1

r1 +

1

r2 +

1

r3 +

1

r4 + · · · +

1

rn +

rn+1 + xn+1

but this works better in handwriting than in print so we shall use the compact

(but less suggestive) notation

x = [r1 , r2 , r3 , r4 , . . . , rn , rn+1 + xn+1 ].

We are interested in what happens when we consider the ˜truncation™

yn+1 = [r1 , r2 , r3 , r4 , . . . , rn , rn+1 ].

438 A COMPANION TO ANALYSIS

Write out the formula for yn in the style of . Show by induction, using

the formula xn+1 = x1n ’ rn+1 , or otherwise, that

pn+1 + pn xn+1

x= ,

qn+1 + qn xn+1

where

pn+1 = rn+1 pn + pn’1 , qn+1 = rn+1 qn + qn’1 ,

and p0 = 0, q0 = 1, p1 = 1, q1 = r1 . If you have met Euclid™s algorithm, you

should note that the recurrence equations for pn and qn are precisely those

which occur in that process. By considering the appropriate interpretation

of right hand side of equation when xn+1 = 0, show that

pn

yn =

qn

for n ≥ 1.

By using the recurrence equations for pn and qn , show that

pn’1 qn ’ pn qn’1 = (’1)n

for all n ≥ 1. Use this fact together with equation to show that

(’1)n xn+1

pn+1

x’ =

qn+1 qn+1 (qn+1 + qn xn+1 )

for all n ≥ 0. Hence show that x ’ pn /qn is alternately positive and negative

and that

1 pn 1

< x’ <

qn’1 qn qn qn+1 qn

for all n ≥ 1. By looking at the recurrence relations for qn , show that qn ≥ n

and deduce that pn /qn ’ x as n ’ ∞. It is thus natural to say that we have

expressed x as an in¬nite continued fraction

1

x= .

1

r1 +

1

r2 +

1

r3 +

r4 + . . .

439

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

We know that the process of constructing successive xn cannot terminate

if x is irrational, but we did not discuss the case x rational. Show that if r,

s, p, q are integers with s > r ≥ 0, q > p ≥ 0, then either r/s = p/q or

rp 1

’ ≥.

sq qs

By considering equation , conclude that, if x is rational, the process

of forming xn will indeed terminate. It is easy to see (but I leave it up the

reader to decide whether to check the details) that we can then express any

rational x with 0 < x < 1 as a continued fraction

x = [r1 , r2 , r3 , r4 , . . . , rN ]

for appropriate rj and N .

In this question we have shown how to express any real number x with

0 < x < 1 as a continued fraction. (So any real number can be written in the

form m or m + y where m is an integer and y can be expressed as a continued

fraction.) We have not discussed whether a general continued fraction

1

,

1

s1 +

1

s2 +

1

s3 +

s4 + . . .

with sn a strictly positive integer, converges. (A fairly simple adaptation

of the argument above shows that it does. Of course, we need to use the

fundamental axiom of analysis at some point.)

Exercise K.14. [1.5, T, ‘ ] In this exercise we continue the discussion of

Exercise K.13. We shall suppose, as before, that 0 < x < 1 and x is irra-

tional. One major advantage of continued fractions is that they give very

good rational approximations, as was shown in , which gave us the

estimate

pn 1 1

x’ ¤ 2.

< (†)

qn qn+1 qn qn

Conclude that, if x is a real number, we can ¬nd integers p and q with q ≥ 1

such that

p 1

x’ < 2.

q q

440 A COMPANION TO ANALYSIS

Since the larger qn is (for ¬xed n), the better the inequality (†), it is

natural to ask how small qn can be.

Use induction and the recurrence relation for qn obtained in Exercise K.13

to show that q(n) ≥ Fn where

Fn+1 = Fn + Fn’1

and F0 = 1, F1 = 1. By directly solving the recurrence relation, if you know

how, or by inductive veri¬cation, if you do not, show that

n+1 n+1

„1 ’ „ 2

√

Fn = ,

5

√ √

where „1 = (1 + 5)/2 and „2 = (1 ’ √ 5)/2. Explain why, if n is su¬ciently

n+1

large Fn is the integer closest to „1 / 5.

It is interesting to look for an example of a ˜worst behaved™ continued

fraction. If qn = Fn then the associated recurrence relation tells us that

rn = 1 and suggests we look at

1

x= .

1

1+

1

1+

1

1+

1 + ...

Since we have not proved the convergence of this continued fraction, our

argument is now merely heuristic. Explain why

1

x=

1+x

√

and deduce that, if 1 > x > 0,√ must have x = ( 5 ’ 1)/2. Returning to

we

full rigour, show that, if x = ( 5 ’ 1)/2, then the process of Exercise K.13

does, indeed, give

1

x= .

1

1+

1

1+

1

1+

1 + ...

as required.

441

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

Hardy and Wright™s delightful The Theory of Numbers [25] contains a dis-

cussion of the problem of approximating real numbers by rationals and shows

√

that ( 5 ’ 1)/2 and its relatives are indeed the ˜worst approximable™ num-

bers. (Chapter X of [25] deals with continued fractions and Chapter XI with

problems of approximation by rationals.) We discussed a related problem in

Exercise K.12.

Exercise K.15. (L´vy™s universal chord theorem.) [1.6, P] Suppose

e

f : [0, 1] ’ R is continuous and f (0) = f (1). By considering the function

g : [0, 1/2] ’ R de¬ned by g(x) = f (x + 1 ) ’ f (x) show that there exist

2

1

a, b ∈ [0, 1] such that b ’ a = 2 and f (b) = f (a). Show, more generally, that

1

there exist ±n , βn ∈ [0, 1] such that βn ’ ±n = n and f (βn ) = f (±n ). (This is

L´vy™s universal chord theorem. For more discussion, including a proof that

e

1

you cannot replace n by a more general h, see [8], page 109.)

