210
IX – Multivariate Differential and Integral Calculus
Having done this, (13) shows that
|z n+1 − z n | = |p(z n ) − p(z n−1 )| ≤ q|z n − z n−1 |
with q < 1. Hence lim z n = z ∈ K(r) exists, with obviously ψ(z) = ζ, qed.
Let us now suppose that the assumptions of theorem 4 hold.
Lemma 2. For all q > 0, there exists r > 0 such that, for x, y ∈ A,
|x − y| ≤ r =⇒ ⇒ϕ
(x) − ϕ
(y) ≤ q/ϕ
(x)
−1
.
(10.14)
The map x → ϕ
(x) being continuous on A and ϕ
(x) being invertible for
all x ∈ A, the map x → ϕ
(x)
−1 is also continuous on A : indeed Cramer’s
formulas
41 tell us how to calculate the entries of the matrix of ϕ
(x)
−1 from
those of the matrix of ϕ
(x). The norm ϕ
(x)
−1
is, therefore, also a continuous function on A and, A being compact, it is bounded on A. Setting
sup ϕ
(x)
−1
= 1/M , M ≤ 1/ϕ
(x)
−1
for all x ∈ A and (14) will hold if
|x − y| ≤ r =⇒ ⇒ϕ
(x) − ϕ
(y) ≤ Mq .
(10.15)
But x → ϕ
(x) is uniformly continuous since A is compact. So for all q > 0,
there exists r satisfying (15), qed.
In the following statement, m(X) denotes the Lebesgue measure of a
measurable set X ⊂ R
n , as it happens a compact set.
Lemma 3. For all q such that 0 < q < 1, there exists r > 0 satisfying the
following property: for all a ∈ U such that K(a, r) = K ⊂ U ,
|m [ϕ(K)] −| J ϕ (a) |.m(K)| ≤ q.m(K) .
(10.16)
Take a point a ∈ U and replace ϕ by
ϕ a : x −→ ϕ
(a)
−1 [ϕ(a + x) − ϕ(a)] ,
41 There is an easier argument in the case of an arbitrary Banach space E; it is based
on the fact that, for any linear operator T with norm < 1, the operator 1 − T has
an inverse, namely
T
n (the series converges absolutely since T
n ≤ q
n , where
q = T ). Let A and X be two continuous linear operators on E; set X = A − Y
and suppose that A is invertible. Then, X = A(1−A
−1 Y ), so that X is invertible
if A
−1 Y = q < 1 ; as A
−1 Y < A
−1 .Y , this is the case if Y < 1/A
−1 ,
i.e. if X is sufficiently near A. Then, X
−1 = (1−A
−1 Y )
−1 A
−1 =
(A
−1 Y )
n A
−1 ,
and so
X
−1 < A
−1 /(1 − q) < 2A
−1
if Y ≤ 1/2A
−1 . As X
−1 − A
−1 = A
−1 (A − X)X
−1 = A
−1 Y X
−1 ,
X
−1 − A
−1 < A
−1 .X − A.X
−1 < 2A
−1
2 .X − A
follows. Hence X → X
−1 is continuous at A.
IX – Multivariate Differential and Integral Calculus
Having done this, (13) shows that
|z n+1 − z n | = |p(z n ) − p(z n−1 )| ≤ q|z n − z n−1 |
with q < 1. Hence lim z n = z ∈ K(r) exists, with obviously ψ(z) = ζ, qed.
Let us now suppose that the assumptions of theorem 4 hold.
Lemma 2. For all q > 0, there exists r > 0 such that, for x, y ∈ A,
|x − y| ≤ r =⇒ ⇒ϕ
(x) − ϕ
(y) ≤ q/ϕ
(x)
−1
.
(10.14)
The map x → ϕ
(x) being continuous on A and ϕ
(x) being invertible for
all x ∈ A, the map x → ϕ
(x)
−1 is also continuous on A : indeed Cramer’s
formulas
41 tell us how to calculate the entries of the matrix of ϕ
(x)
−1 from
those of the matrix of ϕ
(x). The norm ϕ
(x)
−1
is, therefore, also a continuous function on A and, A being compact, it is bounded on A. Setting
sup ϕ
(x)
−1
= 1/M , M ≤ 1/ϕ
(x)
−1
for all x ∈ A and (14) will hold if
|x − y| ≤ r =⇒ ⇒ϕ
(x) − ϕ
(y) ≤ Mq .
(10.15)
But x → ϕ
(x) is uniformly continuous since A is compact. So for all q > 0,
there exists r satisfying (15), qed.
In the following statement, m(X) denotes the Lebesgue measure of a
measurable set X ⊂ R
n , as it happens a compact set.
Lemma 3. For all q such that 0 < q < 1, there exists r > 0 satisfying the
following property: for all a ∈ U such that K(a, r) = K ⊂ U ,
|m [ϕ(K)] −| J ϕ (a) |.m(K)| ≤ q.m(K) .
(10.16)
Take a point a ∈ U and replace ϕ by
ϕ a : x −→ ϕ
(a)
−1 [ϕ(a + x) − ϕ(a)] ,
41 There is an easier argument in the case of an arbitrary Banach space E; it is based
on the fact that, for any linear operator T with norm < 1, the operator 1 − T has
an inverse, namely
T
n (the series converges absolutely since T
n ≤ q
n , where
q = T ). Let A and X be two continuous linear operators on E; set X = A − Y
and suppose that A is invertible. Then, X = A(1−A
−1 Y ), so that X is invertible
if A
−1 Y = q < 1 ; as A
−1 Y < A
−1 .Y , this is the case if Y < 1/A
−1 ,
i.e. if X is sufficiently near A. Then, X
−1 = (1−A
−1 Y )
−1 A
−1 =
(A
−1 Y )
n A
−1 ,
and so
X
−1 < A
−1 /(1 − q) < 2A
−1
if Y ≤ 1/2A
−1 . As X
−1 − A
−1 = A
−1 (A − X)X
−1 = A
−1 Y X
−1 ,
X
−1 − A
−1 < A
−1 .X − A.X
−1 < 2A
−1
2 .X − A
follows. Hence X → X
−1 is continuous at A.