ńņš. 7 |

Necessity Suppose f is continuous and U is an open set in Rp . If x ā f ā’1 (U ),

then f (x) ā U . But U is open, so there exists an > 0 such that B(f (x), ) ā

U . Since f is continuous at x, we can ļ¬nd a Ī“ > 0 such that

f (x) ā’ f (y) < whenever x ā’ y < Ī“.

We thus have B(x, Ī“) ā f ā’1 (U ). It follows that f ā’1 (U ) is open.

Suļ¬ciency Suppose that f ā’1 (U ) is open whenever U is an open subset of Rp .

Let x ā Rm and > 0. Since B(f (x), ) is open, it follows that f ā’1 (B(f (x), ))

is open. But x ā f ā’1 (B(f (x), )), so there exists a Ī“ > 0 such that B(x, Ī“) ā

f ā’1 (B(f (x), )). We thus have

f (x) ā’ f (y) < whenever x ā’ y < Ī“.

It follows that f is continuous.

Exercise 4.2.18. Show that sin((ā’5Ļ, 5Ļ)) = [ā’1, 1]. Give examples of

bounded open sets A in R such that (a) sin A is closed and not open, (b)

sin A is open and not closed, (c) sin A is neither open nor closed, (d) sin A

is open and closed. (Observe that ā… is automatically bounded.)

The reader may object that we have not yet derived the properties of sin.

In my view this does not matter if we are merely commenting on or illustrat-

ing our main argument. (I say a little more on this topic in Appendix C.)

However, if the reader is interested, she should be able to construct a poly-

nomial P such that (a), (b), (c) and (d) hold for suitable A when sin A is

replaced by P (A).

The next exercise gives a simple example of how Lemma 4.2.17 can be

used and asks you to contrast the new ā˜open setā™ method with the old ā˜ , Ī“

method

Exercise 4.2.19. Prove the following result, ļ¬rst directly from Deļ¬nition 4.2.14

and then by using Lemma 4.2.17 instead.

If f : Rm ā’ Rp and g : Rp ā’ Rq are continuous, then so is their

composition g ā—¦ f .

55

Please send corrections however trivial to twk@dpmms.cam.ac.uk

(Recall that we write g ā—¦ f (x) = g(f (x)).)

The reader who has been following carefully may have observed that we

have only deļ¬ned limits of sequences. Here is another notion of limit which

is probably familiar to the reader.

Deļ¬nition 4.2.20. Let E ā Rm , x ā E and a ā Rp . Consider a function

f : E \ {x} ā’ Rp (or4 a function f : E ā’ Rp ). We say that f (y) ā’ a as

y ā’ x through values of y ā E if, given > 0, we can ļ¬nd a Ī“( , x) > 0

such that, whenever y ā E and 0 < x ā’ y < Ī“( , x), we have

f (x) ā’ a < .

(We give a slightly more general deļ¬nition in Exercise K.23.)

Exercise 4.2.21. Let E ā Rm , x ā E and a ā Rp . Consider a function

f : E \ {x} ā’ Rp . Deļ¬ne Ė : E ā’ Rp by Ė(y) = f (y) if y ā E \ {x} and

f f

Ė(x) = a. Show that f (y) ā’ a as y ā’ x through values of y ā E if and only

f

if Ė is continuous at x.

f

Exercise 4.2.22. After looking at your solution of Lemma 4.2.15, state and

prove the corresponding results for limits.

Exercise 4.2.23. [In this exercise you should start from Deļ¬nition 1.7.2]

Let U be an open set in R. Show that a function f : U ā’ R is diļ¬erentiable

at t ā U with derivative f (t) if and only if

f (t + h) ā’ f (t)

ā’ f (t)

h

as h ā’ 0 (through values of h with t + h ā U ).

Exercise 4.2.24. In Chapter 6 we approach the properties of diļ¬erentia-

tion in a more general manner. However the reader will probably already

have met results like the following which can be proved using Exercises 4.2.22

and 4.2.23.