Show that we can ¬nd an , bn ∈ [0, 1] such that an ¤ an+1 < bn+1 ¤ bn ,

bn ’ an = 2’n and f (bn ) = f (an ).

Now suppose that δ > 0, that f is de¬ned on the larger interval (’δ, 1+δ)

and f : (’δ, 1+δ) ’ R is everywhere di¬erentiable. Show that if f (0) = f (1),

then there exists a c ∈ [0, 1] such that f (c) = 0. (A little extra thought gives

the full Rolle™s theorem (Theorem 4.4.4).)

Exercise K.16. [1.6, P] Suppose that f : [a, b] ’ R is increasing and

g : [a, b] ’ R is continuous. Show that, if f (a) ≥ g(a) and f (b) ¤ g(b), then

there exists a c ∈ [a, b] such that f (c) = g(c).

Is it necessarily true that, if f (a) ¤ g(a) and f (b) ≥ g(b), then there

exists a c ∈ [a, b] such that f (c) = g(c). Is the result of the ¬rst paragraph

true if we replace ˜g continuous™ by ˜g decreasing™ ? Is the result of the ¬rst

paragraph true if we replace ˜f increasing™ by ˜f continuous™ ? Give proofs

or counterexamples as appropriate.

Exercise K.17. [3.1, P] Let c ∈ (a, b). Suppose fn : (a, b) ’ R is continu-

ous at c for each n ≥ 1.

Show that, if g : (a, b) ’ R is given by

g(x) = max(f1 (x), f2 (x), . . . , fN (x)),

then g is continuous at c, but show, by means of an example, that, even if

there exists a K such that |fn (x)| ¤ K for all n and all x ∈ (a, b), if we de¬ne

G : (a, b) ’ R by

G(x) = sup(f1 (x), f2 (x), . . . ),

then G need not be continuous at c.

442 A COMPANION TO ANALYSIS

Exercise K.18. [3.1, P] (This requires good command of the notion of

countability)

(i) Suppose that E is a bounded uncountable set of real numbers. Show

that there exist real numbers ± and β with ± < β such that both of the sets

{e ∈ E : e < γ} and {e ∈ E : e > γ}

are uncountable if and only if ± < γ < β.

(ii) State and prove the appropriate version of (ii) if we omit the condition

that E be bounded.

(iii) Give an example of a bounded in¬nite set E of real numbers such

that there does not exist a real γ with both of the sets

{e ∈ E : e < γ} and {e ∈ E : e > γ}

in¬nite.

Exercise K.19. [3.2, P] Consider a sequence of real numbers xn . Which of

the following statements are always true and which are sometimes true and

sometimes false? Give proofs or examples.

(i) There is a strictly increasing subsequence. (That is, we can ¬nd n(j)

with n(j + 1) > n(j) and xn(j+1) > xn(j) for all j ≥ 1.)

(ii) There is an increasing subsequence. (That is, we can ¬nd n(j) with

n(j + 1) > n(j) and xn(j+1) ≥ xn(j) for all j ≥ 1.)

(iii) If there is no increasing subsequence, there must be a strictly de-

creasing subsequence.

(iv) If there is no strictly increasing subsequence and no strictly decreasing

subsequence, then we can ¬nd an N such that xn = xN for all n ≥ N .

Exercise K.20. [3.2, P] (i) Suppose that an is a bounded sequence of real

numbers. Show that, if n(p) ’ ∞ as p ’ ∞ and an(p) ’ a, then

lim sup an ≥ a ≥ lim inf an .

n’∞

n’∞

(ii) Suppose that lim supn’∞ an = 1 and lim inf n’∞ an = 0. Show, by

means of examples, that it may happen that 1 and 0 are the only numbers

which are the limits of convergent subsequences of the an or it may happen

that every a with 0 ¤ a ¤ 1 is such a limit. [You may wish to investigate

precisely what sets can occur, but you will need the notion of a closed set.]

(iii) Suppose that an and bn are bounded sequences of real numbers.

Show, by means of an example, that the equation

?

lim sup(an + bn ) = lim sup an + lim sup bn

n’∞ n’∞ n’∞

443

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

need not hold. By giving a proof or a counterexample, establish whether the

following relation always holds.

?

lim sup(an + bn ) ≥ lim sup an + lim inf bn .

n’∞

n’∞ n’∞

Exercise K.21. (Ptolomey™s inequality.) [4.1, S] We work in a vector

x y

’

space with inner product. By squaring and rewriting the

x2 y2

result, or otherwise, prove Ptolomey™s inequality

x’y z ¤ y’z x + z’x y.

Exercise K.22. [4.2, P] We use the standard Euclidean norm on Rn and

Rn+1 . Which of the following statements are true and which are false? Give

proofs or counterexamples as appropriate.

(i) If E is a closed subset of Rn+1 then

{x ∈ Rn : (0, x) ∈ E}

is closed in Rn .

(ii) If U is an open subset of Rn+1 then

{x ∈ Rn : (0, x) ∈ U }

is open in Rn .

(iii) If E is a closed subset of Rn then

{(0, x) : x ∈ E}

is closed in Rn+1 .

(iv) If U is an open subset of Rn then

{(0, x) : x ∈ U }

is open in Rn+1 .

Exercise K.23. [4.2, T] We can extend De¬nition 4.2.20 as follows.

Let E ⊆ Rm , A ⊆ E, x ∈ E and l ∈ Rp Consider a function f : E \ {x} ’

Rp . We say that f (y) ’ l as y ’ x through values of y ∈ A if, given > 0,

we can ¬nd a δ( , x) > 0 such that, whenever y ∈ A and 0 < x’y < δ( , x),

we have

f (y) ’ l < .

444 A COMPANION TO ANALYSIS

(i) Let E ⊆ Rm , A ⊆ E, x ∈ E and l ∈ Rp Consider a function f :

E \ {x} ’ Rp . Show that if A is ¬nite (this includes the case A = …), then,

automatically, f (y) ’ l as y ’ x through values of y ∈ A.

