In order to tie up loose ends, we need the following lemma.

Lemma 8.4.7. Suppose f : [a, b] ’ R is continuous and

b

f (t)h(t) dt = 0,

a

whenever h : [a, b] ’ R is an in¬nitely di¬erentiable function with h(a) =

h(b) = 0. Then f (t) = 0 for all t ∈ [a, b].

Proof. By continuity, we need only prove that f (t) = 0 for all t ∈ (a, b).

Suppose that, in fact, f (x) = 0 for some x ∈ (a, b). Without loss of generality

we may suppose that f (x) > 0 (otherwise, consider ’f ). Since (a, b) is open

and f is continuous we can ¬nd a δ > 0 such that [x ’ δ, x + δ] ⊆ (a, b) and

|f (t) ’ f (x)| < f (x)/2 for t ∈ [x ’ δ, x + δ]. This last condition tells us that

f (t) > f (x)/2 for t ∈ [x ’ δ, x + δ].

In Example 7.1.6 we constructed an in¬nitely di¬erentiable function E :

R ’ R with E(t) = 0 for t ¤ 0 and E(t) > 0 for t > 0. Setting h(t) =

E(t ’ x + δ)E(’t + x + δ) when t ∈ [a, b], we see that h is an in¬nitely

di¬erentiable function with h(t) > 0 for t ∈ (x ’ δ, x + δ) and h(t) = 0

otherwise (so that, in particular h(a) = h(b) = 0). By standard results on

the integral,

b x+δ x+δ

f (t)h(t) dt ≥

f (t)h(t) dt = (f (x)/2)h(t) dt

a x’δ x’δ

x+δ

f (x)

= h(t) dt > 0,

2 x’δ

so we are done.

Exercise 8.4.8. State explicitly the ˜standard results on the integral™ used in

the last sentence of the previous proof and show how they are applied.

Theorem 8.4.6 is often stated in the following form. If the function y :

[a, b] ’ R minimises J then

‚F d ‚F

= .

‚y dx ‚y

This is concise but can be confusing to the novice6 .

The Euler-Lagrange equation can only be solved explicitly in a small

number of special cases. The next exercise (which should be treated as an

6

It certainly confused me when I met it for the ¬rst time.

197

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

exercise in calculus rather than analysis) shows how, with the exercise of

some ingenuity, we can solve the brachistochrone problem with which we

started. Recall that this asked us to minimise

1/2

b

1 + f (x)2

1

J(f ) = dx.

(2g)1/2 κ ’ f (x)

a

Exercise 8.4.9. We use the notation and assumptions of Theorem 8.4.6.

(i) Suppose that F (u, v, w) = G(v, w) (often stated as ˜t does not appear

explicitly in F =F (t, y, y )™). Show that the Euler-Lagrange equation becomes

d

G,1 (f (t), f (t)) = G,2 (f (t), f (t))

dt

and may be rewritten

d

G(f (t), f (t)) ’ f (t)G,2 (f (t), f (t)) = 0.

dt

Deduce that

G(f (t), f (t)) ’ f (t)G,2 (f (t), f (t)) = c

‚F

where c is a constant. (This last result is often stated as F ’ y = c.)

‚y

(ii) (This is not used in the rest of the question.) Suppose that F (u, v, w) =

G(u, w). Show that

G,2 (t, f (t)) = c

where c is a constant.

(iii) By applying (i), show that solutions of the Euler-Lagrange equation

associated with the brachistochrone are solutions of

1

=c

((κ ’ f (x))(1 + f (x)2 ))1/2

where c is a constant. Show that this equation can be rewritten as

1/2

B + f (x)

f (x) = .

A ’ f (x)

(iv) We are now faced with ¬nding the curve

1/2

dy B+y

= .

A’y

dx

198 A COMPANION TO ANALYSIS

If we are su¬ciently ingenious (or we know the answer), we may be led to

try and express this curve in parametric form by setting

A’B A+B

’

y= cos θ.

