(iv) If f (x) ’ a and g(x) ’ b, with respect to the direction , then,

f (x) + g(x) ’ a + b.

(v) If f (x) ’ a and g(x) ’ b, with respect to the direction , then

f (x)g(x) ’ ab.

(vi) Suppose that f (x) ’ a, with respect to the direction . If f (x) = 0

for each x ∈ X and a = 0, then f (x)’1 ’ a’1 .

(vii) If f (x) ¤ A for each x ∈ X and f (x) ’ a, with respect to the

direction , then a ¤ A. If g(x) ≥ B for each x ∈ X and g(x) ’ b, with

respect to the direction , then b ≥ B.

As one might expect, we can recover Lemma 1.2.2 from Exercise D.8.

Exercise D.9. (i) If N+ is the set of strictly positive integers, show that >

(with its ordinary meaning) is a direction on N+ . Show further that, if f is

a function from N+ to F (an ordered ¬eld) and a ∈ F, then f (n) ’ a, with

respect to the direction >, if and only if f (n) ’ a as n ’ ∞ in the sense of

De¬nition 1.2.1.

(ii) Deduce Lemma 1.2.2 from Exercise D.8.

(iii) Show that (i) remains true if we replace > by ≥. Show that (i)

m if n ≥ 10m + 4. Thus

remains true if we replace > by with n

di¬erent succession relations can produce the same notion of limit.

The real economy of this approach appears when we extend it.

Exercise D.10. (i) Let a, t and b be real with a < t < b. Show that, if we

on (a, b) \ {t} by x y if |x ’ t| < |y ’ t|, then

de¬ne the relation

is a direction. Suppose f : (a, b) ’ R is a function and c ∈ R. Show that

f (x) ’ c with respect to the direction , if and only if f (x) ’ c as x ’ t,

in the traditional sense of De¬nition D.3.

398 A COMPANION TO ANALYSIS

(ii) Deduce the properties of the traditional limit of De¬nition D.3 from

Exercise D.8.

(iii) Give a treatment of the classical ˜limit from above™ de¬ned in De¬-

nition D.4 along the lines laid out in parts (i) and (ii).

Exercise D.11. Obtain a multidimensional analogue of De¬nition D.7 along

the lines of De¬nition 4.1.8 and prove a multidimensional version of Exer-

cise D.8 along the lines of Lemma 4.1.9.

A little thought shows how to bring De¬nition D.2 into this circle of ideas.

De¬nition D.12. If X is a direction on a non-empty set X, Y is a di-

rection on a non-empty set Y and f is a function from X to Y , we say that

f (x) ’ —Y as x ’ —X if, given y ∈ Y , we can ¬nd x0 (y) ∈ X such that

f (x) Y y for all x x0 (y).

Exercise D.13. (i) Show that if we take X = N+ , X to be the usual rela-

tion > on X, Y = R and Y to be the usual relation > on Y , then saying that

a function f : X ’ Y has the property f (x) ’ —Y as x ’ —X is equivalent

to the classical statement f (n) ’ ∞ as n ’ ∞.

(ii) Give a similar treatment for the classical statement that a function

f : R ’ R satis¬es f (x) ’ ’∞ as x ’ ∞.

(iii) Give a similar treatment for the classical statement that a function

f : R ’ R satis¬es f (x) ’ a as x ’ ∞.

In all the examples so far, we have only used rather simple examples of

direction. Here is a more complicated one.

Exercise D.14. This exercise assumes a knowledge of Section 8.2 where we

de¬ned the Riemann integral. It uses the notation of that section. Consider

the set X of ordered pairs (D, E) where D is the dissection

D = {x0 , x1 , . . . , xn } with a = x0 ¤ x1 ¤ x2 ¤ · · · ¤ xn = b,

and

E = {t0 , t1 , . . . , tn } with xj’1 ¤ tj ¤ xj .

If f : [a, b] ’ R is a bounded function, we write

n

σ(f, D, E) = f (tj )(xj ’ xj’1 ).

j=1

399

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

Show that, if we write (D , E ) (D, E) when D ⊇ D, then is a

direction on X. (Note that we place no conditions on E and E .) Show that

not all dissections are comparable.