(ii) Let E ⊆ Rm , A ⊆ E, x ∈ E and l ∈ Rp Consider a function f :

E \ {x} ’ Rp such that f (y) ’ l as y ’ x through values of y ∈ A. Show

that, if B ⊆ A, then f (y) ’ l as y ’ x through values of y ∈ A.

(iii) Let E ⊆ Rm , A ⊆ E, B ⊆ E, x ∈ E and l ∈ Rp . Consider a function

f : E \ {x} ’ Rp such that f (y) ’ l as y ’ x through values of y ∈ A and

f (y) ’ l as y ’ x through values of y ∈ B. Show that f (y) ’ l as y ’ x

through values of y ∈ A ∪ B.

Exercise K.24. [4.2, P] We work in R2 . The projections π1 , π2 : R2 ’ R

are de¬ned by π1 (x, y) = x and π2 (x, y) = y.

(i) Show that E = {(x, 1/x) : x > 0} is a closed set in R2 such that

π1 (E) is not closed.

(ii) Find a closed set E in R2 such that π2 (E) is closed, but π1 (E) is not

closed.

(iii) Show, however, that if E is a closed set in R2 with π2 (E) bounded,

then π1 (E) must be closed.

Exercise K.25. [4.2, P] Let E ⊆ Rn and let f : E ’ Rm be function. We

call

G = {(x, f (x)) : x ∈ E}

the graph of f .

(i) Show that, if G is closed, then E is. Show that, if G is bounded, then

E is.

(ii) Show that, if E is closed and f is continuous, then G is closed.

(iii) Show that G may be closed but f discontinuous. (Look at Exer-

cise K.24 if you need a hint.)

(iv) Show that, if G is closed and bounded, then f is continuous.

Exercise K.26. [4.3, P] Let E ⊆ Rn . A function f : E ’ R is said to be

upper semi-continuous on E if, given any x ∈ E and any > 0, we can ¬nd

a δ( , x) > 0 such that f (y) < f (x) + for all y ∈ E with y ’ x < δ( , x).

Give an example of a function f : [’1, 1] ’ R which is upper semi-continuous

on [’1, 1] but not continuous. If E ⊆ Rn and f : E ’ R is such that both

f and ’f are upper semi-continuous on E, show that f is continuous on E.

If K ⊆ Rn is closed and bounded and f : K ’ R is upper semi-

continuous, show that f is bounded above and attains its least upper bound.

Is it necessarily true that f is bounded below? Is it necessarily true that,

if f is bounded below, it attains its greatest lower bound? Give proofs or

counterexamples as appropriate.

445

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

Exercise K.27. [4.3, H, S] Suppose that f : (a, b) ’ R is a function with

continuous derivative f . The object of this question is to show, using the

mean value inequality (Theorem 1.7.1), that, if f (t) > 0 for all t ∈ (a, b), then

f is strictly increasing on (a, b). (We prove a stronger result in Lemma 4.4.2

using Theorem 4.4.1.)

To this end, suppose a < x < y < b.

(i) Using the fact that a continuous function on a closed bounded interval

is bounded and attains its bounds, show that there exists an m > 0 such

that f (t) ≥ m for all t ∈ [x, y].

(ii) Hence, using the mean value inequality or one of its corollaries, show

that f (y) ’ f (x) ≥ m(y ’ x) > 0.

Exercise K.28. [4.3, T!] Suppose x1 , x2 , . . . are real numbers. Show that

we can ¬nd closed intervals Kn = [an , bn ] (where bn > an ) such that xn ∈ /

Kn for all n ≥ 1 but Kn ⊆ Kn’1 for n ≥ 2. By applying the result of

Exercise 4.3.8 show that there exist a real y with y = xn for all n (i.e. the

reals are uncountable).

[The method of this exercise is quite close to that of Cantor™s original proof.]

Exercise K.29. [4.3, T] This easy question introduces a concept that will

be used in other questions. We say that a non-empty collection A of subsets

of some set X has the ¬nite intersection property if, whenever n ≥ 1 and

A1 , A2 , . . . , An ∈ A, we have n Aj = ….

j=1

(i) Fix N and let A(1) be the collection of subsets A of Z such that

A © {1, 2, 3, . . . , N, N + 1} has at least N members.

If A1 , A2 , . . . , AN ∈ A(1) show that N Aj = …. Show, however, that A(1)

j=1

does not have the ¬nite intersection property.

(ii) Let A(2) be the collection of subsets A of Z such that N \ A is ¬nite.

Show that A(2) has the ¬nite intersection property but A∈A(2) A = ….

(iii) (Requires countability.) Let A(3) be the collection of subsets A of R

such that R\A is ¬nite. Show that if A1 , A2 , · · · ∈ A(3) then ∞ Aj = … (so

j=1

that, in particular, A(3) has the ¬nite intersection property) but A∈A(3) A =

….

(iv) Let A(4) be the collection of subsets A of Z of the form

A = {kr : r ≥ 1}

for some k ≥ 1. Show that A(4) has the ¬nite intersection property but

A∈A(4) A = ….

446 A COMPANION TO ANALYSIS

Exercise K.30. [4.3, T] Suppose ± : Rn ’ Rn is a linear map which is self

adjoint, that is to say, such that

±(x) · y = x · (±y)

for all x, y ∈ Rn . (We use the standard inner product notation.)

(i) Show that the map f : Rn ’ R given by f (x) = x · ±x is continuous.

Hence, stating carefully any theorem that you use, show that there is an

e ∈ Rn with e = 1 such that

e · (±e) ≥ x · (±x),

whenever x = 1.

(ii) If e is as in (i), deduce that if e · h = 0 then

(1 + δ 2 )e · (±e) ≥ (e + δh).(±(e + δh))

for all real δ. By considering what happens when δ is small, or otherwise,

show that

e · (±h) + h · (±e) = 0

and so h · (±e) = 0.

