L1(Rn)L^1(\R^n)上的Fourier变换

\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fL1(Rn)f\in L^1(\R^n),记(Ff)(ξ)=f^(ξ)=Rnf(x)e2πixξdx,ξRn(\F f)(\xi)=\hat{f}(\xi)=\intrn f(x)e^{-2\pi ix\cdot\xi}dx,\forall\xi\in\R^n,称Ff(ξ)\F f(\xi)f^\hat{f}ff的Fourier变换.

例1 R1\R^1上, f(x)=X[a,b](x)f(x)=\Chi_{[a,b]}(x),则f^(ξ)=Rf(x)e2πixξdx=bae2πixξdx\hat f(\xi)=\int_\R f(x)e^{-\ip x\cdot\xi}dx=\int^a_be^{-\ip x\xi}dx f^(ξ)={baξ=0e2πiaξe2πibξ2πiξξ0 \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 R1\R^1上,f(x)=eπx2 (xR)f(x)=e^{-\pi x^2}~(x\in\R) f^(ξ)=Reπx2e2πixξdx=Reπ(x2+2ixξ)dx=Reπ(x+iξ)2+π(iξ)2=eπξ2+eπ(x+iξ)2dx=eπξ2iξiξ+eπz2dz=eπξ2+eπx2dx=eπξ2 \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()=eπ()2f(\cdot)=e^{-\pi(\cdot)^2}f^()=eπ()2\hat{f}(\cdot)=e^{-\pi(\cdot)^2}

例3 f(x)=e2πx,xR,fL1(R)f(x)=e^{-2\pi|x|},x\in\R,f\in L^1(\R) f^(ξ)=Re2πxe2πixξdx=0e2πx2πixξdx+0+e2πx2πixξdx=0e2π(1iξ)xdx+0+e2π(1+iξ)xdx=12π(1iξ)e2π(1iξ)x0+12π(1+iξ)e2π(1+iξ)x0+=12π(1iξ)+12π(1+iξ)=1π11+ξ2 \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(x1...xn)g(xn+1...xn+m)F=f(x_1...x_n)g(x_{n+1}...x_{n+m}),有F^=f^+g^\hat F=\hat f+\hat g

例4 f(x)=eπx,xRn,x=(x1,...,xn),x2=x12+...+xn2f(x)=e^{-\pi|x|},x\in\R^n,x=(x_1,...,x_n),|x|^2=x^2_1+...+x_n^2.
f^=i=1neπxi2^(ξi)=i=1neπξi2=eπξ2\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)=e2πx,xRn,f^(ξ)=cn1(1+ξ2)n+12,ξRnf(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:
11^\circ 平移 (τyf)(x)=f(xy),x,yRn(\tau_yf)(x)=f(x-y),\forall x,y\in\R^n
22^\circ 伸缩 (δaf)(x)=f(ax),xRn(\delta_af)(x)=f(ax), x\in\R^na>0a>0给定
33^\circ 反射 f~(x)=f(x),xRn\widetilde{f}(x)=f(-x),\forall x\in\R^n
44^\circ 展缩 fε(x)=εnf(ε1x)=εn(δε1f)(x)f_\varepsilon(x)=\varepsilon^{-n}f(\varepsilon^{-1}x)=\varepsilon^{-n}(\delta_{\varepsilon^{-1}}f)(x)
55^\circ 多重指标α,β,α=(α1,...,αn)Nn,α1,...,αnN={0,1,...}\alpha,\beta,\alpha=(\alpha_1,...,\alpha_n)\in\N^n,\alpha_1,...,\alpha_n\in\N=\lrb{0,1,...},长度α=α1+...+αn|\alpha|=\alpha_1+...+\alpha_n.
αβαiβi,i=1,...,n\alpha\leq\beta\lrArr\alpha_i\leq\beta_i,\forall i=1,...,n.
βα=(β1α1,...,βnαn)\beta-\alpha=(\beta_1-\alpha_1,...,\beta_n-\alpha_n)
xα=x1α1...xnαn\partial^\alpha_x=\partial^{\alpha_1}_{x_1}...\partial^{\alpha_n}_{x_n}xα=x1α1...xnαn(ax)α=aαxαx^\alpha=x_1^{\alpha_1}...x_n^{\alpha_n},(ax)^\alpha=a^{|\alpha|}x^\alpha
66^\circ Mhf(ξ)=eihf(ξ)M_hf(\xi)=e^{-ih}f(\xi)

