|
|
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. |
|
战巡
发表于 2020-4-5 03:48
hbghlyj
发表于 2022-2-3 03:30
。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。