(iii) In (ii) we showed that h · (±e) = 0 whenever e · h = 0. Explain why

this tells us that ±e = »e for some real » and so the vector e found in (i) is

an eigenvector.

(iv) Let U = {h : e · h = 0}. Explain why U is an n ’ 1 dimensional

subspace of V and ±(U ) ⊆ U . If β = ±|U , the restriction of ± to U , show that

β is self adjoint. Hence use induction to show that Rn has an orthonormal

basis e1 , e2 , . . . , en of eigenvectors of ±.

Exercise K.31. [4.3, T!] (i) Suppose that E is a subspace of Rn and y ∈ E.

/

Show that the map g : E ’ R given by

g(x) = x ’ y

is continuous. Hence, show carefully (note that, except in the trivial case

E = {0}, E itself is not bounded) that there exists a x0 ∈ E such that

x ’ y ≥ x0 ’ y

for all x ∈ E.

(ii) Show that x0 ’ y is perpendicular to E (that is (x0 ’ y) · x = 0

for all x ∈ E). (See Exercise K.30 (ii), if you need a hint.) Show that

x ’ y > x0 ’ y for all x ∈ E with x = x0 (in other words, x0 is unique).

447

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

(iii) Suppose that ± : Rm ’ R is linear and ± = 0. By setting

E = {x : ±x = 0}

and choosing an appropriate y, show that there exists an b, perpendicular to

E, with ±b = 0. Hence show that there exists a unit vector a, perpendicular

to E, with ±a = 0. By considering the e¬ect of ± on the vector x ’ (a · x)a,

show that

±x = a · x

for all x ∈ Rm .

Exercise K.32. (Hahn-Banach for Rn . [4.3, T!, ‘ ] This exercise extends

some of the ideas of Exercise K.31. Recall that a set E in Rm is called convex

if, whenever e1 , e2 ∈ E, and » ∈ [0, 1] we have »e1 + (1 ’ »)e2 ∈ E (in other

words, if two points lie in E, so does the chord joining them).

(i) Suppose that E is a non-empty closed convex set Rm and y ∈ E. Show

/

that there exists a x0 ∈ E such that

x ’ y ≥ x0 ’ y

for all x ∈ E.

(ii) Suppose that x0 = 0. Show that there exists a b such that x · b ¤ 0

for all x ∈ E but b · y > 0.

(iii) Returning to the general case, when all we know is that E is a non-

empty closed convex set Rm and y ∈ E, show that there exists an a ∈ Rm

/

and a real number c such that x · a ¤ c for all x ∈ E but a · y > c. (Results

like this are called Hahn-Banach type theorems.)

Exercise K.33. (Extreme points.) [4.3, T!, ‘ ] Here is an application of

Exercise K.32.

(i) Throughout this question K is a non-empty closed bounded convex

set in Rn . We say that a point z ∈ K is an extreme point for K, if, whenever

x, y ∈ K, 1 > » > 0 and »x + (1 ’ »)y = z, then x = y = z.

Show that the extreme points of the closed unit ball {x ∈ Rn : x ¤ 1}

are precisely those points z ∈ Rn with z = 1. Show that the extreme

points of the closed cube [’1, 1]n are precisely the 2n points (z1 , z2 , . . . , zn )

with |zj | = 1.

(ii) Suppose w ∈ K. By using Exercise K.32, show that there exists an

/

a ∈ R and a real number c such that x · a ¤ c for all x ∈ K but a · w > c.

n

Quoting carefully any theorems that you use, show that there exists a x0 ∈ K

such that x · a ¤ x0 · a for all for all x ∈ K.

448 A COMPANION TO ANALYSIS

Show that

K = {u : x0 + u ∈ K and u · a = 0}

is a non-empty closed convex subset of an n ’ 1 dimensional subspace of Rn .

Show further that, if u0 is an extreme point of K , then x0 + u0 is an extreme

point of K.

(iii) By using induction on the dimension n, or otherwise, show that every

non-empty closed bounded convex set in Rn contains an extreme point.

(iv) Suppose that T : Rn ’ R. Explain why there exists an x0 ∈ K such

that T (x0 ) ≥ (x) for all x ∈ K. By considering extreme points of the set

{x ∈ K : T (x) = T (x0 )},

or otherwise, show that there exists an extreme point e of K with T (e) =

T (x0 ). Thus every linear map T : Rn ’ R attains its maximum on a closed

bounded convex set at some extreme point of that set. This observation is

the foundation of the theory of linear optimisation.

(v) Suppose that L is a non-empty closed convex subset of K. If L = K

(so that there exists a w ∈ K with w ∈ L) use the ideas of (ii) to show that

/

there exists an extreme point of K which does not lie in L.

Exercise K.34. (Krein-Milman for Rn .) [4.3, T!, ‘ ] We continue with

the ideas of Exercise K.33.

(i) If L is a collection of convex sets in Rn , show that L∈L L is convex.

If M is a collection of closed convex sets in Rn , show that M ∈M M is a

closed convex set.

(ii) Explain why Rn is a closed convex set. If A is a subset of Rn , we

write

{L ⊇ A : L is closed and convex}

hull(A) =

and call hull(A) the closed convex hull of A. Show that hull(A) is indeed

closed and convex and that hull(A) ⊇ A. Show also that, if B is a closed

convex set with B ⊇ A, then B ⊇ hull(A).

[This argument parallels the standard approach to the closure and interior

set out in Exercise K.179.]

(iii) Use (ii) and part (v) of Exercise K.33 to show that the closed convex

hull of the set E of extreme points of a non-empty bounded closed convex

subset K of Rn is the set itself. (More brie¬‚y hull(E) = K.) Results of this

kind are known as Krein-Milman theorems.

449

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

Exercise K.35. [4.3, T, ‘ ] We work in R. Let K be a collection of closed

sets lying in some closed interval [a, b] and suppose that K has the ¬nite

intersection property (see Exercise K.29).

(i) If a ¤ c ¤ b and we write

K1 = {K © [a, c] : K ∈ K} and K2 = {K © [c, b] : K ∈ K},

show that at least one of K1 and K2 must have the ¬nite intersection property.

