[proofplan] The proof proceeds in three stages. First, we show that for any null sequence $(a_n)$ in $\mathbb{Q}_p$, the Mahler series $f_a(x) = \sum a_n \binom{x}{n}$ converges uniformly on $\mathbb{Z}_p$ to a continuous function, with $\|f_a\| \leq \sup_n |a_n|_p$. Second, we compute that the Mahler coefficients of $f_a$ recover $(a_n)$, using the Pascal shift identity $\Delta \binom{x}{n} = \binom{x}{n-1}$. Third, we invoke the previously established results that $f \mapsto (a_n(f))$ is injective and norm-decreasing to promote the construction to an isometric bijection, using the [Formal Isometry Lemma](/theorems/???). [/proofplan] [step:Show Mahler series converge uniformly and define continuous functions] Let $(a_n)_{n \geq 0}$ be a sequence in $\mathbb{Q}_p$ with $a_n \to 0$. For each $x \in \mathbb{Z}_p$, the binomial coefficient $\binom{x}{n} = \frac{x(x-1)\cdots(x-n+1)}{n!}$ lies in $\mathbb{Z}_p$ (since it is a continuous $p$-adic integer-valued function agreeing with the usual binomial coefficient on $\mathbb{Z}_{\geq 0}$). In particular, $\left|\binom{x}{n}\right|_p \leq 1$ for all $x \in \mathbb{Z}_p$ and $n \geq 0$. The $n$-th term of the series satisfies $\left|a_n \binom{x}{n}\right|_p \leq |a_n|_p \to 0$, uniformly in $x$. By the non-archimedean convergence criterion (a series in a complete non-archimedean field converges if and only if its terms tend to zero), the series $\sum_{n=0}^\infty a_n \binom{x}{n}$ converges for each $x \in \mathbb{Z}_p$. The convergence is uniform in $x$ because the bound $|a_n|_p$ is independent of $x$. Define \begin{align*} f_a: \mathbb{Z}_p &\to \mathbb{Q}_p \\ x &\mapsto \sum_{n=0}^\infty a_n \binom{x}{n}. \end{align*} Since each $x \mapsto a_n \binom{x}{n}$ is continuous and the convergence is uniform, $f_a$ is continuous. The strong triangle inequality gives the norm bound: \begin{align*} \|f_a\| = \sup_{x \in \mathbb{Z}_p} |f_a(x)|_p \leq \sup_{n \geq 0} |a_n|_p = \|(a_n)\|_{c_0}. \end{align*} [guided] **Why is $\binom{x}{n} \in \mathbb{Z}_p$ for $x \in \mathbb{Z}_p$?** For $x = m \in \mathbb{Z}_{\geq 0}$, $\binom{m}{n}$ is a non-negative integer, hence in $\mathbb{Z}_p$. The map $x \mapsto \binom{x}{n} = \frac{x(x-1)\cdots(x-n+1)}{n!}$ is a polynomial of degree $n$ with rational coefficients, but it takes integer values on $\mathbb{Z}_{\geq 0}$. Since $\mathbb{Z}_{\geq 0}$ is dense in $\mathbb{Z}_p$ and $x \mapsto \binom{x}{n}$ is continuous (as a polynomial in $x$), the image of $\mathbb{Z}_p$ lies in $\overline{\mathbb{Z}} = \mathbb{Z}_p$ (the closure of $\mathbb{Z}$ in $\mathbb{Q}_p$). So $|\binom{x}{n}|_p \leq 1$ for all $x \in \mathbb{Z}_p$. **The construction $a \mapsto f_a$ defines a norm-decreasing map $G: c_0 \to C(\mathbb{Z}_p, \mathbb{Q}_p)$.