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Solution to Principles of Mathematical Analysis Chapter 7 Part B


Chapter 7 Sequences and Series of Functions

Exercise 13

(By analambanomenos)

(a) Following the steps of the hint: (i) Show that some subsequence {fni} converges at all rational points r, say, to f(r).

Let {rm} be an enumeration of the rational points of R. Since {fn(r1)}[0,1], a compact set, there is a subsequence {f1,nj(r1)} which converges to f(r1), by Theorem 2.41. Similarly, the {f1,nj(r2)} has a subsequence {f2,nj(r2)} which converges to f(r2). Continuing in this fashion, we get a sequence of subsequences {fm,nj} such that each {fm,nj} is a subsequence of {fm1,nj} and fm,nj(rm) converges to f(rm). Letting fni=fi,ni, we get a subsequence of {fn} which converges on all rational points r of R to f(r).

(ii) Define f(x), for any xR, to be supf(r), the sup being taken over all rational points rx.

First we need to show that this new definition of f(r) agrees with the previous one; let f(r) be the new definition of f(r). It is clear from the definition that f(r)f(r). Suppose f(r)>f(r), that is, there is a rational point q<r and a real number a such that f(q)>a>f(r). Since f(q)=limfni(q), and f(r)=limfni(r), there is an integer N such that for all iN we have fni(q)>a>fni(r), but this contradicts the fact that these functions are monotonically increasing.

It is clear from the definition that f is monotonically increasing.

(iii) Show that fni(x)f(x) at every x at which f is continuous.

Suppose f is continuous at x and let ε>0. Then there are rational numbers r1<x<r2 such that f(x)ε2<f(r1)f(x)f(r2)<f(x)+ε2. Since fni(r1)f(r1) and fni(r2)f(r2), there is an integer N such that for all i>N we have |fni(r1)f(r1)|<ε2and|fni(r2)f(r2)|<ε2. And since fni is monotonically increasing, we have for all i>N f(x)ε<f(r1)ε2<fni(r1)fni(x)fni(r2)<f(r2)+ε2<f(x)+ε. That is, |f(x)fni(x)|<ε, so fni(x)f(x) as i.

(iv) Show that a subsequence of {fni} converges at every point of discontinuity of f since there are at most countably many such points.

Since f is monotonically increasing, f has at most countably many simple points of discontinuity by Theorems 4.29 and 4.30. Let {xm} be an enumeration of the points of discontinuity of f. Repeating the argument of step (i), there is a subsequence {f1,nj(x1)} of {fni(x1) which converges to a value a between 0 and 1. We can continue this way for all the xm, getting a sequence of subsequences {fm,nj} of {fni} such that each {fm,nj} is a subsequence of {fm1,nj} for which fm,nj(xm) converges. Letting fnk=fk,nk, we get a subsequence of {fn} which converges at all the points {xm}. We can redefine f(xi) at these points to be those values.

Since we have already shown that {fnk} converges at the points where f is continuous, we have fnk(x) converging to f(x) for all real values x.

(b) Let K be a compact subset of R and let ε>0. It was shown in step (iii) of part (a) that, since f is everywhere continuous, f(x)=limfnk(x) for all points x. Also, we are using the initial definition of f, so that is it monotonically increasing. For each xK there is a δx>0 such that for all y[xδx,x+δx] we have
|f(y)f(x)|<ε4.Also, there is an integer Nx such that for all k>Nx |fnk(xδx)|<ε2and|fnk(x+δx)|<ε4. Since fnk is monotonically increasing, we must have, for all y(xδx,x+δx) and all k>Nx, f(x)ε2<f(xδx)ε4<fnk(xδx)fnk(y)fnk(x+δx)+ε4<f(x)+ε2, that is, |fnk(y)f(x)|<ε/2. Hence, for y(xδx,x+δx) and k>Nx, |fnk(y)f(y)||fnk(y)f(x)|+|f(x)f(y)|<ε. Since the open neighborhoods (xδx,x+δx) cover the compact set K, a finite subset of these neighborhoods, centered at x1,,xM, also covers K. Letting N=max(Nx1,,NxM), we have for all yK and all k>N that |fnk(y)f(y)|ε, that is, {fnk} converges uniformly to f on K.


Exercise 14

(By analambanomenos) Since for each n we have |2nf(32n1t)|2n and |2nf(32nt)|2n, the series defining x(t) and y(t) converge uniformly since 2n converges, by Theorem 7.10. And since the partial sums are continuous functions, x(t) and y(t) are continuous functions, by Theorem 7.12. Hence Φ is continuous.

