Machine-translated from Chinese.
极限
对某基本函数空间的广义函数列$\{T_k\},$
若对其基本函数空间上的任意元素$\varphi,$
有$\left<{}T_k,\varphi\right>\rightarrow 0,$ 则称$T_k$(弱)收敛于$0.$
若$T_k-T$收敛于$0,$ 则称$T_k$的极限是$T,$ 记为$T_k\rightarrow T.$
对常义函数, 我们可以做正则化$J_\varepsilon.$ 下面我们说明,
对广义函数也有类似的正则化.
定理 1.1. 若$T\in \mathscr{D}'(\mathbb{R}_y^n),$ $T_\varepsilon(x):=\left<{}T_y,\alpha_\varepsilon(x-y)\right>\in C^\infty(\mathbb{R}_x^n),$ $\alpha_\varepsilon$如上节所示.
证: 不难说明$x\rightarrow x_0$时,
$\alpha_\varepsilon(x-y)\xrightarrow{\mathscr{D}(\mathbb{R}_y^n)} \alpha_\varepsilon(x_0-y)$(等度连续),
因此$\left<{}T,\alpha_\varepsilon(x-y)\right>\rightarrow\left<{}T,\alpha_\varepsilon(x_0-y)\right>,$
$T_\varepsilon(x)$为连续函数. 类似地,
$\frac{\alpha_\varepsilon(x+\Delta x_k,y)-\alpha_\varepsilon(x,y)}{\Delta x_k}\xrightarrow{\mathscr{D}(\mathbb{R}_y^n)} \partial_k \alpha_\varepsilon(x,y),$
从而有
$$\frac{T_\varepsilon(x+\Delta x_k)-T_\varepsilon(x)}{\Delta x_k}=\left<{}T,\frac{\alpha_\varepsilon(x+\Delta x_k,y)-\alpha_\varepsilon(x,y)}{\Delta x_k}\right>\rightarrow\left<{}T,\partial_k \alpha_\varepsilon(x,y)\right>,$$
因此$T_\varepsilon(x)$关于$x_k$可导. 同理可证它有任意阶导数,
因此是光滑函数.
当$T$为常义函数时, 泛函即为积分,
因此对广义函数的正则化是对常义函数正则化的推广.
当然有了下面的定理才让正则化有意义:
定理 1.2. $T\in \mathscr{D}'(\mathbb{R}^n)$时, $T_\varepsilon\xrightarrow{\mathscr{D}'} T.$
证:
即证$\left<{}T_\varepsilon,\varphi\right>=\int_{\mathbb{R}^n} T_\varepsilon\varphi dx\rightarrow \left<{}T,\varphi\right>,$
$\,\forall\,\varphi\in \mathscr{D}(\mathbb{R}^n).$
由于$\varphi$具紧支集, 可以用Riemann积分的方式来逼近式中的积分.
然而将积分写成求和形式后,
$\alpha_\varepsilon(x-y)$可以与$\varphi(x)$先行结合, 即
$$\sum_i\alpha_\varepsilon(x_i-y)\varphi(x_i)\Delta x_i\xrightarrow{\mathscr{E} }\int_{\mathbb{R}_x^n}\alpha_\varepsilon(x-y)\varphi(x)dx,$$
从而$\left<{}T_\varepsilon,\varphi\right>=\left<{}T,\varphi_\varepsilon\right>\rightarrow \left<{}T,\varphi\right>,$
命题得证.
定理 1.3. $T\in \mathscr{E}'(\mathbb{R}^n)$时, $T_\varepsilon\xrightarrow{\mathscr{E}'} T.$
证:
即证$\left<{}T_\varepsilon,\varphi\right>=\int_{\mathbb{R}^n} T_\varepsilon\varphi dx\rightarrow \left<{}T,\varphi\right>,$
$\,\forall\,\varphi\in \mathscr{E}(\mathbb{R}^n).$ 由于$T$具紧支集,
不难证明$T_\varepsilon$也具紧支集, 且充分靠近$\operatorname{supp}T.$
取紧支撑的$\zeta$在$\operatorname{supp}T$某邻域上恒为$1,$ 由前面的定理,
$\left<{}T_\varepsilon,\zeta\varphi\right>\rightarrow \left<{}T,\zeta\varphi\right>.$
由于在$\operatorname{supp}T$邻域上$\zeta\varphi=\varphi,$
且对充分小的$\varepsilon,$ 该邻域包含了$T,T_\varepsilon$的支集, 因此
$$\left<{}T_\varepsilon,\zeta\varphi\right>=\left<{}T_\varepsilon,\varphi\right>\rightarrow \left<{}T,\varphi\right>=\left<{}T,\zeta\varphi\right>,$$
命题得证.
