Since k = 1, we have k = 0 and thus k > 0. Let us set L = k — .

—

If x = 0, then x ’1 x ∈ S(0, 1) so

’1 ’1 ’1

x) ≥ f (k) = k

x x = x x = f( x = L,

— — —

≥ L x . The case x = 0 is trivial, so we have established

and so x —

Kx ≥ x ≥L x

—

for all x ∈ Rn as required.

Exercise 10.4.7. (i) Let aj ∈ R for 1 ¤ j ¤ n and write

n

aj |xj |.

x =

w

j=1

State and prove necessary and su¬cient conditions for w to be a norm on

R.n

be the usual Euclidean norm on Rn . Show that, if n ≥ 2, then

(ii) Let

given any K we can ¬nd a norm — and points y1 and y2 such that

y1 > K y1 and y2 > K y2 — .

—

Does this result remain true if n = 1? Give reasons.

The fact that all norms on a ¬nite dimensional space are, essentially,

the same meant that, in Chapters 4 and 6 and elsewhere, we did not really

need to worry about which norm we used. By the same token, it obscured

the importance of the appropriate choice of norm in approaching problems in

analysis. However, once we consider in¬nite dimensional spaces, the situation

is entirely di¬erent.

250 A COMPANION TO ANALYSIS

Exercise 10.4.8. Consider s00 the space of real sequences a = (an )∞ such

n=1

that all but ¬nitely many of the an are zero.

(i) Show that if we use the natural de¬nitions of addition and scalar

multiplication

(an ) + (bn ) = (an + bn ), »(an ) = (»an )

then s00 is a vector space.

(ii) Show that the following de¬nitions all give norms on s00 .

= max |an |,

a ∞

n≥1

= max |nan |,

a w

n≥1

∞

|an |,

a =

1

n=1

1/2

∞

|an |2

a = ,

2

n=1

∞

n|an |.

a =

u

n=1

(iii) For each of the twenty ¬ve possible pairs of norms A and B from

part (ii), establish whether or not there exists a K such that K a A ≥ a B

for all a ∈ s00 . Show that none of the norms are Lipschitz equivalent.

(iv) Find a family of norms ± on s00 [± > 0] such that ± and β

are not Lipschitz equivalent if ± = β.

When we study an algebraic structure we also study those maps which

preserve algebraic structure. Thus, if we have a map θ : G ’ H be-

tween groups, we want θ to preserve group multiplication, that is, we want

θ(xy) = θ(x)θ(y). Such maps are called homomorphisms. Similarly, when

we study an analytic structure, we also study those maps which preserve an-

alytic structure. Consider for example a map f between metric spaces which

preserves ˜sequences tending to limits™. We then have

xn ’ x0 implies f (xn ) ’ f (x0 )

and study continuous functions.

If we study objects which have linked analytic and algebraic structures,

we will thus wish to study maps between them which preserve both algebraic

and analytic structures. For normed vector spaces this means that we wish

to study continuous linear maps. This fact is obscured by the fact that linear

maps between ¬nite dimensional vector spaces are automatically continuous.

251

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

Exercise 10.4.9. Let U and V be ¬nite dimensional vector spaces with norms

V . Show that any linear map T : U ’ V is automatically con-

U and

tinuous.

However, as we shall see in Exercise 10.4.14, when we deal with in¬nitely

dimensional vector spaces, not all linear maps need be continuous.

Here is a simple but important characterisation of those linear maps which

are continuous.

Lemma 10.4.10. Let (U, U ) and (V, V ) be normed vector spaces.

Then a linear map T : U ’ V is continuous if and only if there exists a K

such that T u V ¤ K u U for all u ∈ U .

Exercise 10.4.11. Prove Lemma 10.4.10. [One possible proof follows that

of Lemma 10.4.2 very closely.]

Exercise 10.4.12. By observing that the space of polynomials of degree n

or less has ¬nite dimension, or otherwise, prove the following result. There

exists a constant Cn such that

sup |P (t)| ¤ Cn sup |P (t)|

t∈[0,1] t∈[0,1]

for all real polynomials P of degree n or less.

