The theorem has been summarised by Conway as follows:-

If E is closed and bounded, say Heine-Borel,

And also Euclidean, then we can tell

That, if it we smother

With a large open cover,

There™s a ¬nite re¬nement as well!

450 A COMPANION TO ANALYSIS

Suppose that E is a non-empty set in F which is bounded above. By

considering

K = {[a, b] : a ∈ E and b ≥ e for all e ∈ E},

or otherwise, deduce that E has a supremum. Conclude that statement (A)

is equivalent to the fundamental axiom of analysis.

Exercise K.38. [4.4, T!] In this question you may assume the standard

properties of sin and cos but not their power series expansion.

(i) By considering the sign of f1 (x), when f1 (t) = t ’ sin t, show that

t ≥ sin t

for all t ≥ 0.

(ii) By considering the sign of f2 (x), when f2 (t) = cos t ’ 1 + t2 /2!, show

that

t2

cos t ≥ 1 ’

2!

for all t ≥ 0.

(iii) By considering the sign of f3 (x), when f3 (t) = sin t ’ t + t3 /3!, show

that

t3

sin t ≥ t ’

3!

for all t ≥ 0.

(iv) State general results suggested by parts (i) to (iii) and prove them

by induction. State and prove corresponding results for t < 0.

(v) Using (iv), show that

N

(’1)n t2n+1

’ sin t

(2n + 1)!

n=0

as N ’ ∞ for all t ∈ R. State and prove a corresponding result for cos.

[This question could be usefully illustrated by computer graphics.]

Exercise K.39. (Jensen™s inequality.) [4.4, T] Good inequalities are

the diamonds of analysis, valuable but rare. Jensen earned his living in the

telephone industry and pursued mathematics in his spare time. He has two

marvelous inequalities named after him. This exercise discusses one of them.

451

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

We call a function f : (a, b) ’ R convex if, whenever x1 , x2 ∈ (a, b) and

1 ≥ » ≥ 0, we have

»f (x1 ) + (1 ’ »)f (x2 ) ≥ f (»x1 + (1 ’ »)x2 ).

(The centre of mass of a particle of mass » at (x1 , f (x1 )) and a particle of

mass 1 ’ » at (x2 , f (x2 )) lies above the graph of the curve y = f (x).)

(i) Suppose g : (a, b) ’ R is twice di¬erentiable with g (t) ≥ 0 for all

t ∈ (a, b). Suppose further that a < x1 < x2 < b and g(x1 ) = g(x2 ) = 0 .

Sketch g. By using the mean value theorem twice, or otherwise, show that

g(t) ¤ 0 for all t ∈ [x1 , x2 ].

(ii) Suppose f : (a, b) ’ R is twice di¬erentiable with f (t) ≥ 0 for all

t ∈ (a, b). Sketch f . By using (i), or otherwise, show that f is convex.

We have now obtained a good supply of convex functions and proceed to

discuss a form of Jensen™s inequality.

(iii) Suppose f : (a, b) ’ R is convex. By writing

n+1

»j xj = »n+1 xn+1 + (1 ’ »n+1 )yn+1

j=1

and using induction, or otherwise, show that if

n

x1 , x2 , . . . , xn ∈ (a, b), »1 , »2 , . . . , »n ≥ 0 and »j = 1,

j=1

then

n n

»j f (xj ) ≥ f »j xj .

j=1 j=1

This is Jensen™s inequality.

(iv) The importance of Jensen™s inequality lies in the fact that it has many

classical inequalities as special cases. By taking f (t) = ’ log t and »j = 1/n

obtain the arithmetic-geometric inequality

x1 + x 2 + · · · + x n

≥ (x1 x2 . . . xn )1/n

n

whenever the xj are strictly positive real numbers.

[We give a more sophisticated account of these matters in Exercise K.128]

452 A COMPANION TO ANALYSIS

Exercise K.40. [4.4, P] (This uses Exercise K.39.) The four vertices A,

B, C, D of a quadrilateral lie in anti-clockwise order on a circle radius a

and center O. We write 2θ1 = ∠AOB, 2θ2 = ∠BOC, 2θ3 = ∠COD, 2θ4 =

∠DOA. Find the area of the quadrilateral and state the relation that θ1 , θ2 ,

θ3 and θ4 must satisfy.