导数
设$T\in\mathscr{D}'(\mathbb{R}^n),$
定义$T$关于$x_k$的(广义)导数$\frac{\partial {}T}{\partial {}x_k}\in \mathscr{D}'(\mathbb{R}^n),$
满足
$$\left<{}\frac{\partial {}T}{\partial {}x_k},\varphi\right>=-\left<{}T,\frac{\partial {}\varphi}{\partial {}x_k}\right>.$$
由分部积分公式, 当$T$为$C^1$函数时, 广义导数即为一般意义下的导数.
因此广义导数的概念是通常导数概念的推广. 特别地, 对于某些常义函数,
它无法按照通常概念求导. 但将其视为广义函数,
就可以得到广义函数空间中的广义导数, 这往往是十分有用的.
类似地定义高阶导数
$$\left<{}\partial^\alpha T,\varphi\right>=(-1)^{|\alpha|}\left<{}T,\partial^\alpha \varphi\right>.$$
可见广义函数任意阶导数存在, 且混合偏导与顺序无关.
一个有意思的例子是, 若$\operatorname{supp}T=\{0\},$
则$T$只能是$\delta$及其导数的有限线性组合.
证明通过对$\varphi$进行Taylor展开给出即可.
乘子
对$T\in \mathscr{D}'(\mathbb{R}^n),$ $\alpha\in C^\infty(\mathbb{R}^n),$
定义$\alpha T\in \mathscr{D}'(\mathbb{R}^n)$为
$$\left<{}\alpha T,\varphi\right>=\left<{}T,\alpha\varphi\right>.$$
称$\alpha$为$\mathscr{D}'(\mathbb{R}^n)$的乘子.
类似地它也是$\mathscr{E}'(\mathbb{R}^n)$的乘子,
但一般不见得是$\mathscr{S}(\mathbb{R}^n)$的乘子.
类似于常义函数, 我们有Leibniz公式:
$$\partial^r(\alpha T)=\sum_{p+q=r} \frac{r!}{p!q!}(\partial^p \alpha \cdot \partial^q T).$$
自变量变换
仿照$C_c^\infty$函数的自变量变换, 对$T\in \mathscr{D}'(\Omega_x),$
给光滑同胚$\psi:\Omega_x\rightarrow \Omega_y,$
定义$S\in \mathscr{D}'(O_y):$
$$\left<{}S,\varphi\right>=\left<{}T,\det|\psi'|\varphi\circ\psi\right>.$$
当$T$为常义函数时, $S(y)=T(\psi^{-1}(y)).$ 该定义自然也是通常定义的推广.
卷积
对$f,g\in C_c^\infty(\mathbb{R}^n),$ 考虑$f\ast g$在$\varphi$上的作用,
类比地定义广义积分间的卷积. 设$S,T$为广义函数, 定义卷积$S\ast T:$
$$\left<{}S\ast T,\varphi\right>=\left<{}S_x,\left<{}T_y,\varphi(x+y)\right>\right>,$$
$\,\forall\,\varphi\in \mathscr{D}(\mathbb{R}^n).$ 和常义函数一样,
广义函数的卷积也不总是有意义的.