2 2

Show that

dx A+B

= (1 + cos θ),

dθ 2

and conclude that our curve is (in parametric form)

x = a + k(θ ’ sin θ), y = b ’ k cos θ

for appropriate constants a, b and k. Thus any curve which minimises the

time of descent must be a cycloid.

It is important to observe that we have shown that any minimising func-

tion satis¬es the Euler-Lagrange equation and not that any function sat-

isfying the Euler-Lagrange equation is a minimising function. Exactly the

same argument (or replacing J by ’J), shows that any maximising function

satis¬es the Euler-Lagrange equation. Further, if we re¬‚ect on the simpler

problem discussed in section 7.3, we see that the Euler-Lagrange equation

will be satis¬ed by functions f such that

Gh (·) = J(f + ·h)

has a minimum at · = 0 for some h ∈ E and a maximum at · = 0 for others.

Exercise 8.4.10. With the notation of this section show that, if f satis¬es

the Euler-Lagrange equations, then Gh (0) = 0.

To get round this problem, examiners ask you to ˜¬nd the values of f

which make J stationary™ where the phrase is equivalent to ˜¬nd the val-

ues of f which satisfy the Euler-Lagrange equations™. In real life, we use

physical intuition or extra knowledge about the nature of the problem to

¬nd which solutions of the Euler-Lagrange equations represent maxima and

which minima.

Mathematicians spent over a century seeking to ¬nd an extension to the

Euler-Lagrange method which would enable them to distinguish true maxima

and minima. However, they were guided by analogy with the one dimensional

(if f (0) = 0 and f (0) > 0 then 0 is minimum) and ¬nite dimensional

case and it turns out that the analogy is seriously defective. In the end,

199

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

Figure 8.1: A problem for the calculus of variations

Weierstrass produced examples which made it plain what was going on. We

discuss a version of one of them.

Consider the problem of minimising

1

(1 ’ (f (x))4 )2 + f (x)2 dx

I(f ) =

0

where f : [0, 1] ’ R is once continuously di¬erentiable and f (0) = f (1) = 0.

Exercise 8.4.11. We look at

Gh (·) = I(·h).

Show that Gh (·) = 1 + Ah · 2 + Bh · 4 + Ch · 8 where Ah , Bh , Ch depend on

h. Show that, if h is not identically zero, Ah > 0 and deduce that Gh has a

strict minimum at 0 for all non-zero h ∈ E.

We are tempted to claim that ˜I(f ) has a local minimum at f = 0™.

Now look at the function gn [n a strictly positive integer] illustrated in

Figure 8.1 and de¬ned by

2r 2r 1

gn (x) = x ’ for x ’ ¤ ,

2n 2n 4n

2r + 1 2r + 1 1

’x for x ’ ¤

gn (x) = ,

2n 2n 4n

whenever r is an integer and x ∈ [0, 1]. Ignoring the ¬nite number of points

where gn is not di¬erentiable, we see that gn (x) = ±1 at all other points,

and so

1

gn (x)2 dx ’ 0 as n ’ ∞.

I(gn ) =

0

200 A COMPANION TO ANALYSIS

Figure 8.2: The same problem, smoothed

It is clear that we can ¬nd a similar sequence of functions fn which is con-

tinuously di¬erentiable by ˜rounding the sharp bits™ as in Figure 8.2.

The reader who wishes to dot the i™s and cross the t™s can do the next

exercise.

> 0 and let k : [0, 1] ’ R be the function

Exercise 8.4.12. (i) Let 1/2 >

such that

k(x) = ’1 x for 0 ¤ x ¤ ,

for ¤ x ¤ 1 ’ ,

k(x) = 1

k(x) = ’1 (1 ’ x) for 1 ’ ¤ x ¤ 1.

Sketch k.

(ii) Let kn (x) = (’1)[2nx] k(2nx ’ 2[nx]) for x ∈ [0, 1]. (Here [2nx] means

the integer part of 2nx.) Sketch the function kn .

