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On continuous functions on the p-adic integers, define the forward difference operator and the Mahler coefficients byWriting $Sf(x)=f(x+1)$, the explicit iterate follows from ${\Delta }^n=(S-I{)}^n$. The ultrametric inequality and integral binomial coefficients imply $\Vert \Delta f{\Vert }_{\infty }\le \Vert f{\Vert }_{\infty }$ and $|a_n(f){|}_p\le \Vert f{\Vert }_{\infty }$, where the supremum norm is taken over ${\mathbb{Z}}_p$.

The Mahler theorem states that every such $f$ has a unique expansion with uniform convergenceConversely, every sequence in ${\mathbb{Q}}_p$ tending to zero yields a continuous function by this expansion. The binomial polynomials form an orthonormal expansion in the non-Archimedean sense: $\Vert f{\Vert }_{\infty }={\sup }_n|a_n(f){|}_p$.

Here is the requested proof under the permitted coefficient-decay assumption. For $n\ge 1$, the binomial polynomial $B_n(x)=x(x-1)\cdots (x-n+1)/n!$ is a continuous function on ${\mathbb{Z}}_p$. Its values on nonnegative integers are integral, and those integers are dense in the p-adic integers. Since ${\mathbb{Z}}_p$ is a closed set in ${\mathbb{Q}}_p$, $B_n({\mathbb{Z}}_p)\subseteq {\mathbb{Z}}_p$. Also $B_n(n)=1$, so $\Vert B_n{\Vert }_{\infty }=1$, including $B_0=1$.

If $a_n\to 0$, the ultrametric inequality gives the uniform tail boundCompleteness of ${\mathbb{Q}}_p$ gives a uniform limit $F$, and the uniform limit theorem makes $F$ continuous. At any nonnegative integer $m$, all $B_n(m)$ with $n>m$ vanish. Finite binomial inversion givessince the inner sum is $(1-1{)}^{m-j}$. Hence $F=f$ on a dense subset and therefore on all of ${\mathbb{Z}}_p$. The values at $m=0,1,2,\dots$ recover each coefficient recursively because $B_m(m)=1$; this proves uniqueness. The expansion bounds $\Vert f{\Vert }_{\infty }$ by $\sup |a_n{|}_p$, and the earlier coefficient inequality proves equality. This also proves the converse statement.

Although the problem allows us to assume decay, it can be established independently. By compactness and uniform continuity, approximate $f$ uniformly by $h$ constant on residue classes modulo $p^r$. On this finite-dimensional space, $S^{p^r}=I$ andThe matrix of ${\Delta }^{p^r}$ has entries divisible by $p$, so its operator norm is at most $p^{-1}$; consequently $\Vert {\Delta }^{j{p}^r+s}h{\Vert }_{\infty }\le p^{-j}\Vert h{\Vert }_{\infty }$ for $0\le s<p^r$. Thus $a_n(h)\to 0$. Since $|a_n(f)-a_n(h){|}_p\le \Vert f-h{\Vert }_{\infty }$, arbitrary uniform approximation proves automatic decay of Mahler coefficients.

For the last claim, construct the discrete antidifferentiation on the p-adic integersIts coefficient sequence is $0,a_0,a_1,\dots$, still tending to zero, so it is continuous. By Pascal's identity, $\Delta B_{n+1}=B_n$. The boundedness of $\Delta$ permits applying it to the uniform limit, giving $\Delta G=f$. For the stated linear map, invariance under translation by one now givesThus translation-invariant linear forms on p-adic continuous functions vanish:No continuity of $T$ has been assumed or used. In particular, one must not justify this by applying $T$ termwise to an infinite Mahler expansion; it is the existence of a continuous discrete antiderivative that makes the conclusion valid.