Mahler’s theorem on continuous p-adic maps via generating functions

Noam D. Elkies

Fix a prime p. Let a_{0}, a_{1}, a_{2}, … be a sequence of p-adic numbers. By induction we may find b_{0}, b_{1}, b_{2}, … ∈ Q_{p} such that

a_{n} = ∑_{i=0}^{∞} \binom{n}{i}b_{i}.

(The sum is actually finite because \binom{n}{i} vanishes once i > n.) If b_{i}→0 as i→∞ then the sum

A(x) := ∑_{i=0}^{∞} \binom{x}{i}b_{i} (x ∈ Z_{p})

is a uniform limit of polynomials and thus converges to a continuous function from Z_{p} to Q_{p} such that A(n) = a_{n} for each n = 0, 1, 2, . . .. Mahler proved that conversely if the function n ↦ a_{n} extends to a continuous function from Z_{p} to Q_{p} then b_{i}→0 in Q_{p} as i→∞. We give a direct and simple proof of this using generating functions.

Let f(t), g(u) be the formal power series

f(t) = ∑_{n=0}^{∞} a_{n}t^{n}, g(u) = ∑_{i=0}^{∞} b_{i}u^{i}

in Q_{p}[[t]] and Q_{p}[[u]]. We have

f(t) = ∑_{n=0}^{∞} ∑_{i=0}^{n} \binom{n}{i}b_{i}t^{n} = ∑_{i=0}^{∞} b_{i} ∑_{n=i}^{∞} \binom{n}{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(\frac{t}{1 − t}).

If u = t/(1 − t) then t = u/(1 + u). We can thus solve for g:

g(u) = \frac{1}{1 + u} f(\frac{u}{1 + u}).

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} = ∑_{m=0}^{i}(−1)^{m−i}\binom{i}{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 : Z_{p}→Q_{p} that extends the function n ↦ a_{n}. Since Z_{p} is compact, the map is uniformly continuous.

Restricting A to {0, 1, 2, …} and unwinding the definition of uniform continuity in the p-adic metric, we conclude that for each integer k there exists an integer r such that a_{n} − a_{n′} has p-adic valuation at least k for all positive integers n, n′ such that n ≡ n′ mod 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} ∈ Z_{p} for all n.) Since k is arbitrary, this will verify that b_{i}→0.

Our assumption that v_{p}(a_{n} − a_{n′}) ≥ k when v_{p}(n − n′) ≥ r means that

f(t) = \frac{P (t)}{1 − t^{p^{r}}} + p^{k}α(t)

for some polynomial P ∈ Q_{p}[t] of degree less than p^{r} and some power series α ∈ Z_{p}[[t]]. (For instance, we may take P (t) = ∑_{n=0}^{p^{r}−1} a_{n}t^{n} and α(t) = ∑_{n=p^{r}}^{∞} 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(\frac{u}{1 + u}) = \frac{Q(u)}{(1 + u)^{p^{r}} − u^{p^{r}}} + \frac{p^{k}}{1 + u} α(\frac{u}{1 + u}),

where

Q(u) = (1 + u)^{p^{r}−1}P (\frac{u}{1 + u}) ∈ Q_{p}[u],

a polynomial of degree less than p^{r}, and the remainder p^{k} α(u/(1 + u)) / (1 + u) is again a power series in p^{k}Z_{p}[[u]].

Now the key point is that the denominator (1 + u)^{p^{r}} − u^{p^{r}} of g(u) is congruent to 1 mod p. (It is well known that (X + Y)^{p} ≡ X^{p} + Y^{p} mod p in Z[X, Y]; it follows by induction on r that (X + Y)^{p^{r}} ≡ X^{p^{r}} + Y^{p^{r}} mod p; now take X = u and Y = 1.) That is,

(1 + u)^{p^{r}} − u^{p^{r}} = 1 + pR(u)

for some polynomial R with coefficients in Z_{p} and degree p^{r} − 1. Therefore

\frac{Q(u)}{(1 + u)^{p^{r}} − u^{p^{r}}} = Q(u)(1 + pR(u) + p^{2}R^{2}(u) + p^{3}R^{3}(u) + ⋯).

Let s be a nonnegative integer such that p^{s}Q(u) ∈ Z_{p}[u]. Then

\frac{Q(u)}{(1 + u)^{p^{r}} − u^{p^{r}}} = Q(u) ∑_{j=0}^{k+s−1}(pR(u))^{j} + β(u)

where

β(u) = Q(u) ∑_{j=k+s}^{∞}(pR(u))^{j} ∈ p^{k}Z_{p}[[u]]

while Q(u) ∑_{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 p-adic valuation at least k. Thus v_{p}(b_{i}) ≥ k for all i > N_{k}. This establishes our claim and completes the proof of Mahler’s theorem.