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Solution to Understanding Analysis Exercise 4.4


Exercise 4.4.1

(a) It is already shown in Example 4.3.5.

(b) Let xn=n+1n and yn=n, then we have
|xnyn|0.On the other hand, we have
|f(xn)f(yn)|= |(n+1n)3n3|= |n3+3n+3n+1n3n3| 3n1.Hence we can choose ϵ0=1 in Theorem 4.4.5. It follows from Theorem 4.4.5 that f is not uniformly continuous on R.

Lemma: If a function f is uniformly continuous on a set A, then it is also uniformly continuous on any subset B of A.

Proof: Since f is uniformly continuous on A, for any ϵ>0, there exists δ>0 such that for all x,yA, |xy|<δ, we have |f(x)f(y)|<ϵ.

Now for any x,yB, |xy|<δ, as B is a subset of A, we also have x,yA, |xy|<δ, therefore we have |f(x)f(y)|<ϵ.

(c) Let A be a bounded subset of R. Then there exists M>0 such that A[M,M]. The closed interval [M,M] is a compact set. As f(x) is continuous on [M,M], by Theorem 4.4.7, it follows that f(x) is uniformly continuous on [M,M]. Now it follows from the lemma above that f(x) is uniformly continuous on A.


Exercise 4.4.2

(a) No, let xn=1n and yn=12n, then|xnyn|=12n0,|f(xn)f(yn)|=n1.Hence by Theorem 4.4.5, f(x) is not uniformly continuous on (0,1).

(b) Yes. Note that g(x) is continuous on [0,1]. Since [0,1] is bounded and closed, [0,1] is compact. By Theorem 4.4.7, g(x) is uniformly continuous on [0,1]. Therefore, g(x) is also uniformly continuous on (0,1).

(c) Yes. We extend the function h(x) by defining h(0)=0, then it is clear that h(x) is continuous on [0,1]. Since [0,1] is bounded and closed, [0,1] is compact. By Theorem 4.4.7, h(x) is uniformly continuous on [0,1]. Therefore, h(x) is also uniformly continuous on (0,1).


Exercise 4.4.3

See Understanding Analysis Instructors’ Solution Manual Exercise 4.4.2.


Exercise 4.4.4

(a) See Understanding Analysis Instructors’ Solution Manual Exercise 4.4.4.

(b) True. We argue it by contradiction. Suppose f(A) is not bounded. Then there exists a sequence (xn) such that xnA and |f(xn+1)||f(xn)|1 for all n1 (this sequence can be obtained inductively). Since A is bounded, by Theorem 2.5.5, then we have a convergent subsequence  (xnk). Let ynk=xnk+1, then|xnkynk|0, but|f(xnk)f(ynk)||f(ynk)||f(xnk)|1.Hence by Theorem 4.4.5, f is not uniformly continuous, which is a contradiction. Therefore, we are done.

(c) False. Define a function f(x) as follows,f(x)={0, if xQ;1, otherwise. Then f(K) is always a finite set and hence compact. But f is not continuous on R.


Exercise 4.4.5

See Understanding Analysis Instructors’ Solution Manual Exercise 4.4.7.


Exercise 4.4.6

(a) Let f(x)=1/x for x(0,1). Then f is continuous on (0,1). Let xn=1/n, then (xn) is Cauchy. But f(xn)=n, so the sequence (f(xn)) is not Cauchy.

(b) Impossible. Since f is uniformly continuous on (0,1), for any ϵ>0, there exists δ>0 such that |f(x)f(y)|<ϵ provided |xy|<δ. Since (xn) is Cauchy, there exists N>0 such that |xnxm|<δ whenever n,m>N. Hence for n,m>N, we have|f(xn)f(xm)|<ϵ.This implies that (f(xn)) is Cauchy.

(c) Impossible. Since (xn) is Cauchy, (xn) is bounded by Lemma 2.6.3. Let |xn|M where M>0 for all n1. Then we can consider the restriction of f on the interval [0,M]. Because [0,M] is compact, by Theorem 4.4.7, f is uniformly continuous on [0,M]. Using the same argument as in (b), you can show that (f(xn)) is Cauchy.


Exercise 4.4.7

See Understanding Analysis Instructors’ Solution Manual Exercise 4.4.8 (b).


Exercise 4.4.8

(a) Impossible. Because f is continuous and [0,1] is compact, by Theorem 4.4.1, the range is compact as well. But (0,1) is not compact.

(b) Let f(x)=|sin1x|, see Example 4.2.6.

(c) Letf(x)=|sin1x|+x1+2x.


Exercise 4.4.9

See Understanding Analysis Instructors’ Solution Manual Exercise 4.4.9.


Exercise 4.4.10

The functions f(x)+g(x) and f(g(x)) are necessarily uniformly continuous.

f(x)+g(x) is uniformly continuous. This follows from the triangle inequality. Let ϵ>0. Because f(x) and g(x) are uniformly continuous on A, there exist δ1>0 and δ2>0 such that |f(x1)f(y1)|<ϵ/2,|g(x2)g(y2)|<ϵ/2for all |x1y1|<δ1 and |x2y2|<δ2 on A. Let δ=min{δ1,δ2}. Then for all x,yA and |xy|<δ, we have|(f(x)+g(x))(f(y)+g(y))||f(x)f(y)|+|g(x)g(y)|<ϵ2+ϵ2=ϵ.Thus f(x)+g(x) is uniformly continuous.

f(x)g(x) may bot be uniformly continuous. Let f(x)=g(x)=x, then f(x)g(x)=x2 is not uniformly continuous. Because we can choose xn=n and yn=n+1/n, then|xnyn|0,|xn2yn2|=|2+1n2|2.By Theorem 4.4.5, x2 is not uniformly continuous.

f(x)/g(x) may bot be uniformly continuous. Let f(x)=1 and g(x)=x defined on (0,1), then f(x)/g(x)=1/x is not uniformly continuous, see Exercise 4.4.2.

f(g(x)) is uniformly continuous. Let ϵ>0. Since f(x) is uniformly continuous, there exists δ1>0 such that for all x,yA and |xy|<δ1 we have |f(x)f(y)|<ϵ. Since g(x) is uniformly continuous, there exists δ>0 such that for all x,yA and |xy|<δ we have |g(x)g(y)|<δ. Therefore, if x,yA and |xy|<δ, we have |g(x)g(y)|<δ1 and therefore |f(g(x))f(g(y))|<ϵ.Thus f(g(x)) is uniformly continuous.


Exercise 4.4.11

See Understanding Analysis Instructors’ Solution Manual Exercise 4.4.11.


Exercise 4.4.12

(a) (b) (c) may be false. Consider the example f(x)=1 for all xR. Let B=K=A, then f1(B)=f1(K)=f1(A)=R which is clear not finite, not compact, not bounded.

(d) is true. To show f1(F) is closed, it suffices to show that f1(F)c is open. We show that f1(F)c=f1(Fc). Then because Fc is open, by definition of Exercise 4.4.11, f1(F)c=f1(Fc) is also open.

Now we show f1(F)c=f1(Fc). First we show f1(F)cf1(Fc). Let af1(F)c, then af1(F), i.e. f(a)F. Therefore f(a)Fc, hence af1(Fc).

Then we show f1(Fc)f1(F)c. Let af1(Fc), then f(a)Fc. Thus f(a)F and af1(F). Therefore af1(F)c, this implies f1(Fc)f1(F)c.

Now we have f1(F)c=f1(Fc).


Exercise 4.4.13

See Understanding Analysis Instructors’ Solution Manual Exercise 4.4.13.

Linearity

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