Machine-translated from Chinese.
基本概念
多重指标
$\alpha\in \mathbb{N}^n,$ $|\alpha|=\sum \alpha_i,$
$\alpha!=\prod \alpha_i !.$ 作为上标, $x^\alpha=\prod x_i^{\alpha_i}$
$\partial^\alpha=\frac{\partial^{|\alpha|} }{\partial x_1^{\alpha_1}\cdots\partial x_n^{\alpha_n} },$
$f^{(\alpha)}(x)=\partial^\alpha f(x).$
$f(x)$的Taylor展开为$\sum_\alpha \frac{f^{(\alpha)}(0)}{\alpha!}x^\alpha.$
$\partial^\alpha(fg)=\sum_{\beta+\gamma=\alpha}\binom{\alpha}{\beta \:\gamma}\partial^\beta f\partial^\gamma g.$
当$f,g$满足适当条件时, 一个有趣的证明是考虑Fourier变换,
利用二项式展开先证明$\xi^\alpha (\widehat{f}\ast \widehat{g})=\sum_{\beta+\gamma=\alpha}\binom{\alpha}{\beta \:\gamma}(\xi^\beta \widehat{f})\ast (\xi^\gamma \widehat{g}),$
再做Fourier逆变换即可.
基本空间
取光滑函数空间$\mathscr{E}(\mathbb{R}^n)=C^\infty(\mathbb{R}^n),$
拓扑规定为,
若序列$\{\varphi_v\}\in C^\infty(\mathbb{R}^n)$满足$\,\forall\,K\Subset \mathbb{R}^n, \alpha \in \mathbb{N}^n,$
$\sup_{x\in K}|\partial^\alpha \varphi_v|\rightarrow 0,$
则称$\varphi_v\rightarrow 0(C^\infty(\mathbb{R}^n)).$
这样的空间是拓扑线性空间, 可以看作有多个(半)模.
元素趋于零需要求所有(可列多个)模均趋于零. 由于难以找到一个统一的模,
因此它和Banach空间有所区别.
取紧支撑光滑函数空间$\mathscr{D}(\mathbb{R}^n)=C_c^\infty(\mathbb{R}^n),$
拓扑规定为,
若序列$\{\varphi_v\}\in C_c^\infty(\mathbb{R}^n)$支集落在共同紧集$K$内,
且$\,\forall\,\alpha\in \mathbb{N}^n,$
$\sup_{x\in K}|\partial^\alpha \varphi_v|\rightarrow 0,$
则称$\varphi_v\rightarrow 0(C_c^\infty(\mathbb{R}^n)).$
取速降函数空间$\mathscr{S}(\mathbb{R}^n),$
为满足$\lim\limits_{|x|\rightarrow \infty}x^\alpha\partial^p\varphi(x)=0,$
$\,\forall\,\alpha,p\in \mathbb{N}^n$的光滑函数全体. 拓扑规定为,
若序列$\{\varphi_v\}\in \mathscr{S}(\mathbb{R}^n)$满足$\,\forall\,\alpha,p\in \mathbb{N}^n,$
$\sup_{x\in \mathbb{R}^n}|x^\alpha\partial^p \varphi_v(x)|\rightarrow 0,$
则称$\varphi_v\rightarrow 0(\mathscr{S}(\mathbb{R}^n)).$
速降函数的条件也可等价地改为有界, 或将$x^\alpha$替换为$(1+|x|^2)^k,$
以便于处理. 速降函数空间的拓扑也类似. 和前两个空间一样,
速降函数空间也关于卷积封闭.
取$\varphi(x)=\begin{cases}
e^{\frac{1}{|x|^2-1} },&|x|<1\\
0,&|x|\ge 1.
\end{cases},$
$\alpha(x)=\frac{1}{\int_{\mathbb{R}^n} \varphi dx}\varphi(x).$
令$\beta_R=1_{B_R}\ast \alpha,$
则$\operatorname{supp}\beta_R\subset B_{R+1},$
且$\beta_R|_{B_{R-1} }\equiv 1.$
从而$\,\forall\,\varphi\in C^\infty(\mathbb{R}^n),$
取$\varphi_\nu=\beta_\nu\varphi\in C_c^\infty(\mathbb{R}^n),$
满足$\varphi_\nu\rightarrow \varphi(C^\infty(\mathbb{R}^n)).$
因此$C_c^\infty(\mathbb{R}^n)$在$C^\infty(\mathbb{R}^n)$中稠密. 类似地,
它在$\mathscr{S}(\mathbb{R}^n)$中也稠密.
对于$\mathscr{D}(\mathbb{R}^n)\subset \mathscr{S}(\mathbb{R}^n)\subset \mathscr{E}(\mathbb{R}^n),$
前面的空间在后面的空间中稠密, 而前面的拓扑比后面的拓扑强.
正则化
对$\Omega$上的函数$u(x),$ 称其局部可积,
若其在任意紧集$K\Subset \Omega$上Lebesgue可积.
取$\alpha_\varepsilon=\frac{1}{\varepsilon^n}\alpha(\frac{x}{\varepsilon})\in C_c^\infty(\mathbb{R}^n),$
令$u_\varepsilon=u\ast \alpha_\varepsilon.$
若$u$局部可积则$u_\varepsilon$由定义, 满足如下定理:
定理 1.1. 若$u\in L_{loc}^1(\mathbb{R}^n),$ 则$u_\varepsilon\in C^\infty(\mathbb{R}^n).$ 当$\varepsilon\rightarrow 0$时, 若$u\in C^0(\mathbb{R}^n),$ 则$u_\varepsilon\rightarrow u(C^0(\mathbb{R}^n));$ 若$u\in L^p(\mathbb{R}^n),$ 则$u_\varepsilon\rightarrow u(L^p(\mathbb{R}^n)).$
前两个性质是易证的. 对$u\in L^p(\mathbb{R}^n)$的情况, 运用Lusin定理,
找$v\in C_c(\mathbb{R}^n)$使得$\parallel u-v\parallel_{L^p}$充分小即可.
