bN k bN k f (’k) = 0

’π ’π

k=’N k=’N

for all N .

Since f is continuous on [’π, π], it is bounded so there exists a K such

that |f (t)| ¤ K for all t ∈ [’π, π]. Since P (0) = · + 1 we can ¬nd an > 0

with > such that P (t) ≥ 1 + ·/2 for all |t| ¤ . Finally we observe

that |P (t)| ¤ 1 ’ · for ¤ |t| ¤ π. Putting all our information together, we

obtain

f (t)P (t)n ≥ f (0)(1 + ·/2)N /2 for all |t| ¤ ,

f (t)P (t)n ≥ 0 for all ¤ |t| ¤ ,

|f (t)P (t)n | ¤ K(1 ’ ·)N for all ¤ |t| ¤ π.

Thus

π

f (t)P (t)N dt

0=

’π

f (t)P (t)N dt + f (t)P (t)N dt + f (t)P (t)N dt

=

|t|¤ ¤|t|¤ ¤|t|¤π

≥ f (0)(1 + ·/2)N + 0 ’ 2πK(1 ’ ·)N ’ ∞.

The assumption that f (0) = 0 has led to a contradiction and the required

result follows by reductio ad absurdum.

301

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Exercise 11.6.5. Draw sketches illustrating the proof just given.

Exercise 11.6.6. (i) If g : R ’ C is continuous and periodic with period

2π and a ∈ R, we write ga (t) = g(t ’ a). Show that ga (n) = exp(ina)ˆ(n).

ˆ g

(ii) By translation, or otherwise, prove the following result. If f : R ’ R

ˆ

is continuous and periodic with period 2π and f (n) = 0 for all n, then f = 0.

Exercise 11.6.7. (i) If g : R ’ C is continuous and periodic with period

2π show that g — (n) = (ˆ(’n))— .

g

(ii) By considering f +f — and f ’f — , or otherwise, prove Theorem 11.6.3.

We can now state and prove our promised result on Fourier sums.

Theorem 11.6.8. If f : R ’ C is continuous and periodic with period 2π

ˆ

and ∞n=’∞ |f (n)| converges, then

N

ˆ

f (n) exp(int) ’ f (t)

n=’N

uniformly as N ’ ∞.

ˆ ˆ ˆ ˆ

Proof. Since |f (n) exp(int)+ f (’n) exp(’int)| ¤ |f (n)|+|f (’n)|, the Weier-

ˆ

strass M-test tells us that N n=’N f (n) exp(int) converges uniformly to g(t),

say. Since the uniform limit of continuous functions is continuous, g is con-

tinuous. We wish to show that g = f .

Observe that, since | exp(’imt)| = 1, we have

N N

ˆ ˆ

f (n) exp(i(n ’ m)t) = exp(’imt) f (n) exp(int) ’ exp(’imt)g(t)

n=’N n=’N

uniformly as N ’ ∞, so by Theorem 11.4.10, we have

N N

π π

1 1

ˆ ˆ

exp(i(n ’ m)t) dt = f (n) exp(i(n ’ m)t) dt

f (n)

2π 2π

’π ’π n=’N

n=’N

π

1

’ exp(’imt)g(t) dt = g (m).

ˆ

2π ’π

π

1

Now 2π ’π exp(irt) dt takes the value 1 if r = 0 and the value 0 otherwise,

ˆ ˆ

so we have shown that f (m) ’ g (m) as N ’ ∞. Thus f (m) = g (m) for all

ˆ ˆ

m and, by the uniqueness of Fourier series (Theorem 11.6.3), we have f = g

as required.

302 A COMPANION TO ANALYSIS

Exercise 11.6.9. If f : R ’ C periodic with period 2π and has continuous

second derivative, show, by integrating by parts twice, that

1

ˆ

f (n) = ’ 2 f (n)

n

for all n = 0. Deduce that

1

ˆ

|f (n)| ¤ sup |f (t)|

n2 t∈[’π,π]

for all n = 0, and that

N

ˆ

f (n) exp(int) ’ f (t)

n=’N

uniformly as N ’ ∞.