(ii) Use lion hunting to show that K∈K K = ….

Exercise K.36. (The Heine-Borel theorem.) [4.3, T, ‘ ] We work in

Rm . Consider the following two statements.

(A) Let R > 0. If a collection K of closed sets lying in [’R, R]m has the

¬nite intersection property then K∈K K = ….

(B) Let E be a bounded closed set. If a collection U of open sets is such

that U ∈U U ⊇ E, then we can ¬nd an n ≥ 1 and U1 , U2 , . . . , Un ∈ U such

that n Un ⊇ E.

j=1

(i) By considering sets of the form K = E \ U deduce statement (B) from

statement (A). Deduce statement (A) from statement (B).

(ii) Prove statement (A) by a lion hunting argument along the lines of

Exercises K.35 and 4.1.14. It follows from (i) that (B) holds. This result is

known as the theorem of Heine-Borel1 .

(iii) Explain why Exercise 4.3.8 is a special case of statement (A).

(iv) Suppose that E is a set with the properties described in the second

sentence of statement (B). Show that E is closed and bounded.

[Hint. To show E is closed, suppose x ∈ E and consider sets U of the form

/

U = {y : r > x ’ y > 1/r}

with r > 1.]

Exercise K.37. [4.3, H, ‘ ] (This complements Exercise K.35.) Suppose

that F is an ordered ¬eld with the following property.

(A) Whenever K is a collection of closed sets lying in some closed interval

[a, b] and K has the ¬nite intersection property it follows that K∈K K = ….

1

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

|g(t)| ¤ 1 and |g (t)| ¤ 1 for all t ∈ R and g(0) = 1.

(e) Given M0 , M2 > 0 we can ¬nd an in¬nitely di¬erentiable function

G : R ’ R with |G(t)| ¤ M0 and |G (t)| ¤ M2 for all t ∈ R and G(0) =

(M0 M2 )1/2 .

(f) There exists a constant A such that, if |f (t)| ¤ M0 and |f (t)| ¤ M2

for all t ∈ R, then |f (t)| ¤ A(M0 M2 )1/4 for all t ∈ R.

Exercise K.53. [4.5, P] We consider functions f : (0, 1) ’ R. Give a proof

or a counterexample for each of these statements.

(i) If f is uniformly continuous, then f is bounded.

(ii) If f is uniformly continuous, then f is bounded and attains its bounds.

(iii) If f is continuous and bounded and attains its bounds, then f is

uniformly continuous.

Exercise K.54. [4.5, P] Give a proof or a counterexample for each of these

statements.

(i) If f : R ’ R is continuous and |f (x)| tends to a limit as |x| ’ ∞,

then f is uniformly continuous.

(ii) If f : R ’ R is uniformly continuous, then |f (x)| tends to a limit as

|x| ’ ∞.

459

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

(iii) If f : R ’ R is continuous and bounded then f is uniformly contin-

uous.

(iv) If f : R ’ R is uniformly continuous, then f is bounded.

(v) If f : C ’ C is continuous and |f (z)| tends to a limit as |z| ’ ∞,

then f is uniformly continuous.

(vi) If f : C ’ C is uniformly continuous, then so is |f | : C ’ R.

(vii) If f : R ’ C is continuous and |f | : R ’ R is uniformly continuous,

then f is uniformly continuous.

(vii) If f, g : R ’ R are uniformly continuous, then so is their product

f — g.

(viii) If f, g : R ’ R are uniformly continuous, then so is their composi-

tion f —¦ g.

Exercise K.55. [4.5, P] If E is a non-empty subset of Rn we de¬ne f :

Rn ’ R by

f (x) = inf{ x ’ y : y ∈ E}.

Show that f is uniformly continuous.

Show also that E is closed if and only if given x ∈ Rn we can ¬nd y ∈ E

such that x ’ y = f (x).

Exercise K.56. [4.5, P] Suppose that f : Q ’ R is uniformly continuous

on Q. The object of this question is to show that f has a unique continuous

extension to R.

(i) By using the general principle of convergence, show that, if x ∈ R,

xn ∈ Q, [n ≥ 1] and xn ’ x as n ’ ∞, then f (xn ) tends to a limit.

(ii) Show that if x ∈ R, xn , yn ∈ Q [n ≥ 1] and xn ’ x is such that

xn ’ x and yn ’ x, then f (xn ) and f (yn ) tend to the same limit.

(iii) Conclude that there is a unique F : R ’ R de¬ned by f (xn ) ’ F (x)

whenever xn ∈ Q and xn ’ x.

(iv) Explain why F (x) = f (x) whenever x ∈ Q.

(v) Show that F is uniformly continuous.

[See also Exercise K.303.]

Exercise K.57. [4.6, P] Suppose that the power series ∞ an z n has ra-

n=0

∞ n

dius of convergence R and the power series n=0 bn z has radius of conver-

gence S.

(i) Show that, if R = S, then ∞ (an + bn )z n has radius of convergence

n=0

min(R, S).

(ii) Show that, if R = S, then ∞ (an + bn )z n has radius of convergence

n=0

T ≥ R.

460 A COMPANION TO ANALYSIS

(iii) Continuing with the notation of (i) show, by means of examples, that

T can take any value with T ≥ R.

(iv) If » = 0 ¬nd the radius of convergence of ∞ »an z n . What happens

n=0

if » = 0?

(v) Investigate what, if anything, we can say about the radius of con-

∞ n

if |cn | = max(|an |, |bn |). Do the same if |cn | =

vergence of n=0 cn z

min(|an |, |bn |).

Exercise K.58. [4.6, P] We work in C. Consider a series of the form

∞ nz

n=0 bn e . Show that there exists an X, which may be ∞, ’∞ (with

appropriate conventions) or any real number, such that the series converges

whenever z < X and diverges whenever z > X. Show, by means of

examples, that any such value of X may occur.

