反例（周民强调和分析讲义P25）：
$\gdef\leq{\leqslant}$ $\gdef\geq{\geqslant}$ $\gdef\lrb#1{\lbrace#1\rbrace}$ $\gdef\norm#1{\Vert#1\Vert}$ $\gdef\F{\mathcal{F}}$ $\gdef\T{\mathcal{T}}$ $\gdef\intrn{\int_{\R^n}}$ $\gdef\ip{2\pi i}$
定义 设$f\in L^1(\R^n)$，记$(\F f)(\xi)=\hat{f}(\xi)=\intrn f(x)e^{-2\pi ix\cdot\xi}dx,\forall\xi\in\R^n$，称$\F f(\xi)$或$\hat{f}$为$f$的Fourier变换.
例1 $\R^1$上, $f(x)=\Chi_{[a,b]}(x)$，则$\hat f(\xi)=\int_\R f(x)e^{-\ip x\cdot\xi}dx=\int^a_be^{-\ip x\xi}dx$ $$ \hat f(\xi)=\begin{cases} b-a &\xi=0\\ \frac{e^{-\ip a\xi}-e^{-\ip b\xi}}{\ip\xi}&\xi\neq0 \end{cases} $$ 例2 $\R^1$上，$f(x)=e^{-\pi x^2}~(x\in\R)$ $$ \begin{aligned} \hat{f}(\xi)=\int_\R e^{-\pi x^2}e^{-\ip x\xi}dx &=\int_\R e^{-\pi(x^2+2ix\xi)}dx\\ &=\int_\R e^{-\pi(x+i\xi)^2+\pi(i\xi)^2}\\ &=e^{-\pi\xi^2}\int^{+\infty}_{-\infty}e^{-\pi(x+i\xi)^2}dx\\ &=e^{-\pi\xi^2}\int_{i\xi-\infty}^{i\xi+\infty}e^{-\pi z^2}dz\\ &=e^{-\pi\xi^2}\int^{+\infty}_{-\infty}e^{-\pi x^2}dx\\ &=e^{-\pi\xi^2} \end{aligned} $$ 注：$f(\cdot)=e^{-\pi(\cdot)^2}$则$\hat{f}(\cdot)=e^{-\pi(\cdot)^2}$
例3 $f(x)=e^{-2\pi|x|},x\in\R,f\in L^1(\R)$ $$ \begin{aligned} \hat{f}(\xi)&=\int_\R e^{-2\pi|x|}e^{-\ip x\xi}dx\\ &=\int^0_{-\infty}e^{2\pi x-\ip x\xi}dx+\int^{+\infty}_0e^{-2\pi x-\ip x\xi}dx\\ &=\int^0_{-\infty}e^{2\pi(1-i\xi)x}dx+\int^{+\infty}_0e^{-2\pi(1+i\xi)x}dx\\ &=\frac{1}{2\pi(1-i\xi)}e^{2\pi(1-i\xi)x}\Big|^0_{-\infty}+\frac{1}{-2\pi(1+i\xi)}e^{-2\pi(1+i\xi)x}\Big|^{+\infty}_0\\ &=\frac{1}{2\pi(1-i\xi)}+\frac{1}{2\pi(1+i\xi)}\\ &=\frac{1}{\pi}\frac{1}{1+\xi^2} \end{aligned} $$ 注：分离变量，$F=f(x_1...x_n)g(x_{n+1}...x_{n+m})$，有$\hat F=\hat f+\hat g$
例4 $f(x)=e^{-\pi|x|},x\in\R^n,x=(x_1,...,x_n),|x|^2=x^2_1+...+x_n^2$.