Exercise 11.6.10. Suppose f : R ’ R is a 2π periodic function with f (x) =

’f (’x) for all x and f (x) = x(π ’ x) for 0 ¤ x ¤ π. Show that

∞

8 sin(2m + 1)x

f (x) =

(2m + 1)3

π m=0

for all x and, by choosing a particular x, show that

∞

(’1)m π3

=.

(2m + 1)3 32

m=0

Exercise 11.6.11. Suppose f : R ’ R is a 2π periodic continuous function

ˆ

with ∞n=’∞ |nf (n)| convergent. Show that f is di¬erentiable and

∞

ˆ

f (t) = inf (n) exp(int).

n=’∞

Chapter 12

Contraction mappings and

di¬erential equations

12.1 Banach™s contraction mapping theorem

This chapter and the next depend on the famous contraction mapping theo-

rem by which Banach transformed a ˜folk-technique™ into a theorem.

De¬nition 12.1.1. Let (X, d) be a metric space and T : X ’ X a mapping.

We say that w ∈ X is a ¬xed point of T if T w = w. We say that T is a

contraction mapping if there exists a positive number K < 1 with d(T x, T y) ¤

Kd(x, y) for all x, y ∈ X.

The next exercise is easy but helps suggest the proof of the theorem that

follows.

Exercise 12.1.2. Let (X, d) be a metric space and T : X ’ X a contrac-

tion mapping with a ¬xed point w. Suppose that x0 ∈ X and we de¬ne xn

inductively by xn+1 = T xn . Show that d(xn , w) ’ 0 as n ’ ∞.

Theorem 12.1.3. (The contraction mapping theorem.) A contraction

mapping on a non-empty complete metric space has a unique ¬xed point.

Proof. Suppose 1 > K > 0, (X, d) is a non-empty complete metric space and

T : X ’ X has the property d(T x, T y) ¤ Kd(x, y) for all x, y ∈ X.

We show ¬rst that, if T has a ¬xed point, it is unique. For suppose

T w = w and T z = z. We have

d(z, w) = d(T z, T w) ¤ Kd(z, w)

so, since K < 1, d(z, w) = 0 and z = w.

303

304 A COMPANION TO ANALYSIS

To prove that a ¬xed point exists, choose any x0 ∈ X and de¬ne xn

inductively by xn+1 = T xn . (The preceding exercise shows this is a good

idea.) By induction,

d(xn+1 , xn ) = d(T xn , T xn’1 ) ¤ Kd(xn , xn’1 ) ¤ · · · ¤ K n d(x1 , x0 )

and so, by the triangle inequality, we have, whenever m > n

m’1 m’1

Kn

j

d(xm , xn ) ¤ d(xj+1 , xj ) ¤ K d(x1 , x0 ) ¤ d(x1 , x0 ) ’ 0

1’K

j=n j=n

as n ’ ∞. Thus the sequence xn is Cauchy. Since (X, d) is complete, we

can ¬nd a w such that d(xn , w) ’ 0 as n ’ ∞.

We now show that w is indeed a ¬xed point. To do this, we observe that

d(T w, w) ¤ d(T w, xn+1 ) + d(xn+1 , w) = d(T w, T xn ) + d(xn+1 , w)

¤ Kd(w, xn ) + d(xn+1 , w) ’ 0 + 0 = 0

as n ’ ∞. Thus d(T w, w) = 0 and T w = w.

Wide though the conditions are, the reader should exercise caution before

attempting to widen them further.

Example 12.1.4. (i) If X = {’1, 1}, d is ordinary distance and the map

T : X ’ X is given by T x = ’x, then (X, d) is a complete metric space and

d(T x, T y) = d(x, y) for all x, y ∈ X, but T has no ¬xed point.

(ii) If X = [1, ∞), d is Euclidean distance and

T x = 1 + x + exp(’x),

then (X, d) is a complete metric space and d(T x, T y) < d(x, y) for all x, y ∈

X, but T has no ¬xed point.

(iii) If X = (0, ∞), d is ordinary distance and the map T : X ’ X is

given by T x = x/2, then (X, d) is a metric space and T is a contraction

mapping, but T has no ¬xed point.

