In distribution theory one often has to replace a function by one with compact support without changing it on a large compact set. This is done by multiplication with a “cutoff function” as constructed in the following
在分布理论中，人们经常不得不用具有紧凑支座的函数替换函数，而不在大型紧凑集合上更改它。这是通过使用 “截止函数” 进行乘法来完成的，如下所示

Theorem 1.4.1. If $X$$X$XX is an open set in ${\mathbb{R}}^n$${\mathbb{R}}^n$R^(n)\mathbb{R}^{n} and $K$$K$KK is a compact subset, then one can find $\varphi \in C_0^{\mathrm{\infty }}(X)$$\varphi \in C_0^{\mathrm{\infty }}(X)$phi inC_(0)^(oo)(X)\phi \in C_{0}^{\infty}(X) with $0\leqq \varphi \leqq 1$$0\leqq \varphi \leqq 1$0 <= phi <= 10 \leqq \phi \leqq 1 so that $\varphi =1$$\varphi =1$phi=1\phi=1 in a neighborhood of $K$$K$KK.
定理 1.4.1.如果 $X$$X$XX 是 中的 ${\mathbb{R}}^n$${\mathbb{R}}^n$R^(n)\mathbb{R}^{n} 开集 并且 $K$$K$KK 是紧子集，则可以在 $\varphi \in C_0^{\mathrm{\infty }}(X)$$\varphi \in C_0^{\mathrm{\infty }}(X)$phi inC_(0)^(oo)(X)\phi \in C_{0}^{\infty}(X) 中找到 ， $0\leqq \varphi \leqq 1$$0\leqq \varphi \leqq 1$0 <= phi <= 10 \leqq \phi \leqq 1 因此 $\varphi =1$$\varphi =1$phi=1\phi=1 在 $K$$K$KK 的邻域中。

Proof. Choose $\epsilon >0$$\epsilon >0$epsi > 0\varepsilon>0 so small that
证明。选择 $\epsilon >0$$\epsilon >0$epsi > 0\varepsilon>0 如此小的

and let $v$$v$vv be the characteristic function of
，设 $v$$v$vv 为 的特征函数

According to Lemma 1.2 .3 we can find a non-negative function $\chi \in C_0^{\mathrm{\infty }}(B)$$\chi \in C_0^{\mathrm{\infty }}(B)$chi inC_(0)^(oo)(B)\chi \in C_{0}^{\infty}(B) where $B$$B$BB is the unit ball, such that $\int \chi dx=1$$\int \chi dx=1$int chi dx=1\int \chi d x=1. Then ${\chi }_{\epsilon }(x)$${\chi }_{\epsilon }(x)$chi_(epsi)(x)\chi_{\varepsilon}(x) $={\epsilon }^{-n}\chi (x/\epsilon )$$={\epsilon }^{-n}\chi (x/\epsilon )$=epsi^(-n)chi(x//epsi)=\varepsilon^{-n} \chi(x / \varepsilon) has support in the ball $\{x;|x|<\epsilon \}$$\{x;|x|<\epsilon \}${x;|x| < epsi}\{x ;|x|<\varepsilon\} and $\int {\chi }_{\epsilon }dx=1$$\int {\chi }_{\epsilon }dx=1$intchi_(epsi)dx=1\int \chi_{\varepsilon} d x=1, so
根据引理 1.2 .3，我们可以找到一个非负函数 $\chi \in C_0^{\mathrm{\infty }}(B)$$\chi \in C_0^{\mathrm{\infty }}(B)$chi inC_(0)^(oo)(B)\chi \in C_{0}^{\infty}(B) ，其中 $B$$B$BB 是单位球，使得 $\int \chi dx=1$$\int \chi dx=1$int chi dx=1\int \chi d x=1 。then ${\chi }_{\epsilon }(x)$${\chi }_{\epsilon }(x)$chi_(epsi)(x)\chi_{\varepsilon}(x) $={\epsilon }^{-n}\chi (x/\epsilon )$$={\epsilon }^{-n}\chi (x/\epsilon )$=epsi^(-n)chi(x//epsi)=\varepsilon^{-n} \chi(x / \varepsilon) 在球 $\{x;|x|<\epsilon \}$$\{x;|x|<\epsilon \}${x;|x| < epsi}\{x ;|x|<\varepsilon\} 和 $\int {\chi }_{\epsilon }dx=1$$\int {\chi }_{\epsilon }dx=1$intchi_(epsi)dx=1\int \chi_{\varepsilon} d x=1 中具有支撑，因此

