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a < x1 < x2 < · · · < xn < b.
Suppose that P is a polynomial of degree n ’ 1 with P (xj ) = f (xj ). (We say
that P is an interpolating polynomial for f .) We are interested in the error
E(t) = f (t) ’ P (t)
at some point t ∈ [a, b]. Since we already know that E(xj ) = 0, we may also
assume that t, x1 , x2 , . . . , xn are distinct.
We consider the function
n
x ’ xj
F (x) = f (x) ’ P (x) ’ E(t) .
t ’ xj
j=1
64 A COMPANION TO ANALYSIS

(i) Show that F vanishes at t, x1 , x2 , . . . , xn and so vanishes at n + 1
distinct points in (a, b).
(ii) By using Rolle™s theorem (Theorem 4.4.4), show that F vanishes at
n distinct points in (a, b).
(iii) By repeated use of Rolle™s theorem show that F (n) vanishes at some
point c ∈ (a, b).
(iv) By computing F (n) explicitly, deduce that
n
1
(n)
(c) ’ n!E(t)
0=f ,
t ’ xj
j=1


and so
n
f (n) (c)
(t ’ xj ).
E(t) =
n! j=1

Of course, we know nothing about c, but, if we know that |f (n) (x)| ¤ A for
all x ∈ [a, b], we can deduce that
n
A
|f (t) ’ P (t)| ¤ (t ’ xj )
n! j=1

for all t ∈ [a, b]. (We discuss this matter further in Exercise K.48.)
(v) Deduce the weaker inequality

(b ’ a)n
|f (t) ’ P (t)| ¤ A
n!
for all t ∈ [a, b].
A similar argument to the one just given is used in Exercise K.49 to prove
a version of Taylor™s theorem.

A very sharp-eyed reader may observe that we cannot prove Lemma 4.4.2 (i)
from the mean value inequality6 .


4.5 Uniform continuity
The mean value inequality (Theorem 1.7.1) is an excellent example of the
way that many theorems of analysis convert local information into global
6
Having gone to all the bother of proving Theorem 4.4.1 from which Lemma 4.4.2 (i)
follows, we might as well use it. However, Exercise K.27 provides an alternative proof.
65
Please send corrections however trivial to twk@dpmms.cam.ac.uk

information. We know that f (x) ¤ K so that locally the rate of increase of
f is no greater than K. We deduce that f (u) ’ f (v) ¤ K(u ’ v) so that
globally the rate of increase of f is no greater than K. I remind the reader,
once again, that this conversion of local to global fails for Q and depends on
the fundamental axiom of analysis.
The main theorem of this section (Theorem 4.5.5 on uniform continuity) is
another excellent example of the conversion of local information to global. We
need a couple of de¬nitions and examples. Recall ¬rst from De¬nition 4.2.14
what it means for a function to be continuous on a set
De¬nition 4.5.1. Let E ⊆ Rm . We say that a function f : E ’ Rp is
continuous on E if, given any point x ∈ E and any > 0, we can ¬nd a
δ( , x) > 0 such that, whenever y ∈ E and x ’ y < δ( , x), we have

f (x) ’ f (y) < .

Now compare De¬nition 4.5.1 with our de¬nition of uniform continuity.
De¬nition 4.5.2. Let E ⊆ Rm . We say that a function f : E ’ Rp is
uniformly continuous on E if, given any > 0, we can ¬nd a δ( ) > 0 such
that, whenever x, y ∈ E and x ’ y < δ( ), we have

f (x) ’ f (y) < .

Example 4.5.4 and Theorem 4.5.5 depend on understanding what it means
for a function not to be uniformly continuous.
Exercise 4.5.3. Let E ⊆ Rm . Write down the de¬nition of what it means
for a function f : E ’ Rp not to be uniformly continuous7 .
Example 4.5.4. The following three functions are continuous but not uni-
formly continuous.
(i) f1 : R ’ R given by f1 (x) = x2 .
(ii) f2 : (0, 1) ’ R given by f2 (x) = x’1 .
(iii) f3 : (0, 1) ’ R given by f3 (x) = sin(x’1 ).
Proof. (i) Suppose that δ > 0. If we take x > δ ’1 and y = x + δ/2, we have
|x ’ y| < δ, yet

|x2 ’ y 2 | = y 2 ’ x2 = (y + x)(y ’ x) > 2δ ’1 δ/2 = 1.