(i) If f, g : (a, b) ā’ R are diļ¬erentiable at x ā (a, b), then so is the sum

f + g and we have (f + g) (x) = f (x) + g (x).

(ii) If f, g : (a, b) ā’ R are diļ¬erentiable at x ā (a, b), then so is the

product f Ć— g and we have (f Ć— g) (x) = f (x)g(x) + f (x)g (x). [Hint: f (x +

h)g(x + h) ā’ f (x)g(x) = (f (x + h) ā’ f (x))g(x + h) + f (x)(g(x + h) ā’ g(x)).]

(iii) If f : (a, b) ā’ R is nowhere zero and f is diļ¬erentiable at x ā (a, b),

then so is 1/f and we have (1/f ) (x) = ā’f (x)/f (x)2 .

4

Thus it does not matter whether f is deļ¬ned at x or not (and, if it is deļ¬ned, it does

not matter what the value of f (x) is).

56 A COMPANION TO ANALYSIS

(iv) State accurately and prove a result along the lines of (ii) and (iii)

dealing with the derivative of f /g.

(v) If c ā R, c = 0 and f : R ā’ R is diļ¬erentiable at x, show that the

function fc deļ¬ned by fc (t) = f (ct) [t ā R] is diļ¬erentiable at cā’1 x and we

have fc (cā’1 x) = cf (x). What happens if c = 0?

(vi) Use part (ii) and induction on n to show that if rn (x) = xn , then rn

is everywhere diļ¬erentiable with rn (x) = nrnā’1 (x) [n ā„ 1]. Hence show that

every polynomial is everywhere diļ¬erentiable. If P and Q are polynomials

and Q(t) = 0 for all t ā (a, b) show that P/Q is everywhere diļ¬erentiable on

(a, b).

Exercise 4.2.25. (i) Use part (ii) of Exercise 4.2.24 to show that, if f, g :

(a, b) ā’ R satisfy the equation f (t)g(t) = 1 for all t ā (a, b) and are diļ¬er-

entiable at x ā (a, b) then g (x) = ā’f (x)/f (x)2 .

(ii) Explain why we can not deduce part (iii) of Exercise 4.2.24 directly

from part (i) of this exercise. Can we deduce the result of part (i) of this

exercise directly from part (iii) of Exercise 4.2.24?

(iii) Is the following statement true or false? If f, g : (a, b) ā’ R are

diļ¬erentiable at x ā (a, b) and f (x)g(x) = 1 then g (x) = ā’f (x)/f (x)2 .

Give a proof or counterexample.

Exercise 4.2.26. From time to time the eagle eyed reader will observe state-

ments like

ā˜f (x) ā’ ā as x ā’ ā’āā™

which have not been formally deļ¬ned. If this really bothers her, she is probably

reading the wrong book (or the right book but too early). It can be considered

a standing exercise to ļ¬ll in the required details.

In Appendix D, I sketch a method used in Beardonā™s elegant treatment [2]

which avoids the need for such repeated deļ¬nitions.

4.3 A central theorem of analysis

In this section we prove Theorem 4.3.4 which says that a real-valued con-

tinuous function on a closed bounded set in Rm is bounded and attains its

bounds. This result together with the intermediate value theorem (proved as

Theorem 1.6.1) and the mean value inequality (proved as Theorem 1.7.1 and

later in a more general context as Theorem 6.3.1) are generally considered

to be the central theorems of elementary analysis.

Our next result looks a little abstract at ļ¬rst.

57

Please send corrections however trivial to twk@dpmms.cam.ac.uk

Theorem 4.3.1. Let K be a closed bounded subset of Rm and f : K ā’ Rp

a continuous function. Then f (K) is closed and bounded.

Thus the continuous image of a closed bounded set is closed and bounded.

Proof. By Theorem 4.2.2 (ii), we need only prove that any sequence in f (K)

has a subsequence converging to a limit in f (K).

