Discrete antidifferentiation on the p-adic integers ($Jf(x)={\sum }_{n\ge 0}{a}_n(f)\left(\genfrac{}{}{0ex}{}{x}{n+1}\right)$)

The Mahler coefficients tend to zero, so the displayed series defines a continuous function and has value zero at zero. The Pascal's identity and uniform convergence give $\Delta Jf=f$. Thus the forward difference operator is surjective on continuous functions on the p-adic integers. Its kernel consists of constants: period one implies agreement on the dense nonnegative integers, and continuity then implies constancy. This selects the unique discrete antiderivative vanishing at zero.

There is also a direct construction on locally constant functions. If $h$ has period $M=p^r$, put $a_j=h(j)$, $S={\sum }_{i=0}^{M-1}{a}_i$, and $P_j={\sum }_{i<j}{a}_i$. For $x=j+My$ with $0\le j<M$ and $y\in {\mathbb{Z}}_p$, setThis is a continuous function on each residue class. Increasing $j$ gives $Jh(x+1)-Jh(x)=a_j$; at $j=M-1$, the next point has residue zero and quotient $y+1$, giving the same identity. The ultrametric inequality gives $\Vert Jh{\Vert }_{\infty }\le \Vert h{\Vert }_{\infty }$, and $Jh(0)=0$. The construction is independent of the chosen period, since two normalized discrete antiderivatives agree on the nonnegative integers and then on the p-adic integers by continuity. It is linear on the locally constant functions, which form a dense subset in the supremum norm. Completeness therefore extends it to every continuous function, retaining $\Delta J=I$ and the norm bound. This proves surjectivity without first using the Mahler theorem.

Let $T:\mathcal{C}({\mathbb{Z}}_p,{\mathbb{Q}}_p)\to {\mathbb{Q}}_p$ be linear and invariant under translation by one. For any $f$, choose its continuous discrete antiderivative $Jf$. Then $T(f)=T(\Delta Jf)=T(Jf(\cdot +1))-T(Jf)=0$. No continuity or boundedness of $T$ is assumed. In particular this applies to a form invariant under every translation by a p-adic integer.