510 CHAPTER 24. IMPLICIT FUNCTION THEOREM*

Consider fi and leth(t)≡ fi (x+ t (z−x) ,y) .

Then h(1) = h(0) and so by the mean value theorem, h′ (ti) = 0 for some ti ∈ (0,1) . There-fore, from the chain rule and for this value of ti,

h′ (ti) =n

∑j=1

∂

∂x jfi (x+ ti (z−x) ,y)(z j − x j) = 0. (24.4)

Then denote by xi the vector, x+ ti (z−x) . It follows from 24.4 that

J(x1, · · · ,xn,y

)(z−x) = 0

and so from 24.3 z−x= 0. (The matrix, in the above is invertible since its determinantis nonzero.) Now it will be shown that if η is chosen sufficiently small, then for all y ∈B(y0,η) , there exists a unique x(y) ∈ B(x0,δ ) such that f (x(y) ,y) = 0.

Claim: If η is small enough, then the function, x→ hy (x) ≡ |f (x,y)|2 achieves itsminimum value on B(x0,δ ) at a point of B(x0,δ ) . This is Proposition 24.0.2.

Choose η < η0 and also small enough that the above claim holds and let x(y) denotea point of B(x0,δ ) at which the minimum of hy on B(x0,δ ) is achieved. Since x(y) is aninterior point, you can consider hy (x(y)+ tv) for |t| small and conclude this function of thas a zero derivative at t = 0. Now

hy (x(y)+ tv) =n

∑i=1

f 2i (x(y)+ tv,y)

and so from the chain rule,

ddt

hy (x(y)+ tv) =n

∑i=1

n

∑j=1

2 fi (x(y)+ tv,y)∂ fi (x(y)+ tv,y)

∂x jv j.

Therefore, letting t = 0, it is required that for every v,

n

∑i=1

n

∑j=1

2 fi (x(y) ,y)∂ fi (x(y) ,y)

∂x jv j = 0.

In terms of matrices this reduces to 0 = 2f (x(y) ,y)T D1f (x(y) ,y)v for every vector v.Therefore, 0 = f (x(y) ,y)T D1f (x(y) ,y) . From 24.3, it follows f (x(y) ,y)=0. Mul-tiply by D1f (x(y) ,y)−1 on the right. This proves the existence of the function y → x(y)such that f (x(y) ,y) = 0 for all y ∈ B(y0,η) .

It remains to verify this function is a C1 function. To do this, let y1 and y2 be points ofB(y0,η) . Then as before, consider the ith component of f and consider the same argumentusing the mean value theorem to write

0 = fi (x(y1) ,y1)− fi (x(y2) ,y2)= fi (x(y1) ,y1)− fi (x(y2) ,y1)+ fi (x(y2) ,y1)− fi (x(y2) ,y2)= D1 fi

(xi,y1

)(x(y1)−x(y2))+D2 fi

(x(y2) ,y

i)(y1 −y2) .

(24.5)

where yi is a point on the line segment joining y1 and y2 and xi is a point on the linesegment joining x(y1) and x(y2) . Thus

(x(y1)−x(y2)) =−J(x1, · · · ,xn,y1

)−1M (y1 −y2)