The Unapologetic Mathematician

Mathematics for the interested outsider

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 jth partial derivative of the ith component function and evaluate it at the ith 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 | Analysis, Calculus

2 Comments »

  1. [...] 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

  2. [...] 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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