[proofplan] We show that the Mahler coefficients $a_n(f) = \Delta^n f(0)$ tend to zero by establishing $\|\Delta^n f\| \to 0$. The key input is the contraction lemma: for every $f \in C(\mathbb{Z}_p, \mathbb{Q}_p)$ with $\|f\| = 1$, there exists $k$ such that $\|\Delta^{p^k} f\| \leq p^{-1}\|f\|$. Iterating this contraction along a subsequence shows $\|\Delta^n f\| \to 0$, using the monotonicity $\|\Delta^{n+1} f\| \leq \|\Delta^n f\|$. Injectivity follows from the density of $\mathbb{Z}_{\geq 0}$ in $\mathbb{Z}_p$. [/proofplan]

[step:Establish monotonicity of $\|\Delta^n f\|$ and the bound $|a_n(f)|_p \leq \|\Delta^n f\|$] The forward difference operator is defined by $\Delta f(x) = f(x+1) - f(x)$. Its iterates satisfy \begin{align*} \Delta^n f(x) = \sum_{j=0}^n (-1)^{n-j} \binom{n}{j} f(x+j), \end{align*} and the Mahler coefficients are $a_n(f) = \Delta^n f(0)$. **Monotonicity.** Since $\Delta^{n+1} f = \Delta(\Delta^n f)$, we have \begin{align*} \|\Delta^{n+1} f\| = \sup_{x \in \mathbb{Z}_p} |\Delta^n f(x+1) - \Delta^n f(x)|_p \leq \sup_{x \in \mathbb{Z}_p} \max(|\Delta^n f(x+1)|_p, |\Delta^n f(x)|_p) = \|\Delta^n f\|, \end{align*} using the strong triangle inequality. So the sequence $(\|\Delta^n f\|)_{n \geq 0}$ is non-increasing. **Coefficient bound.** Since $a_n(f) = \Delta^n f(0)$, we have $|a_n(f)|_p \leq \|\Delta^n f\|$. [/step]

[step:Prove the contraction lemma: $\|\Delta^{p^k} f\| \leq p^{-1}\|f\|$ for $k$ sufficiently large]We may assume $f \neq 0$. By scaling, assume $\|f\| = 1$. We claim there exists $k \geq 1$ such that $\|\Delta^{p^k} f\| \leq p^{-1}$. Using the explicit formula for $\Delta^{p^k}$: \begin{align*} \Delta^{p^k} f(x) = \sum_{i=0}^{p^k} (-1)^{p^k - i} \binom{p^k}{i} f(x+i). \end{align*} For $0 < i < p^k$, the binomial coefficient $\binom{p^k}{i}$ satisfies $v_p\!\left(\binom{p^k}{i}\right) \geq 1$, since $p^k$ divides the numerator $p^k!$ to a higher power of $p$ than the denominator $i!(p^k - i)!$ when $0 < i < p^k$. Therefore, modulo $p$, only the terms $i = 0$ and $i = p^k$ survive: \begin{align*} \Delta^{p^k} f(x) \equiv f(x + p^k) + (-1)^{p^k} f(x) \pmod{p\mathcal{O}_K}. \end{align*} Since $(-1)^{p^k} = -1$ for odd $p$ and $(-1)^{2^k} = 1 \equiv -1 \pmod{2}$ for $p = 2$ (as $1 = -1$ in $\mathbb{F}_2$), this simplifies to \begin{align*} |\Delta^{p^k} f(x)|_p \leq \max\!\left(p^{-1}, |f(x+p^k) - f(x)|_p\right). \end{align*} Since $\mathbb{Z}_p$ is compact and $f$ is continuous, $f$ is uniformly continuous. Choose $k$ large enough that $|x - y|_p \leq p^{-k}$ implies $|f(x) - f(y)|_p \leq p^{-1}$. Since $|p^k|_p = p^{-k}$, we have $|f(x+p^k) - f(x)|_p \leq p^{-1}$ for all $x \in \mathbb{Z}_p$. Therefore $\|\Delta^{p^k} f\| \leq p^{-1} = p^{-1}\|f\|$.[/step]