Using the results of Section 8.2, show that f is Riemann integrable, in

the sense of Section 8.2, with integral I if and only if

σ(f, D, E) ’ I

with respect to the direction .

Here is another example, this time depending on the discussion of metric

spaces in Section 10.3. If we wish to de¬ne the notion of a limit for a func-

tion between two metric spaces the natural classical procedure is to produce

something along the lines of De¬nition 10.3.22

De¬nition D.15. Let (X, d) and (Z, ρ) be metric spaces and f be a map

from X to Z. Suppose that x ∈ X and z ∈ Z. We say that f (y) ’ z as

y ’ x if, given > 0, we can ¬nd a δ( , x) > 0 such that, if y ∈ X and

d(x, y) < δ( , x), we have

ρ(f (y), z) < .

Here is an alternative treatment using direction.

Exercise D.16. Let (X, d) and (Z, ρ) be metric spaces and x ∈ X and z ∈

Z. We take X to be the collection of open sets in X which contain x and Z

to be the collection of open sets in Z which contain z.

Show that if we de¬ne a relation on X by U X V if V ⊇ U , then X is

a direction on X . Is it always true that two elements of X are comparable?

Let Z be de¬ned similarly. If f is a map from X to Z, show that

f (y) ’ z as y ’ x in the sense of De¬nition D.15 if and only if f (y) ’ —Z

as x ’ —X in the sense of De¬nition D.12.

The advantage of the approach given in Exercise D.16 is that it makes

no reference to the metric and raises the possibility of of doing analysis on

more general objects than metric spaces.

We close with a couple of interesting observations (it will be more conve-

nient to use De¬nition D.7 than our more general De¬nition D.12).

Exercise D.17. Suppose that is a direction on a non-empty set X and f

is a function from X to R.

Suppose that there exists an M ∈ R such that f (x) ¤ M for all x ∈ X and

suppose that f is ˜increasing™ in the sense that x y implies f (x) ¤ f (y).

Show that there exists an a ∈ F such that that f (x) ’ a with respect to the

direction .

[Hint. Think about the supremum.]

400 A COMPANION TO ANALYSIS

If the reader thinks about the matter she may recall points in the book

where such a result would have been useful.

Exercise D.18. Prove Lemma 9.2.2 using using Lemma D.17.

Exercise D.19. The result of Lemma D.17 is the generalisation of the state-

ment that every bounded increasing sequence in R has a limit. Find a similar

generalisation of the general principle of convergence. Use it to do Exer-

cise 9.2.10.

If the reader wishes to see more, I refer her to the elegant and e¬cient

treatment of analysis in [2].

Appendix E

Traditional partial derivatives

One of the most troublesome culture clashes between pure mathematics and

applied is that to an applied mathematician variables like x and t have mean-

ings such as position and time whereas to a pure mathematician all variables

are ˜dummy variables™ or ˜place-holders™ to be interchanged at will. To a pure

mathematician, v is an arbitrary function de¬ned by its e¬ect on a variable

so that v(t) = At3 means precisely the same thing as v(x) = Ax3 whereas,

to an applied mathematician who thinks of v as a velocity, the statements

v = At3 and v = Ax3 mean very di¬erent (indeed incompatible) things.

dv δv

The applied mathematician thinks of as representing ˜when every-

dt δt

thing is so small that second order quantities can be neglected™. Since

δv δt δv

= ,

δt δx δx

it is obvious to the applied mathematician that

dv dv dx

= , (A)

dt dx dt

but the more rigid notational conventions of the pure mathematicians prevent

them from thinking of v as two di¬erent functions (one of t and one of x) in

the same formula. The closest a pure mathematician can get to equation (A)

is the chain rule

d

v(x(t)) = v (x(t))x (t).