Use Jensen™s inequality to ¬nd the form of a quadrilateral of greatest area

inscribed in a circle of radius a.

Use Jensen™s inequality to ¬nd the form of an n-gon of greatest area

inscribed in a circle [n ≥ 3].

Use Jensen™s inequality to ¬nd the form of an n-gon of least area circum-

scribing a circle [n ≥ 3].

[Compare Exercise K.295.]

Exercise K.41. [4.4, P, S] Suppose that g : [0, 1] ’ R is a continuous

function such that, for every c ∈ (0, 1), we can ¬nd a k with

1

0 < c ’ k < c < c + k < 1 and g(c) = 2 (g(c + k) + g(c ’ k)).

Show that, if g(0) = g(1) = 0, then g(t) = 0 for all t ∈ [0, 1]. What can you

say if we drop the conditions on g(0) and g(1)?

(In one dimension this is just a brain teaser, but it turns out that the

generalisation to higher dimensions is a branch of mathematics in itself.)

Exercise K.42. [4.4, P] De¬ne f (x) = x2 sin x’4 for x = 0, f (0) = 0.

Show that (f (h) ’ f (0))/h ’ 0 as h ’ 0 and deduce that f is di¬erentiable

at 0. Use the chain rule to show that f is di¬erentiable at all x = 0. Show

that, although f is everywhere di¬erentiable, its derivative is not continuous.

[This example is extended in Exercise K.165.]

Exercise K.43. (Darboux™s theorem.) [4.4, T] Exercise K.42 shows

that the derivative of a di¬erentiable function need not be continuous. Oddly

enough, it still satis¬es a form of the intermediate value theorem. Suppose

that (±, β) ⊃ [a, b], and that f : (±, β) ’ R is di¬erentiable. Darboux™s

theorem asserts that if k lies between f (a) and f (b) then there is a c ∈ [a, b]

with f (c) = k.

Explain why there is no loss in generality in supposing that f (a) > k >

f (b). Set g(x) = f (x) ’ kx. By looking at g (a) and g (b) show that g cannot

have a maximum at a or b. Use the method of the proof of Rolle™s theorem

(Theorem 4.4.4) to show that there exists a c ∈ (a, b) with g (c) = 0 and

deduce Darboux™s theorem.

Use Darboux™s theorem to show that, if (±, β) ⊇ [a, b] and f : (±, β) ’ R

is twice di¬erentiable with a local maximum at a and a local minimum at b,

then f (c) = 0 for some c ∈ [a, b].

453

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

Darboux™s theorem shows that not all functions can be derivatives. It is

a very hard problem to characterise those functions f : R ’ R for which

there exists an F : R ’ R with F = f .

Exercise K.44. [4.4, P] (i) Suppose f : R ’ R is such that for each n ≥ 1

we can ¬nd xn and yn with |xn |, |yn | ¤ 1 such that |xn ’ yn | < n’1 and

|f (xn ) ’ f (yn )| ≥ 1. Consider the following argument.

˜Pick zn ∈ [’1, 1] lying between xn and yn . By the Bolzano-Weierstrass

theorem, we can ¬nd a sequence n(j) ’ ∞ and a z ∈ [’1, 1] such that

zn(j) ’ z and so f must jump by at least 1 at z. Thus f cannot be continuous

at z and in particular cannot be everywhere continuous.™

Which parts of the argument are correct and which incorrect? Is the

conclusion correct? Give a proof or counterexample.

(ii) Suppose f : R ’ R is continuous and for each n ≥ 1 we can ¬nd xn

and yn with |xn |, |yn | ¤ 1 such that |xn ’yn | < n’2 and |f (xn )’f (yn )| ≥ n’1 .

Consider the following argument.

˜By the mean value theorem, we can ¬nd zn ∈ [’1, 1] with |f (zn )| ≥ n.

