di¬erentiation to problems on integration with the advantages pointed out

on page 306.

In the terminology of more advanced work, we have shown how di¬erential

operators like y ’ Sy, where

Sy(t) = y (t) + a(t)y (t) + b(t)y(t),

can be linked with better behaved integral operators like

1

f ’ T f with T f (t) = G(s, t)f (s) ds.

0

Note that we have shown ST f = f for f continuous, but note also that, if f

is merely continuous, Sf need not be de¬ned. The Green™s function G is an

example of an integral kernel1 More formally, if we write

1

Jf (s) = K(s, t)f (s) ds,

0

then J is called an integral operator with kernel K.

We end with an example to show that things really do go awry if our key

assumption fails.

Example 12.4.8. The equation

y (t) + π 2 y(t) = t

has no solution satisfying y(0) = y(1) = 0.

Proof. Suppose that y satis¬es the equation

y (t) + π 2 y(t) = t

1

We shall not discuss these matters much further but most of the new words in this

last paragraph are well worth dropping.

You must lie among the daisies and discourse in novel phrases of your com-

plicated state of mind,

The meaning doesn™t matter if it™s only idle chatter of a transcendental kind.

And everyone will say,

As you walk your mystic way,

˜If this young man expresses himself in terms too deep for me

Why, what a very singularly deep young man this deep young man must be!™

. (Gilbert and Sullivan Patience)

327

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

and satis¬es y(0) = 0. If we set

y (0) ’ π ’2

’2

w(t) = π t + sin πt,

π

then, by direct calculation,

w (t) + π 2 w(t) = t

and w(0) = 0 = y(0), w (0) = y (0) so, by the uniqueness results of the

previous section, y = w. In particular, y(1) = w(1) = 0.

Chapter 13

Inverse and implicit functions

13.1 The inverse function theorem

We start with an exercise giving a very simple example of a technique that

mathematicians have used for centuries.

Exercise 13.1.1. We work in R.

(i) Suppose x and y are close to 1. If (x + ·)2 = y show that

· ≈ (y ’ x2 )/2.

(ii) This suggests the following method for ¬nding the positive square root

of y. Take x0 close to the square root of y (for example x0 = 1) and de¬ne

xn inductively by xn = xn’1 + (y ’ x2 )/2. We expect xn to converge to the

n’1

positive square root of y.

(a) Try this for y = 1.21 and for y = 0.81.

(b) Try this for y = 100.

(iii) Sketch x ’ x2 /2 for 1/2 ¤ x ¤ 3/2. Show that if |y ’ 1| ¤ 1/2 and

we de¬ne T x by

T x = x + (y ’ x2 )/2

then |T x ’ 1| ¤ 1/2 whenever |x ’ 1| ¤ 1/2. Show that T : [1/2, 3/2] ’

[1/2, 3/2] is a contraction mapping and deduce that the proposed method for

¬nding square roots works if |y ’ 1| ¤ 1/2.

(iv) Find a reasonable approximation to the positive square root of 150 by

observing that

(150)1/2 = 12(150/144)1/2

and using the method proposed in part (ii).

329

330 A COMPANION TO ANALYSIS

The discussion that follows may be considered as a natural extension of

the ideas above. We work in Rn with the usual Euclidean norm.

Lemma 13.1.2. Consider a function f : Rm ’ Rm such that f (0) = 0.

Suppose there exists a δ > 0 and an · with 1 > · > 0 such that

(f (x) ’ f (y)) ’ (x ’ y) ¤ · (x ’ y)

for all x , y ¤ δ. Then, if y ¤ (1 ’ ·)δ, there exists one and only one

solution of the equation

f (x) = y

with x < δ. Further, if we denote this solution by g(y), we have

g(y) ’ y ¤ ·(1 ’ ·)’1 y .

Since Lemma 13.1.2 is most important to us when · is small there is no

harm in concentrating our attention on this case. Lemma 13.1.2 is then an

instance of the

Slogan: Anything which is close to the identity has an inverse.

(This is a slogan, not a theorem. See, for example, Exercise 13.1.7.) The

inequality

(f (x) ’ f (y)) ’ (x ’ y) ¤ · (x ’ y)

is then seen as a perturbation of the trivial equality

(I(x) ’ I(y)) ’ (x ’ y) = 0 = 0 (x ’ y) .