To this end, suppose that yn is a sequence in f (K). By deļ¬nition, we can

ļ¬nd xn ā K such that f (xn ) = yn . Since K is closed and bounded subset,

Theorem 4.2.2 (i) tells us that there exist n(j) ā’ ā and x ā K such that

xn(j) ā’ x as j ā’ ā. Since f is continuous,

yn(j) = f (xn(j) ) ā’ f (x) ā f (K)

and we are done.

Exercise 4.3.2. Let N : Rm ā’ R be given by N (x) = x . Show that

N is continuous. Deduce in particular that if xn ā’ x as n ā’ ā, then

xn ā’ x .

Exercise 4.3.3. (i) Let A be the open interval (0, 1). Show that the map

f : A ā’ R given by f (x) = 1/x is continuous but that f (A) is unbounded.

Thus the continuous image of a bounded set need not be bounded.

(ii) Let A = [1, ā) = {x ā R : x ā„ 1} and f : A ā’ R be given by

f (x) = 1/x. Show that A is closed and f is continuous but f (A) is not

closed. Thus the continuous image of a closed set need not be closed.

(iii) Show that the function Ļ : R2 ā’ R given by Ļ(x, y) = x is continu-

ous. (The function Ļ is called a projection.) Show that the set

A = {(x, 1/x) : x > 0}

is closed in R2 but that Ļ(A) is not.

We derive a much more concrete corollary.

Theorem 4.3.4. Let K be a closed bounded subset of Rm and f : K ā’ R a

continuous function. Then we can ļ¬nd k1 and k2 in K such that

f (k2 ) ā¤ f (k) ā¤ f (k1 )

for all k ā K.

Proof. Since f (K) is a non-empty bounded set, it has a supremum M say.

Since f (K) is closed, M ā f (K), that is M = f (k1 ) for some k1 ā K. We

obtain k2 similarly.

58 A COMPANION TO ANALYSIS

In other words, a real-valued continuous function on a closed bounded

set is bounded and attains its bounds. Less usefully we may say that, in this

case, f actually has a maximum and a minimum. Notice that there is no

analogous result for vector-valued functions. Much popular economic writing

consists of attempts to disguise this fact (there is unlikely to be a state of

the economy in which everybody is best oļ¬).

Exercise 4.3.5. When I was an undergraduate, we used another proof of

Theorem 4.3.4 which used lion hunting to establish that f was bounded and

then a clever trick to establish that it attains its bounds.

(i) We begin with some lion hunting in the style of Exercise 4.1.14. As

in that exercise, we shall only consider the case m = 2, leaving the general

case to the reader. Suppose, if possible, that f (K) is not bounded above (that

is, given any Īŗ > 0, we can ļ¬nd a x ā K such that f (x) > Īŗ).

Since K is closed and bounded, we can ļ¬nd a rectangle S0 = [a0 , b0 ] Ć—

[a0 , b0 ] ā K. Show that we can ļ¬nd a sequence of pairs of intervals [an , bn ]

and [an , bn ] such that

f (K ā© [an , bn ] Ć— [an , bn ]) is not bounded,

anā’1 ā¤ an ā¤ bn ā¤ bnā’1 , anā’1 ā¤ an ā¤ bn ā¤ bnā’1 ,

and bn ā’ an = (bnā’1 ā’ anā’1 )/2, bn ā’ an = (bnā’1 ā’ anā’1 )/2,

for all n ā„ 1.

Show that an ā’ c as n ā’ ā for some c ā [a0 , b0 ] and an ā’ c as n ā’ ā

for some c ā [a0 , b0 ]. Show that c = (c, c ) ā K. Use the fact that f is

continuous at c to show that there exists an > 0 such that, if x ā K and

x ā’ c < , then f (x) < f (c) + 1. Show that there exists an N such that

[an , bn ] Ć— [an , bn ] ā B(c, )

for all n ā„ N and derive a contradiction.

Hence deduce that f (K) is bounded above. Show also that f (K) is bounded

below.