By the Bolzano-Weierstrass theorem, we can ¬nd a sequence n(j) ’ ∞ and

a z ∈ [’1, 1] such that zn(j) ’ z and so f (z) (if it exists) must be arbitrarily

large. Thus f cannot be di¬erentiable at z and in particular cannot be

everywhere di¬erentiable.™

Which parts of the argument are correct and which incorrect? Is the

conclusion correct? Give a proof or counterexample.

Exercise K.45. [4.4, H] In this question we examine our proof of Rolle™s

theorem (Theorem 4.4.4) in detail.

(i) Obtain Rolle™s theorem as the consequence of the following three state-

ments.

(a) If g : [a, b] ’ R is continuous we can ¬nd k1 , k2 ∈ [a, b] such that

g(k2 ) ¤ g(x) ¤ g(k1 ) for all x ∈ [a, b].

(b) If g : [a, b] ’ R is a function with g(a) = g(b) and we can ¬nd

k1 , k2 ∈ [a, b] such that g(k2 ) ¤ g(x) ¤ g(k1 ) for all x ∈ [a, b], then at least

one of the following three statements is true:- a < k1 < b, a < k2 < b or g is

constant.

(c) If g : [a, b] ’ R is di¬erentiable at a point c with a < c < b such that

g(x) ¤ g(k1 ) for all x ∈ [a, b] (or g(c) ¤ g(x) for all x ∈ [a, b], then g (c) = 0.

(ii) Suppose we now work over Q as we did in Example 1.1.3. Prove that

the following results still hold.

(b™) If b > a and f : Q ’ Q is a function with f (a) = f (b) and we can

¬nd a ¤ k1 , k2 ¤ b such that f (k2 ) ¤ f (x) ¤ f (k1 ) for all a ¤ x ¤ b, then at

least one of the following three statements is true:- a < k1 < b, a < k2 < b

or f is constant.

454 A COMPANION TO ANALYSIS

(c™) If b > a and f : Q ’ Q is di¬erentiable at a point c with a < c < b

such that f (x) ¤ f (c) for all a ¤ x ¤ b (or f (c) ¤ f (x) for all a ¤ x ¤ b,

then f (c) = 0.

This shows that the key to the mean value theorem is the fact that if

we work with the reals a continuous function on a closed bounded interval is

bounded and attains its bounds. All the rest is mere algebra.

Exercise K.46. [4.4, H, S] (This is more of a comment than an exercise.)

Occasionally even a respected mathematician2 may fall into the trap of

believing that ˜Theorem 4.4.4 is subtle only because we do not ask the deriva-

tive to be continuous and do not include the endpoints™. It is worth pointing

out that, even in this case, there is no simple proof. If you ever ¬nd yourself

believing that there is a simple proof, check it against the following example.

Consider the function f : Q ’ Q given in Example 1.1.3. Set

g(t) = 1 ’ t + f (t).

Show that the function g : Q ’ Q has continuous derivative g and that g

satis¬es the conclusions of the intermediate value theorem. (More formally,

if a, b ∈ Q with a < b and γ ∈ Q is such that either g(a) ¤ γ ¤ g(b) or

g(b) ¤ γ ¤ g(a), then there exists a c ∈ Q with a ¤ c ¤ b such that g(c) = γ.

The statement is complicated but the veri¬cation is trivial.)

Show that g(0) = g(2) = 0 but that there does not exist a c ∈ Q with

g (c) = 0.

Exercise K.47. (Tchebychev polynomials.) [4.4, G, P] Prove that

(cos θ + i sin θ)n = cos nθ + i sin nθ,

whenever n is a positive integer. (This is just to get you started so, provided

you make clear what you are assuming, you may make any assumptions you

wish.)

By taking real and imaginary parts, show that there is a real polynomial

of degree n such that

Tn (cos θ) = cos nθ for all real θ.

Write down T0 (t), T1 (t), T2 (t), and T3 (t) explicitly.

Prove that, if n ≥ 1.

(a) Tn+1 (t) = 2tTn (t)’Tn’1 (t), (why can we omit the restriction |t| ¤ 1?)