当$S,T\in \mathscr{D}'(\mathbb{R}^n)$中至少一者在$\mathscr{E}'(\mathbb{R}^n)$中时,
卷积有意义, $S\ast T\in \mathscr{D}'(\mathbb{R}^n).$
证明过程分为两部分, 首先只有当$T\in \mathscr{D}'/\mathscr{E}'$时,
$\psi:=\left<{}T_y,\varphi(x+y)\right>\in \mathscr{E}/\mathscr{D},$
才能使定义有意义; 其次当$\varphi_\nu\xrightarrow{\mathscr{D} } 0$时,
有$\psi_\nu\xrightarrow{\mathscr{E}/\mathscr{D} }0,$
因此线性泛函是连续的. 证明方法与先前类似, 在此略过.
可以证明,
对$R,S,T\in \mathscr{D}'(\mathbb{R}^n)$且其中至少两个具紧支集时,
广义函数的卷积(同常义卷积一样)满足如下性质:
$\operatorname{supp}(S\ast T)\subset \operatorname{supp}(S)+\operatorname{supp}(T);$
$(R\ast S)\ast T=R\ast (S\ast T);$
$S\ast T=T\ast S;$
$\delta\ast T=T\ast \delta=T;$
$\alpha_1+\alpha_2=\alpha$时,
$\partial^\alpha(S\ast T)=(\partial^{\alpha_1})\ast (\partial^{\alpha_2});$
$T\in \mathscr{D}'/\mathscr{E}',$
$\varphi\in \mathscr{D}/\mathscr{E}$时,
$T\ast \varphi=\left<{}T_y,\varphi(x-y)\right>\in C^\infty(\mathbb{R}_x^n).$
前四个性质说明$\mathscr{E}'(\mathbb{R}^n)$关于通常的线性运算与卷积运算构成有单位元$\delta$的交换代数,
称为卷积代数. 对函数的正则化有$f_\varepsilon=f\ast \alpha_\varepsilon,$
类似地对广义函数的正则化也有$T_\varepsilon=T\ast \alpha_\varepsilon.$
文章最后更新于 2021-10-18 15:02:37
limit
Generalized function sequence $\{T_k\},$ for a certain basic function space
If for any element on its basic function space $\varphi,$
If there is $\left<{}T_k,\varphi\right>\rightarrow 0,$, it is said that $T_k$ (weakly) converges to $0.$
If $T_k-T$ converges to $0,$, then the limit of $T_k$ is said to be $T,$ and is recorded as $T_k\rightarrow T.$
For normal functions, we can do regularization $J_\varepsilon.$. Below we explain,
There are similar regularizations for generalized functions.
Theorem 1.1. If $T\in \mathscr{D}'(\mathbb{R}_y^n),$ $T_\varepsilon(x):=\left<{}T_y,\alpha_\varepsilon(x-y)\right>\in C^\infty(\mathbb{R}_x^n),$ $\alpha_\varepsilon$ is as shown in the previous section.
Certificate: It is not difficult to explain that when $x\rightarrow x_0$,
$\alpha_\varepsilon(x-y)\xrightarrow{\mathscr{D}(\mathbb{R}_y^n)} \alpha_\varepsilon(x_0-y)$(isocratic continuous),
Therefore $\left<{}T,\alpha_\varepsilon(x-y)\right>\rightarrow\left<{}T,\alpha_\varepsilon(x_0-y)\right>,$
$T_\varepsilon(x)$ is a continuous function. Similarly,
$\frac{\alpha_\varepsilon(x+\Delta x_k,y)-\alpha_\varepsilon(x,y)}{\Delta x_k}\xrightarrow{\mathscr{D}(\mathbb{R}_y^n)} \partial_k \alpha_\varepsilon(x,y),$
Thereby there is
$$\frac{T_\varepsilon(x+\Delta x_k)-T_\varepsilon(x)}{\Delta x_k}=\left<{}T,\frac{\alpha_\varepsilon(x+\Delta x_k,y)-\alpha_\varepsilon(x,y)}{\Delta x_k}\right>\rightarrow\left<{}T,\partial_k \alpha_\varepsilon(x,y)\right>,$$
Therefore $T_\varepsilon(x)$ is differentiable with respect to $x_k$. In the same way, it can be proved that it has derivatives of any order,
Therefore it is a smooth function.