(iii) Let

x

Kn (x) = kn (t) dt

0

for x ∈ [0, 1]. Sketch Kn . Show that 0 ¤ Kn (x) ¤ 1/(2n) for all x. Show that

Kn is once di¬erentiable with continuous derivative. Show that |Kn (x)| ¤ 1

for all x and identify the set of points where |Kn (x)| = 1.

(iv) Show that there exists a sequence of continuously di¬erentiable func-

tions fn : [0, 1] ’ R, with fn (0) = fn (1) = 0, such that

I(fn ) ’ 0 as n ’ ∞.

[This result is slightly improved in Example K.134.]

This example poses two problems. The ¬rst is that in some sense the fn

are close to f0 = 0 with I(fn ) < I(f0 ), yet the Euler-Lagrange approach of

201

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

Exercise 8.4.11 seemed to show that I(f0 ) was smaller than those I(f ) with

f close to f0 . One answer to this seeming paradox is that, in Exercise 8.4.11,

we only looked at Gh (·) = I(·h) as · became small, so we only looked at

certain paths approaching f0 and not at all possible modes of approach. As

· becomes small, not only does ·h become small but so does ·h . However,

as n becomes large, fn becomes small but fn does not. In general when

the Euler-Lagrange method looks at a function f it compares it only with

functions which are close to f and have derivative close to f . This does

not a¬ect the truth of Theorem 8.4.6 (which says that the Euler-Lagrange

equation is a necessary condition for a minimum) but makes it unlikely that

the same ideas can produce even a partial converse.

Once we have the notion of a metric space we can make matters even

clearer. (See Exercise K.199 to K.201.)

Exercise 8.4.13. This exercise looks back to Section 7.3. Let U be an open

subset of R2 containing (0, 0). Suppose that f : U ’ R has second order

partial derivatives on U and these partial derivatives are continuous at (0, 0).

Suppose further that f,1 (0, 0) = f,2 (0, 0) = 0. If u ∈ R2 we write Gu (·) =

f (·u).

(i) Show that Gu (0) = 0 for all u ∈ R2 .

(ii) Let e1 = (1, 0) and e2 = (0, 1) Suppose that Ge1 (0) > 0 and Ge2 (0) >

0. Show, by means of an example, that (0, 0) need not be a local minimum

for f . Does there exist an f with the properties given which attains a local

minimum at (0, 0)? Does there exist an f with the properties given which

attains a local maximum at (0, 0)?

(iii) Suppose that Gu (0) > 0 whenever u is a unit vector. Show that f

attains a local minimum at (0, 0).

The second problem raised by results like Exercise 8.4.12 is also very

interesting.

Exercise 8.4.14. Use Exercise 8.3.4 to show that I(f ) > 0 whenever f :

[0, 1] ’ R is a continuously di¬erentiable function.

Conclude, using the discussion above, that the set

{I(f ) : f continuously di¬erentiable}

has an in¬mum (to be identi¬ed) but no minimum.

Exercise 8.1. Here is a simpler (but less interesting) example of a varia-

tional problem with no solution, also due to Weierstrass. Consider the set

202 A COMPANION TO ANALYSIS

E of functions f : [’1, 1] ’ R with continuous derivative and such that

f (’1) = ’1, f (1) = 1. Show that

1

x2 f (x)2 dx = 0

inf

f ∈E ’1

1

x2 f0 (x)2 dx = 0.

but there does not exist any f0 ∈ E with ’1

The discovery that that they had been talking about solutions to prob-

lems which might have no solutions came as a severe shock to the pure

mathematical community. Of course, examples like the one we have been

discussing are ˜arti¬cial™ in the sense that they have been constructed for

the purpose but unless we can come up with some criterion for distinguish-

ing ˜arti¬cial™ problems from ˜real™ problems this takes us nowhere. ˜If we

have actually seen one tiger, is not the jungle immediately ¬lled with tigers,

and who knows where the next one lurks.™ The care with which we proved

Theorem 4.3.4 (a continuous function on a closed bounded set is bounded

and attains its bounds) and Theorem 4.4.4 (Rolle™s theorem, considered as

the statement that, if a di¬erentiable function f on an open interval (a, b)

attains a maximum at x, then f (x) = 0) are distant echos of that shock. On

the other hand, the new understanding which resulted revivi¬ed the study

of problems of maximisation and led to much new mathematics.

