8.1.1 Sequences of Complex-Valued Functions on Metric Spaces

Chapter 8

Sequences and Series of Functions

Given a set A, a sequence of elements of A is a function F : M A rather than using the notation Fnfor the elements that have been selected from A, since the domain is always the natural numbers, we use the notational convention a_n Fn and denote sequences in any of the following forms:

a_n^*_n1 a_nn+M or a₁a₂a₃a₄ .

Given any sequenceck^*_k1of elements of a set A, we have an associated sequence of nth partial sums

sn^*_n1 where sn

;n k1

ck

the symbol 3_*

k1c_k is called a series (or in¿nite series). Because the function gx x 1 is a one-to-one correspondence fromM intoMC 0, i.e., g : M ¹¹ MC 0, a sequence could have been de¿ned as a function on MC 0. In our dis- cussion of series, the symbolic descriptions of the sequences of nth partial sums usually will be generated from a sequence for which the ¿rst subscript is 0. The notation always makes the indexing clear, when such speci¿city is needed.

Thus far, our discussion has focused on sequences and series of complex (and real) numbersi.e., we have taken AFand A U. In this chapter, we take A to be the set of complex (and real) functions onF(andU).

325

8.1 Pointwise and Uniform Convergence

The¿rst important thing to note is that we will have different types of convergence to consider because we have “more variables.” The ¿rst relates back to numerical sequences and series. We start with an example for which the work was done in Chapter 4.

Example 8.1.1 For each n + M, let f_nz zⁿ where z + F. We can use results obtained earlier to draw some conclusions about the convergence offnz^*_n1. In Lemma 4.4.2, we showed that, for any¿xed complex number z₀such thatz₀1,

nlim*zⁿ₀ 0. In particular, we showed that for z₀, 0 z₀ 1, if 0, then taking

M Mz₀ !

!

!!

1 , foro1

zln lnz₀

{

, for 1

.

yields that nnz₀ⁿ0nn for all n M. When z₀ 0, we have the constant sequence. In offering this version of the statement of what we showed, I made a

“not so subtle” change in formatnamely, I wrote the former Mand Mz0. The change was to stress that our discussion was tied to the ¿xed point. In terms of our sequence f_nz^*_n1, we can say that for each ¿xed point z₀ + P z +F:z 1,fnz0^*_n1 is convergent to 0. This gets us to some new termi- nology: For this example, if f z 0 for all z + F, then we say thatf_nz₀^*_n1 is pointwise convergent to f onP.

It is very important to keep in mind that our argument for convergence at each

¿xed point made clear and de¿nite use of the fact that we had a point for which a known modulus was used in¿nding an Mz₀. It is natural to ask if the pointwise dependence was necessary. We will see that the answer depends on the nature of the sequence. For the sequence given in Example 8.1.1, the best that we will be able to claim over the setPis pointwise convergence. The associated sequence of nth partial sums for the functions in the previous example give us an example of a sequence of functions for which the pointwise limit is not a constant.

Example 8.1.2 For a / 0 and each k + MC 0, let f_kz az^k where z + F. In Chapter 4, our proof of the Convergence Properties of the Geometric Series

8.1. POINTWISE AND UNIFORM CONVERGENCE 327 Theorem showed that the associated sequence of nth partial sumss_nz^*_n0 was given by

s_nz

;n k0

f_kz

;n k0

az^k ab

1zⁿ¹c 1z .

In view of Example 8.1.1, we see that for each ¿xed z₀ + P z +F:z1, s_nz₀^*_n1is convergent to a

1z₀. Thus,s_nz^*_n0is pointwise convergent on P. In terminology that is soon to be introduced, we more commonly say that “the series3_*

k0az^k is pointwise convergent onP.”

Our long term goal is to have an alternative way of looking at functions. In par- ticular, we want a view that would give promise of transmission of nice properties, like continuity and differentiability. The following examples show that pointwise convergence proves to be insuf¿cient.

