1 Introduction
In what follows, p is a prime number, $\mathbb {Q}_p$ is the field of p-adic numbers and $\mathbb {Z}_p$ is the ring of p-adic integers. Let $(u_n)_{n\geq 0}$ be a sequence of integers. If there exists a continuous function $f:\mathbb {Z}_p\rightarrow \mathbb {Q}_p$ such that $f(n)=u_n$ for all nonnegative integers n, we say that f is a p-adic interpolation of $(u_n)_{n\geq 0}$ . In addition, if f is analytic, we say that it is a p-adic analytic interpolation of this sequence. Since the set of nonnegative integers is a dense subset of $\mathbb {Z}_p$ , any given sequence of integers admits at most one such interpolation, which will only exist under certain strong conditions on the sequence (for more details, see [Reference Schikhof17]).
Many authors have studied the problem of p-adic interpolation. Bihani et al. [Reference Bihani, Sheppard and Young2] considered the problem of p-adic interpolation of the Fibonacci sequence, they proved that the sequence $(2^nF_n)_{n\geq 0}$ can be interpolated by a p-adic hypergeometric function on $\mathbb {Z}_5$ . Rowland and Yassawi in [Reference Rowland and Yassawi16] studied p-adic properties of sequences of integers (or p-adic integers) that satisfy a linear recurrence with constant coefficients. For such a sequence, they obtained an explicit approximate twisted interpolation to $\mathbb {Z}_p$ . In particular, they proved that for any prime $p\neq 2$ , there is a twisted interpolation of the Fibonacci sequence by a finite family of p-adic analytic functions with coefficients in some finite extension of $\mathbb {Q}_p$ . Inspired by the Skolem–Mahler–Lech theorem on linear recurrent sequences, Bell [Reference Bell1] proved that for a suitable choice of a p-adic analytic function f and a starting point $\overline {x}$ , the iterate-computing map $n\mapsto f^n(\overline {x})$ extends to a p-adic analytic function g defined for all $x\in \mathbb {Z}_p$ . That is, the sequence $f^n(\overline {x})$ can be interpolated by the p-adic analytic function g.
Mahler [Reference Mahler7] states that the polynomial functions
with $n\geq 0$ integer, form an orthonormal basis, called the Mahler basis, for the space of p-adic continuous functions $\mathcal {C}(\mathbb {Z}_p\rightarrow \mathbb {Q}_p)$ . More precisely, he showed that every continuous function $f:\mathbb {Z}_p\rightarrow \mathbb {Q}_p$ has a unique uniformly convergent expansion
where $a_n\rightarrow 0$ and $\|f\|_{\scriptsize \mbox {sup}}=\max _{n\geq 0}\|a_n\|_p$ . Conversely, every such expansion defines a continuous function. Furthermore, if $f\in \mathcal {C}(\mathbb {Z}_p\rightarrow \mathbb {Q}_p)$ has a Mahler expansion given by (1.1), then the Mahler coefficients $a_n$ can be reconstructed from f by the inversion formula
Using the Mahler expansion (1.1) and the inversion formula (1.2), we conclude that the sequence $(u_n)_{n\geq 0}$ of integers can be p-adically interpolated if and only if
We became interested in studying the p-adic analytic interpolation of sequences of integers with polynomial growth while studying a problem about p-adic Liouville numbers. Based on the classic definition of complex Liouville numbers, Clark [Reference Clark3] called a p-adic integer $\lambda $ a p-adic Liouville number if
It is easily seen that all p-adic Liouville numbers are transcendental p-adic numbers. Moreover, if $\lambda $ is a p-adic Liouville number and $a,b$ are integers, with $a>0$ , then $a\lambda +b$ is also a p-adic Liouville number.
In his book, Maillet [Reference Maillet10, Ch. III] discusses some arithmetic properties of complex Liouville numbers. One of them states that given a nonconstant rational function f with rational coefficients, if $\xi $ is a Liouville number, then so is $f(\xi )$ . Motivated by this fact, Mahler [Reference Mahler9] posed the following question.
Question 1.1 (Mahler [Reference Mahler9])
Are there transcendental entire functions $f:\mathbb {C}\rightarrow \mathbb {C}$ such that if $\xi $ is any Liouville number, then $f(\xi )$ is also a Liouville number?
