第一段原文
8.4. FAMILIES OF FUNCTIONS
The following theorem gives us a characterization for local uniform boundedness when the metric space is a subset ofor.
Theorem 8.4.12 A family of complex valued functionson a subsetofis locally uniformly bounded inif and only ifis uniformly bounded on every compact subset of .
Proof. () This is an immediate consequence of the observation that the closure of a neighborhood inoris compact.
() Supposeis locally uniformly bounded on a domainandandis a compact subset of. Then, for eachthere exists a neighborhood of , and a positive real number, , such that
，
Sincecovers, we know that there exists a finite subcover, say . Then, for,,for all , and we conclude thatis uniformly bounded on.
Remark 8.4.13 Note that Theorem 8.4.12 made specific use of the Heine-Borel Theorem; i.e., the fact that we were in a space where compactness is equivalent to being closed and bounded.
Remark 8.4.14 In our text, an example is given to illustrate that a uniformly bounded
sequence of real-valued continuous functions on a compact metric space need not yield a subsequence that converges (even) pointwise on the metric space. Because
the veriiquest;cation of the claim appeals to a theorem given in Chapter 11 of the text, at
this point we accept the example as a reminder to be cautious.
Remark 8.4.15 Again by way of example, the author of our text illustrates that it
is not the case that every convergent sequence of functions contains a uniformly
convergent subsequence. We offer it as our next excursion, providing space for you
to justify the claims.
Excursion 8.4.16 Let and
.
- Show thatis uniformly bounded in.
- Find the pointwise limit offor.
- Justify that no subsequence ofcan converge uniformly on .
***For (a), observing thatforandfor each yields thatfor. In (b), since the only occurrence ofis in the denominator of each, for each fixed, the corresponding sequence of real goes to 0 as. For (c), in view of the negation of the definition of uniform convergence of a sequence, the behavior of the sequenceat the points allows us to conclude that no subsequences ofwill converge uniformly on.***
Now we know that we donrsquo;t have a “straight” analog for the Bolzano-Weierstrass Theorem when we are in the realm of families of functions in . This poses the challenge of finding an additional property (or set of properties) that will yield such an analog. Towards that end, we introduce define a property that requires “local and global” uniform behavior over a family.
Definition 8.4.17 A familyof complexvalued functions defined on a metric space
is equicontinuous onif and only if
.
Remark 8.4.18 If is equicontinuous on , then eachis clearly uniformly
continuous in.
Excursion 8.4.19 On the other hand, for, show that each function in is uniformly continuous on thoughis not equicontinuous on .
Excursion 8.4.20 Use the Mean-Value Theorem to justify that
is equicontinuous in
The next result is particularly useful when we can designate a denumerable subset of the domains on which our functions are defined. When the domain is an open connected subset of or, then the rationals or points with the real and imaginary parts as rational work very nicely. In each case the denumerable subset is dense in the set under consideration.
Lemma 8.4.21 Ifis a pointwise bounded sequence of complex-valued functions
on a denumerable set, thenhas a subsequencethat converges
pointwise on.
Excursion 8.4.22 Finish the following proof.
Proof. Letbe sequence of complexvalued functions that is pointwise bounded on a denumerable set. Then the setcan be realized as a sequenceof distinct points. This is a natural setting for application of the Cantor diagonalization process that we saw earlier in the proof of the denumerability of the rationals. From the Bolzano–Weierstrass Theorem,bounded implies that there exists a convergent subsequence. The process can be applied to to obtain a subsequence that is convergent.
In general, is such that is convergent and is a
subsequence of each of for. Now consider
*** For, there exists an such that. Then is a
subsequence of from which it follows that is convergent
at. ***
The next result tells us that if we restrict ourselves to domainsthat are compact metric spaces that any uniformly convergent sequence inis also an equicontinuous family.
Theorem 8.4.23 Suppose that is a compact metric space and the sequence
of functions is such that. If converges
uniformly on, then is equicontinuous on.
