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\begin{large}
\centerline{
Mahler's theorem on continuous $p$-adic maps
via generating functions
}
\end{large}
\centerline{
Noam D. Elkies
}

Fix a prime $p$.
Let $a_0, a_1, a_2,\ldots$ be a sequence of \hbox{$p$-adic} numbers.
By induction we may find $b_0, b_1, b_2, \ldots \in \Qp$ such that
$$
a_n = \sum_{i=0}^\infty {n\choose i} b_i.
$$
(The sum is actually finite because $n\choose i$ vanishes once $i>n$.)
If $b_i \ra 0$ as $i\ra\infty$ then the sum
$$
A(x) := \sum_{i=0}^\infty {x\choose i} b_i
\quad(x\in\Zp)
$$
is a uniform limit of polynomials
and thus converges to a continuous function from~$\Zp$ to $\Qp$
such that $A(n) = a_n$ for each $n=0,1,2,\ldots$.
Mahler proved that conversely if the function $n \mapsto a_n$
extends to a continuous function from~$\Zp$ to $\Qp$
then $b_i \ra 0$ in~$\Qp$ as $i\ra\infty$.
We give a direct and simple proof of this using generating functions.

Let $f(t)$, $g(u)$ be the formal power series
$$
f(t) = \sum_{n=0}^\infty a_n t^n,
\quad
g(u) = \sum_{i=0}^\infty b_i u^i
$$
in $\Qp[[t]]$ and $\Qp[[u]]$.  We have
$$
f(t) = \sum_{n=0}^\infty \sum_{i=0}^n {n\choose i} b_i t^n
= \sum_{i=0}^\infty b_i \sum_{n=i}^\infty {n\choose i} t^n.
$$
The inner sum is the Taylor series for $t^n/(1-t)^{n+1}$.  Hence
$$
f(t) = \frac{1}{1-t} \, g\Bigl(\frac{t}{1-t}\Bigr).
$$
If $u=t/(1-t)$ then $t=u/(1+u)$.  We can thus solve for $g$:
$$
g(u) = \frac{1}{1+u} \, f\Bigl(\frac{u}{1+u}\Bigr).
$$
We may now obtain the $b_i$ explicitly in terms of the $a_n$
by expanding each term $u^n/(1+u)^{n+1}$ in the sum defining
$(1+u)^{-1}f(u/(1+u))$:
$$
b_i = \sum_{m=0}^i (-1)^{m-i} {i\choose m} a_i.
$$
But for our purposes it is more convenient to use
the formula for the generating function $g(u)$.

Suppose now that there exists a continuous map $A: \Zp \ra \Qp$
that extends the function $n \mapsto a_n$.  Since $\Zp$ is compact,
the map is uniformly continuous.
Restricting $A$\/ to $\{0,1,2,\ldots\}$ and unwinding the definition
of uniform continuity in the \hbox{$p$-adic} metric, we conclude that
for each integer $k$\/ there exists an integer $r$ such that
$a_n - a_{n'}$ has \hbox{$p$-adic} valuation at least~$k$\/
for all positive integers $n,n'$ such that $n\equiv n' \bmod p^r$.
We shall show that this implies the existence of an integer $N(k)$
such that $b_i$ is a multiple of $p^k$ for all $i > N(k)$.
(Specifically, we shall obtain $N(k) = (s+k+1) p^r$
where $s$ is a nonnegative integer such that
$p^s a_n \in \Z_p$ for all~$n$.) 
Since $k$ is arbitrary, this will verify that $b_i \ra 0$.

Our assumption that
$v_p(a_n - a_{n'}) \geq k$ when $v_p(n-n') \geq r$ means that
$$
f(t) = \frac{P(t)}{1-t^{p^r}\9} + p^k \alpha(t)
$$
for some polynomial $P \in \Qp[t]$ of degree less than $p^r$
and some power series $\alpha \in \Zp[[t]]$.  (For instance,
we may take $P(t) = \sum_{n=0}^{p^r-1} a_n t^n$ and
$\alpha(t) = \sum_{n=p^r}^\infty p^{-k} (a_n-a_{n'}) t^n$
where $n'$ is the remainder of~$n$ when dividied by~$p^r$.)
Then
$$
g(u) = \frac{1}{1+u} \, f\Bigl(\frac{u}{1+u}\Bigr)
= \frac{Q(u)}{(1+u)^{p^r}\9 - u^{p^r}\9}
+ \frac{p^k}{1+u} \, \alpha\Bigl(\frac{u}{1+u}\Bigr),
$$
where
$$
Q(u) = (1+u)^{p^r-1}\9 P\Bigl(\frac{u}{1+u}\Bigr) \in \Qp[u],
$$
a polynomial of degree less than $p^r$, and the remainder
$p^k \, \alpha(u/(1+u)) \, / \, (1+u)$
is again a power series in $p^k \Zp[[u]]$.

Now the key point is that the denominator $(1+u)^{p^r}\9 - u^{p^r}\9$
of $g(u)$ is congruent to~$1$ mod~$p$.  (It is well known that
$(X+Y)^p\9 \equiv X^p\9 + Y^p\9 \bmod p$ in $\Z[X,Y]$;
it follows by induction on~$r$ that
$(X+Y)^{p^r}\9 \equiv X^{p^r}\9 + Y^{p^r}\9 \bmod p$;
now take $X=u$ and $Y=1$.)  That is,
$$
(1+u)^{p^r}\9 - u^{p^r}\9 = 1 + p R(u)
$$
for some polynomial $R$\/ with coefficients in $\Zp$ and degree~$p^r-1$.
Therefore
$$
\frac{Q(u)}{(1+u)^{p^r}\9 - u^{p^r}\9}
= Q(u) \bigl( 1 + pR(u) + p^2 R^2(u) + p^3 R^3(u) + \cdots \bigr).
$$
Let $s$ be a nonnegative integer such that $p^s Q(u) \in \Zp[u]$.  Then
$$
\frac{Q(u)}{(1+u)^{p^r}\9 - u^{p^r}\9}
= Q(u) \sum_{j=0}^{k+s-1} (pR(u))^j + \beta(u)
$$
where 
$$
\beta(u) = Q(u) \sum_{j=k+s}^\infty (pR(u))^j \in p^k \Zp[[u]]
$$
while $Q(u) \sum_{j=0}^{k+s-1} (pR(u))^j$ is a polynomial,
say of degree~$N_k$.
We have shown that $g(u)$ differs from this polynomial
by a power series all of whose coefficients have
\hbox{$p$-adic} valuation at least~$k$.
Thus $v_p(b_i) \geq k$ for all $i>N_k$.
This establishes our claim and completes the proof of Mahler's theorem.
\end{document}