by Theorem 1.3.1 and (1.3.11), and $1-\varphi =(1-v)\ast {\chi }_{\epsilon }$$1-\varphi =(1-v)\ast {\chi }_{\epsilon }$1-phi=(1-v)**chi_(epsi)1-\phi=(1-v) * \chi_{\varepsilon} vanishes in $K_{\epsilon }$$K_{\epsilon }$K_(epsi)K_{\varepsilon} by (1.3.11). This proves the theorem.
由定理 1.3.1 和 （1.3.11） 完成，并在 （1.3.11） 中 $1-\varphi =(1-v)\ast {\chi }_{\epsilon }$$1-\varphi =(1-v)\ast {\chi }_{\epsilon }$1-phi=(1-v)**chi_(epsi)1-\phi=(1-v) * \chi_{\varepsilon} $K_{\epsilon }$$K_{\epsilon }$K_(epsi)K_{\varepsilon} 消失。这证明了这个定理。

For future reference we also note that
为了将来参考，我们还注意到

Thus  因此

where $C_{\alpha }$$C_{\alpha }$C_(alpha)C_{\alpha} only depends on $\alpha ,n$$\alpha ,n$alpha,n\alpha, n and the choice of the norm. Using Theorem 1.3.5 it is possible to give a still more precise result which we mention for the sake of completeness since it is sometimes useful. However, the reader can jump to Theorem 1.4 .4 without loss of continuity.
其中 $C_{\alpha }$$C_{\alpha }$C_(alpha)C_{\alpha} 仅取决于 $\alpha ,n$$\alpha ,n$alpha,n\alpha, n 范数的选择。使用 Theorem 1.3.5 可以给出更精确的结果，为了完整起见，我们提到它，因为它有时很有用。但是，读者可以跳转到 Theorem 1.4 .4 而不会丢失连续性。

Theorem 1.4.2. Let $X$$X$XX be an open set in the $n$$n$nn dimensional vector space $V$$V$VV with norm $\Vert \Vert$$\Vert \Vert$||||\|\| and let $K$$K$KK be a compact subset. If
定理 1.4.2.设 $X$$X$XX 为 $n$$n$nn 维向量空间中 $V$$V$VV 具有 norm 的开集 $\Vert \Vert$$\Vert \Vert$||||\|\| ，设 $K$$K$KK 为紧子集。如果

and $d_j$$d_j$d_(j)d_{j} is a positive decreasing sequence with $\sum _1^{\mathrm{\infty }}{d}_j<d$$\sum _1^{\mathrm{\infty }} d_j<d$sum_(1)^(oo)d_(j) < d\sum_{1}^{\infty} d_{j}<d, then one can find $\varphi \in C_0^{\mathrm{\infty }}(X)$$\varphi \in C_0^{\mathrm{\infty }}(X)$phi inC_(0)^(oo)(X)\phi \in C_{0}^{\infty}(X) with $0\leqq \varphi \leqq 1$$0\leqq \varphi \leqq 1$0 <= phi <= 10 \leqq \phi \leqq 1, equal to 1 in a neighborhood of $K$$K$KK, so that
和 $d_j$$d_j$d_(j)d_{j} 是 的正递减序列，则 $\sum _1^{\mathrm{\infty }}{d}_j<d$$\sum _1^{\mathrm{\infty }} d_j<d$sum_(1)^(oo)d_(j) < d\sum_{1}^{\infty} d_{j}<d 可以在 $K$$K$KK 的邻域中找到 $\varphi \in C_0^{\mathrm{\infty }}(X)$$\varphi \in C_0^{\mathrm{\infty }}(X)$phi inC_(0)^(oo)(X)\phi \in C_{0}^{\infty}(X) ，等于 $0\leqq \varphi \leqq 1$$0\leqq \varphi \leqq 1$0 <= phi <= 10 \leqq \phi \leqq 1 1，因此