Proof. Suppose that is a compact metric space, the sequence of functions
converges uniformly onandis given. By Theorem 8.2.3,
is uniforml
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附录X 译文
函数的族
下面的定理给出了当度量空间是R或C的子集时，局部一致有界性的一个刻画。
定理8.4.12一个子集上的复值函数族是局部一致有界的，当且仅当它在每个致密子集上一致有界。
这是观察到在复数域和实数域邻域闭包是紧的直接结果()假设域上局部一致有界且是其中的致密子集。那么对于每一个存在一个邻域，并且存在一个正实数，这样由于包含，我们知道存在一个有限子复盖，，那么对于，，对于所有的，我们证明了在上是一致有界的
注意到定理8.4 . 12具体利用了Heine - Borel定理；也就是说，我们处在一个紧致性等价于封闭和有界的空间中.
在我们的文中，举例说明在紧度量空间上实值连续函数的一致有界序列不必产生在度量空间上逐点收敛(甚至)的子序列.
因为要求上的检定适用于案文第11章中给出的定理.再举例说明，我们文章的作者说明，并非函数的每一个收敛序列都包含一个一致收敛子序列的情况。
我们提供它作为我们的下一段，为您提供空间来证明观点。
证明在上是一致有界的
找到对于上的的逐点极限
证明的任何子序列都不可能在上一致收敛。
***对于（a），观察到有对于，且有对于每一个域（，）
在（b）中，由于n的唯一出现是在每个的分母中，对于每个固定，对应的实数列当n到正无穷趋于0
对于（c），鉴于对一个序列一致收敛定义的否定，该序列的表现在点
允许我们得出结论，n的子序列不会一致收敛在***
现在我们知道，当我们处在函数族的范畴内时，对于博尔扎诺 - Weierstrass定理，我们并没有一个'直'的类比。这就提出了寻找一个额外的属性( 或属性集合 )来产生这样的类似物的挑战。为此，我们引入定义一个属性，该属性要求对一个家庭进行lsquo;局部和全局rsquo;的统一行为.
定义8.4 . 17度量空间上定义的复值函数族是等连续的当且仅当
.
注8.4.18如果在上是等连续的，则每个都是明显一致连续的.
定理8.4.19另一方面，对于，表明每个函数在上是一致连续的，尽管在上不是等连续的.
Excursion8.4.20利用中值定理证明了这一点
是等度连续的在
下一个结果特别有用，当我们可以指定我们定义函数的域的一个可数子集。当域是或的开连通子集时，以实部和虚部为有理的有理或点非常好地工作。在每一种情况下，可数子集在所考虑的集合中是稠密的.
引理8.4.21如果是可数集合上复值函数的点有界序列，则有一个子序列点收敛。
执行部分8.4 . 22完成以下证明.
证明。设是复值函数序列在可数集上点有界.然后该集合可以实现为一系列不同的点.
这是一个应用Cantor对角化过程的自然设置，我们在对有理函数可枚举性的证明中较早看到了这一点。从博尔扎诺 - Weierstrass定理，有界意味着存在收敛子序列。该过程可应用于，以获得收敛的子序列.
一般来说，使得收敛，且是其中的每个子序列。现在考虑
*** 对于,存在一个 使得t.那么 是
的子序列。由此可知 收敛于. ***
下一个结果告诉我们，如果我们把自己限制在紧度量空间的域上，那么中的任意一致收敛序列也是等连续族.
定理8.4.23假设是紧度量空间，函数序列是这样的。如果一致收敛，则在上是等连续的.
证明。假设是紧度量空间，给出了函数序列一致收敛于且满足。由定理8.2 . 3，是一致收敛于上的.因此，存在一个正整数使得，这意味着。尤其是，对于
因为每个是紧集上连续的，从一致连续性定理来看，对于每个，都是一致连续的在上。因此，对于每一个
，存在这样一个使得，并暗示着
让 . 那么
. (8.7)
对于和使得t，我们也知道.( 8.8 )
由( 8.7 )和( 8.8 )以及任意的事实，我们得出结论。
i在上等度连续.
我们现在已经准备好为其中的一个子族提供条件，这将使我们类似于博尔扎诺 - Weierstrass定理.
定理8.4.24假设是紧度量空间，函数序列是这样的。如果是点有界且等连续的在K上，那么
在K上一致有界
（b）包含一个一致收敛的子序列在上
8.4.25将缺失的内容填入，以便完成定理8.4.24的下列证明.
证明。假设是一个紧度量空间，函数序列使得，的族是点有界的，并且是等连续的。部分( a )的证明：
部分( a )的证明：
给定，因为在上等度连续，那么存在使得.(8.9)
我们知道因为得到一个且是致密的，那么存在有限个点，为，使得。另一方面，是逐点有界的；因此，对于每个，这里存在一个正实数使得
对于，由此可见
. (8.10)
设，因为，存在一个使得.
