4.8. IMPLICIT FUNCTION THEOREM 101

Definition 4.8.4 Let δ ,η > 0 satisfy: B(x0,δ )×B(y0,η)⊆U where f : U ⊆ Rn×Rm→ Rp is given as

f(x,y) =

f1 (x,y)f2 (x,y)

...fp (x,y)

and for

(x1 · · · xn

)∈ B(x0,δ )

pand y ∈ B(y0, η̂) define

J(x1, · · · ,xp,y

)≡

 f1,x1

(x1,y

)· · · f1,xn

(x1,y

)...

...fp,x1 (x

p,y) · · · fp,xn (xp,y)

 . (*)

Thus, its ith row is D1 fi(xi,y

). Let K,r be constants.

By Lemma 4.8.1, and (x,y) ∈ B(x0,δ )× B(y0,η) ⊆ U, and h,k sufficiently small,there are xi on the line segment between x and x+h such that

f(x+h,y+k)− f(x,y) = f(x+h,y+k)− f(x,y+k)+ f(x,y +k)− f(x,y)

= J(x1, · · · ,xp,y+k

)h+D2f(x,y)k+o(k) (4.11)

= D1f(x,y)h+D2f(x,y)k+o(k)+(J(x1, · · · ,xp,y+k

)−D1f(x,y)

)h (4.12)

Now by continuity of the partial derivatives, if√|h|2 + |k|2 is sufficiently small,∥∥J

(x1, · · · ,xp,y+k

)−D1f(x,y)

∥∥< ε

and so ∣∣(J (x1, · · · ,xp,y+k)−D1f(x,y)

)h∣∣√

|h|2 + |k|2≤ ε |h|√

|h|2 + |k|2≤ ε

and so the last term in 4.12 is o((h,k)) . Thus f(x+h,y+k)− f(x,y) is of the form

D1f(x,y)h+D2f(x,y)k+o(h,k)

which shows that f is differentiable and its derivative is the p× (n+m) matrix,(D1f(x,y) D2f(x,y)

).

Proposition 4.8.5 Suppose g : B(x0,δ )×B(y0,η0)→ [0,∞) is continuous and

g(x0,y0) = 0

and if x ̸= x0,g(x,y0) > 0. Then there exists η < η0 such that if y ∈ B(y0,η) , then thefunction x→ g(x,y) achieves its minimum on the open set B(x0,δ ).