102 CHAPTER 4. THE DERIVATIVE

Proof: If not, then there is a sequence yk → y0 but the minimum of x→ g(x,yk) forx ∈ B(x0,δ ) happens on ∂B(x0,δ )≡ ∂B≡ {x : |x−x0|= δ} at xk. Now ∂B is closed andbounded and so compact. Hence there is a subsequence, still denoted with subscript k suchthat xk→ x ∈ ∂B and yk→ y0. Let

0 < 2ε < min{g(x̂,y0) : x̂ ∈ ∂B}

Then for k large,

|g(xk,yk)−g(x,y0)|< ε, |g(xk,yk)−g(xk,y0)|< ε

the second inequality from uniform continuity. Then from these inequalities, for k large,

g(x0,yk) ≥ g(xk,yk)> g(xk,y0)− ε

> min{g(x̂,y0) : x̂ ∈ ∂B}− ε > 2ε− ε = ε

Now let k→ ∞ to conclude that g(x0,y0)≥ ε , a contradiction. ■Here is the implicit function theorem. It is based on the mean value theorem from one

variable calculus, the extreme value theorem from calculus, and the formula for the inverseof a matrix in terms of the transpose of the cofactor matrix divided by the determinant.

Theorem 4.8.6 (implicit function theorem) Suppose U is an open set in Rn×Rm.Let f : U → Rn be in C1 (U) and suppose

f(x0,y0) = 0, D1f(x0,y0)−1 exists. (4.13)

Then there exist positive constants δ ,η , such that for every y ∈ B(y0,η) there exists aunique x(y) ∈ B(x0,δ ) such that

f(x(y) ,y) = 0. (4.14)

Furthermore, the mapping, y→ x(y) is in C1 (B(y0,η)).

Proof: Letf(x,y) =

(f1 (x,y) f2 (x,y) · · · fn (x,y)

)T.

Define for(x1, · · · ,xn

)∈ B(x0,δ )

nand y ∈ B(y0,η) the following matrix.

J(x1, · · · ,xn,y

)≡

 f1,x1

(x1,y

)· · · f1,xn

(x1,y

)...

...fn,x1 (x

n,y) · · · fn,xn (xn,y)

 . (*)

Then by the assumption of continuity of all the partial derivatives, there exists r > 0 andδ 0,η0 > 0 such that if δ ≤ δ 0 and η ≤ η0, it follows that for all

(x1, · · · ,xn

)∈ B(x0,δ )

n ≡B(x0,δ )×B(x0,δ )×·· ·×B(x0,δ ), and y ∈ B(y0,η),

detJ(x1, · · · ,xn,y

)/∈ (−r,r). (4.15)

and B(x0,δ 0)× B(y0,η0) ⊆U . Therefore, from the formula for the inverse of a matrixand continuity of all entries of the various matrices, there exists a constant K such that all