Here $C$$C$CC depends only on the dimension $n$$n$nn.
此处 $C$$C$CC 仅取决于 维度 $n$$n$nn 。
Proof. Assume first that $V={\mathbb{R}}^n$$V={\mathbb{R}}^n$V=R^(n)V=\mathbb{R}^{n} and that $\Vert x\Vert =max\left|x_j\right|$$\Vert x\Vert =max\left|x_j\right|$||x||=max|x_(j)|\|x\|=\max \left|x_{j}\right|. Let $u$$u$uu be the function in Theorem 1.3.5 with $a_j=d_{j+1}$$a_j=d_{j+1}$a_(j)=d_(j+1)a_{j}=d_{j+1} and set $h(t)=u(t+\sum d_j/2)$$h(t)=u\left(t+\sum d_j/2\right)$h(t)=u(t+sumd_(j)//2)h(t)=u\left(t+\sum d_{j} / 2\right). Then we have $|t|\leqq \sum \left|d_j\right|/2$$|t|\leqq \sum \left|d_j\right|/2$|t| <= sum|d_(j)|//2|t| \leqq \sum\left|d_{j}\right| / 2 if $t\in \mathrm{supp}h$$t\in \mathrm{supp}h$t in supp ht \in \operatorname{supp} h, and for every $j$$j$jj
证明。首先假设 that $V={\mathbb{R}}^n$$V={\mathbb{R}}^n$V=R^(n)V=\mathbb{R}^{n} 和 that $\Vert x\Vert =max\left|x_j\right|$$\Vert x\Vert =max\left|x_j\right|$||x||=max|x_(j)|\|x\|=\max \left|x_{j}\right| 。设 $u$$u$uu 为定理 1.3.5 中的函数，其中 $a_j=d_{j+1}$$a_j=d_{j+1}$a_(j)=d_(j+1)a_{j}=d_{j+1} 和 集 $h(t)=u(t+\sum d_j/2)$$h(t)=u\left(t+\sum d_j/2\right)$h(t)=u(t+sumd_(j)//2)h(t)=u\left(t+\sum d_{j} / 2\right) 。然后我们有 $|t|\leqq \sum \left|d_j\right|/2$$|t|\leqq \sum \left|d_j\right|/2$|t| <= sum|d_(j)|//2|t| \leqq \sum\left|d_{j}\right| / 2 if $t\in \mathrm{supp}h$$t\in \mathrm{supp}h$t in supp ht \in \operatorname{supp} h ，对于每个 $j$$j$jj

We can now apply the proof of Theorem 1.4 .1 with $\epsilon =1$$\epsilon =1$epsi=1\varepsilon=1 and
我们现在可以用 $\epsilon =1$$\epsilon =1$epsi=1\varepsilon=1 和 来应用定理 1.4 .1 的证明

taking for $v$$v$vv the characteristic function of
$v$$v$vv 取 为特征函数

It follows that  因此

so introducing the differentials instead we have
因此，引入差分，我们有

which proves Theorem 1.4.2 in this case. The passage to a Euclidean norm and from there to an arbitrary norm can be made by the following lemma which gives (1.4.3) with $C=2n^2$$C=2n^2$C=2n^(2)C=2 n^{2}.
这证明了 Theorem 1.4.2 在这种情况下。到欧几里得范数的段落，以及从那里到任意范数的段落，可以通过以下引理来完成，它给出 （1.4.3） 和 $C=2n^2$$C=2n^2$C=2n^(2)C=2 n^{2} 。

Lemma 1.4.3. If $K$$K$KK is a convex symmetric body in the $n$$n$nn dimensional vector space $V$$V$VV, then one can find an ellipsoid $B$$B$BB with center at 0 and
引理 1.4.3.如果 $K$$K$KK 是 $n$$n$nn 维向量空间 $V$$V$VV 中的凸对称体，则可以找到一个 $B$$B$BB 中心为 0 且