Exercise 12.1.5. Verify the statements made in Example 12.1.4. In each

case, state the hypothesis in Theorem 12.1.3 which is not satis¬ed. In each

case, identify the point at which the proof of Theorem 12.1.3 fails.

The contraction mapping theorem is not the only important ¬xed point

theorem in mathematics. Exercise 1.6.5 gives another ¬xed point result which

can be generalised substantially. (For example, if

B = {x : x ¤ 1}

is the unit ball in Rn , then any continuous map of B into itself has a ¬xed

point.) However, the standard proofs involve algebraic topology and are

beyond the scope of this book.

305

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

12.2 Existence of solutions of di¬erential equa-

tions

We use the contraction mapping theorem to show that a wide class of di¬er-

ential equations actually have a solution.

We shall be looking at equations of the form

y = f (t, y).

Our ¬rst, simple but important, result is that this problem on di¬erential

equations can be turned into a problem on integral equations. (We shall

discuss why this may be expected to be useful after Exercise 12.2.2.)

Lemma 12.2.1. If f : R2 ’ R is continuous, t0 , y0 ∈ R and δ > 0, then

the following two statements are equivalent.

(A) The function y : (t0 ’ δ, t0 + δ) ’ R is di¬erentiable and satis¬es the

equation y (t) = f (t, y(t)) for all t ∈ (t0 ’ δ, t0 + δ) together with the boundary

condition y(t0 ) = y0 .

(B) The function y : (t0 ’ δ, t0 + δ) ’ R is continuous and satis¬es the

condition

t

y(t) = y0 + f (u, y(u)) du

t0

for all t ∈ (t0 ’ δ, t0 + δ).

Proof. We show that (A) implies (B). Suppose that y satis¬es condition (A).

Since y is di¬erentiable, it is continuous. Thus, since f is continuous, y is

continuous and one of the standard forms of the fundamental theorem of the

calculus (Theorem 8.3.11) gives

t

y(t) ’ y(t0 ) = f (u, y(u)) du

t0

so, since y(t0 ) = y0 ,

t

y(t) = y0 + f (u, y(u)) du

t0

for all t ∈ (t0 ’ δ, t0 + δ) as required.

The fact that (B) implies (A) is an immediate consequence of the funda-

mental theorem of the calculus in the form Theorem 8.3.6.

306 A COMPANION TO ANALYSIS

Exercise 12.2.2. If f : R2 ’ R is n times di¬erentiable then any solution

of y (t) = f (t, y(t)) is n + 1 times di¬erentiable.

Remark: Most mathematicians carry in their minds a list of operations which

are or are not likely to to be troublesome. Such a list will probably contain

the following entries.

less troublesome more troublesome

multiplication division

interpolation extrapolation

averaging di¬erencing

integration di¬erentiation

direct calculation ¬nding inverses

Integration produces a better behaved function, di¬erentiation may well pro-

duce a worse behaved function. The integral of an integrable function is an

integrable function, the derivative of a di¬erentiable function need not be dif-

ferentiable. The contraction mapping theorem concerns a map T : X ’ X,

so to apply it we must be sure that our operation T does not take us out

of our initial space. This is much easier to ensure if T involves integration

rather than di¬erentiation.

Theorem 12.2.3. Suppose f : R2 ’ R is continuous, t0 , y0 ∈ R and δ > 0.

Suppose further that there exists a K > 0 such that Kδ < 1 and

|f (t, u) ’ f (t, v)| ¤ K|u ’ v|

for all t ∈ [t0 ’ δ, t0 + δ] and all u and v. Then there exists a unique y :

[t0 ’ δ, t0 + δ] ’ R which is continuous and satis¬es the condition

t

y(t) = y0 + f (u, y(u)) du

t0

for all t ∈ [t0 ’ δ, t0 + δ].