过程中用到如下引理:
引理 1.2 (Minkowski积分不等式). $\parallel\parallel f(x,y)\parallel_{L_x^1}\parallel_{L_y^p}\le \parallel\parallel f(x,y)\parallel_{L_y^p}\parallel_{L_x^1}.$
证:
只需证明(可能需要限制在子空间上)线性算子$\mathrm{id}:L_x^1L_y^p\rightarrow L_y^pL_x^1$是有界的,
且$\parallel\mathrm{id}\parallel\le 1.$ 由泛函分析知识,
赋范线性空间间的线性算子$A:X\rightarrow Y$是有界的当且仅当$A^\ast :Y^\ast \rightarrow X^\ast $是有界的,
且有$\parallel A\parallel=\parallel A^\ast \parallel.$ 左推右是经典的命题,
右推左只需利用$A^{\ast \ast }$即可, 它是$A$在$X^{\ast \ast }$上的延拓.
因此考虑线性算子$\mathrm{id}^\ast :L_y^qL_x^\infty\rightarrow L_x^\infty L_y^q,$
说明$\parallel\mathrm{id}^\ast \parallel\le 1$即可,
即$\operatorname{ess}\operatorname{sup}_x\parallel f(x,y)\parallel_{L_y^q}\le \parallel\operatorname{ess}\operatorname{sup}_x|f(x,y)|\parallel_{L_y^q},$
然而这是显然的.
若$\parallel\parallel f(x,y)\parallel_{L_x^1}\parallel_{L_y^p}=\infty,$
考虑用$g_n=(|f|\wedge n)\cdot 1_{B_n}$逼近,
利用Levi单调收敛引理说明即可.
推论 1.3. $\parallel f\ast g\parallel_{L^p}\le \parallel f\parallel_{L^p}\parallel g\parallel_{L^1}.$
证:
$\parallel f\ast g\parallel_{L^p} \le \parallel\parallel f(x-y)g(y)\parallel_{L^1_y}\parallel_{L^p_x}\le \parallel\parallel f(x-y)g(y)\parallel_{L^p_x}\parallel_{L^1_y}=\parallel f\parallel_{L^p}\parallel g\parallel_{L^1}.$
由该推论,
$\parallel u_\varepsilon-v_\varepsilon\parallel\le \parallel u-v\parallel$也充分小,
而$\parallel v-v_\varepsilon\parallel$利用已有的性质也可以充分小,
定理得证.
称线性算子$J_\varepsilon:L_{loc}^1(\mathbb{R}^n)\rightarrow C^\infty(\mathbb{R}^n),$
$u\mapsto u_\varepsilon$为磨光算子,
得到$u_\varepsilon$的过程称为正则化.
定理说明$C^\infty(\mathbb{R}^n)$在$C^0(\mathbb{R}^n),L^p(\mathbb{R}^n)$中稠密,
从而$C_c^\infty(\mathbb{R}^n), \mathscr{S}(\mathbb{R}^n)$也在其中稠密.
一些简单的推论是,
$J_\varepsilon\in \mathfrak{B}(L^p(\mathbb{R}^n)\rightarrow L^p(\mathbb{R}^n));$
$u\in \mathscr{D}/\mathscr{S}/\mathscr{E}(\mathbb{R}^n)$时,
$u_\varepsilon\rightarrow u(\mathscr{D}/\mathscr{S}/\mathscr{E}(\mathbb{R}^n)).$
磨光算子的一个基本应用是对开集$\Omega,$ $\,\forall\,K\Subset \Omega,$
$\,\exists\,\varphi\in C_c^\infty(\mathbb{R}^n),$
$\operatorname{supp}\varphi\subset \Omega,$ $\varphi \in [0,1],$
且$\varphi|_K\equiv 1.$ 取$\varepsilon=d(K,\Omega^c)>0,$
取$J_{\frac{\varepsilon}{4} } (1_{d(x,K)\le \frac{\varepsilon}{2} })$即可.
广义函数
定义
称$\mathscr{D}/\mathscr{S}/\mathscr{E}(\mathbb{R}^n)$上的线性连续泛函为$\mathscr{D}'/\mathscr{S}'/\mathscr{E}'(\mathbb{R}^n)$广义函数.
我们有关系式$\mathscr{D}'\supset \mathscr{S}'\supset \mathscr{E}',$
这里包含通过嵌入来看.
如$\mathrm{id}:\mathscr{S}'\rightarrow \mathscr{D}'$不仅有意义,
而且是单射.
有意义是因为在$\mathscr{D}\subset \mathscr{S}\subset \mathscr{E}$中,
前者拓扑强于后者; 单射通过前面的空间在后面的空间稠密保证.
类似地, 可以在开集$\Omega$上定义$\mathscr{D}',\mathscr{E}'$广义函数,
但$\mathscr{S}'$广义函数只在$\mathbb{R}^n$上讨论.
容易说明$L_{loc}^1\subset \mathscr{D}',$
而Dirac函数$\delta\in \mathscr{E}'.$
我们知道在赋范线性空间上, 线性泛函有界$\Leftrightarrow$连续,
但在拓扑线性空间上, 没有统一的模, 因此并没有”有界”的概念.