** Indeed, $\|G(a)\| = \|f_a\| \leq \|a\|_{c_0}$. This map $G$ will serve as the right inverse to the Mahler coefficient map $F: f \mapsto (a_n(f))$. [/guided] [/step] [step:Compute the Mahler coefficients of $f_a$ using the Pascal shift identity] We verify that the Mahler coefficients of $f_a$ are $(a_n)$ itself. The key identity is the Pascal recurrence for binomial coefficients: for $n \geq 1$, \begin{align*} \binom{x+1}{n} - \binom{x}{n} = \binom{x}{n-1}. \end{align*} Applying the difference operator $\Delta$ to $f_a$: \begin{align*} \Delta f_a(x) = f_a(x+1) - f_a(x) = \sum_{n=0}^\infty a_n \left[\binom{x+1}{n} - \binom{x}{n}\right] = \sum_{n=1}^\infty a_n \binom{x}{n-1} = \sum_{n=0}^\infty a_{n+1} \binom{x}{n}. \end{align*} This shows $\Delta f_a = f_{a^{(1)}}$, where $a^{(1)} = (a_1, a_2, a_3, \ldots)$ is the shifted sequence. Iterating $k$ times: $\Delta^k f_a = f_{a^{(k)}}$ where $a^{(k)} = (a_k, a_{k+1}, \ldots)$. Evaluating at $x = 0$ with $\binom{0}{n} = \delta_{n,0}$ (the Kronecker delta): \begin{align*} a_n(f_a) = \Delta^n f_a(0) = f_{a^{(n)}}(0) = \sum_{m=0}^\infty a_{n+m} \binom{0}{m} = a_n. \end{align*} Therefore the Mahler coefficient map $F: f \mapsto (a_n(f))$ satisfies $F(G(a)) = a$, i.e., $F \circ G = \mathrm{id}_{c_0}$. [/step] [step:Apply the formal isometry lemma to conclude the isometric bijection and norm identity] We have established three maps and properties: - $F: C(\mathbb{Z}_p, \mathbb{Q}_p) \to c_0$ defined by $F(f) = (a_n(f))_{n \geq 0}$ is injective and norm-decreasing (by the [Mahler Coefficients Tend to Zero](/theorems/???) theorem). - $G: c_0 \to C(\mathbb{Z}_p, \mathbb{Q}_p)$ defined by $G((a_n)) = f_a$ is norm-decreasing (shown in the first step). - $F \circ G = \mathrm{id}_{c_0}$ (shown in the second step). By the [Formal Isometry Lemma](/theorems/???), $G \circ F = \mathrm{id}_{C(\mathbb{Z}_p, \mathbb{Q}_p)}$, and both $F$ and $G$ are isometric. In concrete terms: - **Existence and uniqueness of the expansion:** Every $f \in C(\mathbb{Z}_p, \mathbb{Q}_p)$ satisfies $f = G(F(f)) = f_{(a_n(f))}$, i.e., \begin{align*} f(x) = \sum_{n=0}^\infty a_n(f) \binom{x}{n} \quad \text{for all } x \in \mathbb{Z}_p, \end{align*} with $a_n(f) = \Delta^n f(0)$. Uniqueness follows from the injectivity of $F$: if $f = \sum b_n \binom{x}{n}$ for another null sequence $(b_n)$, then $(b_n) = F(f) = (a_n(f))$. - **Norm identity:** Since $F$ is isometric, \begin{align*} \|f\| = \|F(f)\|_{c_0} = \sup_{n \geq 0} |a_n(f)|_p = \max_{n \geq 0} |a_n(f)|_p, \end{align*} where the supremum is a maximum because $a_n(f) \to 0$. [guided] The formal isometry lemma is a clean abstract device that spares us from having to prove surjectivity and the norm equality directly. Let us verify why its hypotheses are met. **$F$ is injective:** Proved in the [Mahler Coefficients Tend to Zero](/theorems/???) theorem. The argument uses $f(0) = a_0(f) = 0$, then $f(1) = 0$ by induction, and density of $\mathbb{Z}_{\geq 0}$ in $\mathbb{Z}_p$. **$F$ is norm-decreasing:** $\|F(f)\|_{c_0} = \sup_n |a_n(f)|_p \leq \|f\|$, since $|a_n(f)|_p = |\Delta^n f(0)|_p \leq \|\Delta^n f\| \leq \|f\|$. **$G$ is norm-decreasing:** $\|G(a)\| = \|f_a\| \leq \sup_n |a_n|_p = \|a\|_{c_0}$. **$F \circ G = \mathrm{id}$:** Verified by the Pascal shift computation. The lemma then gives $G \circ F = \mathrm{id}$ (so every continuous function has a Mahler expansion) and $\|F(f)\| = \|f\|$ (the norm identity). The maximum in $\max_n |a_n|_p$ is attained because $a_n \to 0$, so $|a_n|_p$ achieves its supremum at some finite index. [/guided] [/step]