He pointed out: ‘The difficulty of this problem lies of course in the fact that the set of all Liouville numbers is nonenumerable.’ We are interested in studying the analogous question for p-adic Liouville numbers.
Question 1.2. Are there p-adic transcendental analytic functions $f:\mathbb {Z}_p\rightarrow \mathbb {Q}_p$ such that if $\lambda $ is a p-adic Liouville number, then so is $f(\lambda )$ ?
It is important to note that the analogue of Maillet’s result is not true for p-adic Liouville numbers. In fact, Lelis and Marques [Reference Lelis and Marques5] proved that the analogue of Maillet’s result is true for a class of p-adic numbers called weak p-adic Liouville numbers, but not for all p-adic Liouville numbers.
Inspired by an argument presented by Marques and Moreira in [Reference Marques and Moreira11] and discussed by Lelis and Marques in [Reference Lelis and Marques6], we approached Question 1.2 as follows. If there were a positive integer sequence $(u_n)_{n\geq 0}$ satisfying $u_n\rightarrow \infty $ and $u_n=O(n)$ that could be interpolated by a p-adic transcendental analytic function $f:\mathbb {Z}_p\rightarrow \mathbb {Q}_p$ , then f would answer Question 1.2 affirmatively. Indeed, assuming all that is true, if we get any p-adic Liouville number $\lambda \in \mathbb {Z}_p$ , by definition there would be a sequence of integers $(n_{k})_{k\geq 0}$ such that
The function f being analytic would satisfy a Lipschitz condition (see [Reference Robert15, Ch. 5, Section 3]). Thus, there would be a constant $c>0$ such that
and so
where $u_{n_k}\rightarrow \infty $ and $u_{n_k}=O(n_k)$ . So $f(\lambda )$ would also be a p-adic Liouville number.
In light of this, it is natural to try to characterise the p-adic analytic functions which interpolate sequences of integers $(u_n)_{n \geq 0}$ of linear growth. There are other reasons for seeking such characterisations. Indeed, one may ask whether there exists a p-adic interpolation of some arithmetic function (many of which have linear growth) or, more generally, if polynomials with integer coefficients are the only p-adic analytic functions that take positive integers into positive integers with polynomial order.
Theorem 1.3. Let $(u_n)_{n\geq 0}$ be a sequence of positive integers such that $u_n=O(n^d)$ for some fixed $d\geq 0 \ (d\in \mathbb {R})$ . Assume there exists a p-adic analytic function $f:\mathbb {Z}_p\rightarrow \mathbb {Q}_p$ which interpolates the sequence $(u_n)_{n\geq 0}$ .
-
(i) If $d\leq 1$ , then f is a polynomial function of degree at most one.
-
(ii) If $d>1$ and the Mahler expansion of f converges for all $x\in \mathbb {Q}_p$ , then f is a polynomial function of degree at most $\lfloor d\rfloor $ .
We remark that the condition ‘f is a p-adic analytic function on $\mathbb {Z}_p$ ’ is fundamental in the result above. Indeed, if we write $n=\sum _{i=0}^{k}a_ip^i$ in base p, then the function $f:\{0\}\cup \mathbb {N}\to \mathbb {Q}_p$ given by
clearly can be extended in a unique way to a continuous function $\overline {f}:\mathbb {Z}_p\to \mathbb {Q}_p$ such that $\overline {f}(n)=O(n)$ . However, $\overline {f}$ is nonanalytic and it is clearly not a polynomial function.
Moreover, consider the p-adic function $f_d:\mathbb {Z}_p\to \mathbb {Q}_p$ defined by
where $z=\sum _{k=0}^{\infty }a_kp^k$ is the p-adic expansion of $z\in \mathbb {Z}_p$ . Then it is well known that $f_d$ is a continuous function for all integers $d\geq 2$ . In fact, if $d\geq 2$ is an integer, then
In particular, we have $f^{\prime }_d(x)=0$ for all $x\in \mathbb {Q}_p$ and $f_d\in \mathcal {C}^1(\mathbb {Z}_p\to \mathbb {Q}_p)\subset \mathcal {C}(\mathbb {Z}_p\to \mathbb {Q}_p)$ . Note that $f_d(n)=O(n^d)$ , but $f_d$ is not a polynomial function. However, since $f_d$ is not a p-adic analytic function, its Mahler expansion does not converge for all $x\in \mathbb {Q}_p$ .