不过类似的命题依然成立, 可以由”一些模”控制.
定理 1.4. 若$T\in \mathscr{D}'(\Omega),$ 则$\,\forall\,K\Subset \Omega,$ $\,\exists\,C(K)>0,$ $m(K)\in \mathbb{N},$ 使得 $$|\left<{}T,\varphi\right>|\le C(K)\sup_{x\in \Omega,|\alpha|\le m}|\partial^\alpha \varphi(x)|, \,\forall\,\varphi\in C_c^\infty(K)\subset C_c^\infty(\Omega).$$ 反之若$T$为满足上述条件的线性泛函, 则它是连续的, $T\in \mathscr{D}'(\Omega).$
证:
定义$|\varphi|_\nu=\sup\limits_{x\in \Omega,|\alpha|\le \nu} |\partial^\alpha \varphi(x)|,$
满足$|\varphi|_\nu\le |\varphi|_\mu,$ $\,\forall\,\nu\le \mu.$
若$T\in \mathscr{D}'(\Omega)$不满足命题性质,
则$\,\exists\,K\Subset \Omega,$ $\{\varphi_\nu\}\in C_c^\infty(K),$
使得$|\left<{}T,\varphi_\nu \right>|>\nu|\varphi_\nu|_\nu.$
对$\varphi_\nu$做乘子使$|\left<{}T,\varphi_\nu\right>|=1,$
则$|\varphi_\nu|_\nu<\frac{1}{\nu}.$
容易看出这样有$\varphi_\nu\xrightarrow{\mathscr{D} } 0,$
但$|\left<{}T,\varphi_\nu\right>|\equiv 1,$
$\left<{}T,\varphi_\nu\right>\not\rightarrow 0,$ 矛盾.
反之若$T$为满足条件的线性泛函,
$\,\forall\,\varphi_\nu\xrightarrow{\mathscr{D} } 0,$
$\,\exists\,K$使得$\{\varphi_\nu\}\in C_c^\infty(K).$
由条件有对应的$C(K),m(K),$
使得$|\left<{}T,\varphi_\nu\right>|\le C|\varphi_\nu|_m\rightarrow 0.$
因此$\left<{}T,\varphi_\nu\right>\rightarrow 0,$
$T\in \mathscr{D}'(\Omega).$
定理 1.5. 若$T\in \mathscr{E}'(\Omega),$ 则$\,\exists\,K\Subset \Omega,$ $C>0,$ $m\in \mathbb{N},$ 使得 $$|\left<{}T,\varphi\right>|\le C\sup_{x\in K,|\alpha|\le m}|\partial^\alpha \varphi(x)|, \,\forall\,\varphi\in C^\infty(\Omega).$$ 反之若$T$为满足上述条件的线性泛函, 则它是连续的, $T\in \mathscr{E}'(\Omega).$
证: 取紧集$K_\nu=B_\nu\cap \{x:d(x,\Omega^c)\ge \frac{1}{\nu}\},$
则$K_\nu$递增且$\bigcup K_\nu=\Omega.$ $\,\forall\,K\Subset \Omega,$
$\,\exists\,N>0,$ $\,\forall\,\nu>N,$ $K\subset K_\nu.$
定义$|\varphi|_\nu=\sup\limits_{x\in K_\nu,|\alpha|\le \nu} |\partial^\alpha \varphi(x)|,$
满足$|\varphi|_\nu\le |\varphi|_\mu,$ $\,\forall\,\nu\le \mu.$
同上一定理可证明$T\in \mathscr{E}'(\Omega)$时满足命题性质.
反之若$T$为满足条件的线性泛函,
$\,\forall\,\varphi_\nu\xrightarrow{\mathscr{E} } 0,$ 取使性质成立的$K,$
那么$\sup_{x\in K}|\partial^\alpha\varphi(x)|\rightarrow 0,$
$\,\forall\,\alpha\in \mathbb{N}^n.$
因此$|\left<{}T,\varphi_\nu\right>|\le C\sup\limits_{x\in K,|\alpha|\le m}|\partial^\alpha\varphi_\nu(x)|\rightarrow 0.$
因此$\left<{}T,\varphi_\nu\right>\rightarrow 0,$
$T\in \mathscr{E}'(\Omega).$
可以看出, $\mathscr{D}',\mathscr{E}'$满足某种对偶关系:
在$\mathscr{D}$中, 函数有紧支集, 收敛性要求有一共同紧集,
因此其上的线性泛函连续性条件适当放宽,
对不同紧集上的紧支撑函数分别满足某种有界性即可; 而在$\mathscr{E}$中,
函数无紧支集, 收敛性要求对任意紧集满足一定条件即可,
这导致其上线性泛函连续性条件变严格, 有界性需由一紧集控制,
且要求对所有光滑函数一致地满足有界性.
这便是$\mathscr{D}\subset \mathscr{E}\rightarrow \mathscr{D}'\supset \mathscr{E}'.$
这种对偶关系在下一节会有更强的表现.
支集
对广义函数来说, 它在特定点的取值意义不明确, 但可以谈论它在开集上的取值.
对$T\in \mathscr{D}'(\Omega),$ 给一开集$\Omega'\subset \Omega,$
若$\,\forall\,\varphi\in C_c^\infty(\Omega'),$
$\left<{}T,\varphi\right>=0,$ 则称$T$在$\Omega'$上取零值.
进一步称$T_1,T_2$在$\Omega'$上取值相同, 若$T_1-T_2$在$\Omega'$上取零值.