It is always possible to claim that Nature (with a capital N) will never set

˜arti¬cial™ problems and so the applied mathematician need not worry about

these things. ˜Nature is not troubled by mathematical di¬culties.™ However,

a physical theory is not a description of nature (with a small n) but a model

of nature which may well be troubled by mathematical di¬culties. There

are at least two problems in physics where the model has the characteristic

features of our ˜arti¬cial™ problem. In the ¬rst, which asks for a description

of the electric ¬eld near a very sharp charged needle, the actual experiment

produces sparking. In the second, which deals with crystallisation as a system

for minimising an energy function not too far removed from I, photographs

reveal patterns not too far removed from Figure 8.1!

8.5 Vector-valued integrals

So far we have dealt only with the integration of functions f : [a, b] ’ R.

The general programme that we wish to follow would direct us to consider

the integration of functions f : E ’ Rm where E is a well behaved subset of

Rn . In this section we shall take the ¬rst step by considering the special case

of a well behaved function f : [a, b] ’ Rm . Since C can be identi¬ed with R2 ,

203

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

our special case contains, as a still more special (but very important case),

the integration of well behaved complex-valued functions f : [a, b] ’ C.

The de¬nition is simple.

De¬nition 8.5.1. If f : [a, b] ’ Rm is such that fj : [a, b] ’ R is Rie-

mann integrable for each j, then we say that f is Riemann integrable and

b

f (x) dx = y where y ∈ Rm and

a

b

yj = fj (x) dx

a

for each j.

In other words,

b b

f (x) dx = fj (x) dx.

a a

j

It is easy to obtain the properties of this integral directly from its de¬ni-

tion and the properties of the one dimensional integral. Here is an example.

Lemma 8.5.2. If ± : Rm ’ Rp is linear and f : [a, b] ’ Rm is Riemann

integrable, then so is ±f and

b b

(±f )(x) dx = ± f (x) dx .

a a

Proof. Let ± have matrix representation (aij ). By Lemma 8.2.11,

m

(±f )i = aij fj

j=1

is Riemann integrable and

bm m b

aij fj (x) dx = aij fj (x) dx.

a a

j=1 j=1

Comparing this with De¬nition 8.5.1, we see that we have the required result.

Taking ± to be any orthogonal transformation of Rm to itself, we see

that our de¬nition of the integral is, in fact, coordinate independent. (Re-

member, it is part of our programme that nothing should depend on the

particular choice of coordinates we use. The reader may also wish to look at

Exercise K.137.)

Choosing a particular orthogonal transformation, we obtain the following

nice result.

204 A COMPANION TO ANALYSIS

Theorem 8.5.3. If f : [a, b] ’ Rm is Riemann integrable then

b

f (x) dx ¤ (b ’ a) sup f (x) .

x∈[a,b]

a

This result falls into the standard pattern

size of integral ¤ length — sup.

Proof. If y is a vector in Rm , we can always ¬nd a rotation ± of Rm such

that ±y lies along the x1 axis, that is to say, (±y)1 ≥ 0 and (±y)j = 0 for

b

2 ¤ j ¤ m. Let y = a f (x) dx. Then

b b

f (x) dx = ± f (x) dx

a a

b

= ± f (x) dx

a 1

b

= (±f (x))1 dx

a

¤ (b ’ a) sup |(±f (x))1 |

x∈[a,b]

¤ (b ’ a) sup ±f (x)

x∈[a,b]

= (b ’ a) sup f (x) .

x∈[a,b]