290
III - Convergence: Continuous variables
24 - Implicit functions
The last classical results, and palpably more difficult to establish than the
preceding ones, are the "local inversion" or "implicit function" theorems. In
the second case, one aims to show that, given a real function F of class C1
on an open set U C JR.2, and a point (a, b) E U where F(a, b) = 0, then _
subject to an apparently inoffensive condition - one can find a real function
y = J(x) of class C 1 on a neighbourhood of a, with J(a) = b and such that
(24.1)
F[x,J(x)] = 0
on a neighbourhood of a. In the first case, one is given a function J : G ---+ JR.2
of class e 1 on an open set G C JR.2 and tries to provide an "inverse" map, at
least on a neighbourhood of a given point (a, b).
It is generally a good idea to think before attempting a proof. In the
simplest case, F(x, y) = x-g(y) where 9 is real and of class e 1 , with a = g(b)j
the solution J of (1) must then satisfy x - g[J(x)] = 0, so that the problem
consists of constructing an inverse map of g, at least locally, the one -variable
version of the local inversion problem. Clearly there cannot be a solution
unless 9 is injective on a neighbourhood of b. By nO 4, Theorem 7 bis, this
forces 9 to be strictly monotone on a neighbourhood of b, and so its derivative
g'(y) must have constant sign. If J is differentiable then g'[J(x)].f'(x) = 1,
which again forbids g' from vanishing on a neighbourhood of b, so forces it to
be either everywhere> 0, or everywhere < 0 on a neighbourhood of bj since g'
is continuous, it is enough that g'(b) =I- o. In this case, 9 is strictly monotone
on an open interval I with centre b, so admits an inverse J on J = J(I), an
open interval containing a = g(b), and rule (D 5) of nO 14 shows that it is
of class e 1 , since f'(x) = 1/g'[J(x)], the reciprocal of a continuous function
everywhere =I- o. We emphasis the fact that, if one knows only that g'(b) =I- 0,
these arguments are valid only on a neighbourhood oj b - to be precise, on the
largest open interval containing b where g' does not vanish. If for example
one considers the function g(x) = x 3 on JR., which maps JR. bijectively onto
JR., it admits a e 1 inverse on the interval x > 0 (namely x 1 / 3 ) or on x < 0
(namely -lxI 1 / 3 ), but not on any neighbourhood of 0 where the graph of the
inverse map has a vertical tangent. One can clearly not hope for more when
one progresses to functions of several variables.
In the general case of the equation (1), the existence of a continuous
derivative for the desired solution J would entail
(24.2)
D1F[x, J(x)] + D2F[x, J(x)Jf'(x) = 0
by (21.1) applied to F[x,J(x)]. If D2F vanishes at (a, b) but DIF does not,
there is again no hope of obtaining a derivative f' (a) even under the hypothesis that J exists. This is what happens in the simplest cases, for example
that of the function F(x, y) = x 2 + y2 -1 at the point (a, b) = (1, O)j the two
III - Convergence: Continuous variables
24 - Implicit functions
The last classical results, and palpably more difficult to establish than the
preceding ones, are the "local inversion" or "implicit function" theorems. In
the second case, one aims to show that, given a real function F of class C1
on an open set U C JR.2, and a point (a, b) E U where F(a, b) = 0, then _
subject to an apparently inoffensive condition - one can find a real function
y = J(x) of class C 1 on a neighbourhood of a, with J(a) = b and such that
(24.1)
F[x,J(x)] = 0
on a neighbourhood of a. In the first case, one is given a function J : G ---+ JR.2
of class e 1 on an open set G C JR.2 and tries to provide an "inverse" map, at
least on a neighbourhood of a given point (a, b).
It is generally a good idea to think before attempting a proof. In the
simplest case, F(x, y) = x-g(y) where 9 is real and of class e 1 , with a = g(b)j
the solution J of (1) must then satisfy x - g[J(x)] = 0, so that the problem
consists of constructing an inverse map of g, at least locally, the one -variable
version of the local inversion problem. Clearly there cannot be a solution
unless 9 is injective on a neighbourhood of b. By nO 4, Theorem 7 bis, this
forces 9 to be strictly monotone on a neighbourhood of b, and so its derivative
g'(y) must have constant sign. If J is differentiable then g'[J(x)].f'(x) = 1,
which again forbids g' from vanishing on a neighbourhood of b, so forces it to
be either everywhere> 0, or everywhere < 0 on a neighbourhood of bj since g'
is continuous, it is enough that g'(b) =I- o. In this case, 9 is strictly monotone
on an open interval I with centre b, so admits an inverse J on J = J(I), an
open interval containing a = g(b), and rule (D 5) of nO 14 shows that it is
of class e 1 , since f'(x) = 1/g'[J(x)], the reciprocal of a continuous function
everywhere =I- o. We emphasis the fact that, if one knows only that g'(b) =I- 0,
these arguments are valid only on a neighbourhood oj b - to be precise, on the
largest open interval containing b where g' does not vanish. If for example
one considers the function g(x) = x 3 on JR., which maps JR. bijectively onto
JR., it admits a e 1 inverse on the interval x > 0 (namely x 1 / 3 ) or on x < 0
(namely -lxI 1 / 3 ), but not on any neighbourhood of 0 where the graph of the
inverse map has a vertical tangent. One can clearly not hope for more when
one progresses to functions of several variables.
In the general case of the equation (1), the existence of a continuous
derivative for the desired solution J would entail
(24.2)
D1F[x, J(x)] + D2F[x, J(x)Jf'(x) = 0
by (21.1) applied to F[x,J(x)]. If D2F vanishes at (a, b) but DIF does not,
there is again no hope of obtaining a derivative f' (a) even under the hypothesis that J exists. This is what happens in the simplest cases, for example
that of the function F(x, y) = x 2 + y2 -1 at the point (a, b) = (1, O)j the two