$\hat{f}=\displaystyle\prod^n_{i=1}\widehat{e^{-\pi x_i^2}}(\xi_i)=\prod^n_{i=1}e^{-\pi\xi_i^2}=e^{-\pi|\xi|^2}$
例5 $f(x)=e^{-2\pi|x|},x\in\R^n,\hat{f}(\xi)=c_n\frac{1}{(1+|\xi|^2)^\frac{n+1}{2}},\forall\xi\in\R^n$
一些记号：
$1^\circ$ 平移 $(\tau_yf)(x)=f(x-y),\forall x,y\in\R^n$
$2^\circ$ 伸缩 $(\delta_af)(x)=f(ax), x\in\R^n$，$a>0$给定
$3^\circ$ 反射 $\widetilde{f}(x)=f(-x),\forall x\in\R^n$
$4^\circ$ 展缩 $f_\varepsilon(x)=\varepsilon^{-n}f(\varepsilon^{-1}x)=\varepsilon^{-n}(\delta_{\varepsilon^{-1}}f)(x)$
$5^\circ$ 多重指标$\alpha,\beta,\alpha=(\alpha_1,...,\alpha_n)\in\N^n,\alpha_1,...,\alpha_n\in\N=\lrb{0,1,...}$，长度$|\alpha|=\alpha_1+...+\alpha_n$.
$\alpha\leq\beta\lrArr\alpha_i\leq\beta_i,\forall i=1,...,n$.
$\beta-\alpha=(\beta_1-\alpha_1,...,\beta_n-\alpha_n)$
$\partial^\alpha_x=\partial^{\alpha_1}_{x_1}...\partial^{\alpha_n}_{x_n}$，$x^\alpha=x_1^{\alpha_1}...x_n^{\alpha_n}，(ax)^\alpha=a^{|\alpha|}x^\alpha$
$6^\circ$ $M_hf(\xi)=e^{-ih}f(\xi)$
基本性质：$f,g\in L^1(\R^n),y\in\R^n,b\in\mathbb C,\alpha$为多重指标
(1) $\norm{\hat f}_{L^\infty}\leq\norm{f}_{L^1}$
(2) $\widehat{f+g}=\hat f+\hat g$
(3) $\widehat{bf}=b\hat f$
(4) $\hat{\widetilde{f}}=\widetilde{\hat f}$
(5) $\hat{\overline{f}}=\overline{\widetilde{\hat{f}}}$
(6) $\widehat{\tau_yf}(\xi)=e^{-\ip y\cdot\xi}\hat{f}(\xi)=M_{2\pi y\xi}\hat{f}(\xi)$
(7) $\widehat{e^{\ip xy}f(x)}(\xi)=(\tau_y\hat{f})(\xi)$
(8) $\widehat{\delta_tf}(\xi)=t^{-n}\hat{f}(t^{-1}\xi)=(\hat{f})_t(\xi)=\widehat{f(t\cdot)}(\xi)$
(9) $\widehat{\partial^\alpha_xf}(\xi)=(\ip\xi)^\alpha\hat{f}(\xi)=(\ip)^{|\alpha|}\xi^\alpha\hat{f}(\xi),\partial^\alpha_xf\in L^1$
(10) $\widehat{(-\ip x)^\alpha f(x)}(\xi)=\partial^\alpha_\xi\hat{f}(\xi)$
(11) $\widehat{f\ast g}(\xi)=\hat{f}(\xi)\hat{g}(\xi)$
$$
\begin{aligned}
\widehat{f\ast g}(\xi)&=\intrn (f\ast g)(x)e^{-\ip x\xi}dx\\
&=\intrn \intrn f(x-y)g(y)dye^{-\ip x\xi}dx~~(Fubini)\\
&=\intrn (\intrn f(x-y)e^{-\ip(x-y)\xi}e^{-\ip y\xi}dx)g(y)dy\\
&=\intrn (\intrn f(x-y)e^{-\ip(x-y)\xi}dx)e^{-\ip y\xi}g(y)dy\\
&=\intrn \hat{f}(\xi)e^{-\ip y\xi}g(y)dy\\
&=\hat{f}(\xi)\hat{g}(\xi)
\end{aligned}
$$
定理 若$f\in L^1(\R^n)$，则$\hat{f}\in C_0(\R^n)=\lrb{f\in C(\R^n)|\lim\limits_{|x|\rarr\infty}f(x)=0}$ ($\norm{f}_{C_0}=\sup_{x\in\R^n}|f(x)|$，$f$一致连续)
证明：$1^\circ$ 先证$\hat{f}$在$\R^n$上一致连续.
