5
$\begingroup$

Hi there! First of all i'm not a matematician, i'm just mechanical engineer who interested in some math.

Probably trivial question. Suppose I have a mapping $F: \mathbb{R}^n \to \mathbb{R}^n$ with Jacobian $J(x)$. And at some point $x_0$ Jacobian degenerate $\det J(x_0) = 0$. Is there general condition, for the existence of continuous inverse function near $x_0$, which is weaker than inverse function theorem. For example $y=x^3$ has continuous inverse $x = \sqrt[3]{y}$ near $x = 0$, but this fact is not provided by IFT.

Some usefull thinks I found in book "The implicit function theorem: history, theory, and applications" by Steven George Krantz, Harold R. Parks, but there are only special cases.

Thanks for all inputs.

$\endgroup$
0

1 Answer 1

7
$\begingroup$
  • One result is that small Lipschitz perturbations of the identity on a Banach space are homeomorphisms onto an open set: see here for a more precise statement. Note that the classic Inverse Mapping Theorem is usually proved as a consequence of this.

  • Another one deals with a strongly monotone map on a real Hilbert space, $F:H\to H$, that is, a map satisfying $(F(x)-F(y),x-y)\ge c |x-y|^2$ for all $x$ and $y$. A continuous strongly monotone map on $H$ is a homeomorphism onto $H$ (see e.g.Thm 11.2 in Klaus Deimling's book Nonlinear functional analysis, where you can also find more advanced inversion theorems). For one variable, real-valued functions this reduces to the well-known criterium.

  • In the case of $\mathbb{R}^n$ let's recall the theorem of Invariance of Domain. Injectivity is usually simpler to check than surjectivity, and in the case of a continuous map $f:U\to \mathbb{R}^n$ on an open set $U$ of $\mathbb{R}^n$ guarantees that $f$ is a homeomorphism onto an open set $f(U)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.