Example 8.1.3 For each n + M, let f_nz n²z

1n²z where z +F. For each¿xed z we can use our properties of limits to ¿nd the pointwise limit of the sequences of functions. If z 0, then f_n0^*_n1 converges to 0 as a constant sequence of zeroes. If z is a¿xed nonzero complex number, then

nlim*

n²z

1n²z lim

n*

z 1 n² z

z z 1.

Therefore, f_n f where f z

1 , for z +F 0 0 , for z 0

.

Remark 8.1.4 From Theorem 4.4.3(c) or Theorem 3.20(d) of our text, we know that p 0 and: + U, implies that lim

n*

n^:

1 pⁿ 0. Letting ? 1 1 p for p 0 leads to the observation that

nlim*n^:?ⁿ 0 (8.1)

whenever 0 n ? 1 and for any : + R. This is the form of the statement that is used by the author of our text in Example 7.6 where a sequence of functions for which the integral of the pointwise limit differs from the limit of the integrals of the functions in the sequence is given.

Example 8.1.5 (7.6 in our text) Consider the sequencef_n^*_n1of real-valued func- tions on the interval I [01] that is given by f_nx nxb

1x²cn

for n + M. For ¿xed x + I 0, taking : 1 and ? b

1x²c

in (8.1) yields that nb

1x²cn

0 as n *. Hence, fn

I0 0. Because fn0 0 for all n +M, we see that for each x + I ,

nlim*f_nx lim

n*nx r

1x² s_n

0.

In contrast to having the Riemann integral of the limit function over I being 0, we have that

nlim*

= ₁

0

f_nx d x lim

n*

n

2n2 1 2.

Note that, since : in Equation (8.1) can be any real number, the sequence of real functions g_nxn²xb

1x²cn

for n +Mconverges pointwise to 0 on I with

= ₁

0

g_nx d x n²

2n2 *as n *.

This motivates the search for a stronger sense of convergencenamely, uniform convergence of a sequence (and, in turn, of a series) of functions. Remember that our application of the term “uniform” to continuity required much nicer behavior of the function than continuity at points. We will make the analogous shift in going from pointwise convergence to uniform convergence.

De¿nition 8.1.6 A sequence of complex functionsf_n^*_n1converges pointwise to a function f on a subsetPofF, written f_n f or f_n

z+P f , if and only if the sequencef_nz₀^*_n1 f z₀for each z₀+Pi.e., for each z₀+ P

1 0 2M Mz₀+M n M>z₀" f_nz₀ f z₀ .

De¿nition 8.1.7 A sequence of complex functionsf_nconverges uniformly to f on a subsetPofF, written fn f , if and only if

1 0 2M M M + MF1n 1z n M Fz+ P" f_nz f z .

8.1. POINTWISE AND UNIFORM CONVERGENCE 329 Remark 8.1.8 Uniform convergence implies pointwise convergence. Given a se- quence of functions, the only candidate for the uniform limit is the pointwise limit.

Example 8.1.9 The sequence considered in Example 8.1.1 exhibits the stronger sense of convergence if we restrict ourselves to compact subsets of

P z+ F:z1. For each n + M, let f_nz zⁿ where z + F. Then f_nz^*_n1is uniformly convergent to the constant function f z0 on any com- pact subset ofP.

Suppose K tPis compact. From the Heine-Borel Theorem, we know that K is closed and bounded. Hence, there exists a positive real number r such that r 1 and1z z + K " z nr. LetPr z+ F:z nr. For 0, let

M M

!

!

!!

1 , for o1

zln ln r

{

, for 1

.

Then n M " n ln

ln r "n ln r lnbecause 0 r 1. Consequently, rⁿ and it follows that

f_nz0 nnzⁿnn zⁿ nrⁿ . Since 0 was arbitrary, we conclude that f_n

Pr

f . Because K t Pr, f_n

K

f as claimed.