事实上，$\forall x,h\in\R^n,$ $$
\begin{aligned}
&\hat{f}(x+h)-\hat{f}(x)\\
&=\intrn f(y)e^{-\ip y(x+h)}dy-\intrn f(y)e^{-\ip yx}dy\\
&=\intrn f(y)e^{-\ip yx}(e^{-\ip yh}-1)dy\\
&=(\int_{|r|\leq r}+\int_{|y|>r})f(y)e^{-\ip yx}(e^{-\ip yh}-1)dy\\
&=I_1(h,x)+I_2(h,x)
\end{aligned}
$$ 由$|I_2|\leq\int_{|y|>r}|f(y)|2dy$. $\forall\varepsilon>0,\exist r_0>0~s.t.~\forall0<r\leq r_0$有$\int_{|y|>r}|f(y)|dy<\varepsilon$（因为$f\in L^1,\int_{|y|>r}f(y)dy\rarr0~(r\rarr+\infty)$）. 令$r=r_0$则$|I_2(r_0,h)|<\varepsilon,\forall h,x\in\R^n$.
对于$I_1=I_1(r_0,h)=\int_{|y|\leq r_0}f(y)e^{-\ip xy}(e^{-\ip hy}-1)dy$
在$\lrb{|y|\leq r_0}$，对上面的$\varepsilon>0,\exist\delta_0>0~s.t.~\forall|h|<\delta_0,|e^{-\ip hy}-1|<\varepsilon,\forall |y|\leq r_0~i.e.~e^{-\ip hy}-1\rightrightarrows0~ in~ \lrb{|y|\leq r_0}$
$\rArr|I_1|\leq\int_{|y|\leq r_0}|f(y)||e^{-\ip hy}-1|\leq\varepsilon\int_{|y|\leq r_0}|f(y)|dy<\varepsilon\norm{f}_{L^1}$
$\rArr|I_1|+|I_2|<(1+\norm{f}_{L^1})\varepsilon~(\forall x\in\R^n)$
$i.e.~|\hat{f}(x+h)-\hat{f}(x)|\rightrightarrows0~(|h|\rarr0,x\in\R^n)$
$2^\circ$ 证明Riemann-Lebsgue引理：设$f\in L^1(\R^n)$，则$\lim\limits_{|\xi|\rarr\infty}\hat{f}(\xi)=0$
(1) 对$f(x)=\Chi_{[a,b]}(x)~on~\R^n,\hat{f}(\xi)\begin{cases}\frac{e^{-\ip \xi a}-e^{-\ip \xi b}}{\ip\xi}&\xi\neq0\\b-a &\xi=0\end{cases}$
$\forall|\xi|\geq1,0<|\hat{f}(\xi)|\leq\frac{2}{2\pi|\xi|}=\frac{1}{\pi|\xi|}\rarr0~(|\xi|\rarr\infty)$.
(2) 阶梯函数$g(x_1,...,x_n)=\displaystyle\prod^n_{j=1}\Chi_{[a_j,b_j]}(x)~on~\R^n$
$\rArr\hat{g}(\xi_1...\xi_n)=\displaystyle\prod^n_{j=1}\widehat{\Chi_{[a_j,b_j]}}(\xi_j)\rArr|\hat{g}(\xi)|\rarr0(|\xi|\rarr\infty)$.