Very strict conditions must be satisfied for a sequence $(u_n)_{n\geq 0}$ to be interpolated by a p-adic analytic function. However, if the set $A=\{u_0,u_1,\ldots \}\subseteq \mathbb {Z}$ is a dense subset of $\mathbb {Z}_p$ , one may ask whether there is some re-enumeration $\sigma :\{0\}\cup \mathbb {N}\rightarrow \{0\}\cup \mathbb {N}$ such that $(u_{\sigma (n)})_{n\geq 0}$ can be interpolated by a p-adic analytic function.
In the complex case, Georg [Reference Georg4] established that for each countable subset $X\subset \mathbb {C}$ and each dense subset $Y\subseteq \mathbb {C}$ , there exists a transcendental entire function f such that $f(X)\subset Y$ . In 1902, Stäckel [Reference Stäckel18] used another construction to show that there is a function $f(z)$ , analytic in a neighbourhood of the origin and with the property that both $f(z)$ and its inverse function assume, in this neighbourhood, algebraic values at all algebraic points. Based on these results, Mahler [Reference Mahler8] suggested the following question about the set of algebraic numbers $\overline {\mathbb {Q}}$ .
Question 1.4 (Mahler, [Reference Mahler8])
Are there transcendental entire functions $f(z)=\sum c_nz^n$ with rational coefficients $c_n$ and such that $f(\overline {\mathbb {Q}})\subset \overline {\mathbb {Q}}$ and $f^{-1}(\overline {\mathbb {Q}})\subset \overline {\mathbb {Q}}$ ?
This question was answered positively by Marques and Moreira [Reference Marques and Moreira12]. Moreover, in a more recent paper [Reference Marques and Moreira13], they proved that if X and Y are countable subsets of $\mathbb {C}$ satisfying some conditions necessary for analyticity, then there are uncountably many transcendental entire functions $f(z)=\sum a_nz^n$ with rational coefficients such that $f(X)\subset Y$ and $f^{-1}(Y)\subset X$ . Keeping these results in mind, we prove the following theorem.
Theorem 1.5. Let X and Y be subsets of $\mathbb {Z}$ dense in $\mathbb {Z}_p$ . Then there are uncountably many p-adic analytic injective functions $f:\mathbb {Z}_p\to \mathbb {Q}_p$ with
such that $f(X)=Y$ .
Note that by Theorem 1.5, if $Y=\{y_0,y_1,y_2,\ldots \}\subset \mathbb {Z}$ is a dense subset of $\mathbb {Z}_p$ , that is, if Y contains a complete system of residues modulo any power of p, then there is a p-adic analytic function
and a bijection $\sigma :\{0\}\cup \mathbb {N}\rightarrow \{0\}\cup \mathbb {N}$ such that $f(n)=u_{\sigma (n)}$ , where we take $X=\{0\}\cup \mathbb {N}$ . Moreover, the series above converges for all $x\in \mathbb {Z}_p$ . Thus, if we consider the Mahler expansion, then we immediately obtain the following result.
Corollary 1.6. Let $Y=\{y_0,y_1,y_2,\ldots \}$ be a subset of $\mathbb {Z}$ dense in $\mathbb {Z}_p$ . Then there are $a_0,a_1,a_2,\ldots \in \mathbb {Z}$ and a bijection $\sigma :\{0\}\cup \mathbb {N}\rightarrow \{0\}\cup \mathbb {N}$ such that
for all integers $n\geq 0$ , where $v_p(a_n/n!)\to \infty $ as $n\to \infty $ .
We end this section by presenting some questions which we are still unable to answer. One may ask whether Theorem 1.5 is still true if X and Y are free to contain elements outside $\mathbb {Z}$ . What could one do to guarantee rational coefficients in f in a situation like that? Moreover, if we consider the algebraic closure of $\mathbb {Q}_p$ , denoted by $\overline {\mathbb {Q}}_p$ , and its completion $\mathbb {C}_p$ , we may ask a probably more difficult question.