Proof. Let $B$$B$BB be an ellipsoid of maximal volume contained in $K$$K$KK and choose coordinates so that $B$$B$BB is the unit ball. We have to show that $K$$K$KK $\subset B\sqrt{n}$$\subset B\sqrt{n}$sub Bsqrtn\subset B \sqrt{n}. To do so we assume that $K$$K$KK contains a point $(t,0,\dots ,0)$$(t,0,\dots ,0)$(t,0,dots,0)(t, 0, \ldots, 0) with $t>\sqrt{n}$$t>\sqrt{n}$t > sqrtnt>\sqrt{n} and prove that $B$$B$BB cannot be maximal then. The tangent cone to $B$$B$BB with vertex at $(t,0,\dots ,0)$$(t,0,\dots ,0)$(t,0,dots,0)(t, 0, \ldots, 0) touches $B$$B$BB where $x_1=1/t$$x_1=1/t$x_(1)=1//tx_{1}=1 / t, so the part of $B$$B$BB where $\left|x_1\right|>1/t$$\left|x_1\right|>1/t$|x_(1)| > 1//t\left|x_{1}\right|>1 / t is in the interior of $K$$K$KK because of the convexity and symmetry with respect to 0 . Now consider the ellipsoid
证明。设 $B$$B$BB 是一个包含最大体积的椭球 $K$$K$KK 体，并选择坐标， $B$$B$BB 以便作为单位球。我们必须证明 $K$$K$KK $\subset B\sqrt{n}$$\subset B\sqrt{n}$sub Bsqrtn\subset B \sqrt{n} .为此，我们假设 contains $K$$K$KK a point $(t,0,\dots ,0)$$(t,0,\dots ,0)$(t,0,dots,0)(t, 0, \ldots, 0) with $t>\sqrt{n}$$t>\sqrt{n}$t > sqrtnt>\sqrt{n} 并证明 $B$$B$BB that 不能是最大值。与顶点接触 $(t,0,\dots ,0)$$(t,0,\dots ,0)$(t,0,dots,0)(t, 0, \ldots, 0) $B$$B$BB where $x_1=1/t$$x_1=1/t$x_(1)=1//tx_{1}=1 / t 的切圆锥 $B$$B$BB 体 ，因此 where $\left|x_1\right|>1/t$$\left|x_1\right|>1/t$|x_(1)| > 1//t\left|x_{1}\right|>1 / t 的部分 $B$$B$BB 位于 的内部 $K$$K$KK ，因为相对于 0 的凸性和对称性。现在考虑椭球体

which has the same volume as $B$$B$BB. The inequality may be written
，其体积与 $B$$B$BB .不等式可以写成
that is,  那是

Fig. 3  图 3
For small $\epsilon >0$$\epsilon >0$epsi > 0\varepsilon>0 the right-hand side is negative when $\left|x_1\right|<(1/t$$\left|x_1\right|<(1/t$|x_(1)| < (1//t\left|x_{1}\right|<(1 / t $+1/\sqrt{n})/2$$+1/\sqrt{n})/2$+1//sqrtn)//2+1 / \sqrt{n}) / 2 so this part of the ellipsoid is in the interior of $K$$K$KK. But this is also true of the remaining part since $\left|x_1\right|\geqq (1/t+1/\sqrt{n})/2>1/t$$\left|x_1\right|\geqq (1/t+1/\sqrt{n})/2>1/t$|x_(1)| >= (1//t+1//sqrtn)//2 > 1//t\left|x_{1}\right| \geqq(1 / t+1 / \sqrt{n}) / 2>1 / t. (See Fig. 3.) Hence $B$$B$BB is not maximal.
对于 small $\epsilon >0$$\epsilon >0$epsi > 0\varepsilon>0 ，右侧为负值，因此 $\left|x_1\right|<(1/t$$\left|x_1\right|<(1/t$|x_(1)| < (1//t\left|x_{1}\right|<(1 / t $+1/\sqrt{n})/2$$+1/\sqrt{n})/2$+1//sqrtn)//2+1 / \sqrt{n}) / 2 椭球体的这一部分位于 的内部 $K$$K$KK 。但其余部分也是如此，因为 $\left|x_1\right|\geqq (1/t+1/\sqrt{n})/2>1/t$$\left|x_1\right|\geqq (1/t+1/\sqrt{n})/2>1/t$|x_(1)| >= (1//t+1//sqrtn)//2 > 1//t\left|x_{1}\right| \geqq(1 / t+1 / \sqrt{n}) / 2>1 / t 。（见图 3。因此 $B$$B$BB 不是最大值。

