= Solution
Let $c_n$ be the <Mahler coefficients> of $f$. Define its <discrete antiderivative> by
$$
\boxed{g(x)=\sum_{n\ge0}c_n\binom{x}{n+1}.}
$$
Because $c_n\to0$ and the <binomial polynomials> have <field absolute value> at most one on $\mathbb Z_p$, this series converges uniformly to a continuous $\mathbb Z_p$-valued function. Every summand vanishes at zero. <Pascal's identity> gives $\binom{x+1}{n+1}-\binom{x}{n+1}=\binom xn$, so taking differences through the uniformly convergent series proves $g(x+1)-g(x)=f(x)$. For nonnegative integers, induction from $g(0)=0$ gives precisely $g(m)=\sum_{j=0}^{m-1}f(j)$. Thus this is the required continuous extension, unique because the nonnegative integers are dense in $\mathbb Z_p$.
The <Mahler expansion> of $g$ has coefficient zero in degree zero and coefficient $c_{m-1}$ in degree $m\ge1$. This <discrete antidifferentiation on the p-adic integers> is the mechanism behind the next part.
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