[proofplan] We prove the norm contraction $\|\Delta^{p^k} f\| \leq p^{-1}\|f\|$ by reducing to the case $\|f\| = 1$ and showing $\Delta^{p^k} f(x) \equiv 0 \pmod{p}$ for all $x \in \mathbb{Z}_p$ when $k$ is sufficiently large. The argument has two ingredients: the binomial expansion of $\Delta^{p^k}$ shows that modulo $p$, only the endpoints survive (since $p \mid \binom{p^k}{i}$ for $0 < i < p^k$), reducing the problem to $|f(x + p^k) - f(x)|_p \leq p^{-1}$; then uniform continuity of $f$ on the compact space $\mathbb{Z}_p$ provides the required $k$. [/proofplan]

[step:Reduce to the case $\|f\| = 1$] If $f = 0$, the conclusion holds for any $k \geq 1$. For $f \neq 0$, write $f = \|f\| \cdot g$ where $g = f/\|f\|$ satisfies $\|g\| = 1$. Since $\Delta^{p^k}$ is a linear operator, \begin{align*} \|\Delta^{p^k} f\| = \|f\| \cdot \|\Delta^{p^k} g\|, \end{align*} so it suffices to find $k \geq 1$ with $\|\Delta^{p^k} g\| \leq p^{-1}$. We therefore assume $\|f\| = 1$ and seek $k$ with $\|\Delta^{p^k} f\| \leq p^{-1}$, i.e., $|\Delta^{p^k} f(x)|_p \leq p^{-1}$ for all $x \in \mathbb{Z}_p$. [/step]

[step:Expand $\Delta^{p^k} f(x)$ and reduce modulo $p$]The $N$-th iterate of the forward difference operator satisfies \begin{align*} \Delta^N f(x) = \sum_{i=0}^{N} (-1)^{N-i} \binom{N}{i} f(x + i) \end{align*} for any $N \geq 0$. Taking $N = p^k$: \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*} [claim:$p \mid \binom{p^k}{i}$ for all $0 < i < p^k$] For $0 < i < p^k$, the binomial coefficient $\binom{p^k}{i} = \frac{p^k}{i} \binom{p^k - 1}{i - 1}$ is divisible by $p$. To see this, write $i = p^a m$ with $\gcd(m, p) = 1$ and $0 \leq a < k$. Then $v_p\left(\binom{p^k}{i}\right) = v_p(p^k!) - v_p(i!) - v_p((p^k - i)!)$. By Kummer's theorem, $v_p\left(\binom{p^k}{i}\right)$ equals the number of carries when adding $i$ and $p^k - i$ in base $p$. Since $0 < i < p^k$, the base-$p$ representations of $i$ and $p^k - i$ must produce at least one carry (the sum is $p^k = (1\underbrace{00\cdots0}_{k})_p$, which has a $1$ in a position where both $i$ and $p^k - i$ have $0$), so $v_p\left(\binom{p^k}{i}\right) \geq 1$. [/claim] [proof] By Kummer's theorem, $v_p\left(\binom{m+n}{m}\right)$ equals the number of carries when adding $m$ and $n$ in base $p$. Here $m = i$ and $n = p^k - i$, so $m + n = p^k$. In base $p$, the number $p^k$ is $1$ followed by $k$ zeros. If $i$ has base-$p$ representation $(d_{k-1}, \ldots, d_1, d_0)$ (with $0 \leq d_j \leq p-1$), then $p^k - i$ has digits that are the $(p-1)$-complements of $(d_{k-1}, \ldots, d_0)$ with a final carry propagation. Since $0 < i < p^k$, at least one digit $d_j$ is nonzero, which forces at least one carry. Hence $v_p\left(\binom{p^k}{i}\right) \geq 1$. [/proof] Since $\left|\binom{p^k}{i}\right|_p \leq p^{-1}$ for $0 < i < p^k$ and $|f(x+i)|_p \leq \|f\| = 1$, each interior term satisfies \begin{align*} \left|(-1)^{p^k-i}\binom{p^k}{i}f(x+i)\right|_p \leq p^{-1}. \end{align*} The boundary terms are $i = 0$ (contributing $(-1)^{p^k}f(x)$) and $i = p^k$ (contributing $f(x + p^k)$). Therefore by the ultrametric inequality: \begin{align*} \left|\Delta^{p^k}f(x) - \left((-1)^{p^k}f(x) + f(x+p^k)\right)\right|_p \leq p^{-1}. \end{align*} Since $(-1)^{p^k} = -1$ for odd $p$, and $(-1)^{2^k} = 1 = -1$ in $\mathbb{Z}_2/2\mathbb{Z}_2$ (i.e., $|1 - (-1)|_2 = |2|_2 = 2^{-1} \leq p^{-1}$ when $p = 2$), in both cases: \begin{align*} \left|\Delta^{p^k}f(x) - (f(x+p^k) - f(x))\right|_p \leq p^{-1}. \end{align*}[/step]

[guided]Let us be more careful with the sign $(-1)^{p^k}$. For $p$ odd, $p^k$ is odd, so $(-1)^{p^k} = -1$ and the boundary contribution is $-f(x) + f(x + p^k) = f(x+p^k) - f(x)$. For $p = 2$, $(-1)^{2^k} = 1$, so the boundary contribution is $f(x) + f(x + 2^k)$. But we want $f(x + 2^k) - f(x)$. The difference between the two is $f(x) + f(x+2^k) - (f(x+2^k) - f(x)) = 2f(x)$, and $|2f(x)|_2 = 2^{-1}|f(x)|_2 \leq 2^{-1}$. So modulo the error bound $p^{-1} = 2^{-1}$, the boundary contribution is still equivalent to $f(x + p^k) - f(x)$. More precisely: \begin{align*} |\Delta^{p^k}f(x)|_p \leq \max\left(|f(x+p^k) - f(x)|_p,\; p^{-1}\right). \end{align*} This holds for all primes $p$.[/guided]

[step:Use uniform continuity on $\mathbb{Z}_p$ to choose $k$] The function $f: \mathbb{Z}_p \to \mathbb{Q}_p$ is continuous on the compact space $\mathbb{Z}_p$, hence uniformly continuous. Choose $k \geq 1$ such that \begin{align*} |x - y|_p \leq p^{-k} \implies |f(x) - f(y)|_p \leq p^{-1}. \end{align*} For any $x \in \mathbb{Z}_p$, the element $x + p^k$ satisfies $|(x + p^k) - x|_p = |p^k|_p = p^{-k}$, so by the choice of $k$: \begin{align*} |f(x + p^k) - f(x)|_p \leq p^{-1}. \end{align*} Combining with the estimate from the previous step: \begin{align*} |\Delta^{p^k}f(x)|_p \leq \max\left(|f(x+p^k) - f(x)|_p,\; p^{-1}\right) \leq p^{-1} \end{align*} for all $x \in \mathbb{Z}_p$. Taking the supremum over $x$ gives $\|\Delta^{p^k}f\| \leq p^{-1} = p^{-1}\|f\|$. [/step]