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Suppose $f$ is a $C^k (1\leq k\leq\infty)$ function from the unit ball $\mathcal{B}$ in $\mathbb{R}^n$ to itself, which is a diffeomorphism from the domain to its image, with the upper and lower bounds of the norm of derivative of $f$: $$ \|df(x)\| \leq \alpha < 1, \|df^{-1}(x)\| \leq \beta, ~\forall x\in \mathcal{B} $$ where $\alpha$ and $\beta$ are real constants.

Could $f$ be extended to a $C^k$ function $F$ from the whole space $\mathbb{R}^n$ to itself, still satisfying $$ \|dF(x)\| \leq \alpha, \|dF^{-1}(x)\| \leq \beta, ~\forall x\in \mathbb{R}^n $$ and additionally $$ F(x) \rightarrow \infty \mathrm{~when~} x \rightarrow \infty ? $$

Hence we can use the Hadamard's global inverse function theorem to show that $F$ is a diffeomorphism.

There are related results such as Kirszbraun’s theorem and Whitney's extension theorem, but they are not enough.

If it is true, this result will lead to the global stable and unstable manifold theorem in the thoeory of dynamical systems.

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  • $\begingroup$ Are you sure? Here $df(x)$ may be singular already at some point of $B$. I think you had in mind an hypothesis on $\|df(x)^{-1}\|$ instead, like in Hadamard's theorem in fact. $\endgroup$
    – Pietro Majer
    Commented Sep 11, 2013 at 8:45
  • $\begingroup$ Thank you, I have corrected the description of the problem. $\endgroup$
    – Horizonto
    Commented Sep 11, 2013 at 12:00
  • $\begingroup$ Of course you also need to assume that $f(B)=B$, as assuming just $f(B)\subset B$ allows non-injective locally invertible maps $f:B\to B$ $\endgroup$
    – Pietro Majer
    Commented Sep 12, 2013 at 9:34
  • $\begingroup$ Thanks very much. The origin condition is that $f$ is a diffeomorphism to its image. I have improperly dropped this condition. $\endgroup$
    – Horizonto
    Commented Sep 12, 2013 at 11:00

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