基本性质:f,gL1(Rn),yRn,bC,αf,g\in L^1(\R^n),y\in\R^n,b\in\mathbb C,\alpha为多重指标
(1) f^LfL1\norm{\hat f}_{L^\infty}\leq\norm{f}_{L^1}
(2) f+g^=f^+g^\widehat{f+g}=\hat f+\hat g
(3) bf^=bf^\widehat{bf}=b\hat f
(4) f~^=f^~\hat{\widetilde{f}}=\widetilde{\hat f}
(5) f^=f^~\hat{\overline{f}}=\overline{\widetilde{\hat{f}}}
(6) τyf^(ξ)=e2πiyξf^(ξ)=M2πyξf^(ξ)\widehat{\tau_yf}(\xi)=e^{-\ip y\cdot\xi}\hat{f}(\xi)=M_{2\pi y\xi}\hat{f}(\xi)
(7) e2πixyf(x)^(ξ)=(τyf^)(ξ)\widehat{e^{\ip xy}f(x)}(\xi)=(\tau_y\hat{f})(\xi)
(8) δtf^(ξ)=tnf^(t1ξ)=(f^)t(ξ)=f(t)^(ξ)\widehat{\delta_tf}(\xi)=t^{-n}\hat{f}(t^{-1}\xi)=(\hat{f})_t(\xi)=\widehat{f(t\cdot)}(\xi)
(9) xαf^(ξ)=(2πiξ)αf^(ξ)=(2πi)αξαf^(ξ),xαfL1\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) (2πix)αf(x)^(ξ)=ξαf^(ξ)\widehat{(-\ip x)^\alpha f(x)}(\xi)=\partial^\alpha_\xi\hat{f}(\xi)
(11) fg^(ξ)=f^(ξ)g^(ξ)\widehat{f\ast g}(\xi)=\hat{f}(\xi)\hat{g}(\xi)
fg^(ξ)=Rn(fg)(x)e2πixξdx=RnRnf(xy)g(y)dye2πixξdx  (Fubini)=Rn(Rnf(xy)e2πi(xy)ξe2πiyξdx)g(y)dy=Rn(Rnf(xy)e2πi(xy)ξdx)e2πiyξg(y)dy=Rnf^(ξ)e2πiyξg(y)dy=f^(ξ)g^(ξ) \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fL1(Rn)f\in L^1(\R^n),则f^C0(Rn)={fC(Rn)limxf(x)=0}\hat{f}\in C_0(\R^n)=\lrb{f\in C(\R^n)|\lim\limits_{|x|\rarr\infty}f(x)=0} (fC0=supxRnf(x)\norm{f}_{C_0}=\sup_{x\in\R^n}|f(x)|ff一致连续)

证明:11^\circ 先证f^\hat{f}Rn\R^n上一致连续.
事实上,x,hRn,\forall x,h\in\R^n, f^(x+h)f^(x)=Rnf(y)e2πiy(x+h)dyRnf(y)e2πiyxdy=Rnf(y)e2πiyx(e2πiyh1)dy=(rr+y>r)f(y)e2πiyx(e2πiyh1)dy=I1(h,x)+I2(h,x) \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} I2y>rf(y)2dy|I_2|\leq\int_{|y|>r}|f(y)|2dy. ε>0,r0>0 s.t. 0<rr0\forall\varepsilon>0,\exist r_0>0~s.t.~\forall0<r\leq r_0y>rf(y)dy<ε\int_{|y|>r}|f(y)|dy<\varepsilon(因为fL1,y>rf(y)dy0 (r+)f\in L^1,\int_{|y|>r}f(y)dy\rarr0~(r\rarr+\infty)). 令r=r0r=r_0I2(r0,h)<ε,h,xRn|I_2(r_0,h)|<\varepsilon,\forall h,x\in\R^n.
对于I1=I1(r0,h)=yr0f(y)e2πixy(e2πihy1)dyI_1=I_1(r_0,h)=\int_{|y|\leq r_0}f(y)e^{-\ip xy}(e^{-\ip hy}-1)dy
{yr0}\lrb{|y|\leq r_0},对上面的ε>0,δ0>0 s.t. h<δ0,e2πihy1<ε,yr0 i.e. e2πihy10 in {yr0}\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}
I1yr0f(y)e2πihy1εyr0f(y)dy<εfL1\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}
I1+I2<(1+fL1)ε (xRn)\rArr|I_1|+|I_2|<(1+\norm{f}_{L^1})\varepsilon~(\forall x\in\R^n)
i.e. f^(x+h)f^(x)0 (h0,xRn)i.e.~|\hat{f}(x+h)-\hat{f}(x)|\rightrightarrows0~(|h|\rarr0,x\in\R^n)