(b) the coe¬cient of tn in Tn (t) is 2n’1 ,

2

Name withheld.

455

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

(c) |Tn (t)| ¤ 1 for all |t| ¤ 1,

(d) T has n distinct roots all in (’1, 1).

Show also that, if n, m ≥ 1, then

1

Tn (x)Tm (x) π

dx = δnm ,

(1 ’ x2 )1/2 2

’1

where δnm = 0 if n = m and δnn = 1. What can you say if m = 0 and n ≥ 0?

The Tn are called the Tchebychev3 polynomials after their discoverer.

Exercise K.48. [4.4, T, ‘ ] (This requires Exercises 4.4.10 and K.47.) Reread

Exercise 4.4.10. Suppose now that a = ’1, b = 1 and we choose the xj of

Exercise 4.4.10 to be the n roots of Tn (see part (d) of Exercise 4.4.10). Using

part (b) of Exercise 4.4.10, show that n (t ’ xj ) = 2’n+1 Tn (t). Conclude

j=1

(n)

that, if |f (x)| ¤ A for all x ∈ [’1, 1], then

A

|f (t) ’ P (t)| ¤

2n’1 n!

for all t ∈ [’1, 1].

Modify the ideas above to deal with a general interval [a, b].

(Note that, whilst it is, indeed, true that, if you are going to interpolate a

function by a polynomial, the zeros of the Tchebychev polynomial are good

points of interpolation, it remains the case that it is rarely a good idea to

interpolate a function by a polynomial of high degree. One way of viewing the

problem is to observe that supx∈[a,b] |f (n) (x)| may grow very fast with n. For

another example involving choosing appropriate points see Exercise K.213.)

Exercise K.49. (The nth mean value theorem.) [4.4, T] The object

of this exercise is to prove a version of Taylor™s theorem corresponding to

Theorem 4.4.1 (the one dimensional mean value theorem).

(i) Reread Exercise 4.4.10.

(ii) Suppose that f : (u, v) ’ R is n + 1 times di¬erentiable and that

[a, b] ⊆ (u, v). Show that we can ¬nd a polynomial P of degree n such that

(f ’ P )(r) (a) = 0 for r = 0, 1, . . . , n. Give P explicitly.

(iii) We are interested in the error

E(b) = f (b) ’ P (b).

To this end, we consider the function

n+1

x’a

F (x) = f (x) ’ P (x) ’ E(b) .

b’a

3

The modern view is that Tchebychev should have called himself Chebychev. He seems

to have preferred Tchebyche¬.

456 A COMPANION TO ANALYSIS

Show that F (a) = F (a) = · · · = F (n) (a) = 0 and F (b) = 0.

(iv) By using Rolle™s theorem show that F (c1 ) = 0 for some c1 with

a < c1 < b. By repeating your argument n times show that that F (n+1) (c) = 0

for some c with a < c < b.

(v) Deduce that

(n + 1)!

0 = f (n+1) (c) ’ E(b)

(b ’ a)n+1

and so

n

f (j) (a) f (n+1) (c)

j

(b ’ a)n+1

f (b) ’ (b ’ a) =

j! (n + 1)!

j=0

for some c with a < c < b.

(vi) Show that

n

f (j) (b) f (n+1) (c )

j

(b ’ a)n+1

f (a) ’ (b ’ a) =

j! (n + 1)!

j=0

for some c with a < c < b.

(vii) Suppose that f, g : (u, v) ’ R are n + 1 times di¬erentiable and

that a ∈ (u, v). Show that if

f (a) = f (a) = · · · = f (n) (a) = g(a) = g (a) = · · · = g (n) (a) = 0

but g (n) (a) = 0, then, if f (n+1) and g (n+1) are continuous at a, it follows that

f (n+1) (a)

f (t)

’ (n+1)

g(t) g (a)

as t ’ a.

Exercise K.50. [4.4, P, ‘] Suppose that f : R ’ R is 2n + 2 times dif-

ferentiable. By considering polynomials of the form xk (1 ’ x)l , or otherwise,

show that there is a unique polynomial p of degree 2n + 1 such that

p(r) (0) = f (r) (0) and p(r) (1) = f (r) (1) for all 0 ¤ r ¤ n.