Find the value of X for the sum

∞

2n enz

.

(n + 1)2

n=0

By using results from the ¬rst paragraph, show that, if ∞ cn converges,

n=0

then there exists a Y , which may be ∞ (with an appropriate convention)

or any positive real number, such that ∞ cn cos nz converges absolutely

n=0

whenever | z| < Y and diverges whenever | z| > Y .

Exercise K.59. [4.6, T] Consider the power series ∞ an z n . Show that,

n=0

∞

1/n n

if the sequence |an | is bounded, then n=0 an z has in¬nite radius of

1/n

convergence if lim supn’∞ |an | = 0, and radius of convergence

R = (lim sup |an |1/n )’1

n’∞

otherwise. What can we say if the sequence |an |1/n is unbounded? Prove

your statement.

This formula for the radius of convergence is of considerable theoretical

importance but is hard to handle as a practical tool. It is usually best to use

the de¬nition directly.

Exercise K.60. [4.6, T] Suppose that the sequence an of strictly positive

real numbers has the property that there exists a K ≥ 1 with K ≥ an+1 /an ≥

K ’1 for all n ≥ 1. Prove that

lim inf (an+1 /an ) ¤ lim inf a1/n ¤ lim sup a1/n ¤ lim sup(an+1 /an ).

n n

n’∞ n’∞ n’∞ n’∞

1/n

Conclude that, if an+1 /an ’ l as n ’ ∞, then an ’ l.

461

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

Give examples of sequences an such that

1/n

(i) an ’ l for some l but lim inf n’∞ (an+1 /an ) < l < lim supn’∞ (an+1 /an ).

1/n

(ii) lim inf n’∞ (an+1 /an ) = lim supn’∞ an < lim supn’∞ (an+1 /an ).

In how many di¬erent ways can you replace the ¤ in the initial formula

by combinations of < and =? Can all these possibilities occur? Justify your

answer.

Comment very brie¬‚y on the connection with formulae for the radius of

convergence of a power series given in Exercise K.59.

Exercise K.61. (Summation methods.) [4.6, T] If bj ∈ R, let us write

b0 + b 1 + · · · + b n

Cn = .

n+1

(i) Let > 0. Show that, if |bj | ¤ for all j ≥ 0, then |Cn | ¤ for all

n ≥ 0.

(ii) Show that, if N ≥ 0 and bj = 0 for all n ≥ N , then Cn ’ 0 as n ’ ∞.

(iii) Show that, if bj ’ 0 as j ’ ∞, then Cn ’ 0 as n ’ ∞.

(iv) By considering bj ’ b, or otherwise, show that, if bj ’ b as j ’ ∞,

then Cn ’ b as n ’ ∞.

(v) Let bj = (’1)j . Show that Cn tends to a limit (to be found) but bj

does not.

(vi) Let b2m +k = (’1)m for 0 ¤ k ¤ 2m ’ 1 and m ≥ 1. Show that Cn

does not tend to a limit as n ’ ∞.

(vii) If 1 > r > 0 and bj ∈ R, let us write

∞

(1 ’ r)r n bn .

Ar =

n=0

Show that, if bj ’ b as j ’ ∞, then Ar ’ b as r ’ 1 through values of

r < 1. Give an example where Ar tends to limit but bj does not. Show that

there exists a sequence bj with |bj | ¤ 1 where Ar does not tend to a limit as

r ’ 1 through values of r < 1.

(viii) Let bj ∈ Rm . Show, that, if bj ’ b as j ’ ∞, then

b0 + b 1 + · · · + b n

’b

n+1

You should give two proofs.

(A) By looking at coordinates and using the result for R.

(B) Directly, not using earlier results.

(ix) Explain brie¬‚y why (vii) can be generalised in the manner of (viii).

462 A COMPANION TO ANALYSIS

Exercise K.62. [4.6, T, ‘ ] We continue with the notation of Exercise K.61.

We call the limit of Cn , if it exists, the Ces`ro limit of the sequence bj . We

a

call the limit of Ar if it exists, the Abel limit of the sequence bj . Now suppose

aj ∈ R and we set bj = j aj . r=0

(i) Show that

n

1

Cn = (n + 1 ’ k)ak

n+1 k=0

and

∞

r k ak .

Ar =

k=0

If Cn tends to a limit, we say that the sequence aj has that limit as a

Ces`ro sum and that the sequence aj is Ces`ro summable. If Ar tends to a

a a

limit we say that the sequence aj has that limit as a Abel sum and that the

sequence aj is Abel summable.

(ii) Explain very brie¬‚y why the results of Exercise K.61 (viii) and (ix)

show that we can extend the de¬nitions of (i) to the case aj ∈ C.

(iii) Let aj = z j with z ∈ C.

(a) Show that ∞ z j converges if and only if |z| < 1 and ¬nd the sum

j=0

when it exists.

(b) Show that the sequence z j is Ces`ro summable if and only if |z| ¤ 1

a

and z = 1. Find the Ces`ro sum when it exists.

a

(c) Show that the sequence z j is Abel summable if and only if |z| ¤ 1

and z = 1. Find the Abel sum when it exists.

Exercise K.63. [4.6, T, ‘ ] (This generalises parts of Exercise K.61) Sup-

pose that ujk ∈ R for j, k ≥ 0. Suppose that

(1) If k is ¬xed, ujk ’ 0 as j ’ ∞.

∞

(2) k=0 ujk is absolutely convergent and there exists a M such that

∞

k=0 |ujk | ¤ M for each j ≥ 0.

(3) ∞ ujk ’ 1 as j ’ ∞.

k=0

(i) Show that, if bj ∈ R and bj ’ b as j ’ ∞, then

∞

ujk bk ’ b

k=0

as j ’ ∞.

(ii) Explain why the result just proved gives part (iv) of Exercise K.61.