Excursion 8.1.10 When we restrict ourselves to consideration of uniformly con- vergent sequences of real-valued functions onU, the de¿nition links up nicely to a graphical representation. Namely, suppose that f_n

[ab]

f . Then corresponding to any 0, there exists a positive integer M such that n M " f_nx f x for all x + [ab]. Because we have real-valued functions on the interval, the in- equality translates to

f x f_nx f xfor all x +[ab] . (8.2) Label the following¿gure to illustrate what is described in (8.2) and illustrate the implication for any of the functions f_nwhen n M.

Remark 8.1.11 The negation of the de¿nition offers us one way to prove that a sequence of functions is not uniformly continuous. Given a sequence of functions f_nthat are de¿ned on a subsetPofF, the convergence off_nto a function f on Pis not uniform if and only if

2 0 1M[M +M"

2nb

2z_Mnc b

n MFz_Mn + PFnnf_nb z_Mnc

f b

z_Mncnnoc ].

Example 8.1.12 Use the de¿nition to show that the sequence

| 1 nz

}_*

n1

is point- wise convergent, but not uniformly convergent, to the function f x 0 on P z +F: 0z1.

Suppose that z₀ is a¿xed element ofP. For 0, let M Mz₀ z 1

z₀ {

. Then n M "n 1

z₀ " 1

nz₀ becausez0 0. Hence, nnnn 1

nz0 0nn nn 1

nz0 . Since 0 was arbitrary, we conclude that

| 1 nz₀

}_*

n1

is convergent to 0 for each z0 +P. Therefore,

| 1 nz

}_*

n1

is pointwise convergent onP. On the other hand, let 1

2 and for each n + M, set zn 1

n1. Then

8.1. POINTWISE AND UNIFORM CONVERGENCE 331 z_n +Pand

nnnn 1 nz_n 0nn

nn nnnn nnnn

1 n

t 1 n1

u nnnn

nnnn1 1 n o.

Hence,

| 1 nz

}_*

n1

is not uniformly convergent onP. Example 8.1.13 Prove that the sequence

| 1 1nz

}_*

n1

converges uniformly for z o2 and does not converge uniformly inP^` z +F:z n2

| 1

n : n +M }

. LetP z +F:z o2and, for each n + M, let fnz 1

1nz. From the limit properties of sequences,fnz^*_n1is pointwise convergent onFto

f z

0 , for z + F 0 1 , for z 0

.

Thus, the pointwise limit offnz^*_n1onPis the constant function 0. For 0,

let M M

z1 2

t1 1

u{

. Then n M "n 1 2

t1 1

u

" 1 2n1 because n 1. Furthermore,z o 2 "nz o 2n "nz 1o2n1 0.

Hence,z o2Fn M "

fnz0 nn nn 1

1nz

nnnnn 1

nz 1 n 1

nz 1 n 1

2n1 .

Because 0 was arbitrary, we conclude that f_n

P 0.

On the other hand, let 1

2 and, corresponding to each n + M, set zn 1

n. Then zn +P^` and f_nz_n0

nnnn nnnn

1 1n

t1 n

u nnnn nnnn 1

2 o.

Hence,f_nz^*_n1is not uniformly convergent inP^`.

Excursion 8.1.14 Use the de¿nition to prove that the sequence zⁿ is not uni- formly convergent inz1.

***Hopefully, you thought to make use of the choices?n t

1 1 n

u

that could be related back to e¹.***

Using the de¿nition to show that a sequence of functions is not uniformly con- vergent, usually, involves exploitation of “bad points.” For Examples 8.1.12 and 8.1.13, the exploitable point was x 0 while, for Example 8.1.14, it was x 1.

Because a series of functions is realized as the sequence of nth partial sums of a sequence of functions, the de¿nitions of pointwise and uniform convergence of series simply make statements concerning the nth partial sums. On the other hand, we add the notion of absolute convergence to our list.

