Which continual features are polynomials?

Intend $f \in C(\mathbb{R}^n)$, the room of continual $\mathbb{R}$-valued features on $\mathbb{R}^n$. Exist problems on $f$ that assure it is the pullback of a polynomial under some homeomorphism? That is, when can I locate $\phi:\mathbb{R}^n \to \mathbb{R}^n$ such that $f \circ \phi \in \mathbb{R}[x_1,\ldots, x_n]$? I have actually attempted experimenting with the implied function theory yet have not obtained much. It seems like I might be missing out on something really noticeable.

Some relevant inquiries:

  • An essential problem when it comes to $n = 1$ is that $f$ can not acquire the very same value definitely sometimes (given that a polynomial has just finitely several origins). Is this enough?
  • What happens if we change $\mathbb{R}$ by $\mathbb{C}$?
  • What happens if we consider smooth features rather?
  • What concerning the facility analytic instance?
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2019-05-05 08:23:11
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Answers: 2

Eric, I had actually asked a similar question on MathOverflow, that includes the problem of differentiablility. Please assess this inquiry and also it's solutions as it will likely be of some usage to your inquiry.

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2019-05-10 06:58:31
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Since I can not leave remarks I'm creating this below. I assume this inquiry is made hard by the problem that $\phi$ is simply called for homeomorphism versus claim a diffeomorphism.

In case n = 1 you can absolutely think of continual features that are not differentiable on a distinct set yet can be drawn back to generate a polynomial. As a child instance take into consideration the function $f$ that is $\sqrt{x}$ on the favorable reals and also x on the adverse reals. Take into consideration the homeomorphism that is $x^2$ on the favorable reals and also x on the adverse reals, after that $f$ draws back to the polynomial $x$.

I do not assume its enough that $f$ does not acquire the very same value definitely sometimes. I do not have a counterexample yet I assume a prospect could be had in this article. The idea is that there are features almost everywhere continual and also purely monotonic yet with acquired 0 virtually anywhere.

I assume you would certainly have more good luck making use of the implied function theorem if you called for $\phi$ to be a diffeomorphism. Additionally I think its real that 'most' continual function from $\mathbb{R} \to \mathbb{R}$ are not really wonderful (nonwhere differential) so a more tractable inquiry could be the very same inquiry yet calling for $f$ to be smooth.

If you change $\mathbb{R}$ with $\mathbb{C}$ and also enforce $f$ and also $\phi$ both be holomorphic after that I assume it is adequate that $f^{(n)}$ disappear for all completely huge $n$ due to the fact that you can recoup $f$ from its taylor collection.

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2019-05-08 09:27:41
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