An equivalent way of stating Lemma 1.4 .3 is of course that for any norm || in $V$$V$VV one can find a Euclidean norm || || such that
引理 1.4 .3 的等效表示方式当然是，对于任何范数 ||可以 $V$$V$VV 找到欧几里得范数 ||||使得

In order to make conclusions about a distribution from local hypotheses, it is necessary to write arbitrary test functions as sums of test functions of small support.
为了从局部假设中得出有关分布的结论，有必要将任意测试函数编写为小支持的测试函数的总和。
Theorem 1.4.4. Let $X_1,\dots ,X_k$$X_1,\dots ,X_k$X_(1),dots,X_(k)X_{1}, \ldots, X_{k} be open sets in ${\mathbb{R}}^n$${\mathbb{R}}^n$R^(n)\mathbb{R}^{n} and let $\varphi \in C_0^{\mathrm{\infty }}\left(\bigcup _1^k{X}_j\right)$$\varphi \in C_0^{\mathrm{\infty }}\left(\bigcup _1^k X_j\right)$phi inC_(0)^(oo)(uuu_(1)^(k)X_(j))\phi \in C_{0}^{\infty}\left(\bigcup_{1}^{k} X_{j}\right). Then one can find ${\varphi }_j\in C_0^{\mathrm{\infty }}\left(X_j\right),j=1,\dots ,k$${\varphi }_j\in C_0^{\mathrm{\infty }}\left(X_j\right),j=1,\dots ,k$phi_(j)inC_(0)^(oo)(X_(j)),j=1,dots,k\phi_{j} \in C_{0}^{\infty}\left(X_{j}\right), j=1, \ldots, k, such that
定理 1.4.4.设 $X_1,\dots ,X_k$$X_1,\dots ,X_k$X_(1),dots,X_(k)X_{1}, \ldots, X_{k} be open 集合 ${\mathbb{R}}^n$${\mathbb{R}}^n$R^(n)\mathbb{R}^{n} in 并设 $\varphi \in C_0^{\mathrm{\infty }}\left(\bigcup _1^k{X}_j\right)$$\varphi \in C_0^{\mathrm{\infty }}\left(\bigcup _1^k X_j\right)$phi inC_(0)^(oo)(uuu_(1)^(k)X_(j))\phi \in C_{0}^{\infty}\left(\bigcup_{1}^{k} X_{j}\right) 。然后可以找到 ${\varphi }_j\in C_0^{\mathrm{\infty }}\left(X_j\right),j=1,\dots ,k$${\varphi }_j\in C_0^{\mathrm{\infty }}\left(X_j\right),j=1,\dots ,k$phi_(j)inC_(0)^(oo)(X_(j)),j=1,dots,k\phi_{j} \in C_{0}^{\infty}\left(X_{j}\right), j=1, \ldots, k ，这样