Question 1.7. Are there p-adic transcendental entire functions $f:\mathbb {C}_p\to \mathbb {C}_p$ given by
such that $f(\overline {\mathbb {Q}}_p)\subset \overline {\mathbb {Q}}_p$ and $f^{-1}(\overline {\mathbb {Q}}_p)\subset \overline {\mathbb {Q}}_p$ ?
Naturally, the main difficulty of this problem lies again in the fact that the set $\overline {\mathbb {Q}}_p$ is uncountable.
2 Proof of Theorem 1.3
We start by introducing the classic Strassmann’s theorem about zeros of p-adic power series. This result says that a p-adic analytic function with coefficients in $\mathbb {Q}_p$ has finitely many zeros in $\mathbb {Z}_p$ and provides a bound for the number of zeros.
Theorem 2.1 (Strassmann, [Reference Murty14])
Let $f(x)=\sum _{n=0}^{\infty }c_nx^n$ be a nonzero power series with coefficients in $\mathbb {Q}_p$ and suppose that $\lim _{n\rightarrow \infty }c_n=0$ so that $f(x)$ converges for all x in $\mathbb {Z}_p$ . Let N be the integer defined by conditions
Then the function $f:\mathbb {Z}_p\rightarrow \mathbb {Q}_p$ defined by $x\mapsto f(x)$ has at most N zeros.
Proof of Theorem 1.3
Let $(u_n)_{n\geq 0}$ be a sequence of integers of linear or sublinear growth, that is, $u_n=O(n)$ . Suppose that $(u_n)_{n\geq 0}$ can be interpolated by some p-adic analytic function
Since $f(x)$ is a p-adic analytic function, $\lim _{n\rightarrow \infty }\|c_n\|_p=0$ . Thus, there exists an integer N defined by the conditions
and Strassman’s theorem guarantees that the function $f:\mathbb {Z}_p\rightarrow \mathbb {Q}_p$ has at most N zeros.
By hypothesis, $u_n=O(n)$ , so there is a $C>0$ such that $0 < u_n\leq Cn$ for all $n\geq 0$ . Taking the subsequence $(u_{p^k})_{k\geq 0}$ ,
Since f is an analytic function, it is easily seen that it satisfies the Lipschitz condition
for all $x,y \in \mathbb {Z}_p$ . In particular,
and it follows that
with $t_k\in \mathbb {Z}_+$ , because $u_{p^k}$ is a positive integer. By (2.1) and (2.2), we conclude that $0\leq t_k\leq C$ . Hence, by the pigeonhole principle, there exists an integer t with $0\leq t\leq C$ such that
for infinitely many $j\geq 0$ . Thus, the function
has infinitely many roots and by Strassman’s theorem, we conclude that $f(x)=u_0+tx$ .
Now suppose that $u_n=O(n^d)$ for some fixed positive real number $d>1$ . Let
be the Mahler expansion of f. By hypothesis, the Mahler expansion of f converges for all $x\in \mathbb {Q}_p$ , so the function $x\mapsto \sum _{n=0}^{\infty }a_n\binom {x}{n}$ is analytic on $\mathbb {C}_p$ and
for all real numbers $r>0$ (see [Reference Schikhof17, Ch. 3]). Taking $r=p^2$ , we find $v_p(a_n)\geq 2n$ for all n sufficiently large. Moreover, $a_n$ is an integer for all $n\geq 0$ . In fact, by the Mahler expansion,
where $u_j\in \mathbb {Z}_+$ for all $j\geq 0$ . Hence, either $a_n=0$ or
However,
Since $\|u_j\|_{\infty }\leq j^d\leq n^d$ for all $j\leq n$ , it follows that
where $D>0$ is a fixed constant. It is easily seen that (2.3) and (2.4) cannot both be true for n sufficiently large. Hence, there exists an $N>0$ such that $a_n=0$ for all $n>N$ . Consequently, f is a polynomial function. Furthermore, $f(n)=O(n^d)$ , so its degree must be at most $\lfloor d\rfloor $ .
