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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.

Every function in continuous functions on the p-adic integers has a unique uniformly convergent expansion in the binomial polynomials, with $a_n\to 0$ and $a_n=({\Delta }^{n}f)(0)$. Conversely, any coefficient sequence tending to zero defines such a continuous function. To prove the expansion once $a_n\to 0$ is known, use $|(\genfrac{}{}{0ex}{}{x}{n}){|}_p\le 1$ for uniform convergence. Finite binomial inversion gives agreement with $f$ at each nonnegative integer, and density gives agreement everywhere. The coefficients recover successively from these integer values, proving uniqueness. Moreover $\Vert f{\Vert }_{\infty }={\sup }_n|a_n{|}_p$, because finite differences bound each coefficient by the norm and the expansion gives the reverse inequality.

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.