The Unapologetic Mathematician

Mathematics for the interested outsider

A Lemma on Nonzero Jacobians

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 f and first ask that it be continuous and injective on the closed ball \overline{K} of radius r around the point a. Then we ask that all the partial derivatives of f exist within the open interior K — note that this is weaker than our existence condition for the differential of f — and that the Jacobian determinant J_f(x)\neq0 on K. Then I say that the image f(K) actually contains a neighborhood of f(a). That is, the image doesn’t “flatten out” near a.

The boundary \partial K of the ball K is the sphere of radius r:

\displaystyle\partial K=\left\{x\in\mathbb{R}^n\vert\lVert x-a\rVert=r\right\}

Now the Heine-Borel theorem says that this sphere, being both closed and bounded, is a compact subset of \mathbb{R}^n. We’ll define a function on this sphere by

\displaystyle g(x)=\lVert f(x)-f(a)\rVert

which must be continuous and strictly positive, since if \lVert f(x)-f(a)\rVert=0 then f(x)=f(a), but we assumed that f is injective on \overline{K}. 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, g(\partial K)=[m,M], and there exists some point x on the sphere where this minimum is actually attained: g(x)=m>0.

Now we’re going to let T be the ball of radius \frac{m}{2} centered at f(a). We will show that T\subseteq f(K), and is thus a neighborhood of f(a) contained within f(K). To this end, we’ll pick y\in T and show that y\in f(X).

So, given such a point y\in T, we define a new function on the closed ball \overline{K} by

\displaystyle h(x)=\lVert f(x)-y\rVert

This function is continuous on the compact ball \overline{K}, so it again has an absolute minimum. I say that it happens somewhere in the interior K.

At the center of the ball, we have h(a)=\lVert f(a)-y\rVert<\frac{m}{2} (since y\in T), so the minimum must be even less. But on the boundary \partial K, we find

\displaystyle\begin{aligned}h(x)&=\lVert f(x)-y\rVert\\&=\lVert f(x)-f(a)-(y-f(a))\rVert\\&\geq\lVert f(x)-f(a)\rVert-\lVert f(a)-y\rVert\\&>g(x)-\frac{m}{2}\geq\frac{m}{2}\end{aligned}

so the minimum can’t happen on the boundary. So this minimum of h happens at some point b in the open ball K, and so does the minimum of the square of h:

\displaystyle h(x)^2=\lVert f(x)-y\rVert^2=\sum\limits_{i=1}^n\left(f^i(x)-y^i\right)^2

Now we can vary each component x^i of x separately, and use Fermat’s theorem to tell us that the derivative in terms of x^i must be zero at the minimum value b^i. That is, each of the partial derivatives of h^2 must be zero (we’ll come back to this more generally later):

\displaystyle\frac{\partial}{\partial x^k}\left[\sum\limits_{i=1}^n\left(f^i(x)-y^i\right)^2\right]\Bigg\vert_{x=b}=\sum\limits_{i=1}^n2\left(f^i(b)-y^i\right)\frac{\partial f^i}{\partial x^k}\bigg\vert_{x=b}=0

This is the product of the vector 2(f(b)-y) by the matrix \left(\frac{\partial f^i}{\partial x^k}\right). And the determinant of this matrix is J_f(b): the Jacobian determinant at b\in K, 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 f(b)-y=0, and so y=f(b)\in f(K).

November 13, 2009 - Posted by | Analysis, Calculus


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