Thus f1 is not uniformly continuous.
(ii) and (iii) are left as exercises for the reader.
7
Mathematical educationists call this sort of thing ˜¬nding the contrapositive™.
66 A COMPANION TO ANALYSIS

Theorem 4.5.5. Let K be a closed bounded subset of Rm . If f : K ’ Rp is
continuous on K, then f is uniformly continuous on K.
Proof. Earlier I said that the words ˜closed and bounded™ should elicit the
response ˜Bolzano-Weierstrass™. Let us see how this slogan works here.
If f is not uniformly continuous, then there must exist an > 0 such that,
given any δ > 0, we can ¬nd x, y ∈ K such that

x ’ y < δ but f (x) ’ f (y) > .

Thus we can ¬nd a sequence of pairs of points xn , yn ∈ K such that

xn ’ yn < 1/n but f (xn ) ’ f (yn ) > .

By the Bolzano-Weierstrass theorem, we can ¬nd a k ∈ K and a sequence
n(1) < n(2) < n(3) < . . . such that xn(j) ’ k as j ’ ∞. Since

yn(j) ’ k ¤ yn(j) ’ xn(j) + xn(j) ’ k ’ 0 + 0 = 0,

it follows that yn(j) ’ k as j ’ ∞. Since f is continuous, it follows that

< f (xn(j) ) ’ f (yn(j) ) ¤ f (xn(j) ) ’ f (k) + f (yn(j) ) ’ f (k) ’ 0 + 0 = 0

as j ’ ∞. This contradiction proves the theorem.
Exercise 4.5.6. Use Theorem 4.5.5 to prove that a real-valued continuous
function on a closed bounded set is bounded. Use the trick given in Exer-
cise 4.3.5 (ii) to deduce that a real-valued continuous function on a closed
bounded set is bounded and attains its bounds (Theorem 4.3.4).
Exercise 4.5.7. If we work in Q rather than R use the function of Exam-
ple 1.1.3 to give an example of a function f : Q ’ Q which is continuous
but not uniformly continuous on [0, 2]. At which stage does our proof Theo-
rem 4.5.5 break down?
We shall use the fact that a continuous function on a closed bounded set
is uniformly continuous when we show that every continuous function on a
closed interval is integrable (Theorem 8.3.1) and it was in this context that
the notion of uniformity ¬rst arose.


4.6 The general principle of convergence
We know that an increasing sequence of real numbers tends to a limit if and
only if the sequence is bounded above. In this section we consider a similarly
useful result which applies to sequences in Rn .
67
Please send corrections however trivial to twk@dpmms.cam.ac.uk

De¬nition 4.6.1. We say that a sequence of points xn ∈ Rm is a Cauchy
sequence if, given any > 0, we can ¬nd n0 ( ) such that xp ’ xq < for
all p, q ≥ n0 ( ).
Our ¬rst lemma is merely8 algebraic.
Lemma 4.6.2. Any convergent sequence in Rm forms a Cauchy sequence.
Proof. Suppose that xn ’ x. Let > 0. By de¬nition, we can ¬nd an n0 ( )
such that

xn ’ x < /2 for all n ≥ n0 ( ).

We observe that, by the triangle inequality,

xp ’ xq ¤ xp ’ x + xq ’ x < /2 + /2 =

for all p, q ≥ n0 ( ), so we are done.
The converse to Lemma 4.6.2 is a powerful theorem of analysis.
Theorem 4.6.3. Any Cauchy sequence in Rm converges.
Proof. Suppose that xn ∈ Rm is a Cauchy sequence. Then, by de¬nition,
given any > 0, we can ¬nd N ( ) such that xp ’ xq < for all p, q ≥ N ( ).
In particular, if n ≥ N (1), we have

xn ¤ xn ’ xN (1) + xN (1) < 1 + xN (1) .

