Chapter 7 Sequences and Series of Functions
Exercise 13
(By analambanomenos)
(a) Following the steps of the hint: (i) Show that some subsequence converges at all rational points , say, to .
Let be an enumeration of the rational points of . Since , a compact set, there is a subsequence which converges to , by Theorem 2.41. Similarly, the has a subsequence which converges to . Continuing in this fashion, we get a sequence of subsequences such that each is a subsequence of and converges to . Letting , we get a subsequence of which converges on all rational points of to .
(ii) Define , for any , to be , the sup being taken over all rational points .
First we need to show that this new definition of agrees with the previous one; let be the new definition of . It is clear from the definition that . Suppose , that is, there is a rational point and a real number such that . Since , and , there is an integer such that for all we have , but this contradicts the fact that these functions are monotonically increasing.
It is clear from the definition that is monotonically increasing.
(iii) Show that at every at which is continuous.
Suppose is continuous at and let . Then there are rational numbers such that Since and , there is an integer such that for all we have And since is monotonically increasing, we have for all That is, , so as .
(iv) Show that a subsequence of converges at every point of discontinuity of since there are at most countably many such points.
Since is monotonically increasing, has at most countably many simple points of discontinuity by Theorems 4.29 and 4.30. Let be an enumeration of the points of discontinuity of . Repeating the argument of step (i), there is a subsequence of which converges to a value between 0 and 1. We can continue this way for all the , getting a sequence of subsequences of such that each is a subsequence of for which converges. Letting , we get a subsequence of which converges at all the points . We can redefine at these points to be those values.
Since we have already shown that converges at the points where is continuous, we have converging to for all real values .
(b) Let be a compact subset of and let . It was shown in step (iii) of part (a) that, since is everywhere continuous, for all points . Also, we are using the initial definition of , so that is it monotonically increasing. For each there is a such that for all we have
Also, there is an integer such that for all Since is monotonically increasing, we must have, for all and all , that is, . Hence, for and , Since the open neighborhoods cover the compact set , a finite subset of these neighborhoods, centered at , also covers . Letting , we have for all and all that , that is, converges uniformly to on .
Exercise 14
(By analambanomenos) Since for each we have and , the series defining and converge uniformly since converges, by Theorem 7.10. And since the partial sums are continuous functions, and are continuous functions, by Theorem 7.12. Hence is continuous.
Following the hint, let , and let be the binary expansions of and , where each is 0 or 1. Let By Exercise 3.19, the set of all such is precisely the Cantor set.
We have for
Note that the last term lies between 0 and Also, the first term is an even integer, so that, since , we have If , then the expression in the argument of lies between 0 and , so that . And if , then the expression in the argument of lies between and 1, so that . In either case, we have . Hence .
Exercise 15
(By analambanomenos) The function must be constant on . Let and let . Then there is a such that for all we have if and . For large enough , we have Hence Since was arbitrary, this shows that .
By the way, it is easy to extend this argument to show that if were equicontinuous on , then would be constant on all of .
Exercise 16
(By analambanomenos) Let and let such that if then for all . Let and let be an integer large enough so that if and , then . Then for all and and all such that , we have
Since is compact, there are a finite number of points in such that the neighborhoods of radius centered at the cover . So if we let , then for and and for all , we have . Hence by Theorem 7.8, converges uniformly on .
Exercise 17
(By analambanomenos) The extensions of the definitions, and the statements and proofs of Theorems 7.8 through 7.11, are trivial, so I will simply copy them from the text, italicizing the changes.
Definition 7.7 We say that a sequence of mappings , into a metric space converges uniformly on to a function if for every there is an integer such that implies for all .
Definition 7.22 A family of mappings defined on a set in a metric space with values in a metric space is said to be equicontinuous on if for every there exists a such that whenever , , , and .
Theorem 7.8 The sequence of mappings into a complete metric space , defined on , converges uniformly on if and only if for every there exists an integer such that , , implies .
Proof: Suppose converges uniformly on , and let be the limit function. There there is an integer such that implies so that if , , .
Conversely, suppose the Cauchy condition holds. Since is a complete metric space, the sequence converges, for every , to a limit which we may call . Thus the sequence converges on , to . We have to prove that the convergence is uniform.
Let be given, and choose such that holds. Fix , and let in . Since as , this gives for every and every , which completes the proof.
Theorem 7.9 Suppose {\it is a sequence of mappings from into a metric space such that} for . Put Then uniformly on if and only if as .
Proof: Since this is an immediate consequence of Definition 7.7, we omit the details of the proof.
Theorem 7.10 Suppose is a sequence of vector-valued functions with values in defined on , and suppose Then converges uniformly on if converges.
Proof: If converges, then, for arbitrary , provided and are large enough. Uniform convergence now follows from Theorem 7.8.
Theorem 7.11 Suppose , a sequence of mappings with values in a complete metric space , converges uniformly on a set in a metric space. Let be a limit point
of , and suppose that Then converges and
Proof: Let be given. By the uniform convergence of , there exists such that , , imply Letting in , we obtain for , , so that is a Cauchy sequence and therefore converges, say to .
Next, We first choose such that for all (this is possible by the uniform convergence), and such that Then, for this we choose a neighborhood of such that if , .
Substituting the inequalities , , into , we see that , provided , . This is equivalent to .
Theorem 7.12 If is a sequence of continuous {\it mappings on into a metric space} , and if uniformly on , then is continuous on .
Proof: (In the text, Theorem 7.12 is an immediately corollary of Theorem 7.11. Here, that would require that be complete, which is not necessarily the case. So the following proof is entirely new.)
By Theorem 4.8, to show that is continuous, it suffices to show that is open for any open set . If , we need to find a neighborhood of which is contained in , for then would be the union of the open neighborhoods , , and so be open.
Let be small enough so that the neighborhood of of radius is contained in . Let be large enough so that for all , which is possible by the assumption of uniform convergence. Let such that for all in the neighborhood of of radius , which is possible since is continuous. Then, for all , That is, , or , so that .
The remaining Theorems, for mappings into , are largely corollaries of their scalar counterparts in the text.
Theorem 7.16 Let be monotonically increasing on . Suppose on , for and suppose uniformly on . Then on , and
Theorem 7.17 Suppose is a sequence of vector-valued functions, differentiable on and such that converges for some point on . If converges uniformly on , to a function , and
Proofs: If , then . Hence convergences uniformly if and only if the component functions converge uniformly. And since the integers or derivatives, and integrability or differentiability, of vector-valued functions are defined by component, the vector-valued versions of the Theorems 7.16 and 7.17 are immediate corollaries of the scalar versions in the text.
Theorem 7.24 If is a compact metric space, if for and if converges uniformly on , then is equicontinuous on .
Proof: If , then . Hence is equicontinuous if and only if each of the , are equicontinuous. Hence the vector-valued version of Theorem 7.24 is an immediate corollary of the scalar version in the text.
Theorem 7.25 If is compact, if is pointwise bounded and equicontinuous on , then
(a) is uniformly bounded on ,
(b) contains a uniformly convergent subsequence.
Proof: Since each of the sequences , , is pointwise bounded and equicontinuous on , by the scalar version of Theorem 7.25 in the text each of them is uniformly bounded on , so is uniformly bounded on . Also, contains a uniformly convergent subsequence, . Since the subsequence is also pointwise bounded and equicontinuous on , it also has a uniformly convergent subsequence such that the corresponding subsequence of also converges uniformly. Continuing in this manner, after steps we have a subsequence whose component functions all converge uniformly, so also converges uniformly.
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