Okay, let’s dive right in with a first step towards proving the inverse function theorem we talked about at the end of yesterday’s post. This is going to get messy.
We start with a function and first ask that it be continuous and injective on the closed ball of radius around the point . Then we ask that all the partial derivatives of exist within the open interior — note that this is weaker than our existence condition for the differential of — and that the Jacobian determinant on . Then I say that the image actually contains a neighborhood of . That is, the image doesn’t “flatten out” near .
The boundary of the ball is the sphere of radius :
Now the Heine-Borel theorem says that this sphere, being both closed and bounded, is a compact subset of . We’ll define a function on this sphere by
which must be continuous and strictly positive, since if then , but we assumed that is injective on . But we also know that the image of a continuous real-valued function on a compact, connected space must be a closed interval. That is, , and there exists some point on the sphere where this minimum is actually attained: .
Now we’re going to let be the ball of radius centered at . We will show that , and is thus a neighborhood of contained within . To this end, we’ll pick and show that .
So, given such a point , we define a new function on the closed ball by
This function is continuous on the compact ball , so it again has an absolute minimum. I say that it happens somewhere in the interior .
At the center of the ball, we have (since ), so the minimum must be even less. But on the boundary , we find
so the minimum can’t happen on the boundary. So this minimum of happens at some point in the open ball , and so does the minimum of the square of :
Now we can vary each component of separately, and use Fermat’s theorem to tell us that the derivative in terms of must be zero at the minimum value . That is, each of the partial derivatives of must be zero (we’ll come back to this more generally later):
This is the product of the vector by the matrix . And the determinant of this matrix is : the Jacobian determinant at , which we assumed to be nonzero way back at the beginning! Thus the matrix must be invertible, and the only possible solution to this system of equations is for , and so .