# The Unapologetic Mathematician

## 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