It follows that

xn ¤ max xr + 1
1¤r¤N (1)

for all n and so the sequence xn is bounded.
By the Bolzano-Weierstrass theorem, it follows that there is an x ∈ Rm
and a sequence n(1) < n(2) < . . . such that xn(j) ’ x as j ’ ∞. Thus,
given any > 0, we can ¬nd J( ) such that xn(j) ’ x < for all j ≥ J( ).
We now observe that if n ≥ N ( /2) and we choose a j such that j ≥
J( /2) and n(j) > N ( /2), then

xn ’ x ¤ xn ’ xn(j) + xn(j) ’ x < /2 + /2 = .

Thus xn ’ x as n ’ ∞ and we are done.
8
Remember that, from the standpoint of this book any argument, however di¬cult and
complicated it may be that does not involve the fundamental axiom is ˜mere algebra™.
68 A COMPANION TO ANALYSIS

Exercise 4.6.4. Show that, if any subsequence of a Cauchy sequence con-
verges, then the Cauchy sequence converges.
Combining Theorem 4.6.3 with Lemma 4.6.2, we get the general principle
of convergence.
Theorem 4.6.5. (General principle of convergence.) A sequence in
Rm converges if and only if it is a Cauchy sequence.
The general principle of convergence is used in the study of in¬nite sums.
De¬nition 4.6.6. If aj ∈ Rm we say that ∞ aj converges to s if
j=1

N
aj ’ s
j=1


as N ’ ∞. We write ∞ aj = s.
j=1
N
j=1 aj does not tend to a limit as N ’ ∞, we say that the sum
If

j=1 aj diverges.

[The de¬nition just given corresponds to how mathematicians talk but is
not how logicians talk. Formally speaking, ∞ aj either exists or it does
j=1
not and cannot converge or diverge. In an e¬ort to get round this some books
use locutions like ˜the series aj converges to the sum ∞ aj ™.]
j=1
We note that any result on sums can be rewritten as a result on sequences
and vice versa.
Exercise 4.6.7. We work in Rm .
(i) If sn = n aj , then ∞ aj converges to s if and only if sn ’ s as
j=1 j=1
n ’ ∞.
(ii) If we write s0 = 0 and aj = sj ’ sj’1 , then sn ’ s as n ’ ∞ if and
only if ∞ aj converges to s.
j=1

Exercise 4.6.8. By writing
N N N
(aj + bj ) = aj + bj ,
j=1 j=1 j=1


show that, if ∞ aj and ∞ ∞
bj converge, then so does j=1 (aj + bj ) and
j=1 j=1
that, in this case,
∞ ∞ ∞
(aj + bj ) = aj + bj .
j=1 j=1 j=1
69
Please send corrections however trivial to twk@dpmms.cam.ac.uk

Exercise 4.6.9. If aj = bj for all j ≥ M and ∞ bj converges, show that
j=1

aj converges. Thus ˜the ¬rst few terms of an in¬nite sum do not a¬ect
j=1
its convergence™.

Exercise 4.6.10. Prove the general principle of convergence for sums, that
is to say, prove the following result. The sum ∞ aj converges if and only
j=1
q
if, given any > 0, we can ¬nd n0 ( ) such that j=p aj < for all
q ≥ p ≥ n0 ( ).

’ 0 as
Exercise 4.6.11. (i) Show that if j=1 aj converges, then an
n ’ ∞.
n ’1/2
≥ n1/2 , or otherwise, show that the
(ii) By observing that j=1 j
converse of part (i) is false.
∞ ’1+·
(iii) Show that, if · > 0, then j=1 j diverges. (We will obtain
stronger results in Exercises 5.1.9 and 5.1.10.)

