§5. Differentiable functions of several variables
293
One can clearly assume that the point c E C considered is the origin (0,0)
and that I maps it to (0,0). Since the linear map D 1(0) = A is by hypothesis
. invertible, one can replace I by A -10 I, the composition of I and of the linear
map A -1. The linear tangent map to the new function at a point z EGis
clearly A- 1 DI(z), the product or composition of two linear maps, so is the
identity map when z = o. If one can invert the function z 1-+ A -1 I (z) locally
there will exist a function 9 such that A- 1 /[g(z)] = z on a neighbourhood
of 0, whence I[g(z)] = Az. Since the invertible linear map A transforms every
!oeighbourhood of 0 into a neighbourhood of 0, one has I [g (A-1z)] = z on
a neighbourhood of 0, so that the required local inverse of I is the function
"%....-. 9 (A-1 z ).
We may thus reduce to inverting I under the hypothesis that 1(0) = 0
and D 1(0) = 1. The proof divides into several parts, the essential argument
being that which allowed us to solve an equation of the form I(x) = x in R.
(Chap. II, nO 16).
, (a) Putting I(z) = z + p(z), one has p(O) = 0, Dp(O) = 0; since p is C 1
: there is thus an r > 0 such that
(24.10)
1
Izl :::; r = } II Dp(z) II :::; 2'
80 that (inequality of the mean)
;
(24.11) {lz'l:::; r & Iz"l:::; r} = }
1
Ip(z') - p(z") I :::; 21z' - z"l
~:
I/(z') - l(z")1 ~ 41z' - z"l·
We conclude that the map I is injective on the closed ball B(r) : Izl :::; r, and
~at the inverse map
",.
9 : I(B(r» ~ B(r)
Is continuous: it satisfies Ig«(') - g«(")1 :::; 21(' - ("1.
(b) Let us show that I(B(r» ::) B(r/2), Le. that the equation ( = I(z) =
z + p(z) has a, necessarily unique, solution z E B(r), for all ( such that
1(1:::; r/2. To do this we write (-p(z) = z and apply the method 01 successive
approximations:
Zo = 0, Z1 = ( - p(zo) = (, Z2 = (- P(Z1) = ( - p«(), ....
First we have to verify that the construction can be continued indefinitely
without leaving the ball B(r). But, by (10) or (11), it is clear that
{lei :::; r/2 & Izl:::; r} = } I( - p(z)1 :::; 1(1 + Ip(z)1 :::; r/2 + r/2 = r.
$0 once we have zo, z!, ... ,Zn E B(r) we find Zn+1 = ( - p(zn) E B(r): there
are no obstructions.
This established, the inequalities
293
One can clearly assume that the point c E C considered is the origin (0,0)
and that I maps it to (0,0). Since the linear map D 1(0) = A is by hypothesis
. invertible, one can replace I by A -10 I, the composition of I and of the linear
map A -1. The linear tangent map to the new function at a point z EGis
clearly A- 1 DI(z), the product or composition of two linear maps, so is the
identity map when z = o. If one can invert the function z 1-+ A -1 I (z) locally
there will exist a function 9 such that A- 1 /[g(z)] = z on a neighbourhood
of 0, whence I[g(z)] = Az. Since the invertible linear map A transforms every
!oeighbourhood of 0 into a neighbourhood of 0, one has I [g (A-1z)] = z on
a neighbourhood of 0, so that the required local inverse of I is the function
"%....-. 9 (A-1 z ).
We may thus reduce to inverting I under the hypothesis that 1(0) = 0
and D 1(0) = 1. The proof divides into several parts, the essential argument
being that which allowed us to solve an equation of the form I(x) = x in R.
(Chap. II, nO 16).
, (a) Putting I(z) = z + p(z), one has p(O) = 0, Dp(O) = 0; since p is C 1
: there is thus an r > 0 such that
(24.10)
1
Izl :::; r = } II Dp(z) II :::; 2'
80 that (inequality of the mean)
;
(24.11) {lz'l:::; r & Iz"l:::; r} = }
1
Ip(z') - p(z") I :::; 21z' - z"l
~:
I/(z') - l(z")1 ~ 41z' - z"l·
We conclude that the map I is injective on the closed ball B(r) : Izl :::; r, and
~at the inverse map
",.
9 : I(B(r» ~ B(r)
Is continuous: it satisfies Ig«(') - g«(")1 :::; 21(' - ("1.
(b) Let us show that I(B(r» ::) B(r/2), Le. that the equation ( = I(z) =
z + p(z) has a, necessarily unique, solution z E B(r), for all ( such that
1(1:::; r/2. To do this we write (-p(z) = z and apply the method 01 successive
approximations:
Zo = 0, Z1 = ( - p(zo) = (, Z2 = (- P(Z1) = ( - p«(), ....
First we have to verify that the construction can be continued indefinitely
without leaving the ball B(r). But, by (10) or (11), it is clear that
{lei :::; r/2 & Izl:::; r} = } I( - p(z)1 :::; 1(1 + Ip(z)1 :::; r/2 + r/2 = r.
$0 once we have zo, z!, ... ,Zn E B(r) we find Zn+1 = ( - p(zn) E B(r): there
are no obstructions.
This established, the inequalities