22^\circ 证明Riemann-Lebsgue引理:设fL1(Rn)f\in L^1(\R^n),则limξf^(ξ)=0\lim\limits_{|\xi|\rarr\infty}\hat{f}(\xi)=0
(1) 对f(x)=X[a,b](x) on Rn,f^(ξ){e2πiξae2πiξb2πiξξ0baξ=0f(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}
ξ1,0<f^(ξ)22πξ=1πξ0 (ξ)\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(x1,...,xn)=j=1nX[aj,bj](x) on Rng(x_1,...,x_n)=\displaystyle\prod^n_{j=1}\Chi_{[a_j,b_j]}(x)~on~\R^n
g^(ξ1...ξn)=j=1nX[aj,bj]^(ξj)g^(ξ)0(ξ)\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) gg为有限个阶梯函数的和,同上有g^(ξ)0 (ξ)|\hat{g}(\xi)|\rarr0~(|\xi|\rarr\infty)
(4) fL1(Rn)\forall f\in L^1(\R^n),由{finite\sum_{finite}阶梯函数}在L1(Rn)L^1(\R^n)中稠密. ε>0,g=\forall\varepsilon>0,\exist g={finite\sum_{finite}阶梯函数}s.t. f=g+hs.t.~f=g+hhL1<ε\norm{h}_{L^1}<\varepsilon. f^(ξ)g^(ξ)+f^(ξ)g^(ξ)+f^(ξ)g^(ξ)+hL1<2ε(ξ) \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} limξ+f^(ξ)=0\rArr\lim\limits_{|\xi|\rarr+\infty}\hat{f}(\xi)=0

问题:gC0(Rn),fL1(Rn) s.t. f^=g\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):1

定理(乘法公式)设f,gL1(Rn)f,g\in L^1(\R^n),则有Rnf^(x)g(x)dx=Rnf(x)g^(x)dx\intrn \hat{f}(x)g(x)dx=\intrn f(x)\hat{g}(x)dx

证明:Rnf^(x)g(x)dx=Rn1Rnf(y)e2πiyxdy g(x)dx=Rn×Rnf(y)e2πiyxg(x)dydx  (Tonelli)=Rn(Rng(x)e2πiyxdx)f(y)dy  (Fubini)=Rng^(y)f(y)dy=Rnf(x)g^(x)dx \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}

2

例 Poisson核. f(x)=e2πx,f^(y)=Γ(n+12)πn+121(1+y2)n+12f(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}}

δaf(y)^=anf^(a1y)=(f^(y))a1\widehat{\delta_af(y)}=a^{-n}\hat{f}(a^{-1}y)=(\hat{f}(y))_{a^{-1}}

定理(Fourier变换的逆)若fL1(Rn),f^L1(Rn)f\in L^1(\R^n),\hat{f}\in L^1(\R^n),则T(Ff)(x)=F(Tf)(x)=f(x) a.e. xRn\T(\F f)(x)=\F(\T f)(x)=f(x)~a.e.~x\in\R^n. 其中Tg(x)=Rng(y)e2πiyxdy=g^(x) (gL1)\T g(x)=\intrn g(y)e^{\ip yx}dy=\hat{g}(-x)~(\forall g\in L^1).

注:TFf=f,fL1 i.e. TF=Id\T\F f=f,\forall f\in L^1~i.e.~\T\F=Id恒等算子.FTf=f i.e. FT=IdT\F\T f=f~i.e.~\F\T=Id\rArr\TF\F的逆,记作F1\F^{-1}

注意,(f^)ˇ=Rnf^(y)e2πiyxdy=Rn(Rnf(z)e2πiyzdz)e2πixydy=Rn(Rne2πiy(zx)dy)f(z)dz \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} e2πiyxL1e^{\ip yx}\notin L^1不能用乘法公式
Rne2πiy(zx)dy\intrn e^{-\ip y(z-x)}dy不可积,没有意义(don't make sense)