We use the general principle of convergence to prove a simple but impor-
tant result.
Theorem 4.6.12. Let an ∈ Rm for each n. If ∞ aj converges, then
j=1
so does ∞ aj .
j=1

Proof. Since ∞ aj converges, the (trivial, algebraic) necessity part of
j=1
the general principle of convergence tells that, given any > 0, we can ¬nd
n0 ( ) such that q for all q ≥ p ≥ n0 ( ). The triangle inequality
j=p aj <
now tells us that
q q
aj ¤ aj <
j=p j=p


for all q ≥ p ≥ n0 ( ) and the (profound, analytic) su¬ciency part of the
general principle of convergence tells that ∞ aj converges.
j=1

Theorem 4.6.12 is often stated as saying that ˜absolute convergence implies
convergence™. We formalise this by making a de¬nition.

De¬nition 4.6.13. If aj ∈ Rm , we say that the sum aj is absolutely
j=1
convergent if ∞ aj converges.
j=1

Theorem 4.6.12 has the natural interpretation that if we wander around
R taking a sequence of steps of ¬nite total length then we must converge
m

on some point. However, although the result appears ˜intuitively evident™,
the proof ultimately depends on the fundamental axiom of analysis.
70 A COMPANION TO ANALYSIS

Exercise 4.6.14. The ¬rst proof of Theorem 4.6.12 that I was given went
as follows.
(i) We work ¬rst in R. Let an ∈ R for each n. If ∞ |aj | converges,
n=1
then set

a+ = a n , a ’ = 0 if an ≥ 0
n n

+
an = 0, an = ’an otherwise.

Use the fact that N a+ is an increasing sequence to show that limN ’∞ N a+
n=1 n n=1 n
N ’
exists. Show similarly that limN ’∞ n=1 an exists. Establish the equality
N N N
a’
a+ ’
an = n n
n=1 n=1 n=1


and use it to show that limN ’∞ N an exists.
n=1
(ii) By taking real and imaginary parts, use (i) to prove that if an ∈ C
for each n and ∞ |aj | converges, then so does ∞ aj .
j=1 j=1
(iii) Use (i) to prove Theorem 4.6.12.
Although this is, if anything, an easier proof than the one I gave, it is
a bit inelegant and does not generalise to to the context of general normed
spaces as discussed in Chapter 10.

Theorem 4.6.12 is often used in tandem with another simple but powerful
tool.

Theorem 4.6.15. (The comparison test.) We work in R. Suppose that
0 ¤ bj ¤ aj for all j. Then, if ∞ aj converges, so does ∞ bj .
j=1 j=1

Proof. Since An = n aj is an increasing sequence tending to a limit A we
j=1
know that An ¤ A for each n. Thus
n n
bj ¤ aj ¤ A
j=1 j=1


for each n and Bn = n bj is an increasing sequence bounded above. By
j=1
the fundamental axiom, Bn tends to a limit and the result follows.

Exercise 4.6.16. We work in R. Spell out in detail the proof that, if A1 ¤
A2 ¤ . . . and An ’ A as n ’ ∞, then Aj ¤ A for all j.

Exercise 4.6.17. (This should be routine for the reader. It will be needed
in the proof of Lemma 4.6.18.) We work in R.
71
Please send corrections however trivial to twk@dpmms.cam.ac.uk

(i) Show that n rj = (1 ’ r n+1 )/(1 ’ r).
j=0
∞ j
j=0 r converges if and only if |r| < 1 and that, if
(ii) Deduce that
|r| < 1, then

1
rj = .
1’r
j=0


We are now in a position to discuss power series ∞ an z n in C. The
n=0
reader may object that we have not discussed convergence in C but from
our point of view C is just R2 with additional algebraic structure. (Alterna-
tively, the reader should be able to supply all the appropriate de¬nitions and
generalizations for herself.)

Lemma 4.6.18. Suppose that an ∈ C. If n
n=0 an z0 converges for some
z0 ∈ C, then ∞ an z n converges for all z ∈ C with |z| < |z0 |.
n=0

Proof. Since ∞ an z0 converges, the general principle of convergence tells
n
n=0
n
us that |an z0 | ’ 0 as n ’ ∞. In particular, we can ¬nd an N such that
n n
|an z0 | ¤ 1 for all n ≥ N . Thus, setting M = 1 + max0¤n¤N |an z0 |, we have
n
|an z0 | ¤ M for all n ≥ 0.
Now suppose that |z| < |z0 |. We observe that

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