(ii) Since any non-empty bounded subset of R has a supremum, we know

that M = sup f (K) and m = inf f (K) exist. We now produce our clever

trick. Suppose, if possible, that f (x) = M for all x ā K. Explain why,

if we set g(x) = 1/(M ā’ f (x)), g : K ā’ R will be a well deļ¬ned strictly

positive continuous function. Deduce that there exists a real number M > 0

such that g(x) ā¤ M for all x ā K and show that f (x) ā¤ M ā’ 1/M for all

x ā K. Explain why this contradicts the deļ¬nition of M and conclude that

there must exist some k1 ā K such that f (k1 ) = M . We obtain k2 similarly.

(The author repeats the remark he made on page 38 that amusing as proofs

59

Please send corrections however trivial to twk@dpmms.cam.ac.uk

like these are, clever proofs should only be used when routine proofs do not

work.)

Our next theorem is just a particular but useful case of Theorem 4.3.4.

Theorem 4.3.6. Let f : [a, b] ā’ R be a continuous function. Then we can

ļ¬nd k1 , k2 ā [a, b] such that

f (k2 ) ā¤ f (x) ā¤ f (k1 )

for all x ā [a, b].

Later we will use this result to prove Rolleā™s theorem (Theorem 4.4.4)

from which in turn we shall obtain the mean value theorem (Theorem 4.4.1).

Theorem 4.3.4 can also be used to prove the fundamental theorem of

algebra which states that every complex polynomial has a root. If the reader

cannot wait to see how this is done then she can look ahead to section 5.8.

Exercise 4.3.7. (i) Prove Theorem 4.3.6 directly from the one-dimensional

version of the Bolzano-Weierstrass theorem. (Essentially just repeat the ar-

guments of Theorem 4.3.1.)

(ii) Give an example of a continuous f : (a, b) ā’ R which is unbounded.

(iii) Give an example of a continuous f : (a, b) ā’ R which is bounded but

does not attain its bounds.

(iv) How does your argument in (i) fail in (ii) and (iii)?

(v) Suppose now we work over Q and write [a, b] = {x ā Q : a ā¤ x ā¤ b}.

Show that f (x) = (1+(x2 ā’2)2 )ā’1 deļ¬nes a continuous function f : [0, 2] ā’ Q

which is continuous and bounded but does not attain its upper bound. How

does your argument in (i) fail?

Deļ¬ne a continuous function g : [0, 2] ā’ Q which is continuous and

bounded but does not attain either its upper bound or its lower bound. Deļ¬ne

a continuous function h : [0, 2] ā’ Q which is continuous but unbounded.

We conclude this section with an exercise which emphasises once again the

power of the hypotheses ā˜closed and boundedā™ combined with the Bolzano-

Weierstrass method. The result is important but we shall not make much

use of it.

Exercise 4.3.8. (i) By picking xj ā Kj and applying the Bolzano-Weierstrass

theorem, prove the following result.

Suppose that K1 , K2 , . . . are non-empty bounded closed sets in Rm such

that K1 ā K2 ā . . . . Then ā Kj = ā…. (That is, the intersection of a

j=1

nested sequence of bounded, closed, non-empty sets is itself non-empty.)

60 A COMPANION TO ANALYSIS

(ii) By considering Kj = [j, ā), show that boundedness cannot be dropped

from the hypothesis.

(iii) By considering Kj = (0, j ā’1 ), show that closedness cannot be dropped

from the hypothesis.

Exercises K.29 to K.36 discuss a substantial generalisation of Exercise 4.3.8

called the Heine-Borel theorem.

4.4 The mean value theorem

Traditionally one of the ļ¬rst uses of the theorem that every continuous func-

tion on a closed interval is bounded and attains its bounds has been to prove

a slightly stronger version of the mean value inequality.

In common with DieudonnĀ“ ([13], page 142) and Boas ([8], page 118), I

e

think that the mean value inequality is suļ¬cient for almost all needs and

that the work required to understand the subtleties in the statement and

proof of Theorem 4.4.1 far outweigh any gain.