证明:1 ε>01^\circ~\forall\varepsilon>0,定义φx(ξ)=e2πixξeπε2ξ2=e2πixξδε(eπξ2),x,ξRn\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
φx^(y)=εneπyxε2=gε(xy)\rArr\widehat{\varphi_x}(y)=\varepsilon^{-n}e^{-\pi|\frac{y-x}{\varepsilon}|^2}=g_\varepsilon(x-y),高斯核g(x)=eπx2g(x)=e^{-\pi|x|^2}.
2 Rnφx(ξ)f^(ξ)dξ=Rnφx^(y)f(y)dy=Rngε(xy)f(y)dy=(gεf)(x)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,Rnφx(ξ)f^(ξ)dξ=Rneπεξ2e2πixξf^(ξ)dξ\intrn\varphi_x(\xi)\hat{f}(\xi)d\xi=\intrn e^{-\pi\varepsilon|\xi|^2}e^{\ip x\xi}\hat{f}(\xi)d\xi limε0Rnφx(ξ)f^(ξ)dξ=Rnlimε0eπεξ2e2πixξf^(ξ)dξ=Rne2πixξf^(ξ)dξ=F1(f^)(x)=F1(Ff)(x)  (xRn) \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} Rnφx(ξ)f^(ξ)dξ=(gεf)(x)\intrn\varphi_x(\xi)\hat{f}(\xi)d\xi=(g_\varepsilon\ast f)(x){gε}ε>0\lrb{g_\varepsilon}_{\varepsilon>0}为恒等逼近,fL1f\in L^1
limε0gεffL1=0 (i.e. gεff in L1)\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)
εj0(j) s.t. gεjff a.e. xRn\rArr\exist\varepsilon_j\rarr0(j\rarr\infty)~s.t.~g_{\varepsilon_j}\ast f\rarr f~a.e.~x\in\R^n
gεfF1(Ff)(x),xRng_\varepsilon\ast f\rarr\F^{-1}(\F f)(x),\forall x\in\R^n.
所以F1(Ff)=f(x) a.e. xRn\F^{-1}(\F f)=f(x)~a.e.~x\in\R^n
33^\circ 类似地,可证F(F1f)(x)=f(x) a.e. xRn\F(\F^{-1}f)(x)=f(x)~a.e.~x\in\R^n

注:(1) f,f^L1(f^)ˇ(x)=(f^)^(x)C0(Rn)fˇC0(Rn),(f^)ˇ(x)=f(x) a.e. xRnf,\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)f(x)在某个零测集上的值,有fC(Rn)f\in C(\R^n)且这时(f^)ˇ(x)=f(x),xRn(\hat f)^{\check{}}(x)=f(x),\forall x\in\R^n
(2) fL1,f^=0 a.e. xRnf^L1(f^)ˇ(x)=f(x) a.e. xRnf\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,(f^)ˇ(x)=f(x)=0f=0 a.e. xRn(\hat f)^{\check{}}(x)=f(x)=0\rArr f=0~a.e.~x\in\R^n

定义(Schwartz空间)S(Rn)={fC(Rn):f(k)<ck,kN}S(\R^n)=\lrb{f\in C^\infty(\R^n):\norm{f}_{(k)}<c_k,\forall k\in\N},其中f(k)=maxα+lksupxRn(1+x2)l2xαf(x) \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+x2)l2~(1+x)l. c=cl s.t. c1(1+x)l(1+x)l2c(1+x2)l(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

S(Rn)S(\R^n)为Schwartz空间(速降函数空间).

fS(Rn)f\in S(\R^n)\rArr
(1+x)lxαf(x)<ck,xRn,α,l,α+lk(1+|x|)^l|\partial^\alpha_xf(x)|<c_k,\forall x\in\R^n,\forall\alpha,l,|\alpha|+l\leq k
i.e. xα(xβf(x))<ck,xRn(α+βk)i.e.~|\partial^\alpha_x(x^\beta f(x))|<c_k,\forall x\in\R^n(|\alpha|+|\beta|\leq k)

定理 若fS(Rn)f\in S(\R^n),则f^S(Rn)\hat f\in S(\R^n).

注:S(Rn)L1(Rn)S(\R^n)\subset L^1(\R^n)且在L1(Rn)L^1(\R^n)中稠密,S(Rn)S(\R^n)是一个Frechet空间.

目录
定义1
例1
例2
例3
例4
性质1
定理1
引理1
例5
定理2
例6
定理3
注1
注2
定义2
注3
定理4
注4
引用1
引用2