因此若某个广义函数和一常义函数(诱导的线性连续泛函)在一开集上取值相同,
则可以说该广义函数在该开集上等于这一常义函数.
若常义函数点点取零则整体为零, 类似地若广义函数点点小邻域上取值为零,
则整体取零. 证明上, 只需利用单位分解定理.
对$\varphi\in C_c^\infty(\Omega),$
取单位分解$\{\varphi_i\}$从属于覆盖$\operatorname{supp}\varphi$的有限个小邻域组成的开覆盖$\{O_i\}$.
那么$\varphi_i \varphi\in C_c^\infty(O_i),$
且在整个$\Omega$上$\sum_i \varphi_i\varphi=\varphi.$
那么广义函数作用在$\varphi$上可分解到$\{O_i\}$上进行,
再线性组合取值即可. 零值作和仍为零, 故广义函数在全空间上取零值.
由这一性质, 广义函数在取零值的全体开集之并上仍取零值.
而全体开集之并仍为开集, 因此可以定义广义函数取零值的最大开集.
那么称该最大开集的补集为该广义函数的支集,
记为$\operatorname{supp}T.$ 对连续函数$f(x),$
容易看出其作为广义函数的支集与原本定义的支集是相同的.
因此广义函数支集的定义是连续函数支集定义的延伸.
定理 1.6. 任意$T\in \mathscr{E}'(\Omega)$具紧支集; 反之任意具紧支集$T\in \mathscr{D}'(\Omega),$ $T$也在$\mathscr{E}'(\Omega)$中.
证: 由上节定理, $\,\forall\,T\in \mathscr{E}'(\Omega),$
$\,\exists\,K\Subset \Omega,$ $C>0,$ $m\in \mathbb{N},$ 使得
$$|\left<{}T,\varphi\right>|\le C\sup\limits_{x\in K,|\alpha|\le m}|\partial^\alpha\varphi(x)|,\,\forall\,\varphi\in C^\infty(\Omega),$$
我们说明该$K$包含$T$的支集, 从而支集是紧的. 而这是容易的,
因为只要$\operatorname{supp}\varphi\subset \Omega\setminus K,$
不等式右端即为零, 从而迫使$T$在$\Omega\setminus K$上取零值, 支集在$K$内.
反之, 若$T\in \mathscr{D}'(\Omega)$,
我们希望将其定义域从$\mathscr{D}(\Omega)$延拓为$\mathscr{E}(\Omega),$
为此需想办法将光滑函数改造为紧支撑的, 且不改变泛函取值.
自然想到利用其支集即可. 记$\operatorname{supp}T=K\Subset \Omega$,
那么取$K\Subset \Omega_1\Subset \Omega,$
作$\zeta\in C_c^\infty(\overline{\Omega}_1),$
$\zeta\in[0,1]$且$\zeta|_{\overline{O}_\delta(K)}\equiv 1.$
对$\varphi\in C^\infty(\Omega),$
定义$\left<{}T,\varphi\right>=\left<{}T,\zeta\varphi\right>,$
$\zeta\varphi\in C_c^\infty(\overline{\Omega}_1).$ 需验证定义的合理性:
若还有$\eta\in C_c^\infty(\overline{\Omega}_2),$
$\eta\in [0,1]$且$\eta|_{\overline{O}_{\varepsilon}(K)}\equiv 1,$
那么$\zeta-\eta\in C_c^\infty(\Omega \setminus K),$
支集落在广义函数支集外,
从而$\left<{}T,\zeta\varphi\right>-\left<{}T,\eta\varphi\right>=\left<{}T,(\zeta-\eta)\varphi\right>=0.$
这样泛函就可以作用在$\varphi$上, 且由上节关于$\mathscr{D}'$的定理,
$$|\left<{}T,\varphi\right>|\le C|\zeta\varphi|_m\le C'\sup_{x\in \Omega_1,|\alpha|\le m}|\partial^\alpha\varphi|,$$
那么由关于$\mathscr{E}'$的定理,
$T$改造为$\mathscr{E}$上线性泛函后也是连续的,
即$T\in \mathscr{E}'(\Omega).$
由此可见$\mathscr{D}'_c=\mathscr{E}',$
这与$\mathscr{D}=\mathscr{E}_c$构成对偶, 十分赏心悦目.
可以看成为了让$\left<{}T,\varphi\right>$合理, 一侧条件弱时,
另一侧条件要强. 证明中后半部分即是在说,
为了让$T\in \mathscr{D}'_c$可以作用在$\mathscr{E}$中的$\varphi$上,
只需考虑在$\varphi|_{\operatorname{supp}T}$部分上的作用即可,
因为其余的部分落在支集外. 就具体操作而言, 不能直接进行限制,
需要利用证明中的方法改造为具紧支集函数.
事实上, 对(用于改造$\varphi$的)辅助函数$\zeta,$
可以将要求替换为$\zeta|_K=1.$
这是因为$\overline{O}_\delta(K)$的$\delta$可任意小,
可以取到$\zeta_\delta\xrightarrow{\mathscr{D} }\zeta,$
再利用线性泛函的连续性即可($T\in \mathscr{D}'$).
