Re: Jacobian determinant and inverse function theorem

On 03.09.2006 11:05, frank wrote:
Hello,

I am stuck right at the beginning of a proof of the inverse function
theorem.

The theorem is stated as follows: Let U be an open set R^n and let f be
a continuously differentiable function mapping U into R^n. Suppose that
c is a point of U where Jf(c)≠0. ... (I omit the rest, because it
will not concern us here).

Now the proof starts by pointing out that, because Jf(c)≠0 and Djfi
is continuous for every i,j=1,2,...,n, one can choose a neighborhood
N(c) contained in U, such that, for every choice of z1, z2,...,zn in
N(c), the following determinant ≠0.
Here is that determinant:
det
[D1f1(z1) D2f1(z1) ... Dnf1(z1)
D1f2(z2) D2f2(z2) ... Dnf2(z2)
D1f3(z3) D2f3(z3) ... Dnf3(z3)
...
D1fn(zn) D2fn(zn) ... Dnfn(zn)]
I hope that the system does not mess up the lay-out, but anyway, this
is kind of the Jacobian determinant, but with the components of the
gradient in each row being evaluated at a different point of the
neighborhood N(c).

Now my question is: why should this determinant be ≠0, how does this
follow from Jf(c)≠0 and the continuity of the Djfi's?
Thank you for explaining me that.

Frank

Hint: The map g: U -> R, x |-> Jf(x) is continuous. Let R* be the set of
real number without 0. If g(c)≠0 then g^{-1}(R*) is an open neighborhood
of c in U.

J.
.

Relevant Pages