De¿nition 8.1.15 Corresponding to the sequence c_kz^*_k0 of complex-valued functions on a setPtF, let

S_nz

;n k0

c_kz denote the sequence of nt h partial sums. Then

(a) the series 3_*

k0c_kz is pointwise convergent onPto S if and only if, for each z₀ +P,S_nz₀^*_n0converges to Sz₀and

(b) the series3_*

k0c_kzis uniformly convergent onPto S if and only if S_n S. P

8.1. POINTWISE AND UNIFORM CONVERGENCE 333 De¿nition 8.1.16 Corresponding to the sequence c_kz^*_k0 of complex-valued functions on a set P t F, the series 3_*

k0c_kzis absolutely convergent on P if and only if3_*

k0ckzis convergent for each z +P.

Excursion 8.1.17 For a /0 and k + MC 0, let c_kz az^k. In Example 8.1.2, we saw that

;* k0

c_kz;^*

k0

az^k

is pointwise convergent for each z₀ +P z +F:z1to a1z₀¹. Show that

(i) 3_*

k0c_kzis absolutely convergent for each z₀+ P

(ii) 3_*

k0ckzis uniformly convergent on every compact subset K ofP

(iii) 3_*

k0c_kzis not uniformly convergent onP.

***For part (i), hopefully you noticed that the formula derived for the proof of the Convergence Properties of the Geometric Series applied to the real series that re- sults from replacing az^k witha z^k. Since 3_n

k0a z^k ab

1 zⁿ¹c

1 z , we

conclude that j3_n

k0a z^kk_*

n0 a

1 z for each z + C such thatz 1 i.e., 3_*

k0c_kz 3_*

k0az^k is absolutely convergent for each z + P. To show part (ii) it is helpful to make use of the fact that if K is a compact subset of P then there exists a positive real number r such that r 1 and K t Pr z +F:z nr. The uniform convergence of3_*

k0c_kzonPr then yields uni- form convergence on K . For S_nz 3_n

k0c_kz 3_n

k0az^k ab

1zⁿ¹c 1z and Sz a

1z, you should have noted that S_nzSz n arⁿ¹

1r for all z + Pr which leads to M max

1 lnb

1ra¹c

ln r 1

as one pos- sibility for justifying the uniform convergence. Finally, with (iii), corresponding to each n +M, let z_n

t

1 1

n1 u

then z_n + Pfor each n and S_nz_nSz_n n1a

t

1 1

n1 un1

can be used to justify that we do not have uniform convergence.***

8.1.1 Sequences of Complex-Valued Functions on Metric Spaces

In much of our discussion thus far and in numerous results to follow, it should become apparent that the properties claimed are dependent on the properties of the codomain for the sequence of functions. Indeed our original statement of the de¿ni- tions of pointwise and uniform convergence require bounded the distance between images of points from the domain while not requiring any “nice behavior relating the points of the domain to each other.” To help you keep this in mind, we state the de¿nitions again for sequences of functions on an arbitrary metric space.

De¿nition 8.1.18 A sequence of complex functions fn^*_n1 converges pointwise to a function f on a subset P of a metric space Xd, written f_n f or

f_n

*+P f , if and only if the sequencef_n*0^*_n1 f *0for each*0 + P i.e., for each*0 +P

1 0 2M M *0+M n M> *0" f_n*0 f *0 .

8.2. CONDITIONS FOR UNIFORM CONVERGENCE 335 De¿nition 8.1.19 A sequence of complex functionsf_nconverges uniformly to f on a subset Pof a metric spaceXd, written f_n f onPor f_n

P f , if and only if

1 0 2M M[M +MF1n 1* n M F* +P

" fn* f * ].

8.2 Conditions for Uniform Convergence

We would like some other criteria that can allow us to make decisions concerning the uniform convergence of given sequences and series of functions. In addition, if can be helpful to have a condition for uniform convergence that does not require knowledge of the limit function.

De¿nition 8.2.1 A sequencef_n^*_n1of complex-valued functions satis¿es the Cauchy Criterion for Convergence onPtFif and only if

1 0 2M +M[1n 1m 1z n M Fm MFz +P

" f_nz f_mz ].