Following the hint, let (x0,y0)I2, and let x0=n=12na2n1,y0=n=12na2n be the binary expansions of x0 and y0, where each ai is 0 or 1. Let t0=i=13i1(2ai). By Exercise 3.19, the set of all such t0 is precisely the Cantor set.

We have for k=1,2,3,,
3kt0=i=13ki1(2ai)=2n=0k23nak1n+23ak+23n=13nak+nNote that the last term lies between 0 and 23n=13n=23(1/31(1/3))=13. Also, the first term is an even integer, so that, since f(t+2)=f(t), we have f(3kt0)=f(23ak+23n=13nak+n). If ak=0, then the expression in the argument of f lies between 0 and 13, so that f(3kt0)=0. And if ak=1, then the expression in the argument of f lies between 23 and 1, so that f(3kt0)=1. In either case, we have f(3kt0)=ak. Hence Φ(t0)=(x(t0),y(t0))=(x0,y0).


Exercise 15

(By analambanomenos) The function f must be constant on [0,]. Let 0x<y and let ε>0. Then there is a δ>0 such that for all n=1,2,3,, we have |fn(z)fn(w)|<ε if 0w,z1 and |wz|<δ. For large enough N, we have |yNxN|=|yx|N<δ,0xN<yN1.Hence |f(y)f(x)|=|fN(yN)fN(xN)|<ε. Since ε>0 was arbitrary, this shows that f(x)=f(y).

By the way, it is easy to extend this argument to show that if {fn} were equicontinuous on [1,1], then f would be constant on all of R.


Exercise 16

(By analambanomenos) Let ε>0 and let δ>0 such that if d(x,y)<δ then |fn(x)fn(y)|<ε/3 for all n. Let xK and let Nx be an integer large enough so that if m>Nx and n>Nx, then |fm(x)fn(x)|<ε. Then for all m>Nx and n>Nx and all yK such that d(x,y)<δ, we have
|fn(y)fm(y)||fn(y)fn(x)|+|fn(x)fm(x)|+|fm(x)fm(y)|<ε. Since K is compact, there are a finite number of points x1,,xM in K such that the neighborhoods of radius δ centered at the xi cover K. So if we let N=max(Nx1,,NxM), then for m>N and n>N and for all yK, we have |fm(y)fn(y)|<ε. Hence by Theorem 7.8, {fn} converges uniformly on K.


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 {fn}, n=1,2,3,, into a metric space F converges uniformly on E to a function f if for every ε>0 there is an integer N such that nN implies (1)d(fn(x),f(x))ε for all xE.

Definition 7.22 A family F of mappings f defined on a set E in a metric space X with values in a metric space Y is said to be equicontinuous on E if for every ε>0 there exists a δ>0 such that dY(f(x),f(y))<ε whenever dX(x,y)δ, xE, yE, and fF.

Theorem 7.8 The sequence of mappings {fn} into a complete metric space F, defined on E, converges uniformly on E if and only if for every ε>0 there exists an integer N such that mN, nN, xE implies d(fn(x),fm(x))ε.

Proof: Suppose {fn} converges uniformly on E, and let f be the limit function. There there is an integer N such that nE implies d(fn(x),f(x))ε/2 so that d(fn(x),fm(x))d(fn(x),f(x))+d(f(x),fm(x))ε if nN, mM, xE.

Conversely, suppose the Cauchy condition holds. Since F is a complete metric space, the sequence {fn(x)} converges, for every x, to a limit which we may call f(x). Thus the sequence {fn} converges on E, to f. We have to prove that the convergence is uniform.

Let ε>0 be given, and choose N such that (1) holds. Fix n, and let m in (1). Since fm(x)f(x) as m, this gives d(fn(x),f(x))ε for every nN and every xE, which completes the proof.

Theorem 7.9 Suppose {fn} {\it is a sequence of mappings from E into a metric space F such that} limfn(x)=f(x) for xE. Put Mn=supxEd(fn(x),f(x)). Then fnf uniformly on E if and only if Mn0 as n.

Proof: Since this is an immediate consequence of Definition 7.7, we omit the details of the proof.

Theorem 7.10 Suppose {fn} is a sequence of vector-valued functions with values in Rk defined on E, and suppose ||fn(x)||Mn(xE,n=1,2,3,). Then fn converges uniformly on E if Mn converges.