However, Theorem 4.4.1 is likely to remain part of the standard analysis

course for many years, so I include it here.

Theorem 4.4.1. (The mean value theorem.) If f : [a, b] ā’ R is a

continuous function with f diļ¬erentiable on (a, b), then we can ļ¬nd a c ā

(a, b) such that

f (b) ā’ f (a) = (b ā’ a)f (c).

Here are some immediate consequences.

Lemma 4.4.2. If f : [a, b] ā’ R is a continuous function with f diļ¬eren-

tiable on (a, b), then the following results hold.

(i) If f (t) > 0 for all t ā (a, b) then f is strictly increasing on [a, b].

(That is, f (y) > f (x) whenever b ā„ y > x ā„ a.)

(ii) If f (t) ā„ 0 for all t ā (a, b) then f is increasing on [a, b]. (That is,

f (y) ā„ f (x) whenever b ā„ y > x ā„ a.)

(iii) If f (t) = 0 for all t ā (a, b) then f is constant on [a, b]. (That is,

f (y) = f (x) whenever b ā„ y > x ā„ a.)

Proof. We prove part (i), leaving the remaining parts to the reader. If b ā„

y > x ā„ a, then the mean value theorem (Theorem 4.4.1) tells us that

f (y) ā’ f (x) = (y ā’ x)f (c)

for some c with y > c > x. By hypothesis f (c) > 0, so f (y) ā’ f (x) > 0.

61

Please send corrections however trivial to twk@dpmms.cam.ac.uk

Exercise 4.4.3. Prove Theorem 1.7.1 from Theorem 4.4.1.

The key step in proving Theorem 4.4.1 is the proof of the special case

when f (a) = f (b).

Theorem 4.4.4. (Rolleā™s theorem.) If g : [a, b] ā’ R is a continuous

function with g diļ¬erentiable on (a, b) and g(a) = g(b), then we can ļ¬nd a

c ā (a, b) such that g (c) = 0.

The next exercise asks you to show that the mean value theorem follows

from Rolleā™s theorem.

Exercise 4.4.5. (i) If f is as in Theorem 4.4.1, show that we can ļ¬nd a real

number A such that, setting

g(t) = f (t) ā’ At,

the function g satisļ¬es the conditions of Theorem 4.4.4.

(ii) By applying Rolleā™s theorem (Theorem 4.4.4) to the function g in (i),

obtain the mean value theorem (Theorem 4.4.1). (Thus the mean value the-

orem is just a tilted version of Rolleā™s theorem.)

Cauchy produced an interesting variant on the argument of Exercise 4.4.5.

which we give as Exercise K.51.

Exercise 4.4.6. The following very easy consequence of Deļ¬nition 1.7.2 will

be used in the proof of Rolleā™s theorem. Let U be an open set in R and let

f : U ā’ R be diļ¬erentiable at t ā U with derivative f (t). Show that if

tn ā U , tn = t and tn ā’ t as n ā’ ā, then

f (tn ) ā’ f (t)

ā’ f (t)

tn ā’ t

as n ā’ ā.

We now turn to the proof of Rolleā™s theorem.

Proof of Theorem 4.4.4. Since the function g is continuous on the closed in-

terval [a, b], Theorem 4.3.6 tells us that it is bounded and attains its bounds.

More speciļ¬cally, we can ļ¬nd k1 , k2 ā [a, b] such that

g(k2 ) ā¤ g(x) ā¤ g(k1 )

for all x ā [a, b]. If both k1 and k2 are end points of [a, b] (that is k1 , k2 ā

{a, b}) then

g(a) = g(b) = g(k1 ) = g(k2 )

62 A COMPANION TO ANALYSIS

and g(x) = g(a) for all x ā [a, b]. Taking c = (a + b)/2, we have g (c) = 0

(the derivative of a constant function is zero) and we are done.