Remark 8.2.2 Alternatively, when a sequence satis¿es the Cauchy Criterion for Convergence on a subset P t Fit may be described as being uniformly Cauchy onPor simply as being Cauchy.

In Chapter 4, we saw that in Uⁿ being convergent was equivalent to being Cauchy convergent. The same relationship carries over to uniform convergence of functions.

Theorem 8.2.3 Let fn^*_n1 denote a sequence of complex-valued functions on a setPtF. Thenf_n^*_n1converges uniformly onPif and only iff_n^*_n1satis¿es the Cauchy Criterion for Convergence onP.

Space for scratch work.

Proof. Suppose that f_n^*_n1 is a sequence of complex-valued functions on a setPtFthat converges uniformly onPto the function f and let 0 be given.

Then there exists M + Msuch that n M implies thatf_nz f z 2 for all z + P. Taking any other m M also yields that f_mz f z

2 for all z +P. Hence, for m MFn M,

f_mz f_nz f_mz f zf_nz f z n fmz f z fnz f z for all z +P. Therefore,f_n^*_n1is uniformly Cauchy onP.

Suppose the sequence f_n^*_n1of complex-valued functions on a setP t Fsatis¿es the Cauchy Criterion for Convergence onPand let 0 be given. For z + P,f_nz^*_n1 is a Cauchy sequence in F becauseF is complete, it follows that f_nz^*_n1 is convergent to some ?z + F. Since z + P was arbitrary, we can de¿ne a function f : P Fby1z z +P" f z?z. Then, f is the pointwise limit off_n^*_n1. Because f_n^*_n1 is uniformly Cauchy, there exists an M +Msuch that m M and n M implies that

f_nz f_mz

2 for all z +P. Suppose that n M is ¿xed and z + P. Since lim

m*fmG f G for each G + P, there exists a positive integer M^{`</sup> M such that m M<sup>`} implies that fmz f z

2. In particular, we have thatfM^`1z f z

2. There- fore,

f_nz f z f_nz f_M`1z f<sub>M</sub>`1z f z n f_nz f_M`1z f<sub>M</sub>`1z f z . But n M and z +Pwere both arbitrary. Consequently,

8.2. CONDITIONS FOR UNIFORM CONVERGENCE 337

1n 1z n MFz +P" f_nz f z . Since 0 was arbitrary, we conclude that fn

P f .

Remark 8.2.4 Note that in the proof just given, the positive integer M^{`</sup>was depen- dent on the point z and thei.e., M<sup>`} M^{`</sup>z. However, the¿nal inequality ob- tained via the intermediate travel through information from M<sup>`},f_nz f z , was independent of the point z. What was illustrated in the proof was a process that could be used repeatedly for each z +P.

Remark 8.2.5 In the proof of both parts of Theorem 8.2.3, our conclusions relied on properties of the codomain for the sequence of functions. Namely, we used the metric on F and the fact that F was complete. Consequently, we could allowP to be any metric space and claim the same conclusion. The following corollary formalizes that claim.

Corollary 8.2.6 Let f_n^*_n1 denote a sequence of complex-valued functions de-

¿ned on a subsetPof a metric spaceXd. Thenf_n^*_n1converges uniformly on Pif and only iff_n^*_n1satis¿es the Cauchy Criterion for Convergence onP. Theorem 8.2.7 Let f_n^*_n1 denote a sequence of complex-valued functions on a setPtFthat is pointwise convergent onPto the function fi.e.,

nlim*f_nz f z and, for each n + M, let M_n sup

z+Pf_nz f z. Then f_n

P f if and only if

nlim*M_n 0.

Use this space to¿ll in a proof for Theorem 8.2.7.

