Lecture 4: The Inverse and the Implicit Function Theorems

Lemma 1: Let be defined and differentiable on a convex open set A. Then for all , one has

where is chosen to be an upper bound on all the for all , , and all .

Proof: Let . For each , the Mean Value Theorem says that there is a between 0 and 1 such that

By the chain rule,

So,

Theorem 1: (Inverse Function Theorem) Let be continuously differentiable on an open subset containing . If , then there are open subsets containing and containing such that and there is an inverse which is differentiable with derivative satisfying:

Exercise 1: Show that Theorem 1 is true for linear functions.

Proof: (of Theorem 1) By replacing with , we can assume that is the identity map (Why?).

Apply Lemma 1 to : for and in some open rectangle containing . But then,

Rearranging gives

Since , and so by choosing the rectangle small enough, we can assume that and so

In particular, it follows that f is one-to-one when restricted to this rectangle, and the inverse will be continuous if it exists. Replacing the rectangle with a smaller one, we can assume the same is true when f is restricted to the closure of the rectangle. Now the boundary of the rectangle is compact and so is also compact and does not contain . Let be the minimum of for . Let be the set .

Now, for every , there is at least one in the rectangle with . In fact, consider the function . The image of under the closure of the rectangle. The minimum of this function does not occur on because . So it must occur where the derivative is zero, i.e. one has:

for . But by taking the rectangle sufficiently small, we can assume that for all in the rectangle. But then the only solution of this system of linear equations is .

If , then maps the open set one-to-one and onto the open set . It remains to check differentiability of the inverse. Since is differentiable, one has for ,

where . Letting and , we get after substitution:

Rearranging gives:

It remains to show that

Since the derivative is just a linear function, it is enough to show that . As , we have because is continuous. So, . But we know that So, the limit of the product is zero, as desired.

Note: It follows from the formula for the derivative of the inverse that the inverse is also continuously differentiable.

Corollary 1: (The Implicit Function Theorem) Let be continuously differentiable in an open set containing the point which satisfies . Suppose that . Then there is a continuously differentiable function mapping an open set containing to an open set containing such that for all .

Proof: We can extend to a function by and apply the Inverse Function Theorem to get an inverse defined in an open subset containing and mapping onto an open subset containing . Let where is the projection onto the last coordinates of . Then one has .

Exercise 2: Show that the Inverse Function Theorem is a Corollary of the Implicit Function Theorem.