= Solution
On <continuous functions on the p-adic integers>, define the <forward difference operator> and the <Mahler coefficients> by
$$
(\Delta f)(x)=f(x+1)-f(x),\qquad
a_n(f)=(\Delta^nf)(0)=\sum_{j=0}^n(-1)^{n-j}\binom nj f(j).
$$
Writing $S f(x)=f(x+1)$, the explicit iterate follows from $\Delta^n=(S-I)^n$. The <ultrametric inequality> and integral <binomial coefficients> imply $\|\Delta f\|_\infty\leq\|f\|_\infty$ and $|a_n(f)|_p\leq\|f\|_\infty$, where the <supremum norm> is taken over $\mathbb Z_p$.
The <Mahler theorem> states that every such $f$ has a unique expansion with <uniform convergence>
$$
\boxed{f(x)=\sum_{n=0}^{\infty}a_n(f)\binom xn,\qquad a_n(f)\longrightarrow0.}
$$
Conversely, 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: $\|f\|_\infty=\sup_n|a_n(f)|_p$.
Here is the requested proof under the permitted coefficient-decay assumption. For $n\geq1$, 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 $\|B_n\|_\infty=1$, including $B_0=1$.
If $a_n\to0$, the <ultrametric inequality> gives the uniform tail bound
$$
\left\|\sum_{n=N}^M a_nB_n\right\|_\infty\leq\max_{N\leq n\leq M}|a_n|_p\longrightarrow0.
$$
Completeness 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 gives
$$
\sum_{n=0}^m\binom mn a_n
=\sum_{j=0}^m f(j)\binom mj\sum_{n=j}^m(-1)^{n-j}\binom{m-j}{n-j}
=f(m),
$$
since 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 $m=0,1,2,\ldots$ recover each coefficient recursively because $B_m(m)=1$; this proves uniqueness. The expansion bounds $\|f\|_\infty$ 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, $S^{p^r}=I$ and
$$
(S-I)^{p^r}\equiv S^{p^r}-I=0\pmod p.
$$
The matrix of $\Delta^{p^r}$ has entries divisible by $p$, so its operator norm is at most $p^{-1}$; consequently $\|\Delta^{jp^r+s}h\|_\infty\leq p^{-j}\|h\|_\infty$ for $0\leq s<p^r$. Thus $a_n(h)\to0$. Since $|a_n(f)-a_n(h)|_p\leq\|f-h\|_\infty$, arbitrary uniform approximation proves <automatic decay of Mahler coefficients>.
For the last claim, construct the <discrete antidifferentiation on the p-adic integers>
$$
G(x)=\sum_{n=0}^{\infty}a_n(f)\binom{x}{n+1}.
$$
Its coefficient sequence is $0,a_0,a_1,\ldots$, 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 gives
$$
T(f)=T(\Delta G)=T(G(\cdot+1))-T(G)=0.
$$
Thus <translation-invariant linear forms on p-adic continuous functions vanish>:
$$
\boxed{T=0.}
$$
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.
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