463

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

(iii) Let 0 < rn < 1 and rn ’ 1. Use (i) to show that, in the notation of

Exercise K.61 (vii), if bj ’ b as j ’ ∞, then Arn ’ b as n ’ ∞. Deduce

Exercise K.61 (vii).

(iv) State and prove an extension of part (i) along the lines of Exer-

cise K.61 (viii).

(v) If » > 0 and bj ∈ C, let us write

∞

»n ’»

B» = e bn .

n!

n=0

Show that, if bj ’ b as j ’ ∞, then B» ’ b as » ’ ∞.

(vi) Much as in the introduction to Exercise K.62 we suppose aj ∈ C and

set bj = j aj . Using the notation of part (v), we say that, if B» tends

r=0

to a limit, then the sequence aj has that limit as a Borel sum and that the

sequence aj is Borel summable.

Show that the sequence z j is Borel summable if and only if z < 1. Find

the Borel sum when it exists.

Exercise K.64. [4.6, P, ‘ ] Let us write G for the set of real double se-

quences U = (ujk )j,k≥0 which satisfy conditions (1), (2) and (3) of Exer-

cise K.63. By providing a proof or counterexample as appropriate, establish

which of the following statements are true and which are false.

(i) If bj is a bounded real sequence, there exists a U ∈ G such that

∞

k=0 ujk bk tends to a limit as j ’ ∞. (Hint. Bolzano-Weierstrass.)

(ii) If bj is a bounded real sequence which does not tend to a limit, there

exist U, V ∈ G such that ∞ ujk bk and ∞ vjk bk converge to di¬erent

k=0 k=0

limits as j ’ ∞.

(iii) If bj is a bounded real sequence, we can ¬nd real numbers ± and β

such that there exists a U ∈ G having the property that ∞ ujk bk tends to

k=0

» if and only if » ∈ [±, β].

(iv) If U ∈ G, we can ¬nd a bounded real sequence bj such that ∞ ujk bk

k=0

does not tend to limit as j ’ ∞.

(v) If bj is a bounded real sequence, we can ¬nd a U ∈ G such that

∞

k=0 ujk bk does not tend to limit as j ’ ∞.

Exercise K.65. [4.6, T!, ‘ ] The object of this exercise is to prove the

converse of part (i) of Exercise K.63. In other words we want to show that,

if ujk ∈ R for j, k ≥ 0 and

∞

ujk bk ’ b

k=0

464 A COMPANION TO ANALYSIS

as j ’ ∞ whenever bj ∈ R and bj ’ b as j ’ ∞, then it follows that the

ujk satisfy conditions (1), (2) and (3) of Exercise K.63. (This is quite hard

work and simpler proofs exist using more advanced techniques.)

(i) By choosing particular sequences bk , show that, if the ujk satisfy the

stated hypothesis, then they must satisfy conditions (1) and (3).

(ii) Show that if ∞ ujk bk exists for each convergent sequence bk , then

k=0

∞

k=0 ujk is absolutely convergent. (Look at Exercise 5.1.11 if you need a

hint.)

(iii) Suppose that bk is given for 0 ¤ k ¤ N ’ 1 with |bk | ¤ L Suppose

that · > 0, > 0, |uk | ¤ · for 1 ¤ k ¤ N ’ 1 and M ’1 |uk | ≥ K for some

k=1

M > N . Show that we can ¬nd bk [N ¤ k ¤ M ’ 1] such that |bk | ¤ for

N ¤ k ¤ M ’ 1 and

M ’1

uk bk ≥ (K ’ N ·) ’ N L·.

k=0

(iii) Suppose that the ujk satisfy conditions (1) and (3) together with

the conclusion of (ii), but do not satisfy condition (2). Show, by induction,

or otherwise, that we can ¬nd a sequence of integers 0 = N (0) < N (1) <

N (2) < . . . , a sequence of integers 0 = j(0) < j(1) < (2) < . . . , a sequence

of real numbers 1 = (0) > (1) > (2) > . . . with 2’r ≥ (r) > 0, and a

sequence bk of real numbers such that

|bk | ¤ (r) for N (r) ¤ k ¤ N (r + 1) ’ 1

N (r+1)’1

uj(r)k bk ≥ 2r + 1

k=0

∞

uj(r)k xk ¤ 1 provided |xk | ¤ (r + 1) for all k ≥ N (r + 1).

k=N (r+1)

Show that | ∞ uj(r)k bk | ≥ 2r and use contradiction to deduce the result

k=0

stated at the beginning of the exercise.

Exercise K.66. [5.2, P] We work in Rn . Show that the following state-

ments are equivalent.

(i) ∞ xn converges absolutely.

n=1

(ii) Given any sequence n with n = ±1, ∞ n xn converges.

n=1

Exercise K.67. [5.2, P] Let S1 and S2 be a partition of Z+ . (That is to

say, S1 ∪ S2 = Z+ and S1 © S2 = ….) Show that, if an ≥ 0 for all n ≥ 1, then

465

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

∞

an converges if and only if n∈S1 an converges (that is n∈S1 , n¤N an

n=1

tends to a limit as N ’ ∞) and n∈S2 an converges.

Establish, using proofs or counterexamples, which parts of the ¬rst para-

graph remain true and which become false if we drop the condition an ≥ 0.

Show that if an , bn ≥ 0, ∞ an and ∞ bn converge, and max(an , bn ) ≥

n=1 n=1

cn ≥ 0 then ∞ cn converges.

n=1

Show that if an ≥ 0 for all n and ∞ an converges then

n=1

∞ 1/2

an

converges.

n

n=1

Give an alternative proof of this result using the Cauchy-Schwarz inequal-

ity.

Suppose that an , bn ≥ 0, ∞ an and ∞ bn diverge, and max(an , bn ) ¤

n=1 n=1

∞

cn . Does it follow that n=1 cn diverges? Give a proof or a counterexample.

Exercise K.68. (Testing for convergence of sums.)[5.2, P] There is

no certain way for ¬nding out if a sum ∞ an converges or not. However,

n=0

there are a number of steps that you might run through.

