210 CHAPTER 8. IMPLICIT FUNCTION THEOREM

Finally consider the claim that this implicitly defined function is C1.

0 = f (x(y+u) ,y+u)−f (x(y) ,y)= D1f (x(y) ,y)(x(y+u)−x(y))+D2f (x(y) ,y)u

+o(x(y+u)−x(y) ,u) (8.9)

Consider the last term. o(x(y+u)−x(y) ,u)/∥u∥ equals{o(x(y+u)−x(y),u)

∥(x(y+u)−x(y),u)∥X×Y

max(∥x(y+u)−x(y)∥,∥u∥)∥u∥ if ∥(x(y+u)−x(y) ,u)∥X×Y ̸= 0

0 if ∥(x(y+u)−x(y) ,u)∥X×Y = 0

Now the Lipschitz condition just established shows that

max(∥x(y+u)−x(y)∥ ,∥u∥)∥u∥

is bounded for nonzero u sufficiently small that y,y+u ∈ B(y0,η). Therefore,

limu→0

o(x(y+u)−x(y) ,u)∥u∥

= 0

Then 8.9 shows that

0= D1f (x(y) ,y)(x(y+u)−x(y))+D2f (x(y) ,y)u+o(u)

Therefore, solving for x(y+u)−x(y) , it follows that

x(y+u)−x(y) = −D1f (x(y) ,y)−1 D2f (x(y) ,y)u+D1f (x(y) ,y)

−1o(u)

= −D1f (x(y) ,y)−1 D2f (x(y) ,y)u+o(u)

and now, the continuity of the partial derivatives D1f,D2f, continuity of the map A→ A−1,along with the continuity of y→ x(y) shows that y→ x(y) is C1 with derivative equal to−D1f (x(y) ,y)

−1 D2f (x(y) ,y). ■The following is a nice result on functional dependence which is an application of the

implicit function theorem. See Widder [59].

Example 8.1.9 Suppose f ,g are C1 near (x0,y0) ∈ R2 andSuppose f ,g are C1 and

1. det(

fx (x,y) fy (x,y)gx (x,y) gy (x,y)

)= 0 near (x0,y0)

2. fx (x0,y0) ̸= 0.

Then there is a C1 function F such that g(x,y) = F ( f (x,y)) for (x,y) near (x0,y0).Consider f (x,y)− z = 0 where z0 ≡ f (x0,y0). By assumption and implicit func-

tion theorem, there is a C1 function (y,z) → φ (y,z) so that for (y,z) near (y0,z0) thex which solves f (x,y)− z = 0 is x = φ (y,z). In particular, for (x,y) close to (x0,y0) ,f (φ (y, f (x,y)) ,y)− f (x,y) = 0 and so

φ (y, f (x,y)) = x. (∗)