(3) $g$为有限个阶梯函数的和，同上有$|\hat{g}(\xi)|\rarr0~(|\xi|\rarr\infty)$
(4) $\forall f\in L^1(\R^n)$，由{$\sum_{finite}$阶梯函数}在$L^1(\R^n)$中稠密. $\forall\varepsilon>0,\exist g=${$\sum_{finite}$阶梯函数}$s.t.~f=g+h$且$\norm{h}_{L^1}<\varepsilon$. $$
\begin{aligned}
|\hat{f}(\xi)|\leq|\hat{g}(\xi)+\hat{f}(\xi)|&\leq|\hat{g}(\xi)|+|\hat{f}(\xi)|\\
&\leq|\hat{g}(\xi)|+\norm{h}_{L^1}\\
&<2\varepsilon\quad(|\xi|\rarr\infty)
\end{aligned}
$$ $\rArr\lim\limits_{|\xi|\rarr+\infty}\hat{f}(\xi)=0$
问题：$\forall g\in C_0(\R^n),\exist f\in L^1(\R^n)~s.t.~\hat{f}=g$ ？
反例（周民强调和分析讲义P25）：
定理（乘法公式）设$f,g\in L^1(\R^n)$，则有$\intrn \hat{f}(x)g(x)dx=\intrn f(x)\hat{g}(x)dx$
证明：$$ \begin{aligned} \intrn \hat{f}(x)g(x)dx&=\intrn 1\intrn f(y)e^{-\ip yx}dy~g(x)dx\\ &=\int_{\R^n\times\R^n}f(y)e^{-\ip yx}g(x)dydx~~(Tonelli)\\ &=\intrn (\intrn g(x)e^{-\ip yx}dx)f(y)dy~~(Fubini)\\ &=\intrn \hat{g}(y)f(y)dy=\intrn f(x)\hat{g}(x)dx \end{aligned} $$
例 Poisson核. $f(x)=e^{-2\pi|x|},\hat{f}(y)=\Gamma(\frac{n+1}{2})\pi^{-\frac{n+1}{2}}\frac{1}{(1+|y|^2)^\frac{n+1}{2}}$
$\widehat{\delta_af(y)}=a^{-n}\hat{f}(a^{-1}y)=(\hat{f}(y))_{a^{-1}}$
定理（Fourier变换的逆）若$f\in L^1(\R^n),\hat{f}\in L^1(\R^n)$，则$\T(\F f)(x)=\F(\T f)(x)=f(x)~a.e.~x\in\R^n$. 其中$\T g(x)=\intrn g(y)e^{\ip yx}dy=\hat{g}(-x)~(\forall g\in L^1)$.
注：$\T\F f=f,\forall f\in L^1~i.e.~\T\F=Id$恒等算子.$\F\T f=f~i.e.~\F\T=Id\rArr\T$是$\F$的逆，记作$\F^{-1}$
注意，$$
\begin{aligned}
(\hat{f})^{\check{}}&=\intrn \hat{f}(y)e^{\ip yx}dy\\
&=\intrn(\intrn f(z)e^{-\ip yz}dz)e^{\ip xy}dy\\
&=\intrn(\intrn e^{-\ip y(z-x)}dy)f(z)dz
\end{aligned}
$$ $e^{\ip yx}\notin L^1$不能用乘法公式
$\intrn e^{-\ip y(z-x)}dy$不可积，没有意义(don't make sense)
证明：$1^\circ~\forall\varepsilon>0$，定义$\varphi_x(\xi)=e^{\ip x\xi}e^{-\pi\varepsilon^2|\xi|^2}=e^{\ip x\xi}\delta_\varepsilon(e^{-\pi|\xi|^2}),\forall x,\xi\in\R^n$
$\rArr\widehat{\varphi_x}(y)=\varepsilon^{-n}e^{-\pi|\frac{y-x}{\varepsilon}|^2}=g_\varepsilon(x-y)$，高斯核$g(x)=e^{-\pi|x|^2}$.