If we try to solve equation by the method of Exercise 13.1.1, we are

led to the observation that, if such a solution exists, with value u, say, then,

if x is close to u,

u ’ x ≈ f (u) ’ f (x) = y ’ f (x),

and so

u ≈ x + (y ’ f (x)).

This suggests that we should start with an x0 close to the solution of equa-

tion (for example x0 = y) and de¬ne xn inductively by xn = T xn’1

where

T x = x + (y ’ f (x)).

If we add the contraction mapping as a further ingredient, we arrive at the

following proof of Lemma 13.1.2.

331

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

Proof of Lemma 13.1.2. We set

¯

X = B(0, δ) = {x ∈ Rm : x ¤ δ}.

Since X is closed in Rm , we know that (X, ) is complete.

Let y ¤ (1 ’ ·)δ. If we set

T x = x + (y ’ f (x)),

then (since f (0) = 0)

T x ¤ y + x ’ f (x)

¤ y + (f (x) ’ f (0)) ’ (x ’ 0)

¤ y +· x’0

¤ (1 ’ ·)δ + · x < δ,

whenever x ∈ X. Thus T is a well de¬ned function T : X ’ X.

If x, x ∈ X, then

Tx ’ Tx = (x ’ x ) ’ (f (x) ’ f (x )) ¤ · x ’ x ,

by the hypotheses of the lemma. Thus T is a contraction mapping and has

a unique ¬xed point u ∈ X such that

u = T u = u + (y ’ f (u)),

or, equivalently,

f (u) = y

as required. To obtain the ¬nal inequality, we return to the original proof

of the contraction mapping theorem (Theorem 12.1.3). Observe, as we did

there, that, if x ∈ X,

T n x ’ T n’1 x ¤ · T n’1 x ’ T n’2 x ¤ · n’1 T x ’ x

and so

n n

· j’1 T x ’ x ¤ (1 ’ ·)’1 T x ’ x .

T nx ’ x ¤ T j x ’ T j’1 x ¤

j=1 j=1

Since T n x ’ u,

u ’ x ¤ u ’ T nx + T nx ’ x

¤ u ’ T n x + (1 ’ ·)’1 T x ’ x ’ (1 ’ ·)’1 T x ’ x

332 A COMPANION TO ANALYSIS

as n ’ ∞, it follows that

u ’ x ¤ (1 ’ ·)’1 T x ’ x .

If we take x = y, the last inequality takes the form

u ’ y ¤ (1 ’ ·)’1 T y ’ y = (1 ’ ·)’1 y ’ f (y)

= (1 ’ ·)’1 (y ’ 0) ’ (f (y) ’ f (0))

¤ ·(1 ’ ·)’1 y ’ 0 = ·(1 ’ ·)’1 y ,

as required.

Exercise 13.1.3. Let (X, d) be a metric space (not necessarily complete)

and T : X ’ X a mapping such that d(T x, T y) ¤ Kd(x, y) for all x, y ∈ X

and some K < 1. Suppose that T has a ¬xed point w. If x0 ∈ X and we

de¬ne xn inductively by xn+1 = T xn , show that d(x0 , w) ¤ (1’K)’1 d(x0 , x1 ).

Lemma 13.1.2 provides the core of the proof of our next result. Here, and

for the rest of the chapter, we use, in addition to the Euclidean norm on Rm ,

the operator norm on the space of linear maps L(Rm , Rm ).

Lemma 13.1.4. Consider a function f : Rm ’ Rm such that f (0) = 0 and

there exists a δ0 > 0 such that f is di¬erentiable in the open ball B(0, δ0 ). If

Df is continuous at 0 and Df (0) = I (the identity map), then we can ¬nd a

δ1 with δ0 ≥ δ1 > 0 and a ρ > 0 such that, if y ¤ ρ, there exists one and

only one solution of the equation

f (x) = y

with x < δ1 . Further, if we denote this solution by g(y), the function g is

di¬erentiable at 0 with Dg(0) = I.

