104 CHAPTER 4. THE DERIVATIVE

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) ,yi)(y1−y2) (4.17)

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

(x(y1)−x(y2)) =−J(x1, · · · ,xn,y1

)−1M (y1−y2)

where M denotes the matrix having the ith row equal to D2 fi(x(y2) ,yi

)all entries being

bounded by K. It follows that

|x(y1)−x(y2)| ≤ Kn |M (y1−y2)| ≤ K2nm |y1−y2|

Thus y→ x(y) is continuous near y0.Now let y2 = y,y1 = y+hek for small h. Then M described above depends on h and

limh→0

M (h) = D2f(x(y) ,y)

thanks to the continuity of y→ x(y) just shown. Also,

x(y+hek)−x(y)h

=−J(x1 (h) , · · · ,xn (h) ,y+hek

)−1M (h)ek

Passing to a limit and using the formula for the inverse of a matrix in terms of the cofactormatrix, and the continuity of y→ x(y) shown above, this yields

∂x∂yk

=−D1f(x(y) ,y)−1 D2 fi (x(y) ,y)ek

Then continuity of y→ x(y) and the assumed continuity of the partial derivatives of fshows that each partial derivative of y→ x(y) exists and is continuous. ■

This theorem implies the inverse function theorem stated next.

Theorem 4.8.7 (inverse function theorem) Let x0 ∈U, an open set in Rn , and letf : U → Rn. Suppose

f is C1 (U) , and Df(x0)−1 exists. (4.18)

Then there exist open sets W, and V such that x0 ∈W ⊆U, f : W → V is one to one andonto, f−1 is C1.

Proof: Apply the implicit function theorem to the function F(x,y) ≡ f(x)− y wherey0 ≡ f(x0). Thus the function y→ x(y) defined in that theorem is f−1 and there is B(y0,η)where this function is defined. Now let W ≡ f−1 (B(y0,η)) and V ≡ B(y0,η) . ■