The Inverse Function Theorem (Proof)
Point of Post: This is a continuation of this post.
Inverse Function Theorem
Let , where is open, we say that is a regular point of if is totally differentiable at and . If is differentiable at and we say that is a critical point. By a basic theorem of linear algebra we know that if we have that a point , for which is differentiable, is regular if is invertible (i.e. ) and critical otherwise (i.e. if ). Thus, with this terminology in mind we state the inverse function theorem as follows:
Theorem (Inverse Function Theorem): Let , where is open, be . Let then be a regular point for , then there exists a neighborhood of such that is open, is a bijection, and the inverse is with .
Proof: Since is a regular point we have that is invertible, call it’s inverse . With a little fiddling one can see that we can choose and such that and
both hold. Now, for each we define . Note then that for each one has that could be equally well described as (where the constant in and is the identity function) and so, using the chain rule,
where we used the fact that and are linear and the total derivative of a linear map is itself. Thus, we see that
We can easily deduce then by the multivariable mean value theorem that
Next note that for each one has that
Thus, is a contraction mapping and so by the Banach fixed point theorem there is a unique solution to in . But, note that writing it out one finds that if and only if . Thus, there is a unique solution to for . So, define the map by sending to the unique such that . It is clear that is an embedding (i.e. that it is a homemorphism onto its image). What we claim though is that is open in . To see this let be arbitrary. Since is continuous we know there exists a neighborhood such that but it’s clear to see that this implies . Thus, since evidently we have that if and then are both open and is a bijection with inverse equal to .
What we lastly claim is that is differentiable and satisfies . So, let be any point of . Let then be some vector in with small enough magnitude such that . We note then that
from this it easily follows that
That said, since is differentiable we know that where . Thus, putting it all together we see that
But, this left side precisely says that exists and .
Corollary: Let be as in the statement of the inverse function theorem, but that for some then the guaranteed is also .
Proof: This follows immediately from the fact that . Indeed, since it’s easy to see that the entries of for any are , but then the entries of are (as can be checked since they are quotients of polynomials of functions and so we can go back quickly to conclude that is .
1. Spivak, Michael. Calculus on Manifolds; a Modern Approach to Classical Theorems of Advanced Calculus. New York: W.A. Benjamin, 1965. Print.
2. Apostol, Tom M. Mathematical Analysis. Reading, MA: Addison-Wesley Pub., 1974. Print.