State, with proof, whether we can ¬nd a C independent of n such that

sup |P (t)| ¤ C sup |P (t)|

t∈[0,1] t∈[0,1]

for all real polynomials P .

State, with proof, whether we can ¬nd a constant An such that of n such

that

sup |P (t)| ¤ An sup |P (t)|

t∈[0,1] t∈[0,1]

for all real polynomials P of degree n or less.

The close link between Lemma 10.4.10 and Lemma 10.4.2 is highlighted

in the next exercise.

Exercise 10.4.13. Let 1 and 2 be two norms on a vector space U .

(i) Show that the following statements are equivalent.

(a) If xn ’ x0 1 ’ 0, then xn ’ x0 2 ’ 0.

1 ) ’ (U,

(b) The identity map I : (U, 2 ) from U with norm 1

to U with norm 2 is continuous.

252 A COMPANION TO ANALYSIS

(c) There exists a K such that K u 1 ≥ u 2 for all u ∈ U .

(ii) Write down, and prove equivalent, three similar statements (a) , (b)

and (c) where (c) is the statement

(c) There exist K and L with K > L > 0 such that K u 1 ≥ u 2 ≥

L u 1 for all u ∈ U .

Here are some examples of continuous and discontinuous linear maps5 .

Exercise 10.4.14. Consider the vector space s00 de¬ned in Exercise 10.4.8

and the norms

= max |an |,

a ∞

n≥1

= max |nan |,

a w

n≥1

∞

|an |,

a =

1

n=1

1/2

∞

|an |2

a = ,

2

n=1

∞

n|an |.

a =

u

n=1

given there.

(i) For each of the twenty ¬ve possible pairs of norms A and B

A) ’

listed above state, with reasons, whether the identity map I : (s00 ,

(s00 , B ) from s00 with norm A to s00 with norm B is continuous.

(ii) Show that the map T : s00 ’ R de¬ned by T a = ∞ aj is linear.

j=1

If we give R the usual Euclidean norm and s00 one of the ¬ve norms listed

above, state, with reasons, whether T is continuous.

(iii) Show that the map S : s00 ’ s00 de¬ned by Sa = b with bj = jaj

is linear. For each of the twenty ¬ve possible pairs of norms A and B

listed above, state, with reasons, whether S, considered as a map from s 00

with norm A to s00 with norm B , is continuous.

Once we have Lemma 10.4.10, the way is open to an extension of the idea

of an operator norm investigated in Section 6.2.

5

If the reader returns to this example after studying the chapter on completeness,

she may note that the normed spaces given here are not complete. The question of the

existence of discontinuous linear maps between complete normed vector spaces involves

more advanced ideas, notably the axiom of choice. However, the consensus is that it

is unreasonable to expect all linear maps between complete normed vector spaces to be

continuous, in the same way, and for much the same reasons, as it is unreasonable to

expect all sets in R3 to have volume (see page 172).

253

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

De¬nition 10.4.15. Let U and V be vector spaces with norms and

U

V . If ± : U ’ V is a continuous linear map, then we set

± = sup ±x .

V

U ¤1

x

Exercise 10.4.16. (i) Let U = V = s00 and let a U = a V = ∞ |an |. n=1

’1

Show that if T : U ’ V is de¬ned by T a = b with bj = (1 ’ j )aj then T is

a continuous linear map. However, there does not exist an a ∈ U with a = 0

such that T a V = T a U . [See also Exercise 11.1.16.]

(ii) If U and V are ¬nite dimensional normed vector spaces and T : U ’

V is linear can we always ¬nd an a ∈ U with a = 0 such that T a V =

T a U ? Give reasons.

In exactly the way we proved Lemma 6.2.6, we can prove the following

results.

Exercise 10.4.17. Let U and V be vector spaces with norms U and V

and let ±, β : U ’ V be continuous linear maps.

(i) If x ∈ U , then ±x V ¤ ± x U .

(ii) ± ≥ 0.

(iv) If ± = 0, then ± = 0.

(v) If » ∈ R, then »± is a continuous linear map and »± = |»| ± .