If $\varphi \geqq 0$$\varphi \geqq 0$phi >= 0\phi \geqq 0 one can take all ${\varphi }_j\geqq 0$${\varphi }_j\geqq 0$phi_(j) >= 0\phi_{j} \geqq 0.
如果 $\varphi \geqq 0$$\varphi \geqq 0$phi >= 0\phi \geqq 0 一个人可以拿走所有 ${\varphi }_j\geqq 0$${\varphi }_j\geqq 0$phi_(j) >= 0\phi_{j} \geqq 0 .
Proof. We can choose compact sets $K_1,\dots ,K_k$$K_1,\dots ,K_k$K_(1),dots,K_(k)K_{1}, \ldots, K_{k} with $K_j\subset X_j$$K_j\subset X_j$K_(j)subX_(j)K_{j} \subset X_{j} so that $\mathrm{supp}\varphi \subset \bigcup _1^k{K}_j$$\mathrm{supp}\varphi \subset \bigcup _1^k K_j$supp phi subuuu_(1)^(k)K_(j)\operatorname{supp} \phi \subset \bigcup_{1}^{k} K_{j}. (In fact, every point in $\mathrm{supp}\varphi$$\mathrm{supp}\varphi$supp phi\operatorname{supp} \phi has a compact neighborhood contained in some $X_j$$X_j$X_(j)X_{j}. By the Borel-Lebesgue lemma a finite number of such neighborhoods can be chosen which cover all of $\mathrm{supp}\varphi$$\mathrm{supp}\varphi$supp phi\operatorname{supp} \phi. The union of those which belong to $X_j$$X_j$X_(j)X_{j} is a compact set $K_j$$K_j$K_(j)K_{j} $\subset X_j$$\subset X_j$subX_(j)\subset X_{j}.) Using Theorem 1.4.1 we now choose ${\psi }_j\in C_0^{\mathrm{\infty }}\left(X_j\right)$${\psi }_j\in C_0^{\mathrm{\infty }}\left(X_j\right)$psi_(j)inC_(0)^(oo)(X_(j))\psi_{j} \in C_{0}^{\infty}\left(X_{j}\right) with $0\leqq {\psi }_j\leqq 1$$0\leqq {\psi }_j\leqq 1$0 <= psi_(j) <= 10 \leqq \psi_{j} \leqq 1 and ${\psi }_j=1$${\psi }_j=1$psi_(j)=1\psi_{j}=1 in $K_j$$K_j$K_(j)K_{j}. Then the functions
证明。我们可以选择紧凑的集合 $K_1,\dots ,K_k$$K_1,\dots ,K_k$K_(1),dots,K_(k)K_{1}, \ldots, K_{k} ， $K_j\subset X_j$$K_j\subset X_j$K_(j)subX_(j)K_{j} \subset X_{j} 以便 $\mathrm{supp}\varphi \subset \bigcup _1^k{K}_j$$\mathrm{supp}\varphi \subset \bigcup _1^k K_j$supp phi subuuu_(1)^(k)K_(j)\operatorname{supp} \phi \subset \bigcup_{1}^{k} K_{j} 。（实际上，中的每个 $\mathrm{supp}\varphi$$\mathrm{supp}\varphi$supp phi\operatorname{supp} \phi 点都有一个紧凑的邻域包含在 some $X_j$$X_j$X_(j)X_{j} 中。通过 Borel-Lebesgue 引理，可以选择有限数量的此类邻域，这些邻域涵盖所有 $\mathrm{supp}\varphi$$\mathrm{supp}\varphi$supp phi\operatorname{supp} \phi 。属于的那些的并集 $X_j$$X_j$X_(j)X_{j} 是一个紧凑的集合 $K_j$$K_j$K_(j)K_{j} $\subset X_j$$\subset X_j$subX_(j)\subset X_{j} 。使用 Theorem 1.4.1，我们现在选择 ${\psi }_j\in C_0^{\mathrm{\infty }}\left(X_j\right)$${\psi }_j\in C_0^{\mathrm{\infty }}\left(X_j\right)$psi_(j)inC_(0)^(oo)(X_(j))\psi_{j} \in C_{0}^{\infty}\left(X_{j}\right) with $0\leqq {\psi }_j\leqq 1$$0\leqq {\psi }_j\leqq 1$0 <= psi_(j) <= 10 \leqq \psi_{j} \leqq 1 和 ${\psi }_j=1$${\psi }_j=1$psi_(j)=1\psi_{j}=1 in $K_j$$K_j$K_(j)K_{j} 。然后函数