因此且，从(8.9),，我们得到对于所有。
但当由此产生对于。从( 8.10 )开始，我们得出结论对所有。既然是任意性的，就这样
is
大部分( b )部分的证明：
如果是有限的，我们将完成。对于无穷大，设其中稠密的一个可数子集。( 这部分证明标题中的lsquo;差不多rsquo;的原因是我们没有做第 45页的练习 # 25作作业。如果or,因此 ,有理函数的密度立即导致满足期望性质的集合 ;在任意度量空间的一般情况下 ,第 45页的演习 # 25指示了如何使用具有合理半径的开复盖来获得这样的集合。 )因为在E上，由引理8.4.21，存在一个的子序列，即对每个，都收敛.
假设是给定的。既然在上是等连续的，那么存在这样一个
使得.
因为在上时稠密的，形成了一个开放的覆盖对。由于由于是致密的，我们得出结论，在上存在有限个元素，为使得
. (8.11)
因为，且是一个复数收敛序列对于每个，由于的完备性，对于每个，柯西收敛，因此，存在一个正正数使得，有推得
设从( 8.11 )式中，存在这样一个使得。然后说明
对于每一个，使，因此，对于和
因为，是任意的，可得;i.e.,
是.根据定理8.4.23，需要在在上一致收敛.
***可接受的响应是：( 1 )开复盖，( 2 )，( 3 )，( 4 )，( 5 )，( 6 )所有，( 7 )一致有界，( 8 )点有界，( 9 )，( 10 )，( 11 )一致Cauchy有界。***
由于我们现在知道，对于函数族，并不是函数的每一个收敛序列都包含一个一致收敛子序列的情况，确实具有该性质的族需要一个特殊的标签.
定义8.4 . 26定义在度量空间上的复值函数族称为正规，只有当每个序列都有一个子序列一致收敛于紧子集上.
注8.4.27根据定理8.4.24，紧度量空间上点有界且等连续的任何族在上.
我们的最后一个定义考虑了来自族的序列的极限在族中的情况.
定义8.4.28复值函数的正规族据说是致密的，当且仅当收敛于的所有序列的一致极限也是其中的成员.
附录Y 外文原文
8.4. FAMILIES OF FUNCTIONS
The following theorem gives us a characterization for local uniform boundedness when the metric space is a subset ofor.
Theorem 8.4.12 A family of complex valued functionson a subsetofis locally uniformly bounded inif and only ifis uniformly bounded on every compact subset of .
Proof. () This is an immediate consequence of the observation that the closure of a neighborhood inoris compact.
() Supposeis locally uniformly bounded on a domainandandis a compact subset of. Then, for eachthere exists a neighborhood of , and a positive real number, , such that
，
Sincecovers, we know that there exists a finite subcover, say . Then, for,,for all , and we conclude thatis uniformly bounded on.
Remark 8.4.13 Note that Theorem 8.4.12 made specific use of the Heine-Borel Theorem; i.e., the fact that we were in a space where compactness is equivalent to being closed and bounded.
Remark 8.4.14 In our text, an example is given to illustrate that a uniformly bounded
sequence of real-valued continuous functions on a compact metric space need not yield a subsequence that converges (even) pointwise on the metric space. Because
the veriiquest;cation of the claim appeals to a theorem given in Chapter 11 of the text, at
this point we accept the example as a reminder to be cautious.
Remark 8.4.15 Again by way of example, the author of our text illustrates that it
is not the case that every convergent sequence of functions contains a uniformly
convergent subsequence. We offer it as our next excursion, providing space for you
to justify the claims.
Excursion 8.4.16 Let and
.
- Show thatis uniformly bounded in.
- Find the pointwise limit offor.
- Justify that no subsequence ofcan converge uniformly on .
***For (a), observing thatforandfor each yields thatfor. In (b), since the only occurrence ofis in the denominator of each, for each fixed, the corresponding sequence of real goes to 0 as. For (c), in view of the negation of the definition of uniform convergence of a sequence, the behavior of the sequenceat the points allows us to conclude that no subsequences ofwill converge uniformly on.***
Now we know that we donrsquo;t have a “straight” analog for the Bolzano-Weierstrass Theorem when we are in the realm of families of
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