文章最后更新于 2021-10-13 12:46:51
Basic concepts
Multiple indicators
$\alpha\in \mathbb{N}^n,$ $|\alpha|=\sum \alpha_i,$
$\alpha!=\prod \alpha_i !.$ as superscript, $x^\alpha=\prod x_i^{\alpha_i}$
$\partial^\alpha=\frac{\partial^{|\alpha|} }{\partial x_1^{\alpha_1}\cdots\partial x_n^{\alpha_n} },$
$f^{(\alpha)}(x)=\partial^\alpha f(x).$
The Taylor expansion of $f(x)$ is $\sum_\alpha \frac{f^{(\alpha)}(0)}{\alpha!}x^\alpha.$
$\partial^\alpha(fg)=\sum_{\beta+\gamma=\alpha}\binom{\alpha}{\beta \:\gamma}\partial^\beta f\partial^\gamma g.$
When $f,g$ satisfies appropriate conditions, an interesting proof is to consider the Fourier transform,
Use binomial expansion to prove $\xi^\alpha (\widehat{f}\ast \widehat{g})=\sum_{\beta+\gamma=\alpha}\binom{\alpha}{\beta \:\gamma}(\xi^\beta \widehat{f})\ast (\xi^\gamma \widehat{g}),$
Then do the inverse Fourier transformation.
basic space
Get the smooth function space $\mathscr{E}(\mathbb{R}^n)=C^\infty(\mathbb{R}^n),$
The topology is defined as,
If the sequence $\{\varphi_v\}\in C^\infty(\mathbb{R}^n)$ satisfies $\,\forall\,K\Subset \mathbb{R}^n, \alpha \in \mathbb{N}^n,$
$\sup_{x\in K}|\partial^\alpha \varphi_v|\rightarrow 0,$
It is called $\varphi_v\rightarrow 0(C^\infty(\mathbb{R}^n)).$
Such a space is a topological linear space, which can be regarded as having multiple (half) modules.
Elements tending to zero require that all (can be listed as multiple) modules tend to zero. Since it is difficult to find a unified module,
Therefore it is different from Banach space.
Take the compactly supported smooth function space $\mathscr{D}(\mathbb{R}^n)=C_c^\infty(\mathbb{R}^n),$
The topology is defined as,
If the support set of sequence $\{\varphi_v\}\in C_c^\infty(\mathbb{R}^n)$ falls within the common compact set $K$,
And $\,\forall\,\alpha\in \mathbb{N}^n,$
$\sup_{x\in K}|\partial^\alpha \varphi_v|\rightarrow 0,$
It is called $\varphi_v\rightarrow 0(C_c^\infty(\mathbb{R}^n)).$
Get the speed-drop function space $\mathscr{S}(\mathbb{R}^n),$
To satisfy $\lim\limits_{|x|\rightarrow \infty}x^\alpha\partial^p\varphi(x)=0,$
The whole set of smooth functions of $\,\forall\,\alpha,p\in \mathbb{N}^n$. The topology is defined as,
If the sequence $\{\varphi_v\}\in \mathscr{S}(\mathbb{R}^n)$ satisfies $\,\forall\,\alpha,p\in \mathbb{N}^n,$
$\sup_{x\in \mathbb{R}^n}|x^\alpha\partial^p \varphi_v(x)|\rightarrow 0,$
It is called $\varphi_v\rightarrow 0(\mathscr{S}(\mathbb{R}^n)).$
The condition of the rapid descent function can also be equivalently changed to bounded, or $x^\alpha$ can be replaced by $(1+|x|^2)^k,$
In order to facilitate processing. The topology of the rapid descent function space is also similar. Like the first two spaces,
The space of rapid descending functions is also closed with respect to convolutions.
Take $\varphi(x)=\begin{cases}
e^{\frac{1}{|x|^2-1} },&|x|<1\\
0,&|x|\ge 1.
\end{cases},$
$\alpha(x)=\frac{1}{\int_{\mathbb{R}^n} \varphi dx}\varphi(x).$
Order$\beta_R=1_{B_R}\ast \alpha,$
Then $\operatorname{supp}\beta_R\subset B_{R+1},$
And $\beta_R|_{B_{R-1} }\equiv 1.$
Thus $\,\forall\,\varphi\in C^\infty(\mathbb{R}^n),$
Take $\varphi_\nu=\beta_\nu\varphi\in C_c^\infty(\mathbb{R}^n),$
Satisfied$\varphi_\nu\rightarrow \varphi(C^\infty(\mathbb{R}^n)).$
Therefore $C_c^\infty(\mathbb{R}^n)$ is dense in $C^\infty(\mathbb{R}^n)$. Similarly,
It is also dense in $\mathscr{S}(\mathbb{R}^n)$.
For $\mathscr{D}(\mathbb{R}^n)\subset \mathscr{S}(\mathbb{R}^n)\subset \mathscr{E}(\mathbb{R}^n),$
The front space is denser than the back space, and the front topology is stronger than the back topology.
regularization
Call the function $u(x),$ on $\Omega$ Locally integrable,
If it is Lebesgue integrable on any compact set $K\Subset \Omega$.
Take $\alpha_\varepsilon=\frac{1}{\varepsilon^n}\alpha(\frac{x}{\varepsilon})\in C_c^\infty(\mathbb{R}^n),$
Order$u_\varepsilon=u\ast \alpha_\varepsilon.$
If $u$ is locally integrable, then $u_\varepsilon$ is defined by and satisfies the following theorem:
Theorem 1.1. If $u\in L_{loc}^1(\mathbb{R}^n),$ then $u_\varepsilon\in C^\infty(\mathbb{R}^n).$ when $\varepsilon\rightarrow 0$, if $u\in C^0(\mathbb{R}^n),$ then $u_\varepsilon\rightarrow u(C^0(\mathbb{R}^n));$ if $u\in L^p(\mathbb{R}^n),$ then $u_\varepsilon\rightarrow u(L^p(\mathbb{R}^n)).$
The first two properties are easy to prove. For the case of $u\in L^p(\mathbb{R}^n)$, using Lusin’s theorem,
Just find $v\in C_c(\mathbb{R}^n)$ to make $\parallel u-v\parallel_{L^p}$ sufficiently small.
