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 John Armstrong | Analysis, Calculus

3 Comments »

[…] yourself. Just like last time we’ve got a messy technical lemma about what happens when the Jacobian determinant of a […]

Pingback by Another Lemma on Nonzero Jacobians « The Unapologetic Mathematician | November 17, 2009 | Reply
[…] a smaller neighborhood on which is injective. We pick some closed ball centered at , and use our first lemma to find that must contain an open neighborhood of . Then we define , which is open since both […]

Pingback by The Inverse Function Theorem « The Unapologetic Mathematician | November 18, 2009 | Reply
[…] that for all . A maximum is similarly defined, except that we require in the neighborhood. As I alluded to recently, we can bring Fermat’s theorem to bear to determine a necessary […]

Pingback by Local Extrema in Multiple Variables « The Unapologetic Mathematician | November 23, 2009 | Reply

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