Proof: If Mn converges, then, for arbitrary ε>0, ||i=nmfi(x)||i=nmMiε(xE), provided m and n are large enough. Uniform convergence now follows from Theorem 7.8.

Theorem 7.11 Suppose {fn}, a sequence of mappings with values in a complete metric space F, converges uniformly on a set E in a metric space. Let x be a limit point
of E, and suppose that limtxfn(t)=An(n=1,2,3,). Then {An} converges and (2)limtxf(t)=limnAn.

Proof: Let ε>0 be given. By the uniform convergence of {fn}, there exists N such that nN, mN, tE imply (3)d(fn(t),fm(t))ε. Letting tx in (3), we obtain d(An,Am)ε for nN, mM, so that {An} is a Cauchy sequence and therefore converges, say to A.

Next, (4)d(f(t),A)d(f(t),fn(t))+d(fn(t),An)+d(An,A) We first choose n such that (5)d(f(t),fn(t))ε3 for all tE (this is possible by the uniform convergence), and such that (6)d(An,A)ε3. Then, for this n we choose a neighborhood V of x such that (7)d(fn(t)An)ε3, if tVE, tx.

Substituting the inequalities (5), (6), (7) into (4), we see that d(f(t),A)ε, provided tVE, tx. This is equivalent to (2).

Theorem 7.12 If {fn} is a sequence of continuous {\it mappings on E into a metric space} F, and if fnf uniformly on E, then f is continuous on E.

Proof: (In the text, Theorem 7.12 is an immediately corollary of Theorem 7.11. Here, that would require that F be complete, which is not necessarily the case. So the following proof is entirely new.)

By Theorem 4.8, to show that fis continuous, it suffices to show that f1(V) is open for any open set VF. If xf1(V), we need to find a neighborhood Nx of x which is contained in f1(V), for then f1(V) would be the union of the open neighborhoods Nx, xf1(V), and so be open.

Let ε>0 be small enough so that the neighborhood My of y=f(x) of radius ε is contained in V. Let n be large enough so that d(fn(z),f(z))ε/3 for all zE, which is possible by the assumption of uniform convergence. Let δ>0 such that d(fn(z),fn(x))<ε/3 for all z in the neighborhood Nx of x of radius δ, which is possible since fn is continuous. Then, for all zNx, d(f(z),f(x))d(f(z),fn(z))+d(fn(z),fn(x))+d(fn(x),f(x))ε. That is, f(z)MyV, or zf1(V), so that Nxf1(V).

The remaining Theorems, for mappings into Rk, are largely corollaries of their scalar counterparts in the text.

Theorem 7.16 Let α be monotonically increasing on [a,b]. Suppose fn=(fn1,,fnk)R(α) on [a,b], for n=1,2,3,, and suppose fnf=(f1,,fk) uniformly on [a,b]. Then fR(α) on [a,b], and abfdα=limnabfndα.

Theorem 7.17 Suppose {fn=(fn1,,fnk)} is a sequence of vector-valued functions, differentiable on [a,b] and such that {f(x0)} converges for some point x0 on [a,b]. If {fn} converges uniformly on [a,b], to a function f=(f1,,fk), and f(x)=limnfn(x)(axb).

Proofs: If x=(x1,,xk)Rk, then |xi|||x||ik|xi|. Hence fn convergences uniformly if and only if the component functions fn1,,fnk 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 K is a compact metric space, if fn=(fn1,,fnk)C(K) for n=1,2,3,, and if {fn} converges uniformly on K, then {fn} is equicontinuous on K.

Proof: If x=(x1,,xk)Rk, then |xi|||x||i|xi|. Hence {fn} is equicontinuous if and only if each of the {fni}, i=1,,k 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 K is compact, if fn=(fn1,,fnk) is pointwise bounded and equicontinuous on K, then

(a) {fn} is uniformly bounded on K,

(b){fn} contains a uniformly convergent subsequence.

Proof: Since each of the sequences {fni}, i=1,,k, is pointwise bounded and equicontinuous on K, by the scalar version of Theorem 7.25 in the text each of them is uniformly bounded on K, so {fn} is uniformly bounded on K. Also, {fn1} contains a uniformly convergent subsequence, {fnj1}. Since the subsequence {fnj2} is also pointwise bounded and equicontinuous on K, it also has a uniformly convergent subsequence such that the corresponding subsequence of {fnj1} also converges uniformly. Continuing in this manner, after k steps we have a subsequence {fnj} whose component functions all converge uniformly, so {fnj} also converges uniformly.

Baby Rudin 数学分析原理完整第七章习题解答


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