The following lemma is used in the process:
Lemma 1.2 (Minkowski integral inequality). $\parallel\parallel f(x,y)\parallel_{L_x^1}\parallel_{L_y^p}\le \parallel\parallel f(x,y)\parallel_{L_y^p}\parallel_{L_x^1}.$
Certificate:
Just prove (possibly restricted to a subspace) that the linear operator $\mathrm{id}:L_x^1L_y^p\rightarrow L_y^pL_x^1$ is bounded,
And $\parallel\mathrm{id}\parallel\le 1.$ is based on functional analysis knowledge,
The linear operator $A:X\rightarrow Y$ between normed linear spaces is bounded if and only if $A^\ast :Y^\ast \rightarrow X^\ast $ is bounded,
And there is $\parallel A\parallel=\parallel A^\ast \parallel.$ pushing left to right is a classic proposition,
To push right to left, just use $A^{\ast \ast }$, which is an extension of $A$ on $X^{\ast \ast }$.
So consider the linear operator $\mathrm{id}^\ast :L_y^qL_x^\infty\rightarrow L_x^\infty L_y^q,$
Just indicate $\parallel\mathrm{id}^\ast \parallel\le 1$,
That is $\operatorname{ess}\operatorname{sup}_x\parallel f(x,y)\parallel_{L_y^q}\le \parallel\operatorname{ess}\operatorname{sup}_x|f(x,y)|\parallel_{L_y^q},$
However this is obvious.
If $\parallel\parallel f(x,y)\parallel_{L_x^1}\parallel_{L_y^p}=\infty,$
Consider using $g_n=(|f|\wedge n)\cdot 1_{B_n}$ to approximate,
Just use Levi's monotonic convergence lemma to illustrate.
Corollary 1.3. $\parallel f\ast g\parallel_{L^p}\le \parallel f\parallel_{L^p}\parallel g\parallel_{L^1}.$
Certificate:
$\parallel f\ast g\parallel_{L^p} \le \parallel\parallel f(x-y)g(y)\parallel_{L^1_y}\parallel_{L^p_x}\le \parallel\parallel f(x-y)g(y)\parallel_{L^p_x}\parallel_{L^1_y}=\parallel f\parallel_{L^p}\parallel g\parallel_{L^1}.$
From this inference,
$\parallel u_\varepsilon-v_\varepsilon\parallel\le \parallel u-v\parallel$ is also small enough,
And $\parallel v-v_\varepsilon\parallel$ can be sufficiently small by using existing properties,
The theorem is proved.
Called linear operator $J_\varepsilon:L_{loc}^1(\mathbb{R}^n)\rightarrow C^\infty(\mathbb{R}^n),$
$u\mapsto u_\varepsilon$ is Polishing operator,
The process of obtaining $u_\varepsilon$ is called regularization.
Theorem states that $C^\infty(\mathbb{R}^n)$ is in $C^0(\mathbb{R}^n),L^p(\mathbb{R}^n)$ Dense,
Therefore $C_c^\infty(\mathbb{R}^n), \mathscr{S}(\mathbb{R}^n)$ is also dense in it.
Some simple corollaries are,
$J_\varepsilon\in \mathfrak{B}(L^p(\mathbb{R}^n)\rightarrow L^p(\mathbb{R}^n));$
When $u\in \mathscr{D}/\mathscr{S}/\mathscr{E}(\mathbb{R}^n)$,
$u_\varepsilon\rightarrow u(\mathscr{D}/\mathscr{S}/\mathscr{E}(\mathbb{R}^n)).$
A basic application of the polishing operator is the open set $\Omega,$ $\,\forall\,K\Subset \Omega,$
$\,\exists\,\varphi\in C_c^\infty(\mathbb{R}^n),$
$\operatorname{supp}\varphi\subset \Omega,$ $\varphi \in [0,1],$
And $\varphi|_K\equiv 1.$ takes $\varepsilon=d(K,\Omega^c)>0,$
Just take $J_{\frac{\varepsilon}{4} } (1_{d(x,K)\le \frac{\varepsilon}{2} })$.
generalized function
definition
On the scale $\mathscr{D}/\mathscr{S}/\mathscr{E}(\mathbb{R}^n)$ linear continuous functional is $\mathscr{D}'/\mathscr{S}'/\mathscr{E}'(\mathbb{R}^n)$ generalized function.
We have the relation $\mathscr{D}'\supset \mathscr{S}'\supset \mathscr{E}',$
This includes viewing via embedding.
For example, $\mathrm{id}:\mathscr{S}'\rightarrow \mathscr{D}'$ is not only meaningful,
And it's a solo shot.
It makes sense because in $\mathscr{D}\subset \mathscr{S}\subset \mathscr{E}$,
The former is topologically stronger than the latter; injectivity through the front space is guaranteed to be dense in the back space.
Similarly, the $\mathscr{D}',\mathscr{E}'$ generalized function can be defined on the open set $\Omega$,
But the $\mathscr{S}'$ generalized function is only discussed on $\mathbb{R}^n$.
Easy to explain$L_{loc}^1\subset \mathscr{D}',$
And the Dirac function $\delta\in \mathscr{E}'.$
We know that on normed linear spaces, linear functionals are bounded $\Leftrightarrow$ continuous,
But in topological linear space, there is no unified module, so there is no concept of "bounded".