[guided]The contraction lemma is the heart of the argument. The key observation is that $\Delta^{p^k}$ is almost the same as the $p^k$-th iterate of the shift operator, modulo terms that are divisible by $p$. **Why $\binom{p^k}{i}$ is divisible by $p$ for $0 < i < p^k$.** Write $\binom{p^k}{i} = \frac{p^k}{i} \binom{p^k - 1}{i - 1}$. Since $0 < i < p^k$, the factor $p^k/i$ contributes at least $k - v_p(i) \geq 1$ to the $p$-adic valuation (because $v_p(i) < k$ when $i < p^k$). So $v_p(\binom{p^k}{i}) \geq 1$, and the corresponding terms in $\Delta^{p^k} f(x)$ have absolute value at most $p^{-1} \|f\| = p^{-1}$. **The surviving terms.** Only $i = 0$ (contributing $(-1)^{p^k} f(x)$) and $i = p^k$ (contributing $f(x + p^k)$) survive modulo $p$. Their difference $f(x + p^k) - f(x)$ can be made small by uniform continuity: $|x - (x + p^k)|_p = p^{-k}$, and for $k$ large enough, this forces $|f(x+p^k) - f(x)|_p \leq p^{-1}$. **Why does this give a contraction and not just a bound?** Because $\|f\| = 1$ and the bound $p^{-1}$ is strictly less than $1$. The point is that $\Delta^{p^k}$ "averages" $f$ over a $p$-adically small neighborhood, and the non-archimedean structure ensures the binomial coefficients carry factors of $p$ that suppress the intermediate terms.[/guided]

[step:Iterate the contraction to show $\|\Delta^n f\| \to 0$] Starting from $f$, apply the contraction lemma to find $k_1$ with $\|\Delta^{p^{k_1}} f\| \leq p^{-1}\|f\|$. Apply the contraction lemma to $g_1 := \Delta^{p^{k_1}} f$ (which is also continuous) to find $k_2$ with $\|\Delta^{p^{k_2}} g_1\| \leq p^{-1}\|g_1\| \leq p^{-2}\|f\|$. Since $\Delta^{p^{k_2}} g_1 = \Delta^{p^{k_1} + p^{k_2}} f$, we obtain \begin{align*} \|\Delta^{p^{k_1} + p^{k_2}} f\| \leq p^{-2} \|f\|. \end{align*} Continuing inductively, after $m$ applications we obtain indices $N_m = p^{k_1} + p^{k_2} + \cdots + p^{k_m}$ with $\|\Delta^{N_m} f\| \leq p^{-m}\|f\|$. Since $N_m \to \infty$ and the sequence $(\|\Delta^n f\|)$ is non-increasing, for any $n \geq N_m$ we have $\|\Delta^n f\| \leq \|\Delta^{N_m} f\| \leq p^{-m}\|f\|$. Therefore $\|\Delta^n f\| \to 0$. [/step]

[step:Conclude that the map $f \mapsto (a_n(f))$ is injective, norm-decreasing, and lands in $c_0$]**Coefficients tend to zero.** From $|a_n(f)|_p \leq \|\Delta^n f\| \to 0$, the sequence $(a_n(f))_{n \geq 0}$ belongs to $c_0$, the space of sequences in $\mathbb{Q}_p$ converging to $0$. **Norm-decreasing.** $\|(a_n(f))\|_{c_0} = \sup_n |a_n(f)|_p \leq \sup_n \|\Delta^n f\| = \|\Delta^0 f\| = \|f\|$, using the monotonicity of $\|\Delta^n f\|$. **Linearity.** The map $f \mapsto a_n(f) = \Delta^n f(0)$ is linear in $f$ since $\Delta$ is linear (it commutes with addition and scalar multiplication). **Injectivity.** Suppose $a_n(f) = 0$ for all $n \geq 0$. Then $a_0(f) = f(0) = 0$. Since $a_1(f) = \Delta f(0) = f(1) - f(0) = 0$, we get $f(1) = 0$. By induction: $\Delta^n f(0) = 0$ for all $n$ implies $f(m) = 0$ for all $m \in \mathbb{Z}_{\geq 0}$ (this follows from the explicit formula $\Delta^n f(0) = \sum_{j=0}^n (-1)^{n-j}\binom{n}{j} f(j)$, which expresses $f(n)$ as a linear combination of $f(0), \ldots, f(n-1)$ and $\Delta^n f(0)$). Since $\mathbb{Z}_{\geq 0}$ is dense in $\mathbb{Z}_p$ and $f$ is continuous, $f = 0$.[/step]

[guided]**Why is $\mathbb{Z}_{\geq 0}$ dense in $\mathbb{Z}_p$?** Every element of $\mathbb{Z}_p$ has a $p$-adic expansion $\sum_{i=0}^\infty a_i p^i$ with $a_i \in \{0, \ldots, p-1\}$. The partial sums $\sum_{i=0}^{n-1} a_i p^i$ are non-negative integers converging to the element in the $p$-adic metric. So $\mathbb{Z}_{\geq 0}$ is dense. **The induction for $f(m) = 0$.** We use the identity $f(n) = \sum_{k=0}^n \binom{n}{k} \Delta^k f(0) = \sum_{k=0}^n \binom{n}{k} a_k(f)$. If $a_k(f) = 0$ for all $k$, then $f(n) = 0$ for all $n \in \mathbb{Z}_{\geq 0}$. (This identity follows from the binomial inversion formula, which relates the values $f(0), f(1), \ldots, f(n)$ to the differences $\Delta^0 f(0), \Delta^1 f(0), \ldots, \Delta^n f(0)$.)[/guided]