Theorem 8.2.8 (Weierstrass M-Test) For each n + M, let u_n* be a complex- valued function that is de¿ned on a subsetPof a metric spaceXd. Suppose that there exists a sequence of real constants Mn^*_n1such thatun* n Mn for all

* + Pand for each n + M. If the series3_*

n1M_n converges, then 3_*

n1u_n* and3_*

n1u_n*converge uniformly onP.

Excursion 8.2.9 Fill in what is missing in order to complete the following proof of the Weierstrass M-Test.

Proof. Suppose thatu_n*^*_n1,P, and M_n^*_n1 are as described in the hy- potheses. For each n + J , let

S_n*

;n k1

u_k* and T_n*

;n k1

u_k* and suppose that 0 is given. Since 3_*

n1M_n converges and M_n^*_n1 t U, j3_n

k1M_kk_*

n1 is a convergent sequence of real numbers. In view of the com- pleteness of the reals, we have thatj3_n

k1Mk

k_*

n1is

1

. Hence, there exists a positive integer K such that n K implies that

n;p kn1

M_k for each p+M.

Sinceu_k* n M_k for all*+ Pand for each k +M, we have that nnT_np*T_n*nnnn

nnn

n;p kn1

u_k*nn nnn

np

;

kn1

u_k* for all*+ P. Therefore,T_n^*_n1is

2

inP. It follows from the

3

that

4

5

n

n;p kn1

u_k* n

n;p kn1

M_k for all* +P. Hence,S_n^*_n1is uniformly Cauchy inP. From Corollary 8.2.6, we conclude that

6

.

8.3. PROPERTY TRANSMISSION AND UNIFORM CONVERGENCE 339

***Acceptable responses include: (1) Cauchy, (2) uniformly Cauchy, (3) triangular inequality, (4)nnS_np*S_n*nn, (5)nnn3np

kn1u_k*nnn, and (6)3_*

n1u_n*and 3_*

n1u_n*converge uniformly onP.***

Excursion 8.2.10 Construct an example to show that the converse of the Weier- strass M-Test need not hold.

8.3 Property Transmission and Uniform Convergence

We have already seen that pointwise convergence was not suf¿cient to transmit the property of continuity of each function in a sequence to the limit function. In this section, we will see that uniform convergence overcomes that drawback and allows for the transmission of other properties.

Theorem 8.3.1 Letfn^*_n1denote a sequence of complex-valued functions de¿ned on a subsetPof a metric spaceXdsuch that f_n

P f . For*a limit point ofP and each n + M, suppose that

tlim*

t+P

f_nt A_n. ThenA_n^*_n1converges and lim

t*f t lim

n*A_n.

Excursion 8.3.2 Fill in what is missing in order to complete the following proof of the Theorem.

Proof. Suppose that the sequencef_n^*_n1of complex-valued functions de¿ned on a subset Pof a metric space Xdis such that f_n

P f ,* is a limit point of P and, for each n + M, lim

t*fnt An. Let 0 be given. Since fn

P f ,

by Corollary 8.2.6, f_n^*_n1 is

1

onP. Hence, there exists a positive integer M such that

2

implies that f_nt f_mt

3 for all

3

. Fix m and n such that m M and n M. Since lim

t*f_kt A_k for each k + M, it follows that there exists a= 0 such that 0dt * =implies that

fmtAm 3 and

4

From the triangular inequality, An Am n An fnt

nnnn

n ₅

nnnn

n fmt Am . Since m and n were arbitrary, for each 0 there exists a positive integer M such that1m 1n n MFm M " A_n A_m i.e., A_n^*_n1 t F is Cauchy. From the completeness of the complex numbers, if follows thatA_n^*_n1is convergent to some complex numberlet lim

n*A_n A.

We want to show that A is also equal to lim

t*

t+P

f t. Again we suppose that 0 is given. From fn

P f there exists a positive integer M1such that n M1

implies that nnnn

n ₆

nnnn n

3 for all t + P, while the convergence ofA_n^*_n1 yields a positive integer M₂such thatA_n A

3 whenever n M₂. Fix n such that n maxM₁M₂. Then, for all t +P,

f t f_nt

3 and A_n A 3. Since lim

t*

t+P

f_nt A_n, there exists a= 0 such that f_nt A_n

3 for all t +N₌* *DP.

