§5. Differentiable functions of several variables
299
and since 9 is continuous by the point (b) of the proof, k tends to 0 with h;
the functions DIG and D 2G being continuous on K and D 2G[x,g(x)] being
#- 1, we can pass to the limit in the preceding relation, whence the existence
and the value of
(24.18)
lim~ = g'(x) = DlG[x,g(x)] .
h
1 - D2G[x,g(x)]
Second method. Since we already know that 9 is continuous, k tends to 0
with h. Since G is C1, (21.12) or (12') shows that
k = DlG(x,y)h + D 2G(x,y)k + o(lhl + Ikl).
Since 1- D2G(x, y) is invertible 47 and independent of hand k, it follows that
(24.19) k =,h + o(lhl + Ikl) where, = [1 - D 2G(x, y)r l DlG(x, y).
For Ihl and so Ikl sufficiently small, we then have
Ik -,hi ~ e(lhl + Ikl),
whence Ikl ~ (ITI + e)lhl + elkl and consequently (1 - e)lkl ~ (ITI + e)lhl·
Choosing e = ~, for example, we deduce that k = O(h). The relation (19)
then shows that k =,h + o(h), whence the existence of limkjh =, where,
very luckily, , has the value already found by the first method ...
(18) shows that 9 is C l , so the solution of F[x,f(x)] = c is so too. That
f is of class CP if F is, follows from the formula
(24.20)
f'{x) = -DlF[x, f{x)JI D2F[x, f(x)]
and from the fact that D2F does not vanish on the square I x I considered
at the beginning of the proof: if F is CP and if one has already shown that f
is C k with k < p, then (20) shows that l' is similarly Ck and so f is C k +!, qed.
Theorem 25 allows one to show that, under certain conditions, the set
C C G of solutions of F{x, y) = c is an excellent "curve" possessing, at each
of its points, a tangent varying in a continuous way as a function of the point
considered. The hypothesis to make is that the derivatives DIF and D2F
are never simultaneously zero at the points of C; example: F(x, y) = x 2 + y2,
with c > 0 arbitrary; examples to the contrary: xy = 0, x 2 + y3 + y2 = 0,
etc., cases where the derivatives are zero at (0,0).
47 Abstract proof: if, on a Banach space, we have a linear map A such that IIA\\ < 1,
then 1 - A is invertible and we even have
(1 - A)-l = 1 + A + A2 + ...
Banal proof in the case in question: 1 - D2G(X, y) # O.
299
and since 9 is continuous by the point (b) of the proof, k tends to 0 with h;
the functions DIG and D 2G being continuous on K and D 2G[x,g(x)] being
#- 1, we can pass to the limit in the preceding relation, whence the existence
and the value of
(24.18)
lim~ = g'(x) = DlG[x,g(x)] .
h
1 - D2G[x,g(x)]
Second method. Since we already know that 9 is continuous, k tends to 0
with h. Since G is C1, (21.12) or (12') shows that
k = DlG(x,y)h + D 2G(x,y)k + o(lhl + Ikl).
Since 1- D2G(x, y) is invertible 47 and independent of hand k, it follows that
(24.19) k =,h + o(lhl + Ikl) where, = [1 - D 2G(x, y)r l DlG(x, y).
For Ihl and so Ikl sufficiently small, we then have
Ik -,hi ~ e(lhl + Ikl),
whence Ikl ~ (ITI + e)lhl + elkl and consequently (1 - e)lkl ~ (ITI + e)lhl·
Choosing e = ~, for example, we deduce that k = O(h). The relation (19)
then shows that k =,h + o(h), whence the existence of limkjh =, where,
very luckily, , has the value already found by the first method ...
(18) shows that 9 is C l , so the solution of F[x,f(x)] = c is so too. That
f is of class CP if F is, follows from the formula
(24.20)
f'{x) = -DlF[x, f{x)JI D2F[x, f(x)]
and from the fact that D2F does not vanish on the square I x I considered
at the beginning of the proof: if F is CP and if one has already shown that f
is C k with k < p, then (20) shows that l' is similarly Ck and so f is C k +!, qed.
Theorem 25 allows one to show that, under certain conditions, the set
C C G of solutions of F{x, y) = c is an excellent "curve" possessing, at each
of its points, a tangent varying in a continuous way as a function of the point
considered. The hypothesis to make is that the derivatives DIF and D2F
are never simultaneously zero at the points of C; example: F(x, y) = x 2 + y2,
with c > 0 arbitrary; examples to the contrary: xy = 0, x 2 + y3 + y2 = 0,
etc., cases where the derivatives are zero at (0,0).
47 Abstract proof: if, on a Banach space, we have a linear map A such that IIA\\ < 1,
then 1 - A is invertible and we even have
(1 - A)-l = 1 + A + A2 + ...
Banal proof in the case in question: 1 - D2G(X, y) # O.