24.1. MORE CONTINUOUS PARTIAL DERIVATIVES 511
where M denotes the matrix having the ith row equal to D2 fi(x(y2) ,y
i)
all entries beingbounded by K. Since the hypothesis of Lemma 24.0.4 hold, it follows that
|x(y1)−x(y2)| ≤ Kn |M (y1 −y2)| ≤ K2nm |y1 −y2| (24.6)
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 implies the inverse function theorem given next.
Theorem 24.0.6 (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. (24.7)
Then there exist open sets W, and V such that
x0 ∈W ⊆U, (24.8)
f : W →V is one to one and onto, (24.9)
f−1 is C1, (24.10)
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. Now let W ≡B(x0,δ )∩f−1 (B(y0,η)) and V ≡ B(y0,η) . This proves the theorem.
24.1 More Continuous Partial DerivativesThe implicit function theorem will now be improved slightly. If f is Ck, it follows thatthe function which is implicitly defined is also Ck, not just C1, meaning all mixed partialderivatives of f up to order k are continuous. Since the inverse function theorem comesas a case of the implicit function theorem, this shows that the inverse function also inheritsthe property of being Ck. First some notation is convenient. Let α = (α1, · · · ,αn) whereeach α i is a nonnegative integer. Then letting |α|= ∑i α i,
Dαf (x)≡ ∂ |α|f
∂ α1∂ α2 · · ·∂ αn(x) , D0f (x)≡ f (x)