Show that the error at y ∈ [0, 1]

f (n+2) (ξ) n+1

y (y ’ 1)n ,

E(y) = f (y) ’ P (y) is given by E(y) =

(2n + 2)!

for some ξ ∈ (0, 1).

457

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

Exercise K.51. (Cauchy™s mean value theorem.) [4.4, P, ‘ ] Suppose

that f, g : [a, b] ’ R are continuous functions with f and g di¬erentiable on

(a, b) and g (x) = 0 for x ∈ (a, b).

(i) Use the mean value theorem (Theorem 4.4.1) to show that g(a) = g(b).

(ii) Show that we can ¬nd a real number A such that, setting

h(t) = f (t) ’ Ag(t),

the function h satis¬es the conditions of Rolle™s theorem (Theorem 4.4.4).

(iii) By applying Rolle™s theorem (Theorem 4.4.4) to the function h in

(ii), show that there exists a c ∈ (a, b) such that

f (b) ’ f (a) f (c)

= .

g(b) ’ g(a) g (c)

(This is Cauchy™s mean value theorem).

(iv) Suppose that f (a) = g(a) = 0 and f (t)/g (t) ’ L as t ’ a through

values of t > a. Prove the version of L™Hˆpital™s rule which states that, under

o

these conditions, f (t)/g(t) ’ L as t ’ a through values of t > a.

(v) Suppose that f, g : (u, v) ’ R are di¬erentiable, that a ∈ (u, v),

f (a) = g(a) = 0 and f (t)/g (t) ’ L as t ’ a. Show that f (t)/g(t) ’ L as

t ’ a.

(vi) Suppose that f, g : (u, v) ’ R are di¬erentiable, that a ∈ (u, v),

f (a) = f (a) = · · · = f (n’1) (a) = g(a) = g (a) = g (a) = · · · = g (n’1) = 0

and f (n) (t)/g (n) (t) ’ L as t ’ a. Show that f (t)/g(t) ’ L as t ’ a.

(vii) Suppose that F : (u, v) ’ R is n times di¬erentiable, that a ∈ (u, v)

and F (n) is continuous at a. By setting

n

F (j) (a)

(t ’ a)j

f (t) = F (t) ’

j!

j=0

and g(t) = (t ’ a)n and applying (vi), show that

n

F (j) (a)

(t ’ a)j + (t)|t ’ a|n

F (t) =

j!

j=0

where (t) ’ 0 as t ’ a. (This is the local Taylor™s theorem which we obtain

by other arguments as Theorem 7.1.3.)

458 A COMPANION TO ANALYSIS

Exercise K.52. [4.4, P] (i) Write down the result of Exercise K.49 in the

speci¬c case n = 2, b ’ a = h.

(ii) Throughout the rest of this question f : R ’ R will be an everywhere

twice di¬erentiable function. Suppose that |f (t)| ¤ M0 and |f (t)| ¤ M2 for

all t ∈ R. Show that

2M0 h

|f (t)| ¤ +

h 2M2

for all h > 0, and deduce, by choosing h carefully, that

|f (t)| ¤ (M0 M2 )1/2

for all t ∈ R.

(iii) Which of the following statements are true and which false? Give a

proof or counterexample as appropriate.

(a) Let a < b. If |f (t)| ¤ M0 and |f (t)| ¤ M2 for all t ∈ [a, b], then

|f (a)| ¤ (M0 M2 )1/2 .

(b) There exists an L > 0 such that, if a + L ≥ b, |f (t)| ¤ M0 and

|f (t)| ¤ M2 for all t ∈ [a, b], then |f (a)| ¤ (M0 M2 )1/2 .

(c) Given M0 , M2 > 0, we can ¬nd an L > 0 such that if, a + L ≥ b,

|f (t)| ¤ M0 and |f (t)| ¤ M2 for all t ∈ [a, b], then |f (a)| ¤ (M0 M2 )1/2 .

(d) There exists an in¬nitely di¬erentiable function g : R ’ R with