(vi) (The triangle inequality) ± + β is a continuous linear map and

±+β ¤ ± + β .

W and γ : V ’ W is a continuous

(vii) If W is vector spaces with norm

linear map, then γ± is a continuous linear map and γ± ¤ γ ± .

We restate part of Exercise 10.4.17 in the language of this chapter.

Lemma 10.4.18. Let U and V be vector spaces with norms U and V.

The space L(U, V ) of continuous linear maps is a vector space and the oper-

ator norm is a norm on L(U, V ).

Although we shall not develop the theme further, we note that we now

have an appropriate de¬nition for di¬erentiation of functions between general

normed vector spaces.

De¬nition 10.4.19. Let U and V be vector spaces with norms U and

V . Suppose that E is a subset of of U and x a point such that there exists

a δ > 0 with B(x, δ) ⊆ E. We say that f : E ’ V is di¬erentiable at x, if

we can ¬nd a continuous linear map ± : U ’ V such that, when h ∈ B(x, δ),

f (x + h) = f (x) + ±h + (x, h) h U

254 A COMPANION TO ANALYSIS

where (x, h) V ’ 0 as h U ’ 0. We write ± = Df (x) or ± = f (x).

If E is open and f is di¬erentiable at each point of E, we say that f is

di¬erentiable on E.

Notice that the only important change from De¬nition 6.1.4 is that we

speci¬cally demand that ± = Df (x) is a continuous linear function.

The best way to understand what is going on is probably to do the next

exercise.

Exercise 10.4.20. State and prove the appropriate extension of the chain

rule given as Lemma 6.2.10.

There is no di¬culty in extending all our work on di¬erentiation from

¬nite dimensional to general normed spaces. The details are set out in

Dieudonn´™s book ([13], Chapter VIII).

e

Exercise 10.4.21. The title of this section was ˜Norms and the interaction

of algebra and analysis™ but much of it was really pure algebra in the sense

that it did not use the fundamental axiom of analysis. Go back through the

section and note which results are genuine results of analysis and which are

just algebra disguised as analysis.

Geodesics ™

10.5

This section introduces an idea which is important in more advanced work

but which will not be used elsewhere in this book. The discussion will be

informal.

Suppose we wish to build a road joining two points of the plane. The cost

per unit length of road will depend on the nature of the terrain. If the cost

of building a short stretch of length δs near a point (x, y) is (to ¬rst order)

g(x, y)δs then, subject to everything being well behaved, the cost of road “

will be

g(x, y) ds.

“

The reader may use whatever de¬nition of line integral she is comfortable

with. In particular she does not need to have read Section 9.5. It is tempting

to de¬ne the distance d(A, B) between two points A and B in the plane as

d(A, B) = inf g(x, y) ds : “ joining A and B .

“

If we include amongst our conditions that g is continuous and g(x, y) > 0

everywhere, we see that d is a metric.

255

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

Plausible statement 10.5.1. (i) d(A, B) ≥ 0 for all A and B.

(ii) d(A, B) = 0 implies A = B.

(iii) d(A, B) = d(B, A) for all A and B.

(iv) d(A, C) ¤ d(A, B) + d(B, C) for all A, B and C.

I have carefully used ˜inf™ rather than ˜min™ in the de¬nition of d(A, B).

One problem is that I have not speci¬ed the kind of “ that is permissible.

(Notice that the argument needed to obtain part (iv) of Statement 10.5.1

means that we cannot just use ˜smooth™.) However, it can be shown that, if

we choose a suitable class for the possible “ and a suitably well behaved g,

then the minimum is attained. If “0 is a path from A to B such that

g(x, y) ds = d(A, B),

“0

we call “0 a geodesic. (The geodesic need not be unique, consider road

building for two towns at diametrically opposite points of a circular marsh.)

If any two points are joined by a geodesic, then d(A, B) is the ˜length of the

geodesic path joining A and B™ where length refers to the metric d and not

to the Euclidean metric.