If at least one of k1 and k2 is not an end point there is no loss in generality

in assuming that k1 is not an end point (otherwise, consider ā’g). Write

c = k1 . Since c is not an end point, a < c < b and we can ļ¬nd a Ī“ > 0 such

that a < c ā’ Ī“ < c + Ī“ < b. Set xn = c ā’ Ī“/n. Since c is a maximum for g,

we have g(c) ā„ g(xn ) and so

g(xn ) ā’ g(c)

ā„0

xn ā’ c

for all n. Since

g(xn ) ā’ g(c)

ā’ g (c),

xn ā’ c

it follows that g (c) ā„ 0. However, if we set yn = c + Ī“/n, a similar argument

shows that

g(yn ) ā’ g(c)

ā¤0

yn ā’ c

for all n and so g (c) ā¤ 0. Since 0 ā¤ g (c) ā¤ 0, it follows that g (c) = 0 and

we are done.

(We look more closely at the structure of the preceeding proof in Exer-

cise K.45.)

In his interesting text [11], R. P. Burn writes

Both Rolleā™s theorem and the mean value theorem are geomet-

rically transparent. Each claims, with slightly more generality

in the case of the mean value theorem, that for a graph of a

diļ¬erentiable function, there is always a tangent parallel to the

chord.

My view is that the apparent geometrical transparency is due to our strong

intuitive feeling a function with positive derivative ought to increase ā” which

is precisely what we are ultimately trying to prove5 . It is because of this strug-

gle between intuition and rigour that the argument of the second paragraph

of the proof always brings to my mind someone crossing a tightrope above

5

This should not be interpreted as a criticism of Burnā™s excellent book. He is writing

a ļ¬rst course in analysis and is trying to persuade the unwilling reader that what looks

complicated is actually simple. I am writing a second course in analysis and trying to

persuade the unwilling reader that what looks simple is actually complicated.

63

Please send corrections however trivial to twk@dpmms.cam.ac.uk

a pool full of crocodiles. Let me repeat to any reader tempted to modify

that argument, we wish to use Theorem 4.4.1 to prove that a function with

positive derivative is increasing and so we cannot use that result to prove

Theorem 4.4.1. If you believe that you have a substantially simpler proof of

Rolleā™s theorem than the one given above, ļ¬rst check it against Exercise K.46

and then check it with a professional analyst. Exercise K.43 gives another

use of the kind of argument used to prove Rolleā™s theorem.

If the reader uses Theorem 4.4.1, it is important to note that we know

nothing about c apart from the fact that c ā (a, b).

Exercise 4.4.7. Suppose that k2 is as in the proof of Theorem 4.4.4. Show

explicitly that, if k2 is not an end point, g (k2 ) = 0.

Exercise 4.4.8. Suppose that g : R ā’ R is diļ¬erentiable, that a < b and

that g(a) = g(b). Suppose k1 and k2 are as in the proof of Theorem 4.4.4.

Show that, if k1 = a, then g (a) ā¤ 0 and show by example that we may have

g (a) < 0. State similar results for the cases b = k1 and a = k2 .

Exercise 4.4.9. (This exercise should be compared with Lemma 4.4.2.)

(i) Suppose that f : (a, b) ā’ R is diļ¬erentiable and increasing on (a, b).

Show that f (t) ā„ 0 for all t ā (a, b).

(ii) If f : R ā’ R is deļ¬ned by f (t) = t3 , show that f is diļ¬erentiable and

everywhere strictly increasing yet f (0) = 0.

Exercise 4.4.10. I said above that the mean value inequality is suļ¬cient

for most purposes. For the sake of fairness here is an example where the

extra information provided by Rolleā™s theorem does seem to make a diļ¬erence.

Here, as elsewhere in the exercises, we assume that reader knows notations

like F (r) for the rth derivative of F and can do things like diļ¬erentiating a

polynomial which have not been explicitly treated in the main text.

Suppose that f : R ā’ R is n times diļ¬erentiable and that

ńņš. 7 |