8.3. PROPERTY TRANSMISSION AND UNIFORM CONVERGENCE 341 From the triangular inequality, for all t + Psuch that 0 dt * =,

f t A n

7

. Therefore,

8

.

***Acceptable responses are: (1) uniformly Cauchy, (2) n MFm M, (3) t + P, (4)f_nt A_n

3, (5) f_nt f_mt, (6) f tf_nt, (7)f t f_nt f_nt A_n A_n A, and (8) lim

t*

t+P

f t A.***

Theorem 8.3.3 (The Uniform Limit of Continuous Functions) Letf_n^*_n1denote a sequence of complex-valued functions that are continuous on a subsetPof a met- ric spaceXd. If fn

P f , then f is continuous onP.

Proof. Suppose that f_n^*_n1 is a sequence of complex-valued functions that are continuous on a subset P of a metric space Xd. Then for each ? + P,

tlim?f_nt f_n?. Taking A_n f_n? in Theorem 8.3.1 yields the claim.

Remark 8.3.4 The contrapositive of Theorem 8.3.3 affords us a nice way of show- ing that we do not have uniform convergence of a given sequence of functions.

Namely, if the limit of a sequence of complex-valued functions that are continuous on a subsetPof a metric space is a function that is not continuous on P, we may immediately conclude that the convergence in not uniform. Be careful about the appropriate use of this: The limit function being continuous IS NOT ENOUGH to conclude that the convergence is uniform.

The converse of Theorem 8.3.3 is false. For example, we know that

| 1 nz

}_*

n1

converges pointwise to the continuous function f z 0 inF 0and the con- vergence is not uniform. The following result offers a list of criteria under which continuity of the limit of a sequence of real-valued continuous functions ensures that the convergence must be uniform.

Theorem 8.3.5 Suppose thatPis a compact subset of a metric space Xdand f_n^*_n1satis¿es each of the following:

(i) f_n^*_n1is a sequence of real-valued functions that are continuous onP (ii) f_n

P f and f is continuous onPand

(iii) 1n 1* n +MF* +P" fn*o fn1*. Then fn

P f .

Excursion 8.3.6 Fill in what is missing in order to complete the following proof of Theorem 8.3.5.

Proof. Forf_n^*_n1 satisfying the hypotheses, set g_n f_n f . Then, for each n + M, g_n is continuous on P and, for each ? + P, lim

n*g_n?

1

. Since f_n*o f_n1*implies that f_n* f *o f_n1* f *, we also have that1n 1*

‚

n +MF*+ P"

2

. To see that g_n

P 0, suppose that 0 is given. For each n +M, let K_n x + P: g_nxo.

Because P and U are metric spaces, g_n is continuous, and * +U:* o is a closed subset ofU, by Corollary 5.2.16 to the Open Set Characterization of Con- tinuous Functions,

3

. As a closed subset of a compact metric space, from Theorem 3.3.37, we conclude that K_n is

4

. If x + K_n1, then gn1x o and gnx o gn1x it follows from the transitivity of o that

5

. Hence, x + K_n. Since x was arbitrary,1x x + K_n1"x + K_n i.e.,

6

. Therefore, Kn^*_n1 is a

7

sequence of compact subsets of7 P. From Corollary 3.3.44 to Theorem 3.3.43, 1n+ M K_n / 3 "

k+M

K_k / 3.

Suppose that*+P. Then lim

n*g_n*0 andg_nxdecreasing yields the existence of a positive integer M such that n M implies that 0ng_n* . In particular, * + K_M1 from which it follows that* + 7

n+MK_n. Because* was

8.3. PROPERTY TRANSMISSION AND UNIFORM CONVERGENCE 343 arbitrary,1* +P

‚

* + 7

n+MKn

i.e., 7

n+MKn 3. We conclude that there exists a positive integer P such that K_P 3. Hence, K_n 3for all

8

that is, for all no P,x + P: gnxo 3. Therefore,

1x 1n x +PFn P "0ng_nx . Since 0 was arbitrary, we have that g_n

P 0 which is equivalent to showing that f_n

P f .