However, similar propositions still hold and can be controlled by "some modules".
Theorem 1.4. If $T\in \mathscr{D}'(\Omega),$ then $\,\forall\,K\Subset \Omega,$ $\,\exists\,C(K)>0,$ $m(K)\in \mathbb{N},$ such that $$|\left<{}T,\varphi\right>|\le C(K)\sup_{x\in \Omega,|\alpha|\le m}|\partial^\alpha \varphi(x)|, \,\forall\,\varphi\in C_c^\infty(K)\subset C_c^\infty(\Omega).$$ On the other hand, if $T$ is a linear functional that satisfies the above conditions, then it is continuous, $T\in \mathscr{D}'(\Omega).$
Certificate:
Definition$|\varphi|_\nu=\sup\limits_{x\in \Omega,|\alpha|\le \nu} |\partial^\alpha \varphi(x)|,$
Meet $|\varphi|_\nu\le |\varphi|_\mu,$ $\,\forall\,\nu\le \mu.$
If $T\in \mathscr{D}'(\Omega)$ does not satisfy propositional properties,
Then $\,\exists\,K\Subset \Omega,$ $\{\varphi_\nu\}\in C_c^\infty(K),$
Make$|\left<{}T,\varphi_\nu \right>|>\nu|\varphi_\nu|_\nu.$
Make a multiplier on $\varphi_\nu$ to make $|\left<{}T,\varphi_\nu\right>|=1,$
Then $|\varphi_\nu|_\nu<\frac{1}{\nu}.$
It is easy to see that $\varphi_\nu\xrightarrow{\mathscr{D} } 0,$
But$|\left<{}T,\varphi_\nu\right>|\equiv 1,$
$\left<{}T,\varphi_\nu\right>\not\rightarrow 0,$ Contradiction.
On the contrary, if $T$ is a linear functional that satisfies the conditions,
$\,\forall\,\varphi_\nu\xrightarrow{\mathscr{D} } 0,$
$\,\exists\,K$ makes $\{\varphi_\nu\}\in C_c^\infty(K).$
$C(K),m(K),$ corresponding to the conditions
Make$|\left<{}T,\varphi_\nu\right>|\le C|\varphi_\nu|_m\rightarrow 0.$
Therefore $\left<{}T,\varphi_\nu\right>\rightarrow 0,$
$T\in \mathscr{D}'(\Omega).$
Theorem 1.5. If $T\in \mathscr{E}'(\Omega),$ then $\,\exists\,K\Subset \Omega,$ $C>0,$ $m\in \mathbb{N},$ such that $$|\left<{}T,\varphi\right>|\le C\sup_{x\in K,|\alpha|\le m}|\partial^\alpha \varphi(x)|, \,\forall\,\varphi\in C^\infty(\Omega).$$ On the other hand, if $T$ is a linear functional that satisfies the above conditions, then it is continuous, $T\in \mathscr{E}'(\Omega).$
Certificate: Take compact set $K_\nu=B_\nu\cap \{x:d(x,\Omega^c)\ge \frac{1}{\nu}\},$
Then $K_\nu$ increases and $\bigcup K_\nu=\Omega.$ $\,\forall\,K\Subset \Omega,$
$\,\exists\,N>0,$ $\,\forall\,\nu>N,$ $K\subset K_\nu.$
Definition$|\varphi|_\nu=\sup\limits_{x\in K_\nu,|\alpha|\le \nu} |\partial^\alpha \varphi(x)|,$
Meet $|\varphi|_\nu\le |\varphi|_\mu,$ $\,\forall\,\nu\le \mu.$
The same theorem as above can prove that $T\in \mathscr{E}'(\Omega)$ satisfies the propositional property.
On the contrary, if $T$ is a linear functional that satisfies the conditions,
$\,\forall\,\varphi_\nu\xrightarrow{\mathscr{E} } 0,$ Take $K,$ which makes the property established
Then $\sup_{x\in K}|\partial^\alpha\varphi(x)|\rightarrow 0,$
$\,\forall\,\alpha\in \mathbb{N}^n.$
Therefore $|\left<{}T,\varphi_\nu\right>|\le C\sup\limits_{x\in K,|\alpha|\le m}|\partial^\alpha\varphi_\nu(x)|\rightarrow 0.$
Therefore $\left<{}T,\varphi_\nu\right>\rightarrow 0,$
$T\in \mathscr{E}'(\Omega).$
It can be seen that $\mathscr{D}',\mathscr{E}'$ satisfies a certain duality relationship:
In $\mathscr{D}$, the function has compact supports, and convergence requires a common compact set.
Therefore, the continuity condition of the linear functional on it is appropriately relaxed,
Compact support functions on different compact sets only need to satisfy a certain boundedness; and in $\mathscr{E}$,
The function has no compact support, and convergence requires that any compact set must meet certain conditions.
This causes the continuity condition of the linear functional on it to become strict, and the boundedness needs to be controlled by a compact set.
And it is required that all smooth functions consistently satisfy boundedness.
This is $\mathscr{D}\subset \mathscr{E}\rightarrow \mathscr{D}'\supset \mathscr{E}'.$
This dual relationship will be stronger in the next section.
support
For a generalized function, the meaning of its value at a specific point is not clear, but we can talk about its value on an open set.
For $T\in \mathscr{D}'(\Omega),$, give an open set $\Omega'\subset \Omega,$
If $\,\forall\,\varphi\in C_c^\infty(\Omega'),$
$\left<{}T,\varphi\right>=0,$ then $T$ is said to take zero value on $\Omega'$.