Proof. We know that C([t0 ’ δ, t0 + δ]) the space of continuous functions on

[t0 ’ δ, t0 + δ] with the uniform norm ∞ is complete. Now consider the

map T : C([t0 ’ δ, t0 + δ]) ’ C([t0 ’ δ, t0 + δ]) given by

t

(T g)(t) = y0 + f (u, g(u)) du.

t0

307

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

If t0 + δ ≥ t ≥ t0 , we have

t

|(T g)(t) ’ (T h)(t)| = f (u, g(u)) ’ f (u, h(u)) du

t0

t

¤ |f (u, g(u)) ’ f (u, h(u))| du

t0

t

¤ K|g(u) ’ h(u)| du

t0

¤ (t ’ t0 )K g ’ h ¤ Kδ g ’ h ∞,

∞

and a similar argument gives

|(T g)(t) ’ (T h)(t)| ¤ Kδ g ’ h ∞

for t0 ≥ t ≥ t0 ’ δ. Thus

Tg ’ Th ¤ Kδ g ’ h

∞ ∞

and T is a contraction mapping.

The contraction mapping theorem tells us that T has a unique ¬xed point,

that is there exists a unique y ∈ C([t0 ’ δ, t0 + δ]) such that

t

y(t) = y0 + f (u, y(u)) du

t0

for all t ∈ [t0 ’ δ, t0 + δ] and this is the required result.

Exercise 12.2.4. Restate Theorem 12.2.3 in terms of di¬erential equations.

Condition is called a Lipschitz condition.

Exercise 12.2.5. (i) Show that, if f : R2 ’ R has continuous partial deriva-

tive f,2 , then given any [a, b] and [c, d] we can ¬nd a K such that

|f (t, u) ’ f (t, v)| ¤ K|u ’ v|

for all t ∈ [a, b] and u, v ∈ [c, d].

(ii) If f : R2 ’ R is given by f (t, y) = |y| show that

|f (t, u) ’ f (t, v)| ¤ K|u ’ v|

for all t, u and v, but f does not have a partial derivative f,2 everywhere.

In the absence of a condition like di¬erential equations can have un-

expected properties.

308 A COMPANION TO ANALYSIS

Exercise 12.2.6. Consider the di¬erential equation

y = 3y 2/3

with y(0) = 0. Show that it has the solution

y(t) = (t ’ a)3 for t < a,

for a ¤ t ¤ b,

y(t) = 0

y(t) = (t ’ b)3 for b < t

whenever a ¤ b.

Exercise 12.2.6 is worth remembering whenever you are tempted to con-

vert the useful rule of thumb ˜¬rst order di¬erential equations involve one

choice of constant™ into a theorem.

Remark: It is easy to write down di¬erential equations with no solution. For

example, there is no real-valued solution to

(y )2 + y 2 + 1 = 0.

However, it can shown that the existence part of Theorem 12.2.3 continues

to hold, even if we drop condition , provided merely that f is continuous.

The reader may wish to ponder on the utility of an existence theorem in the

absence of a guarantee of uniqueness.

There is no di¬culty in extending the proof of Theorem 12.2.3 to higher

dimensions. In the exercise that follows the norm is the usual Euclidean

norm and y (t) = (y1 (t), y2 (t), . . . , yn (t)).

Exercise 12.2.7. (i) Suppose f : Rn+1 ’ Rn is continuous, t0 ∈ R, y0 ∈ Rn

and δ > 0. Suppose, further, that there exists a K > 0 such that Kδ < 1 and

f (t, u) ’ f (t, v) ¤ K u ’ v

for all t ∈ [t0 ’ δ, t0 + δ]. Then there exists a unique y : [t0 ’ δ, t0 + δ] ’ Rn

which is continuous and satis¬es the condition

t

y(t) = y0 + f (u, y(u)) du

t0

for all t ∈ [t0 ’ δ, t0 + δ].

(ii) With the notation and conditions of (i), y is the unique solution of

y (t) = f (t, y(t)), y(t0 ) = y0 .

on (t0 ’ δ, t0 + δ).

309

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

Exercise 12.2.7 is particularly useful because it enables us to deal with

higher order di¬erential equations. To see how the proof below works, observe

that the second order di¬erential equation

y +y =0

can be written as two ¬rst order di¬erential equations