Proof. Since Df is continuous at 0 and Df = I, we can ¬nd a δ1 > 0 such

that δ0 > δ1 > 0 and

Df (w) ’ I < 1/2

for w ¤ δ1 . Applying the mean value inequality to the function h = f ’ I,

we obtain

(f (x) ’ f (y)) ’ (x ’ y) = h(x) ’ h(y) ¤ x ’ y sup Dh(w)

w¤δ1

= x’y Df (w) ’ I ¤ x ’ y /2

sup

w ¤δ1

333

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

for all x , y ¤ δ1 . Setting ρ = δ1 /2, we see, by Lemma 13.1.2, that, if

y ¤ ρ, there exists one and only one solution of the equation

f (x) = y

with x < δ1 . For the rest of the proof we denote this solution by g(y).

To discuss the behaviour of g near 0, we echo the discussion of the ¬rst

paragraph. Let · > 0 be given. Since Df is continuous at 0 and Df = I, we

can ¬nd a δ(·) > 0 such that δ1 > δ(·) > 0 and

Df (w) ’ I < ·

for w ¤ δ(·). By exactly the same reasoning as in the ¬rst paragraph,

f (x) ’ f (y) ’ (x ’ y) ¤· x’y

for all x , y ¤ δ(·). The last sentence of Lemma 13.1.2 now tells us that,

if y ¤ (1 ’ ·)δ(·), the unique solution of the equation

f (x) = y

with x < δ(·), which we have already agreed at the end of the previous

paragraph to denote by g(y), satis¬es

g(y) ’ y ¤ ·(1 ’ ·)’1 y .

In less roundabout language,

g(y) ’ y ¤ ·(1 ’ ·)’1 y

whenever y ¤ (1 ’ ·)δ(·). Since f (0) = 0, we have g(0) = 0 and

g(y) ’ g(0) = Iy + (y) y

(y) ’ 0 as y ’ 0. Thus g is di¬erentiable at 0 with derivative

with

I.

Exercise 13.1.5. In the second paragraph of the proof just given it says ˜By

exactly the same reasoning as in the ¬rst paragraph™. Fill in the details.

Exercise 13.1.6. The only point of proving Lemma 13.1.2 was to expose the

inner workings of the proof of Lemma 13.1.4. Prove Lemma 13.1.4 directly.

(The proof is essentially the same but combining the proofs of the two lemma

is more economical.)

334 A COMPANION TO ANALYSIS

The next exercise shows that simply knowing that f is di¬erentiable at 0

with derivative I is not enough to give the existence of a local inverse.

Exercise 13.1.7. (i) Let f : R ’ R be given by f (x) = x + x2 sin(1/x)

for x = 0, f (0) = 0. Show, by using the de¬nition of di¬erentiability, or

otherwise, that f is di¬erentiable at 0 with f (0) = 1. By considering the

maxima and minima of f , or otherwise, show that there exist a sequence

of non-zero yn ’ 0 such that the equation f (x) = yn has more than one

solution.

(ii) Let f : R ’ R be given by

f (x) = 1 for x > 1,

for 1/n ≥ x > 1/(n + 1), n a strictly positive integer,

f (x) = 1/n

f (0) = 0

f (x) = ’f (’x) for x < 0.

Show that f is di¬erentiable at 0 with f (0) = 1 but that there exist a sequence

of non-zero yn ’ 0 such that the equation f (x) = yn has no solution.

It might be thought that Lemma 13.1.4 refers to a very special situation,

but we can now apply another

Slogan: Anything which works for the identity will work, in a suitable form,

for invertible elements. Moreover, the general case can be obtained from the

special case of the identity.

Lemma 13.1.8. Consider a function f : Rm ’ Rm such that f (0) = 0 and

there exists a δ0 > 0 such that f is di¬erentiable in the open ball B(0, δ0 ).

If Df is continuous at 0 and Df (0) is invertible, then we can ¬nd a δ1 with

δ0 ≥ δ1 > 0 and a ρ > 0 such that, if y ¤ ρ, there exists one and only one

solution of the equation

f (x) = y

with x < δ1 . Further, if we denote this solution by g(y), the function g is

di¬erentiable at 0 with Dg(0) = Df (0)’1 .

Proof. Set ± = Df (0) and

F(x) = ±’1 f (x).

The proof now runs along totally predictable and easy lines.

335

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

By using standard chain rules (or simple direct calculations), the function

F : Rm ’ Rm satis¬es F(0) = 0 and F is di¬erentiable in the open ball

B(0, δ0 ) with derivative DF given by

DF(x) = ±’1 Df (x).