When $T$ is a normal function, the functional is the integral,
Therefore, regularization of generalized functions is a generalization of regularization of normal functions.
Of course, the following theorem makes regularization meaningful:
Theorem 1.2. When $T\in \mathscr{D}'(\mathbb{R}^n)$, $T_\varepsilon\xrightarrow{\mathscr{D}'} T.$
Certificate:
That’s proof $\left<{}T_\varepsilon,\varphi\right>=\int_{\mathbb{R}^n} T_\varepsilon\varphi dx\rightarrow \left<{}T,\varphi\right>,$
$\,\forall\,\varphi\in \mathscr{D}(\mathbb{R}^n).$
Since $\varphi$ has compact support, the Riemann integral can be used to approximate the integral in the formula.
However, after writing the integral in summation form,
$\alpha_\varepsilon(x-y)$ can be combined with $\varphi(x)$ first, that is
$$\sum_i\alpha_\varepsilon(x_i-y)\varphi(x_i)\Delta x_i\xrightarrow{\mathscr{E} }\int_{\mathbb{R}_x^n}\alpha_\varepsilon(x-y)\varphi(x)dx,$$
Thus $\left<{}T_\varepsilon,\varphi\right>=\left<{}T,\varphi_\varepsilon\right>\rightarrow \left<{}T,\varphi\right>,$
The proposition is proved.
Theorem 1.3. When $T\in \mathscr{E}'(\mathbb{R}^n)$, $T_\varepsilon\xrightarrow{\mathscr{E}'} T.$
Certificate:
That’s proof$\left<{}T_\varepsilon,\varphi\right>=\int_{\mathbb{R}^n} T_\varepsilon\varphi dx\rightarrow \left<{}T,\varphi\right>,$
$\,\forall\,\varphi\in \mathscr{E}(\mathbb{R}^n).$ Since $T$ has a compact support,
It is not difficult to prove that $T_\varepsilon$ also has a compact support and is sufficiently close to $\operatorname{supp}T.$
Taking the compact support $\zeta$ in a certain neighborhood of $\operatorname{supp}T$ is always $1,$. According to the previous theorem,
$\left<{}T_\varepsilon,\zeta\varphi\right>\rightarrow \left<{}T,\zeta\varphi\right>.$
Since $\zeta\varphi=\varphi,$ is on $\operatorname{supp}T$ neighborhood
And for sufficiently small $\varepsilon,$, the neighborhood contains the support of $T,T_\varepsilon$, so
$$\left<{}T_\varepsilon,\zeta\varphi\right>=\left<{}T_\varepsilon,\varphi\right>\rightarrow \left<{}T,\varphi\right>=\left<{}T,\zeta\varphi\right>,$$
The proposition is proved.
Derivative
Let $T\in\mathscr{D}'(\mathbb{R}^n),$
Definition $T$ about $x_k$ (generalized) derivative $\frac{\partial {}T}{\partial {}x_k}\in \mathscr{D}'(\mathbb{R}^n),$
satisfy
$$\left<{}\frac{\partial {}T}{\partial {}x_k},\varphi\right>=-\left<{}T,\frac{\partial {}\varphi}{\partial {}x_k}\right>.$$
According to the integral formula by parts, when $T$ is the $C^1$ function, the generalized derivative is the derivative in the general sense.
Therefore, the concept of generalized derivative is an extension of the concept of ordinary derivative. In particular, for some normal functions,
It cannot be differentiated according to the usual concepts. But consider it as a generalized function,
You can get the generalized derivatives in the generalized function space, which is often very useful.
Define higher order derivatives similarly
$$\left<{}\partial^\alpha T,\varphi\right>=(-1)^{|\alpha|}\left<{}T,\partial^\alpha \varphi\right>.$$
It can be seen that derivatives of any order of generalized functions exist, and mixed partial derivatives have nothing to do with order.
An interesting example is if $\operatorname{supp}T=\{0\},$
Then $T$ can only be a finite linear combination of $\delta$ and its derivatives.
