Another Lemma on Nonzero Jacobians

Sorry for the late post. I didn’t get a chance to get it up this morning before my flight.

Brace yourself. Just like last time we’ve got a messy technical lemma about what happens when the Jacobian determinant of a function is nonzero.

This time we’ll assume that $f:X\rightarrow\mathbb{R}^n$ is not only continuous, but continuously differentiable on a region $X\subseteq\mathbb{R}^n$ . We also assume that the Jacobian $J_f(a)\neq0$ at some point $a\in X$ . Then I say that there is some neighborhood $N$ of $a$ so that $f$ is injective on $N$ .

First, we take $n$ points $\{z_i\}_{i=1}^n$ in $X$ and make a function of them

$\displaystyle h(z_1,\dots,z_n)=\det\left(\frac{\partial f^i}{\partial x^j}\bigg\vert_{x=z_i}\right)$

That is, we take the $j$ th partial derivative of the $i$ th component function and evaluate it at the $i$ th sample point to make a matrix $\left(a_{ij}\right)$ , and then we take the determinant of this matrix. As a particular value, we have

$\displaystyle h(a,\dots,a)=J_f(a)\neq0$

Since each partial derivative is continuous, and the determinant is a polynomial in its entries, this function is continuous where it’s defined. And so there’s some ball $N$ of $a$ so that if all the $z_i$ are in $N$ we have $h(z_1,\dots,z_n)\neq0$ . We want to show that $f$ is injective on $N$ .

So, let’s take two points $x$ and $y$ in $N$ so that $f(x)=f(y)$ . Since the ball is convex, the line segment $[x,y]$ is completely contained within $N\subseteq X$ , and so we can bring the mean value theorem to bear. For each component function we can write

$\displaystyle0=f^i(y)-f^i(x)=df^i(\xi_i)(y-x)=\frac{\partial f^i}{\partial x^j}\bigg\vert_{\xi_i}(y^j-x^j)$

for some $\xi_i$ in $[x,y]\subseteq N$ (no summation here on $i$ ). But like last time we now have a linear system of equations described by an invertible matrix. Here the matrix has determinant

$\displaystyle\det\left(\frac{\partial f^i}{\partial x^j}\bigg\vert_{\xi_i}\right)=h(\xi_1,\dots,\xi_n)\neq0$

which is nonzero because all the $\xi_i$ are inside the ball $N$ . Thus the only possible solution to the system of equations is $x^i=y^i$ . And so if $f(x)=f(y)$ for points within the ball $N$ , we must have $x=y$ , and thus $f$ is injective.

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November 17, 2009 - Posted by John Armstrong | Analysis, Calculus

2 Comments »

[...] trying to invert a function which is continuously differentiable on some region . That is we know that if is a point where , then there is a ball around where is one-to-one onto some [...]

Pingback by Cramer’s Rule « The Unapologetic Mathematician | November 17, 2009 | Reply
[...] as a function of , so there is some neighborhood of so that the Jacobian is nonzero within . Our second lemma tells us that there is a smaller neighborhood on which is injective. We pick some closed ball [...]

Pingback by The Inverse Function Theorem « The Unapologetic Mathematician | November 18, 2009 | Reply

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