous (written $\left.f, \hat{f} \in C^{0}(\mathbb{R})\right)$, a fact we justify in the remark below. Writing the Fourier transform explicitly, the expression for the inverse reads
$$
f(t)=\frac{1}{2 \pi} \int\left(\int f(u) e^{-i \omega u} d u\right) e^{i \omega t} d \omega .
$$
It seems tempting to exchange the order of integration in the above expression. However, we cannot do this, as the hypotheses of Fubini's theorem are not satisfied (that is, $f(u) e^{i \omega(t-u)}$ is not in $\left.L^{1}(\mathbb{R} \times \mathbb{R})\right)$. We use the following trick. Define
$$
I_{\varepsilon}(t)=\frac{1}{2 \pi} \iint f(u) e^{i \omega(t-u)} e^{-\varepsilon^{2} \omega^{2} / 2} d u d \omega,
$$
which one can reasonably see as an approximation to the RHS of $(2)$, as $\epsilon \downarrow 0 .$ In this case the integrand is summable and we can invoke Fubini's theorem to change the order of integration. If we integrate over $u$ first we have
$$
I_{\varepsilon}(t)=\frac{1}{2 \pi} \int \hat{f}(\omega) e^{i \omega t} \underbrace{e^{-\varepsilon^{2} \omega^{2} / 2}}_{\rightarrow 1 \text { as } \varepsilon \downarrow 0} d t \quad \stackrel{\varepsilon \downarrow 0}{\longrightarrow} \frac{1}{2 \pi} \int \hat{f}(\omega) e^{i \omega t} d \omega .
$$
The limit taken as $\varepsilon \downarrow 0$ is justified by the Dominated Convergence Theorem (DCT): the integrand is dominated by $|\hat{f}(\omega)|$, which is summable by hypothesis. On the other hand, integrating over $\omega$ first gives
$$
I_{\varepsilon}(t)=\int f(u) g_{\varepsilon}(t-u) d u, \quad \text { where } \quad g_{\varepsilon}(u)=\frac{1}{\varepsilon} g\left(\frac{u}{\varepsilon}\right)
$$
and $g(t)=(\sqrt{2 \pi})^{-1} e^{-t^{2} / 2}$. Note that $g_{\varepsilon}$ is positive, has integral equal to one, and concentrates around the origin as $\varepsilon$ tends to zero. Since $f$ is continuous by assumption, we have $I_{\varepsilon}(t) \rightarrow f(t)$ as $\varepsilon \downarrow 0$. This proves the claim.

Remark: The additional assumption $f, \hat{f} \in C^{0}(\mathbb{R})$ is justified (by the exercise from Lecture 1): $f \in L^{1}(\mathbb{R})$ implies $\hat{f} \in C^{0}(\mathbb{R}) .$ To see why, let $\omega_{0} \in \mathbb{R}$ be fixed. Then
$$
\hat{f}(\omega)-\hat{f}\left(\omega_{0}\right)=\int f(t)\left(e^{-i \omega t}-e^{-i \omega_{0} t}\right) d t
$$
The integrand is dominated by the summable function $2|f(t)|$ so that $\hat{f}(\omega) \rightarrow \hat{f}\left(\omega_{0}\right)$ as $\omega \rightarrow \omega_{0}$. Since $\omega_{0}$ is arbitrary, $\hat{f} \in C^{0}(\mathbb{R}) .$ The same argument can be applied to $f$ when $\hat{f} \in L^{1}(\mathbb{R})$ using $(1) .$
The Fourier inversion formula is valid if both $f$ and $\hat{f}$ are in $L^{1}(\mathbb{R}) .$ However, many functions of interest do not satisfy this hypothesis. For example, the boxcar function $f(t)=\mathbb{I}_{\{-1 / 2 \leq t \leq 1 / 2\}}$ has as its Fourier transform
$$
\hat{f}(\omega)=\frac{\sin (\omega / 2)}{\omega / 2} \notin L^{1}(\mathbb{R})
$$
and thus the above theorem does not apply. From the above remark, one can equally note that the boxcar function is not continuous. We will later discuss what to do in the case that the summability hypotheses are not satisfied, at which time we will extend the inversion formula to a larger class of functions that includes the boxcar function, among others.

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回复 11# hbghlyj

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