The proof is given by performing Taylor expansion on $\varphi$.
multiplier
pair $T\in \mathscr{D}'(\mathbb{R}^n),$ $\alpha\in C^\infty(\mathbb{R}^n),$
Define $\alpha T\in \mathscr{D}'(\mathbb{R}^n)$ as
$$\left<{}\alpha T,\varphi\right>=\left<{}T,\alpha\varphi\right>.$$
Call $\alpha$ $\mathscr{D}'(\mathbb{R}^n)$ multiplier.
Similarly it is also a multiplier of $\mathscr{E}'(\mathbb{R}^n)$,
But it is generally not necessarily a multiplier of $\mathscr{S}(\mathbb{R}^n)$.
Similar to normal functions, we have the Leibniz formula:
$$\partial^r(\alpha T)=\sum_{p+q=r} \frac{r!}{p!q!}(\partial^p \alpha \cdot \partial^q T).$$
independent variable transformation
Modeling the independent variable transformation of $C_c^\infty$ function, for $T\in \mathscr{D}'(\Omega_x),$
Give smooth homeomorphism $\psi:\Omega_x\rightarrow \Omega_y,$
Definition$S\in \mathscr{D}'(O_y):$
$$\left<{}S,\varphi\right>=\left<{}T,\det|\psi'|\varphi\circ\psi\right>.$$
When $T$ is a normal function, the definition $S(y)=T(\psi^{-1}(y)).$ is naturally a generalization of the usual definition.
Convolution
For $f,g\in C_c^\infty(\mathbb{R}^n),$, consider the role of $f\ast g$ on $\varphi$,
Analogously define the convolution between generalized integrals. Let $S,T$ be a generalized function, define Convolution $S\ast T:$
$$\left<{}S\ast T,\varphi\right>=\left<{}S_x,\left<{}T_y,\varphi(x+y)\right>\right>,$$
$\,\forall\,\varphi\in \mathscr{D}(\mathbb{R}^n).$ is the same as the normal function,
Convolution of generalized functions doesn't always make sense either.
When at least one of $S,T\in \mathscr{D}'(\mathbb{R}^n)$ is in $\mathscr{E}'(\mathbb{R}^n)$,
Convolution makes sense, $S\ast T\in \mathscr{D}'(\mathbb{R}^n).$
The proof process is divided into two parts. First, only when $T\in \mathscr{D}'/\mathscr{E}'$,
$\psi:=\left<{}T_y,\varphi(x+y)\right>\in \mathscr{E}/\mathscr{D},$
Only then can the definition be meaningful; secondly, when $\varphi_\nu\xrightarrow{\mathscr{D} } 0$,
Yes $\psi_\nu\xrightarrow{\mathscr{E}/\mathscr{D} }0,$
Therefore, the linear functional is continuous. The proof method is similar to the previous one and is omitted here.
It can be proved that
For $R,S,T\in \mathscr{D}'(\mathbb{R}^n)$ and at least two of them have compact supports,
The convolution of generalized functions (same as normal convolution) satisfies the following properties:
$\operatorname{supp}(S\ast T)\subset \operatorname{supp}(S)+\operatorname{supp}(T);$
$(R\ast S)\ast T=R\ast (S\ast T);$
$S\ast T=T\ast S;$
$\delta\ast T=T\ast \delta=T;$
When $\alpha_1+\alpha_2=\alpha$,
$\partial^\alpha(S\ast T)=(\partial^{\alpha_1})\ast (\partial^{\alpha_2});$
$T\in \mathscr{D}'/\mathscr{E}',$
When $\varphi\in \mathscr{D}/\mathscr{E}$,
$T\ast \varphi=\left<{}T_y,\varphi(x-y)\right>\in C^\infty(\mathbb{R}_x^n).$
The first four property descriptions $\mathscr{E}'(\mathbb{R}^n)$ are about ordinary linear operations and convolution operations forming a commutative algebra with identity elements $\delta$,
called convolution algebra. The regularization of the function is $f_\varepsilon=f\ast \alpha_\varepsilon,$
Similarly, the regularization of generalized functions also has $T_\varepsilon=T\ast \alpha_\varepsilon.$
The article was last updated on 2021-10-18 15:02:37