Let us try to use these ideas to ¬nd a metric on the upper half-plane

H = {z ∈ C : z > 0}

which is invariant under M¨bius transformation. (If you have not met M¨bius

o o

transformations skip to after Exercise 10.5.4, where I restate the problem.)

More precisely, we want a metric d such that, if T is a M¨bius map mapping H

o

bijectively to itself, then d(T z1 , T z2 ) = d(z1 , z2 ) for all z1 , z2 ∈ H. Of course,

no such metric might exist, but we shall see where the question leads.

Lemma 10.5.2. The set H of M¨bius maps T such that T |H : H ’ H is

o

bijective is a subgroup of the group M of all M¨bius transformations. The

o

subgroup H is generated by the transformations Ta with a ∈ R, D» with »

real and » > 0 and J, where

Ta (z) =a + z

D» (z) =»z

J(z) = ’ z ’1

Sketch proof. The fact that H is a subgroup of M follows from easy general

principles (see Exercise 10.5.3). We check directly that (if a is real and » > 0)

Ta , D» and J lie in H. Thus the composition of elements of the stated type

will also lie in H.

256 A COMPANION TO ANALYSIS

We now wish to identify H. If T ∈ H, then T is a well behaved bijective

map on C— = C ∪ {∞} and must take the boundary of H to the boundary

of H. Thus, writing R for the real axis, we know that T takes R ∪ {∞}

to R ∪ {∞}. Thus, if T (∞) = ∞, then T (∞) = a with a real and we set

M1 = JT’a . If T (∞) = ∞ we take M1 to be the identity map. In either case

M1 T ∈ H and M1 T (∞) = ∞. We now observe that M1 T (0) ∈ R ∪ {∞},

by our previous argument, and that M1 T (0) = ∞, since M¨bius maps are

o

bijections. Thus M1 T (0) = b with b real. Set M2 = T’b . We have M2 M1 T ∈

H, M2 M1 T (0) = 0 and M2 M1 T (∞) = ∞. But any M¨bius map M which

o

¬xes 0 and ∞ must have the form M (z) = µz for some µ = 0. If M takes H

to H, then µ is real and positive. Thus M2 M1 T = D» for some » > 0 and

’1 ’1

T = M1 M2 D» . We have shown that any element H is the composition of

maps of the stated type and the lemma follows.

(The remark on page 234 applies here too. We know so much about M¨bius

o

transformations that the argument above is more a calculation than a proof.

However, for more complicated conformal maps than are generally found in

undergraduate courses, arguments involving ˜boundaries™ may be mislead-

ing.)

Exercise 10.5.3. (Most readers will already know the content of this exer-

cise.) Let X be a set and S(X) the collection of bijections of X. Show that

S(X) is a group under the operation of composition. If G is a subgroup of

S(X), Y a subset of X and we de¬ne

K = {f ∈ G : f (Y ) = Y },

show that K is a subgroup of G.

Exercise 10.5.4. If |a| < 1 and Ma is the M¨bius map Ma given by

o

z’a

Ma z = ,

a— z ’ 1

show that |Ma eiθ | = 1 for all real θ. Deduce, stating any properties of M¨bius

o

transforms that you need, that Ma maps the unit disc D = {z : |z| < 1} to

itself and interchanges 0 and a. Show that the set H of M¨bius maps T such

o

that T |H : D ’ D is bijective is a subgroup of the group M of all M¨bius o

transformations generated by the transformations Ma with |a| < 1 and the

rotations.

Lemma 10.5.2 reduces our problem to one of ¬nding a metric d on the half

plane H = {z : z > 0} such that d(T z1 , T z2 ) = d(z1 , z2 ) for all z1 , z2 ∈ H,

257

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

whenever T is one of the transformations Ta with a ∈ R, D» with » > 0 and

J de¬ned by

Ta (z) =a + z

D» (z) =»z

J(z) = ’ z ’1 .

We try using the ideas of this section and de¬ning

d(z1 , z2 ) = inf g(z) ds : “ joining z1 and z2 ,

“

for some appropriate strictly positive function g : H ’ R. (Throughout, we

write z = x + iy and identify the plane R2 with C in the usual manner.)