3 Proof of Theorem 1.5
Suppose that $X=\{x_0,x_1,x_2,\ldots \}$ and $Y=\{y_0,y_1,y_2,\ldots \}$ are subsets of $\mathbb {Z}$ dense in $\mathbb {Z}_p$ . Our proof consists in determining a sequence of polynomial functions $f_0,f_1,\ldots $ such that $f_n\to f$ as $n\to \infty $ , where f is a p-adic analytic injective function on $\mathbb {Z}_p$ with rational coefficients satisfying $f(X)=Y$ . In addition, we will show that there are uncountably many such functions.
To be more precise, we will construct a sequence of polynomial functions $f_0,f_1,f_2,\ldots \in \mathbb {Q}[x]$ of degrees $t_0,t_1,t_2,\ldots \in \mathbb {Z}$ , respectively, such that for all $m\geq 0$ ,
where $c_0=y_0-x_0$ , $c_1=1$ and $\|c_i\|_p\leq p^{-1}$ for all $2\leq i\leq t_m$ . Furthermore, our sequence will obey the recurrence relation
where the polynomial functions $P_m\in \mathbb {Z}[x]$ are given by
with $X_m=\{x_0,\ldots ,x_m\}$ and $Y^{-1}_m=f_m^{-1}(\{y_0,\ldots ,y_m\})$ , and $\delta _m$ and $\epsilon _m$ are rational numbers such that
Finally, our sequence will also satisfy $f_m(x_k)\in Y$ and $f_m^{-1}(\{y_k\})\cap X \neq \emptyset $ for all ${0\leq k\leq m}$ .
We make some remarks regarding such a sequence of polynomials. First, since $f_m$ is a polynomial, $Y^{-1}_m$ must be a finite subset of $\mathbb {Z}_p$ for each m, so the polynomials $P_m$ are well defined. Second, by (3.1), $\|c_1\|_p> \|c_i\|_p$ for all $i \geq 2$ , so each $f_m$ is necessarily injective on $\mathbb {Z}_p$ by Strassmann’s theorem. Lastly, since $f_m$ is injective, there is only one $x_s \in X \cap f_m^{-1}(\{y_k\})$ . The existence of such a sequence is guaranteed by the following lemma.
Lemma 3.1. Suppose that $ f_m(x)=c_0+c_1x+\cdots +c_{t_m}x^{t_m}\in \mathbb {Q}[x] $ is a polynomial with
such that $f_m(X_m)\subset Y$ and $Y_m^{-1}\subset X$ . Then there exist rational numbers $\delta _m$ and $\epsilon _m$ with
such that the function
is a polynomial given by
satisfying $f_{m+1}(X_{m+1})\subset X$ and $Y_{m+1}^{-1}\subset X$ and, moreover, $\|c_i\|_p<\|c_1\|_p$ for all integers i with $2\leq i\leq t_{m+1}$ .
Proof. Suppose that for some $m\geq 0$ , there is a function $f_m$ satisfying the hypotheses of the lemma. We will show that we can choose rational numbers $\delta _m$ and $\epsilon _m$ such that
in such a way that the polynomial $f_{m+1}$ in (3.2) has the desired properties.
First, we will determine $\delta _m\in \mathbb {Q}$ such that $f_{m+1}(x_{m+1})\in Y$ . Suppose that $ f_m(x_{m+1}) \in \{y_0,y_1,\ldots ,y_m\}$ . Since $P_m(x_{m+1})=0$ , we have $f_{m+1}(x_{m+1}) = f_{m}(x_{m+1})\in Y$ . Note that here we did not make direct use of $\delta _m$ to get $f_{m+1}(x_{m+1}) \in Y$ . So we are free to choose any $\delta _m \in \mathbb {Q}$ and we do so by setting $\delta _m = p^m$ . Now, suppose that $f_m(x_{m+1})\notin \{y_0,y_1,\ldots ,y_m\}$ , which implies that $P_m(x_{m+1})\neq 0$ . Since Y is a dense subset of $\mathbb {Z}_p$ , there exists $\hat {y}\in Y$ such that
Then, taking
we obtain $f_{m+1}(x_{m+1})=\hat {y}\in Y$ independently of $\epsilon _m$ . Observe that in both cases just analysed, $\|\delta _m\|_p \leq p^{-m}$ .