$2^\circ~\intrn\varphi_x(\xi)\hat{f}(\xi)d\xi=\intrn\widehat{\varphi_x}(y)f(y)dy=\intrn g_\varepsilon(x-y)f(y)dy=(g_\varepsilon\ast f)(x)$
另一方面，$\intrn\varphi_x(\xi)\hat{f}(\xi)d\xi=\intrn e^{-\pi\varepsilon|\xi|^2}e^{\ip x\xi}\hat{f}(\xi)d\xi$ $$
\begin{aligned}
\lim_{\varepsilon\rarr0}\intrn\varphi_x(\xi)\hat{f}(\xi)d\xi&=\intrn\lim_{\varepsilon\rarr0}e^{-\pi\varepsilon|\xi|^2}e^{\ip x\xi}\hat{f}(\xi)d\xi\\
&=\intrn e^{\ip x\xi}\hat{f}(\xi)d\xi\\
&=\F^{-1}(\hat{f})(x)=\F^{-1}(\F f)(x)~~(\forall x\in\R^n)
\end{aligned}
$$
又$\intrn\varphi_x(\xi)\hat{f}(\xi)d\xi=(g_\varepsilon\ast f)(x)$，$\lrb{g_\varepsilon}_{\varepsilon>0}$为恒等逼近，$f\in L^1$
$\rArr\lim\limits_{\varepsilon\rarr0}\norm{g_\varepsilon\ast f-f}_{L^1}=0~(i.e.~g_\varepsilon\ast f\rarr f~in~L^1)$
$\rArr\exist\varepsilon_j\rarr0(j\rarr\infty)~s.t.~g_{\varepsilon_j}\ast f\rarr f~a.e.~x\in\R^n$
又$g_\varepsilon\ast f\rarr\F^{-1}(\F f)(x),\forall x\in\R^n$.
所以$\F^{-1}(\F f)=f(x)~a.e.~x\in\R^n$
$3^\circ$ 类似地，可证$\F(\F^{-1}f)(x)=f(x)~a.e.~x\in\R^n$
注：(1) $f,\hat{f}\in L^1\rArr(\hat f)^{\check{}}(x)=(\hat f)^{\hat{}}(-x)\in C_0(\R^n)\rArr\check{f}\in C_0(\R^n), (\hat f)^{\check{}}(x)=f(x)~a.e.~x\in\R^n$
$\rArr$改变$f(x)$在某个零测集上的值，有$f\in C(\R^n)$且这时$(\hat f)^{\check{}}(x)=f(x),\forall x\in\R^n$
(2) $f\in L^1,\hat{f}=0~a.e.~x\in\R^n\rArr\hat{f}\in L^1\rArr(\hat f)^{\check{}}(x)=f(x)~a.e.~x\in\R^n$
另一方面，$(\hat f)^{\check{}}(x)=f(x)=0\rArr f=0~a.e.~x\in\R^n$
定义（Schwartz空间）$S(\R^n)=\lrb{f\in C^\infty(\R^n):\norm{f}_{(k)}<c_k,\forall k\in\N}$，其中$$ \norm{f}_{(k)}=\max_{|\alpha|+l\leq k}\sup_{x\in\R^n}(1+|x|^2)^\frac{l}2|\partial^\alpha_xf(x)| $$
注：$(1+|x|^2)^\frac{l}2\text{\textasciitilde}(1+|x|)^l.~\exist c=c_l~s.t.~c^{-1}(1+|x|)^l\leq(1+|x|)^\frac{l}2\leq c(1+|x|^2)^l$
称$S(\R^n)$为Schwartz空间（速降函数空间）.
$f\in S(\R^n)\rArr$
$(1+|x|)^l|\partial^\alpha_xf(x)|<c_k,\forall x\in\R^n,\forall\alpha,l,|\alpha|+l\leq k$
$i.e.~|\partial^\alpha_x(x^\beta f(x))|<c_k,\forall x\in\R^n(|\alpha|+|\beta|\leq k)$
定理 若$f\in S(\R^n)$，则$\hat f\in S(\R^n)$.
注：$S(\R^n)\subset L^1(\R^n)$且在$L^1(\R^n)$中稠密，$S(\R^n)$是一个Frechet空间.