Thus DF is continuous at 0 and DF(0) = I. It follows, by Lemma 13.1.4,

that there exists a δ1 with δ0 ≥ δ1 > 0 and a ρ > 0 such that if y ¤ δ1 ,

˜

˜

there exists one and only one solution of the equation

˜

F(x) = y

with x < ρ. Further, if we denote this solution by G(˜ ), the function G

˜ y

is di¬erentiable at 0 with DG(0) = I.

Equation can be rewritten as

±’1 f (x) = y.

˜

Taking y = ±˜ , and writing

y

g(t) = G(±’1 t),

we can rewrite the conclusion of the last paragraph as follows. If ± ’1 y ¤ ρ

˜

there exists one and only one solution of the equation

f (x) = y

with x < δ1 . Further, if we denote this solution by g(y), the function g is

di¬erentiable at 0 with Dg(0) = ±’1 = Df (0)’1 .

If we set ρ = ±’1 ’1 ρ, then, whenever y ¤ ρ, we have

˜

±’1 y ¤ ±’1 y ¤ρ

˜

and we have obtained all the conclusions of the lemma.

A simple translation argument transforms Lemma 13.1.8 to an apparently

more general form.

Lemma 13.1.9. Consider a function f : Rm ’ Rm and a w ∈ Rm such that

there exists a δ0 > 0 such that f is di¬erentiable in the open ball B(w, δ0 ). If

Df is continuous at w and Df (w), is invertible then we can ¬nd a δ1 with

δ0 ≥ δ1 > 0 and a ρ > 0 such that, if y ’ f (w) ¤ ρ, there exists one and

only one solution of the equation

f (x) = y

with x ’ w < δ1 . Further, if we denote this solution by g(y), the function

g is di¬erentiable at f (w) with Dg(f (w)) = Df (w)’1 .

336 A COMPANION TO ANALYSIS

Exercise 13.1.10. Prove Lemma 13.1.9 from Lemma 13.1.8.

We have now done most of the hard work of this section but Lemma 13.1.9

can be made much more useful if we strengthen its hypotheses.

Slogan: Analysis is done not at points but on open sets.

Our ¬rst two results are preliminary.

Lemma 13.1.11. (i) We write ι for the identity map ι : Rn ’ Rn . If

± : Rn ’ Rn is linear and ι ’ ± < 1, then ± is invertible.

(ii) Suppose β, γ : Rn ’ Rn are linear and β is invertible. If β ’ γ <

β ’1 ’1 , then γ is invertible.

(iii) Consider a function f : Rm ’ Rm which is di¬erentiable on an open

set U . If Df is continuous and invertible at a point u0 ∈ U , then we can ¬nd

an open set B ⊆ U with u0 ∈ B such that Df is invertible at every point of

B.

Proof. (i) Since we are working with ¬nite dimensional vector spaces, ± is

invertible if and only if it is injective and ± is injective if and only if ker ± =

{0}. If x = 0 then

±x = x ’ (ι ’ ±)x ≥ x ’ (ι ’ ±)x ≥ x ’ ι ’ ± x > 0,

and so ±x = 0. The result follows.

(ii) Observe that

ι ’ β ’1 γ = β ’1 (β ’ γ) ¤ β ’1 (β ’ γ) < 1

and so β ’1 γ is invertible. Write θ = (β ’1 γ)’1 and observe that (θβ ’1 )γ =

θ(β ’1 γ) = ι. Thus, since we are dealing with ¬nite dimensional spaces, γ is

invertible with inverse θβ ’1 .

(iii) By the continuity of Df we can ¬nd an open set B ⊆ U with u0 ∈ B

such that

Df (u0 ) ’ Df (u) < Df (u0 )’1 ’1

.

The stated result follows from part (ii).

(Both Lemma 13.1.11 and its proof may be classi¬ed as ˜quick and dirty™. A

more leisurely approach, which reveals much more of what is actually going

on, is given in Exercise K.281.)

Lemma 13.1.12. Consider a function f : Rm ’ Rm which is di¬erentiable

on an open set U . If Df is continuous and invertible at every point of U ,

then f (U ) is open.

337

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

Proof. Suppose w ∈ U . Since U is open, we can ¬nd a δ0 > 0 such that

the open ball B(w, δ0 ) is a subset of U . By hypothesis, Df is continuous at

w and Df (w) is invertible. Thus, by Lemma 13.1.9, we can ¬nd a δ1 with