***Acceptable responses are: (1) 0, (2) g_n* o g_n1*, (3) K_n is closed, (4) compact, (5) g_nxo, (6) K_n1 t K_n, (7) nested, and (8) n o P.***

Remark 8.3.7 Since compactness was referred to several times in the proof of The- orem 8.3.5, it is natural to want to check that the compactness was really needed.

The example offered by our author in order to illustrate the need is

| 1 1nx

}_*

n1

in the segment01.

Our results concerning transmission of integrability and differentiability are for sequences of functions of real-valued functions on subsets ofU.

Theorem 8.3.8 (Integration of Uniformly Convergent Sequences) Let:be a func- tion that is (de¿ned and) monotonically increasing on the interval I [ab]. Sup- pose thatf_n^*_n1is a sequence of real-valued functions such that

1n n +M" f_n + 4: on I and fn

[ab]

f . Then f + 4:on I and

= _b

a

f xd: x lim

n*

= _b

a

f_nxd: x

Excursion 8.3.9 Fill in what is missing in order to complete the following proof of the Theorem.

Proof. For each n + J , letn sup

x+I

f_nx f x. Then fnxn n f xn

1

for a nx nb and if follows that

= _b

a

fnxnd: xn

= _b

a

f xd: xn

= _b

a

f xd: xn

= _b

a f_nxnd: x. (8.3) Properties of linear ordering yield that

0n

= _b

a

f xd: x

= _b

a

f xd: xn

= _b

a

f_nxnd: x

2

. (8.4)

Because the upper bound in equation (8.4) is equivalent to

3

, we conclude that

1n +M

‚

0n5_b

a f xd: x5_b

a f xd: xn

4

. By The- orem 8.2.7,n 0 as n *. Since5_b

a f xd: x5_b

a f xd: xis constant, we conclude that

5

. Hence f + 4:. Now, from equation 8.3, for each n + J ,

= _b

a

f_nxnd: xn

= _b

a

f xd: xn

= _b

a

f_nxnd: x. 6Finish the proof in the space provided.

8.3. PROPERTY TRANSMISSION AND UNIFORM CONVERGENCE 345

***Acceptable responses are:(1) fnxn(2)5_b

a fnxnd: x, (3)5_b

a end: x, (4) 2n[: b: a], (5)5_b

a f xd: x 5_b

a f xd: x, (6) Hopefully, you thought to repeat the process just illustrated. From the modi¿ed inequality it follows that

nnn5_b

a f xd: x5_b

a fnxd: xnnnnn[: b: a]then becausen 0 as n *, given any 0 there exists a positive integer M such that n M implies thatn[: b: a] .***

Corollary 8.3.10 If f_n + 4:on [ab], for each n + M, and

;* k1

f_kxconverges uniformly on [ab] to a function f , then f + 4:on [ab] and

= _b

a

f xd: x;^*

k1

= _b

a

f_kxd: x.

Having only uniform convergence of a sequence of functions is insuf¿cient to make claims concerning the sequence of derivatives. There are various results that offer some additional conditions under which differentiation is transmitted. If we restrict ourselves to sequences of real-valued functions that are continuous on an interval [ab] and Riemann integration, then we can use the Fundamental Theorems of Calculus to draw analogous conclusions. Namely, we have the following two results.

Theorem 8.3.11 Suppose that fn^*_n1 is a sequence of real-valued functions that are continuous on the interval [ab] and f_n

[ab]

f . For c +[ab] and each n+ M, let

Fnx

de f

= _x

c

fntdt . Then f is continuous on [ab] and F_n

[ab]

F where

Fx

= _x

c

f tdt . The proof is left as an exercise.