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If anyone bothered to check on my question history, they would see that I have already been asking several questions related to this theme. Unfortunately I haven't had much success on the specifics, so I might as well as ask this super general question as a last resort.

But fundamentally, I just want to know that if $f \in C^\infty(R^n)$ and all of its derivatives vanish in some sense rapidly enough as $|x| \to \infty$, and $\phi: U \to R^n$ is a diffeomorphism, $U$ an open bounded connected subset of $R^n$, then can we assert that $f\circ \phi: U \to R^n$ has bounded derivatives of all orders?

The issue is of course that since $U$ is a bounded set, $D\phi$ must diverge, and in fact every derivative diverges hence they must do so rapidly (?). $D(f\circ \phi) = (Df)\circ \phi D\phi$, so it comes down to comparing the rate of vanishing of $D^n f$ and the rate of divergence of $D^m \phi$.

This question will probably defy any sort of generalization, but can we at least assert that if $f$ and $D^n f$ vanish say $O(e^{-|x|^2})$ as $|x| \to \infty$, then given $U$, we may find $\phi$ such that $f \circ \phi$ has bounded derivatives of all orders?

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I think your question has a positive answer. Here is an example in one dimension that has the salient features of your question. Let$$\phi = \text{tan},$$so$$\phi:\left(-{\pi\over2}, {\pi\over2}\right) \to \mathbb{R},$$and let$$f(x) = e^{-x^2}.$$Then$$f \circ \phi = {1\over{e^{\text{tan}^2 x}}}.$$You check that it has bounded derivatives. I think that given any $\phi$ one could find $f$ such that $f \circ \phi$ has bounded derivatives.

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