Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 27 / 2

The Mahler theorem states that a function $f:{\mathbb{Z}}_p\to {\mathbb{Q}}_p$ is continuous if and only if it has a unique expansionand the expansion converges uniformly. Moreover $\Vert f{\Vert }_{\infty }={\sup }_n|c_n{|}_p$; in particular $f$ takes values in ${\mathbb{Z}}_p$ exactly when all $c_n$ belong to ${\mathbb{Z}}_p$. Here $\left(\genfrac{}{}{0ex}{}{x}{0}\right)=1$ and $\left(\genfrac{}{}{0ex}{}{x}{n}\right)=x(x-1)\cdots (x-n+1)/n!$. These binomial polynomials are continuous and integer-valued on ${\mathbb{Z}}_p$, since they are integer-valued on the dense nonnegative integers. The coefficient condition $c_n\to 0$ therefore makes the series uniformly convergent.

Evaluate at a nonnegative integer $m$. Terms with $n>m$ vanish, giving the finite triangular relation $f(m)={\sum }_{n=0}^m\left(\genfrac{}{}{0ex}{}{m}{n}\right)c_n$. Binomial inversion gives the Mahler coefficientsTo obtain the generating function, multiply the last finite identity by $T^n/n!$ and compare coefficients. Setting $n=j+k$ yieldsThe Mahler coefficient exponential generating function is an identity in ${\mathbb{Q}}_p[[T]]$: each coefficient is a finite sum. It does not assert convergence for every $p$-adic substitution for $T$.

Let $c_n$ be the Mahler coefficients of $f$. Define its discrete antiderivative byBecause $c_n\to 0$ 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 $\left(\genfrac{}{}{0ex}{}{x+1}{n+1}\right)-\left(\genfrac{}{}{0ex}{}{x}{n+1}\right)=\left(\genfrac{}{}{0ex}{}{x}{n}\right)$, 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 $m\ge 1$. This discrete antidifferentiation on the p-adic integers is the mechanism behind the next part.

Take an arbitrary $f\in C({\mathbb{Z}}_p,{\mathbb{Q}}_p)$. Compactness of ${\mathbb{Z}}_p$ bounds its field absolute value, so some $p^{m}f$ takes values in ${\mathbb{Z}}_p$. Apply part (ii) to obtain a continuous $h$ with $h(x+1)-h(x)=p^{m}f(x)$. Then $g=p^{-m}h$ belongs to the same continuous-function space and satisfies $f=g(\cdot +1)-g$.

Translation invariance by one alone now gives $L(f)=L(g(\cdot +1))-L(g)=0$. Since $f$ was arbitrary,This proves translation-invariant linear forms on p-adic continuous functions vanish without assuming that the linear form itself is continuous. A proof using only density of polynomials would not suffice for an arbitrary, potentially discontinuous linear form; surjectivity of the forward difference operator on the entire function space is what is needed.