# Which continuous functions are polynomials?

Suppose $f \in C(\mathbb{R}^n)$, the space of continuous $\mathbb{R}$-valued functions on $\mathbb{R}^n$. Are there conditions on $f$ that guarantee it is the pullback of a polynomial under some homeomorphism? That is, when can I find $\phi:\mathbb{R}^n \to \mathbb{R}^n$ such that $f \circ \phi \in \mathbb{R}[x_1,\ldots, x_n]$? I have tried playing around with the implicit function theorem but haven't gotten far. It feels like I may be missing something very obvious.

Some related questions:

• A necessary condition in the case of $n = 1$ is that $f$ cannot attain the same value infinitely many times (since a polynomial has only finitely many roots). Is this sufficient?
• What if we replace $\mathbb{R}$ by $\mathbb{C}$?
• What if we look at smooth functions instead?
• What about the complex analytic case?
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I wasn't quite sure how to tag this so feel free to edit them. – Eric O. Korman Aug 2 '10 at 20:35
Why don't infinite polynomials count? – BlueRaja - Danny Pflughoeft Aug 2 '10 at 20:47
@BlueRaja: Because I am interested in what type of loss there is when considering only polynomials (which is what's done in algebraic geometry). I also don't want to worry about issues of convergence, and the theory of Taylor series is pretty well developed. – Eric O. Korman Aug 2 '10 at 21:01
We have $f(x )= f( \phi ( \phi ^{-1}(x))) = g( \phi ^{-1} ( x))$ for some polynomial $g$, and $\phi ^ {-1}$ is a homeomorphism. – user254665 Aug 21 '15 at 19:23

Since I can't leave comments I'm writing this here. I think this question is made difficult by the condition that $\phi$ is just required homeomorphism versus say a diffeomorphism.

In the case n = 1 you can certainly come up with continuous functions that are not differentiable on a discrete set but can be pulled back to yield a polynomial. As a baby example consider the function $f$ that is $\sqrt{x}$ on the positive reals and x on the negative reals. Consider the homeomorphism that is $x^2$ on the positive reals and x on the negative reals, then $f$ pulls back to the polynomial $x$.

I don't think its sufficient that $f$ doesn't attain the same value infinitely many times. I don't have a counterexample but I think a candidate might be contained in this article. The gist is that there are functions everywhere continuous and strictly monotonic but with derivative 0 almost everywhere.

I think you'd have more luck using the implicit function theorem if you required $\phi$ to be a diffeomorphism. Also I believe its true that 'most' continuous function from $\mathbb{R} \to \mathbb{R}$ are not very nice (nonwhere differential) so a more tractable question might be the same question but requiring $f$ to be smooth.

If you replace $\mathbb{R}$ with $\mathbb{C}$ and impose $f$ and $\phi$ both be holomorphic then I think it suffices that $f^{(n)}$ vanish for all sufficiently large $n$ because you can recover $f$ from its taylor series.

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Thanks for the comments. I'll have to check out that paper. – Eric O. Korman Aug 3 '10 at 18:13

Eric, I had asked a similar question on MathOverflow, which includes the condition of differentiablility. Please review this question and it's answers as it will likely be of some use to your question.

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