dt

Now consider a particle moving along the x-axis so its position is x at

time t. Since

δx δt

= 1,

δt δx

401

402 A COMPANION TO ANALYSIS

it is obvious to the applied mathematician that

dx dt

= 1,

dt dx

and so

dt dx

=1 (B)

dx dt

dx

What does equation (B) mean? In the expression we treat x as a function

dt

dt

of t, which corresponds to common sense, but in the expression we treat t

dx

as a function of x, which seems a little odd (how can the position of a particle

in¬‚uence time?). However, if the particle occupies each particular position

at a unique time, we can read o¬ time from position and so, in this sense, t

is indeed a function of x. A pure mathematician would say that the function

x : R ’ R is invertible and replace equation (B) by the inverse function

formula

d ’1 1

x (t) = .

x (x’1 (t))

dt

One of the reasons why this formula seems more complicated than equa-

dx dt

tion (B) is that the information on where to evaluate and has been

dt dx

suppressed in equation (B), which ought to read something like

dt dx

(x0 ) = 1 (t0 ) ,

dx dt

or

dt dx

=1 ,

dx dt

x=x0 t=t0

where the particle is at x0 at time t0 .

The clash between the two cultures is still more marked when it comes

to partial derivatives. Consider a gas at temperature T , held at a pressure

P in a container of volume V and isolated from the outside world. The

applied mathematician knows that P depends on T and V and so writes P =

P (T, V ) or P = P (V, T ). (To see the di¬erence in conventions between pure

and applied mathematics, observe that, to a pure mathematician, functions

f : R2 ’ R such that f (x, y) = f (y, x) form a very restricted class!) Suppose

403

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

that, initially, T = T0 , P = P0 , V = V0 . If we change T0 to T0 + δT whilst

keeping V ¬xed, then P changes to P0 + δP and the applied mathematician

‚P δP

thinks of as representing ˜when everything is so small that

‚T V =V0 ,T =T0 δT

‚P

second order quantities can be neglected™. In other words, is

‚T V =V0 ,T =T0

the rate of change of P when T varies but V is kept ¬xed. Often applied

mathematicians write

‚P ‚P

=

‚T ‚T V =V0 ,T =T0

but you should note this condensed notation suppresses the information that

V (rather than, say, V + T ) should be kept constant when T is varied.

There is good reason to suppose that there is a well behaved function

g : R3 ’ R such that, if a particular gas is at temperature T , held a pressure

P in a container of volume V , then the temperature, pressure, volume triple

(T, P, V ) satis¬es

g(T, P, V ) = 0

(at least for a wide range of values of T , P and V ). We may link the pure

mathematician™s partial derivatives with those of the applied mathematician

by observing that, if (T0 , P0 , V0 ) and (T0 + δT, P0 + δP, V0 ) are possible

temperature, pressure, volume triples, then, to ¬rst order,

0 = g(T0 + δT, P0 + δP, V0 )

= g(T0 , P0 , V0 ) + g,1 (T0 , P0 , V0 )δT + g,2 g(T0 , P0 , V0 )δP

= g,1 (T0 , P0 , V0 )δT + g,2 (T0 , P0 , V0 )δP

and so, to ¬rst order,

δP g,1 (T0 , P0 , V0 )

=’ .

δT g,2 (T0 , P0 , V0 )

It follows that

‚P g,1 (T0 , P0 , V0 )

=’ .

‚T g,2 (T0 , P0 , V0 )

V =V0 ,T =T0

Essentially identical calculations show that

‚T g,3 (T0 , P0 , V0 )

=’

‚V g,1 (T0 , P0 , V0 )

P =P0 ,V =V0

404 A COMPANION TO ANALYSIS

and

‚V g,2 (T0 , P0 , V0 )

=’ .

‚P g,3 (T0 , P0 , V0 )

T =T0 ,P =P0

Putting the last three equations together, we obtain

‚P ‚T ‚V

= ’1.

‚T ‚V ‚P

V =V0 ,T =T0 P =P0 ,V =V0 T =T0 ,P =P0

This is a very beautiful equation. It can be made much more mysterious by

leaving implicit what we have made explicit and writing

‚P ‚T ‚V

= ’1. (C)

‚T ‚V ‚P

If we further neglect to mention that T , P and V are restricted to the surface

g(T, P, V ) = 0, we get ˜an amazing result that common sense could not

possibly have predicted . . . It is perhaps the ¬rst time in our careers, for

most of us, that we do not understand 3-dimensional space™ ([34], page 65).