(1) Is the matter obvious? If an 0 then the sum cannot converge. Does

the ratio test work?

(2) Are the ¬rst few terms untypical? The convergence or otherwise of

∞ ∞

n=N an determines that of n=1 an .

(3) (Mainly applies to examination questions.) Is the sum ∞ an ob-n=0

∞

tained from a sum n=0 bn whose behaviour is known by adding or omitting

irrelevant terms? (For examples see (i) and (ii) below.)

(4) Check for absolute convergence before checking for convergence. It is

generally easier to investigate absolute convergence than convergence and, of

course, absolute convergence implies convergence.

(5) To test for the convergence of a sum of positive terms (such as arises

when we check for absolute convergence) try to ¬nd another sum whose

behaviour is known and which will allow you to use the comparison test.

(6) To test for the convergence of a sum of decreasing positive terms

consider using the integral test (see Lemma 9.2.4) or the Cauchy condensation

test.

(7) If (5) and (6) fail, try to combine them or use some of the ideas of

parts (iii) and (iv) below. Since all that is needed to ¬nd whether the sum

∞

n=1 an of positive terms converges is to discover whether the partial sums

are bounded (that is, there exists a K with N an ¤ K for all N ) it is

n=1

rarely hard to discover whether a naturally occurring sum of positive terms

converges or not.

466 A COMPANION TO ANALYSIS

(8) If the series is not absolutely convergent you may be able to show

that it is convergent using the alternating series test.

(9) If (8) fails, then the more general Abel™s test (Lemma 5.2.4) may

work.

(10) If (8) and (9) fail then grouping of terms (see part (vi)) may help

but you must be careful (see parts (v) and (vii)).

(11) If none of the above is helpful and you are dealing with a naturally

occurring in¬nite sum which is not absolutely convergent but which you hope

is convergent, remember that the only way that such a series can converge

is by ˜near perfect cancellation of terms™. Is there a natural reason why the

terms should (almost) cancel one another out?

(12) If you reach step (12) console yourself with the thought that where

routine ends, mathematics begins4 .

∞

1 + (’1)n

(i) State, with proof, whether converges.

n

n=1

∞

1

(ii) Let pn be the nth prime. State, with proof, whether , converges.

p2

n=1 n

∞

(iii) Suppose an ≥ 0 for each n ≥ 1. Show that n=1 an converges if and

N (j)

only if we can ¬nd a sequence N (j) ’ ∞ such that n=1 an tends to a limit

as j ’ ∞.

(iv) [This is an variation on the integral test described in Lemma 9.2.4.]

Suppose that we can ¬nd a positive continuous function f : [1, ∞] ’ R and

a constant K with K ≥ 1 such that

Kan ≥ f (t) ≥ K ’1 an ≥ 0 for all n ¤ t ¤ n + 1.

∞

Show that ∞ an converges if and only if 0 f (t) dt does.

n=1

(v) Find a sequence an ∈ R such that 2N an tends to a limit as N ’ ∞

n=1

but ∞ an does not converge.

n=1

(vi) Let an ∈ Rm . Suppose that there exists a strictly positive integer M

such that M N an tends to a limit as N ’ ∞ and suppose that an ’ 0 as

n=1

n ’ ∞. Show that ∞ an converges.

n=1

(vii) Find a sequence an ∈ R and a sequence N (j) ’ ∞ as j ’ ∞

N (j)

such that n=1 an tends to a limit as N ’ ∞ and an ’ 0 as n ’ ∞, yet

∞

n=1 an diverges.

Exercise K.69. [5.2, P] Let an , bn be sequences of non-negative real num-

bers.

4

Minkowski was walking through G¨ttingen when he passed a young man lost in deep

o

thought. ˜Don™t worry™ said Minkowski, ˜it is sure to converge.™

467

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

(i) Show that, if ∞ an and ∞ bn converge, so does ∞ (an bn )1/2 .

n=1 n=1 n=1

(ii) Show that, if ∞ an converges, so does ∞ (an an+1 )1/2 .

n=1 n=1

(iii) Show that, if the an form a decreasing sequence, then, if ∞ (an an+1 )1/2

n=1

converges, so does ∞ an .

n=1

(iv) Give an example with ∞ (an an+1 )1/2 convergent and ∞ an di-

n=1 n=1

vergent.

Exercise K.70. [5.2, P] We work in the real numbers. Are the following

true or false? Give a proof or counterexample as appropriate.

(i) If ∞ a4 converges, then ∞ a5 converges.

n=1 n n=1 n

(ii) If n=1 an converges, then ∞ a4 converges.

∞ 5

n=1 n

∞

(iii) If an ≥ 0 for all n and n=1 an converges, then nan ’ 0 as n ’ ∞.

(iv) If an ≥ 0 for all n and ∞ an converges, then n(an ’ an’1 ) ’ 0 as

n=1

n ’ ∞.

(v) If an is a decreasing sequence of positive numbers and ∞ an con-

n=1

verges, then nan ’ 0 as n ’ ∞.

(vi) If an is a decreasing sequence of positive numbers and nan ’ 0 as

n ’ ∞, then ∞ an converges.

n=1

(vii) If n=1 an converges, then ∞ n’3/4 an converges.

∞

n=1

[Hint: Cauchy-Schwarz]

(ix) If ∞ an converges, then ∞ n’1 |an | converges.

n=1 n=1

Exercise K.71. [5.2, T] Show that if an ≥ 0 and k is a strictly positive

integer, the convergence of ∞ ak implies the convergence of ∞ ak+1 .

n=1 n n=1 n

Suppose now that we drop the condition that all the an be positive. By

considering a series of real numbers an with

a3n+1 = 2f (n), a3n+2 = ’f (n), a3n+3 = ’f (n)

for a suitable f (n) show that we may have

∞ ∞

ak divergent for all k ≥ 2.

an convergent but n

n=1 n=1

Exercise K.72. (Euler™s γ.) [5.2, T] Explain why