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Assume $f\colon \mathbb Q\to \mathbb Q$ is a function which admits continuous extensions

$f_0\colon\mathbb R\to \mathbb R$ and
$f_p\colon \mathbb Q_p\to \mathbb Q_p$ for each prime $p$.

Is it true that $f$ is a polynomial?

I guess the answer is no, but I do not see a counterexample.

edited Aug 29 2011 at 17:16

asked Aug 26 2011 at 18:57

ε-δ
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Given two countable dense subsets $A,B\subset \mathbb{R}$, there is a continuous homeomorphism $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $f(A)=B$. You can construct this function by just carefully choosing, one-by-one, the images of points of A (and the preimages of points of B). I believe this problem is exactly the same: just enumerate the rationals $q_{0},q_{1},...$, then choose $f(q_{i})$ so that it is near where you would expect it in $\mathbb{R}$, and $\mathbb{Q_{p})$ for the first $i$ primes $p$. – David Cohen Aug 26 2011 at 19:18

Lavender Honey · Accepted Answer · 2011-08-26 21:41:02Z UTC

The answer is no, and one can essentially use the same construction as in the answer: http://mathoverflow.net/questions/42460

Specifically, enumerate the non-zero rationals $\{r_1,r_2, \ldots\}$ in some way. Now consider the function: $$f(x) = \sum_{n=1}^{\infty} c_n x^{n^2} \prod_{i=1}^{n} (x - r_i).$$ If $c_n \in \mathbf{Q}$, then this is a well defined function from rationals to rationals. On the other hand, $f(x)$ converges to an analytic function in $\mathbf{Q}_v$ if and only if the coefficients of this power series converge to zero fast enough. Since the coefficients of the power series in the range $k = n^2$ to $k < (n+1)^2$ are simply the cofficients of $c_n x^{n^2} \prod_{i=1}^{n} (x - r_i)$, this can be ensured by forcing these coefficients to be very highly divisible by the first $n$ primes, and small in the archimedean sense (by including a very very large prime in the denominator).

Continuous extensions reals and to p-adic numbers

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