Now we will choose $\epsilon _m \in \mathbb {Q}$ to get $f_{m+1}(\hat {x})=y_{m+1}$ for some $\hat {x}\in X$ . Since $f_m$ is injective on $\mathbb {Z}_p$ , there is at most one $\hat {x}\in X$ such that $f_m(\hat {x})=y_{m+1}$ . If there exists $\hat {x}\in X_m$ such that $f_m(\hat {x})=y_{m+1}$ , then $P_m(\hat {x})=0$ and we obtain $f_{m+1}(\hat {x})=y_{m+1}$ . In this case, $\epsilon _m$ does not play a role and we are free to set $\epsilon _m=p^m$ . It remains to consider the case where there is no $\hat {x} \in X_m$ with $f_m(\hat {x})=y_{m+1}$ . Note that if we choose
then $f_{m+1}(x_{m+1})=y_{m+1}$ and we have $\hat {x} = x_{m+1}$ . Since we again did not use $\epsilon _m$ to ensure that $f_{m+1}(x_{m+1})=y_{m+1}$ , we are free to take $\epsilon _m=p^m$ . However, if
we consider the polynomial equation
Since $\|\delta _m\|_p\leq p^{-m}$ and $\|c_i\|_p<p^{-1}$ for $i\geq 2$ ,
for all $m\geq 2$ . Thus, the congruence
has a solution $\overline {x} \equiv y_{m+1}-y_0 \pmod {p\mathbb {Z}_p}$ . Moreover, taking the formal derivative,
Hence, by Hensel’s lemma [Reference Murty14], there exists $b\in \mathbb {Z}_p$ such that
Let $v_p(x)$ be the p-adic valuation of $x\in \mathbb {Z}_p$ and take
Note that $s<+\infty $ , since $P_m(b)(b-x_{m+1})\neq 0$ . Thus, we have a Lipschitz condition on $\mathbb {Z}_p$ , namely
for all $x,y\in \mathbb {Z}_p$ . Since X is a dense subset of $\mathbb {Z}_p$ , there is an integer $\hat {x}\in X$ such that
and $v_p(\hat {x}^{t_m+1}P_m(\hat {x})(\hat {x}-x_{m+1}))=s$ . So,
Taking
we get $\epsilon _m\in \mathbb {Q}$ , $\|\epsilon _m\|_p<p^{-m}$ and $f_{m+1}(\hat {x})=y_{m+1}$ . This completes the proof of the lemma.
Proof of Theorem 1.5
If in Lemma 3.1 we start with $f_0(x)=(x-x_0)+y_0$ , we get a sequence of polynomials as described in the beginning of this section. Furthermore, in each step, we have at least two options for the choice of $\delta _m$ and $\epsilon _m$ so we get uncountably many sequences. We will fix one of these sequences and prove that $f(x)=\lim _{m\rightarrow \infty }f_m(x)$ solves Theorem 1.5. Indeed,
where $\|c_i\|_p\leq p^{-j}$ for $t_{j-1}<i\leq t_j$ and $1\leq j\leq m$ (since $\max \{\|\delta _j\|_p,\|\epsilon _j\|_p\}\leq p^{-j}$ ). Therefore, $\lim _{i\rightarrow \infty }\|c_i\|_p=0$ and
is a p-adic analytic function on $\mathbb {Z}_p$ .
Moreover, $f(X)= Y$ . Indeed, we are assuming that $f_k(x_k)\in Y$ . By (3.3), ${P_m(x_k)=0}$ for all $m \geq k \geq 0$ and, consequently, $f_m(x_k)=f_{m-1}(x_k)=\cdots =f_k(x_k)$ . Thus, we conclude that
However, by hypothesis, given an integer $j\geq 0$ , there exists an integer $s\geq 0$ such that $f_j(x_s)=y_j$ . Similarly,
and we conclude $f(X)= Y$ .
It remains to prove that f is injective. For this, suppose that there are $a_1$ and $a_2$ in $\mathbb {Z}_p$ such that $f(a_1)=f(a_2)=b\in \mathbb {Z}_p$ and note that by (3.1), $c_1=1$ satisfies
Now, consider the function
Note that in the equation above, $c_1=1$ still satisfies the conditions in (3.4). Hence, $f(x)-b$ has at most one zero (by Strassman’s theorem), so we have $a_1=a_2$ .