It is further said that $T_1,T_2$ takes the same value on $\Omega'$, if $T_1-T_2$ takes zero value on $\Omega'$.
Therefore, if a generalized function and a normal function (induced linear continuous functional) have the same value on an open set,
Then it can be said that the generalized function is equal to this normal function on the open set.
If the normal function takes zero at every point, the whole is zero. Similarly, if the generalized function takes zero at a small neighborhood,
Then the whole is zero. To prove, just use partition of unity theorem.
Right $\varphi\in C_c^\infty(\Omega),$
Take the partition of unity $\{\varphi_i\}$ as an open cover $\{O_i\}$ composed of a limited number of small neighbors belonging to the cover $\operatorname{supp}\varphi$.
Then $\varphi_i \varphi\in C_c^\infty(O_i),$
And on the whole $\Omega$ $\sum_i \varphi_i\varphi=\varphi.$
Then the generalized function acting on $\varphi$ can be decomposed into $\{O_i\}$,
Then linear combination can be used to take the value. The sum of zero values is still zero, so the generalized function takes zero value in the whole space.
Due to this property, the generalized function still takes zero value on the union of all open sets that takes zero value.
The sum of all open sets is still an open set, so we can define the largest open set in which the generalized function takes zero value.
Then the complement of the maximum open set is called the generalized function support,
Denote it as $\operatorname{supp}T.$ for the continuous function $f(x),$
It is easy to see that its support as a generalized function is the same as the support originally defined.
Therefore, the definition of generalized function supports is an extension of the definition of continuous function supports.
Theorem 1.6. Any $T\in \mathscr{E}'(\Omega)$ has a compact support; conversely, any set $T\in \mathscr{D}'(\Omega),$ $T$ with a compact support is also in $\mathscr{E}'(\Omega)$.
Certificate: From the theorem of the previous section, $\,\forall\,T\in \mathscr{E}'(\Omega),$
$\,\exists\,K\Subset \Omega,$ $C>0,$ $m\in \mathbb{N},$ such that
$$|\left<{}T,\varphi\right>|\le C\sup\limits_{x\in K,|\alpha|\le m}|\partial^\alpha\varphi(x)|,\,\forall\,\varphi\in C^\infty(\Omega),$$
We show that $K$ contains the support of $T$, so that the support is compact. This is easy,
Because as long as $\operatorname{supp}\varphi\subset \Omega\setminus K,$
The right-hand side of the inequality is zero, thus forcing $T$ to take zero value on $\Omega\setminus K$, and the support is within $K$.
On the contrary, if $T\in \mathscr{D}'(\Omega)$,
We hope to extend its domain from $\mathscr{D}(\Omega)$ to $\mathscr{E}(\Omega),$
To this end, we need to find a way to transform the smooth function into a compact support without changing the value of the functional.
Naturally, I can think of using its support. Note $\operatorname{supp}T=K\Subset \Omega$,
Then take $K\Subset \Omega_1\Subset \Omega,$
Make$\zeta\in C_c^\infty(\overline{\Omega}_1),$
$\zeta\in[0,1]$ and $\zeta|_{\overline{O}_\delta(K)}\equiv 1.$
Right $\varphi\in C^\infty(\Omega),$
Definition$\left<{}T,\varphi\right>=\left<{}T,\zeta\varphi\right>,$
$\zeta\varphi\in C_c^\infty(\overline{\Omega}_1).$ The rationality of the definition needs to be verified:
If there is still $\eta\in C_c^\infty(\overline{\Omega}_2),$
$\eta\in [0,1]$ and $\eta|_{\overline{O}_{\varepsilon}(K)}\equiv 1,$
Then $\zeta-\eta\in C_c^\infty(\Omega \setminus K),$
The support lies outside the support of the generalized function,
Thus $\left<{}T,\zeta\varphi\right>-\left<{}T,\eta\varphi\right>=\left<{}T,(\zeta-\eta)\varphi\right>=0.$
In this way, the functional can act on $\varphi$, and according to the theorem about $\mathscr{D}'$ in the previous section,
$$|\left<{}T,\varphi\right>|\le C|\zeta\varphi|_m\le C'\sup_{x\in \Omega_1,|\alpha|\le m}|\partial^\alpha\varphi|,$$
Then from the theorem about $\mathscr{E}'$,
$T$ is also continuous after being transformed into a linear functional on $\mathscr{E}$,
That is $T\in \mathscr{E}'(\Omega).$
It can be seen that $\mathscr{D}'_c=\mathscr{E}',$
This forms a duality with $\mathscr{D}=\mathscr{E}_c$, which is very pleasing to the eye.
It can be seen that in order to make $\left<{}T,\varphi\right>$ reasonable, when the conditions on one side are weak,
The conditions on the other side must be stronger. The second half of the proof says,
In order to allow $T\in \mathscr{D}'_c$ to act on $\varphi$ in $\mathscr{E}$,
Just consider the effect on the $\varphi|_{\operatorname{supp}T}$ part,
Because the remaining parts fall outside the support set. As far as specific operations are concerned, they cannot be directly restricted.
It needs to be transformed into a compactly supported set function using the method in the proof.
In fact, for the auxiliary function $\zeta,$ (used to transform $\varphi$)
Requirements can be replaced with $\zeta|_K=1.$
This is because $\delta$ of $\overline{O}_\delta(K)$ can be arbitrarily small,
Can get $\zeta_\delta\xrightarrow{\mathscr{D} }\zeta,$
Just use the continuity of linear functionals ($T\in \mathscr{D}'$).
The article was last updated on 2021-10-13 12:46:51
