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Suppose $f:I \rightarrow \Bbb R$ is an analytic function defined on the interval $I\subset \Bbb R$ with the property that for every $q \in \Bbb Q:f(q)\in \Bbb Q$. Does this already imply that $f\in \Bbb Q(X)$, i. e. that $f$ is a rational function with rational coefficients?

If this question is to hard, I would also be glad if one could find an answer to the following presumably easier question: Suppose $f:\Bbb R\rightarrow \Bbb R$ is a function given by an everywhere convergent power series with the property that for every $q \in \Bbb Q:f(q)\in \Bbb Q$. Does this already imply that $f$ is a polynomial?

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1 Answer 1

up vote 3 down vote accepted

No, it doesn't. In fact, given any two dense countable subsets $A$, $B$ of $\mathbb R$, there are uncountably many entire functions that map $A$ one-to-one onto $B$.

See e.g. https://groups.google.com/forum/?hl=en&fromgroups=#!topic/sci.math/UjRgb0y_iBE

EDIT: here's an example of this type of construction.

Let $f_0$ be an entire function taking $\mathbb R$ to $\mathbb R$ with $ f'_0(x) \ge 1$ for all $ x \in \mathbb R$.

Let $a_n$ and $b_n$, $n \in \mathbb N$, be enumerations of $A$ and $B$. I will define inductively sequences $\langle D_n \rangle$ and $\langle R_n \rangle$ of finite subets of $A$ and $B$ respectively and $\langle g_n \rangle$ of entire functions taking $\mathbb R$ into $\mathbb R$, and take $f_n = f_0 + \sum_{j=1}^n g_j$, in such a way that

  1. $|g_n(z)| < 2^{-n}$ for $ |z| \le n$
  2. $|g_n'(x)| < 2^{-n-1}$ for all $ x \in \mathbb R$
  3. $ g_n(x) = 0$ if $ x \in D_{n-1}$
  4. $ f_n(D_n) = R_n$
  5. $ D_n \subseteq D_{n+1}$ and $ R_n \subseteq R_{n+1}$, with $a_n \in D_n$ and $b_n \in R_n$

Once we have such sequences, by (1) the series $ f_0(z) + \sum_{j=1}^\infty g_j(z)$ converges uniformly on compact sets to an entire function $f(z)$. By (2), $ f'(x) \ge 1/2$ for all $x \in \mathbb R$ (and in particular $f$ is one-to-one on $\mathbb R$).
By (3) and (4), $ f(D_n) = R_n$. Using (5), $f$ takes $A$ onto $B$.

Now to construct the sequences. Let $ D_0$ and $R_0$ be empty. Suppose we have $ D_{n-1}$, $ R_{n-1}$ and $ f_{n-1}$.

Let $a_k$ be the first element (in the enumeration of $A$) that is not in $D_{n-1}$. By the induction hypothesis, $k \ge n$. Let $ u(z)$ be an entire function taking $\mathbb R$ into $\mathbb R$, with zeros at all points of $ D_{n-1}$ but $ u(a_k) \ne 0$ and with $ u'_i$ bounded on $ \mathbb R$ (e.g. we could take $u(z) = P(z) \exp(-z^2)$ for a suitable polynomial $P$). For $\alpha$ in some interval $ (-\varepsilon,\varepsilon)$, we have $ |\alpha u(z)| < 2^{-n-1}$ for $|z| \le n$ and $|\alpha u'(x)| < 2^{-n-2}$ for all $ x \in \mathbb R$. For a dense set of $\alpha$'s in this interval, $f_{n-1}(a_k) + \alpha u(a_k) \in B$. Choosing such an $\alpha$, let $v(z) = \alpha u(z)$ and $r = f_{n-1}(a_k) + \alpha u(a_k)$. Note that since $f_{n-1} + \alpha u$ is increasing on $\mathbb R$, $r \notin R_{n-1} = (f_{n-1} + \alpha u)(D_{n-1})$.

Let $b_j$ be the first element (in the enumeration of $B$) that is not in $R_{n-1} \cup \{r\}$. Again, $j \ge n$. Let $t$ be the unique real number with $ f_{n-1}(t) + \alpha u(t) = b_j$. We know that $t \notin D_{n-1}$ and $t \ne a_k$. Let $ w(z)$ be an entire function taking $\mathbb R$ into $\mathbb R$, with zeros at all points of $ D_{n-1} \cup \{a_k\}$ but $w(t) > 0$ and with $ w'_i$ bounded on $ \mathbb R$. For $\beta$ in some interval $(-\delta,\delta)$, we have $ |\beta w(z)| < 2^{-n-1}$ for $|z| \le n$ and $|\beta w'(x)| < 2^{-n-2}$ for all $x \in \mathbb R$. The real solution $ x = s(\beta)$ of $f_{n-1}(x) + \alpha u(x) + \beta w(x) = b_j$ is a continuous function of $\beta$ in this interval, and (since $\beta w(x) > 0$ in a neighbourhood of $t$) strictly decreasing in some subinterval containing $0$. Therefore we can choose $\beta$ so that $s(\beta) \in A$. We let $g_n(z) = \alpha u(z) + \beta w(z)$, $D_n = D_{n-1} \cup \{a_k, s(\beta)\}$, and $R_n = R_{n-1} \cup \{r, b_j\}$. It is easy to see that requirements (1) to (5) are satisfied.

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Could you give an explicit example or a general method to construct such functions? –  Dominik Dec 5 '12 at 19:49

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