§5. Differentiable functions of several variables
297
The interest of this result is not only that it shows the existence of a
solution Y of F(x, y) = z for z = c, which was the original problem; but also
it shows that this solution is a CP function of x and of the right hand side z
of the equation to be solved, granted of course that z is sufficiently close to
c = F(a, b) and that one restricts oneself to looking for solutions for which
(x, y) is sufficiently close to (a, b).
Since Theorem 25 is a direct consequence of the local inversion theorem
it must be possible to adapt its proof to obtain a direct proof of Theorem 25.
Though we will not obtain all the conclusions of Theorem 25, let us show
how one can solve F(x, y) = c.
To simplify the notation, we first reduce to the case where a = b = c = 0
and where D2F(0, 0) = 1, by replacing F by [F(a + x, b + y) - e] / D2F(a, b),
a function which we again denote by F, which vanishes for x = y = 0 and
satisfies D2 F(0,0) = 1. If we put e = D 1 F(0, 0) then
F(x, y) = ex + y + o(lxl + Iyl)
, since D2 F(0, 0) = 1. Now consider the function
G(x, y) = y - F(x, y - ex),
defined and of class C1 on a neighbourhood of (0,0); we have
G(x,y) = y - (ex + y - ex) + o(lxl + Iyl) = o(lxl + Iyl),
on a neighbourhood of the origin, so
(24.13)
D1 G(O, 0) = D2G(0, 0) = o.
This done, suppose we have found a function g(x) of class C1 on a neighbourhood of 0 such that
(24.14)
G[x,g(x)] = g(x);
we then have g(x) - F[x,g(x) - ex] = g(x), so that f(x) = g(x) - ex will be
a Cl solution of F[x, f(x)] = o.
The construction of 9 is then, for x given, to define a sequence of points
tin = Yn(x) by
(24.15)
Yo = 0, Yl = G(x, Yo) = G(x, 0), Y2 = G(x, Y1), ...
and to verify that the Yn converge to the desired solution g(x). This proof
too can be divided into several stages.
(a) Since DG(O,O) = 0, there exists a compact interval I = [-r, r] such
that G is defined on K = I x I and satisfies IIDG(z)11 :-:::; ! on K, i.e.
IIDGIIK :-:::; !. Then