Exercise E.1. Obtain the result corresponding to equation (C) for four vari-

ables. Without going into excessive details, indicate the generalisation to n

variables with n ≥ 2. (Check that if n = 2 you get a result corresponding to

equation (B).)

The di¬erence between the ˜pure™ and ˜applied™ treatment is re¬‚ected in

a di¬erence in language. The pure mathematician speaks of ˜di¬erentiation

of functions on many dimensional spaces™ and the applied mathematician of

˜di¬erentiation of functions of many variables™. Which approach is better de-

pends on the problem at hand. Although (T, P, V ) is a triple of real numbers,

it is not a vector, since pressure, volume and temperature are quantities of

di¬erent types. Treating (T, P, V ) as a vector is like adding apples to pears.

The ˜geometric™ pure approach will yield no insight, since there is no ge-

ometry to consider. On the other hand theories, like electromagnetism and

relativity with a strong geometric component, will bene¬t from a treatment

which does not disguise the geometry.

I have tried to show that the two approaches run in parallel but that, al-

though statements in one language can be translated ˜sentence by sentence™

into the other, there is no word for word dictionary between them. A good

mathematician can look at a problem in more than one way. In particular a

good mathematician will ˜think like a pure mathematician when doing pure

mathematics and like an applied mathematician when doing applied math-

ematics™. (Great mathematicians think like themselves when doing mathe-

matics.)

405

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

Exercise E.2. (In this exercise you should assume that everything is well

behaved.) Rewrite the following statement in pure mathematics notation and

prove it using whichever notation (pure, applied or mixed) you prefer.

Suppose that the equations

f (x, y, z, w) = 0

g(x, y, z, w) = 0

can be solved to give z and w as functions of (x, y). Then

‚f ‚g ‚f ‚g

’

‚z ‚x ‚w ‚w ‚x

=’ ‚f ‚g ‚f ‚g

‚x ’

‚z ‚w ‚w ‚z

‚f ‚g ‚f ‚g

’

‚z ‚y ‚w ‚w ‚y

=’ .

‚f ‚g ‚f ‚g

‚y ’

‚z ‚w ‚w ‚z

Exercise E.3. Each of the four variables p, V , T and S can be regarded as a

well behaved function of any two of the others. In addition we have a variable

U which may be regarded as a function of any two of the four variables above

and satis¬es

‚U ‚U

= ’p.

= T,

‚S ‚V

V S

(i) Show that

‚V ‚T

= .

‚S ‚p

p S

‚2U

(ii) By ¬nding two expressions for , or otherwise, show that

‚p‚V

‚S ‚T ‚S ‚T

’ = 1.

‚V ‚p ‚p ‚V

p V V p

Exercise E.4. You should only do this exercise if you have met the change

of variable formula for multiple integrals. Recall that, if (u, v) : R 2 ’ R2 is

a well behaved bijective map, we write

‚u ‚u

‚(u, v) ‚x ‚y

J= = det ‚v ‚v

‚(x, y) ‚x ‚y

406 A COMPANION TO ANALYSIS

and call J the Jacobian determinant1 . Recall the the useful formula

‚(u, v) ‚(x, y)

= 1.

‚(x, y) ‚(u, v)

Restate it in terms of Df and Df ’1 evaluated at the correct points. (If you

can not see what is going on, look at the inverse function theorem (Theo-

rem 13.1.13 (ii)) and at our discussion of equation (B) in this appendix.)

Restate and prove the formula

‚(u, v) ‚(s, t) ‚(u, v)

=

‚(s, t) ‚(x, y) ‚(x, y)

in the same way.

1

J is often just called the Jacobian but we distinguish between the Jacobian matrix

(see Exercise 6.1.9) and its determinant.

Appendix F

Another approach to the

inverse function theorem

The object of this appendix is to outline a variant on the approach to the

inverse function theorem given in section 13.1.

We begin with a variation on Lemma 13.1.2.