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Let $I$ be a complete discrete valuation ring with a continuous ${\mathbb{Z}}_p$-algebra structure. Every continuous function on ${\mathbb{Z}}_p$ with values in $I$ has a unique uniformly convergent expansion in the binomial polynomials, with $a_n={\Delta }^{n}f(0)$. This is the coefficient-ring version of the Mahler theorem. To see that $a_n\to 0$, approximate the function modulo any power of the maximal ideal by a function of period $p^k$. In the expansion of $(E-1{)}^{p^k}$, all intermediate binomial coefficients are divisible by $p$, while $(E^{p^k}-1)$ kills the periodic function. Iterating makes its differences vanish modulo the prescribed ideal power. Finite differences do not increase the uniform norm, so the same conclusion holds for the original function. Binomial inversion gives agreement at the nonnegative integers, and their density in the p-adic integers gives the expansion everywhere. This gives the dual description of p-adic measures by arbitrary bounded sequences of binomial moments, and hence the Amice transform.

The locally constant function equal to one on $2{\mathbb{Z}}_2$ and minus one on its other coset has Mahler coefficients $a_n=(-2{)}^n$: at nonnegative integers its values are $(-1{)}^j$, so the forward difference operator multiplies it by $-2$. Since $|a_n{|}_2=2^{-n}$ and the binomial polynomials have absolute value at most one on the 2-adic integers, the series converges uniformly. The binomial theorem verifies its values on the nonnegative integers, and continuity extends the identity to the entire domain.

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.

The binomial polynomial $P_r$ equals $1$ when $r=0$. For $n\ge 0$, it has value $\left(\genfrac{}{}{0ex}{}{n}{r}\right)$ if $r\le n$ and value zero otherwise. These are integers. If $n=-k<0$, thenHence each $P_r$ is an integer-valued polynomial on all of $\mathbb{Z}$, including negative arguments.

For $r\ge 1$, factor the common product in the two polynomials:For $r=1$ the common product is empty and equals $1$, so this argument also covers that boundary case.

The polynomial $P_r$ has polynomial degree $r$ and leading coefficient $1/r!$. For a polynomial $F$ of polynomial degree $d$ with leading coefficient $a_d\in \mathbb{Q}$, subtract $d!a_d{P}_d$; the remainder has smaller polynomial degree. Repeating gives the required expansion with rational numbers $c_r$. For uniqueness, suppose two expansions differ and let $s$ be the largest index with different coefficients. Their difference then has a nonzero term of polynomial degree $s$, with leading coefficient equal to that coefficient difference divided by $s!$. It cannot be the zero polynomial. Thus the expansion is unique.

Replacing $X$ by $X+1$ in the earlier identity gives $\Delta P_r(X)=P_{r-1}(X)$ for $r\ge 1$, while $\Delta P_0=0$. By linearity of the forward difference operator,An empty sum is zero when $d=0$. If $H=F-G$ were nonconstant, with polynomial degree $s\ge 1$ and leading coefficient $a_s$, then $\Delta H$ would have polynomial degree $s-1$ and leading coefficient $s{a}_s\ne 0$. Therefore $\Delta F=\Delta G$ forces $H$ to be constant.

There is a direct way to recover every coefficient. Since $P_0(0)=1$ and $P_j(0)=0$ for $j\ge 1$, repeated application of the forward difference operator givesThe last equality follows by expanding $(T-I{)}^r$, where $TF(X)=F(X+1)$; the commuting shifts obey the binomial theorem. If $F$ is an integer-valued polynomial, every summand on the right is an integer, proving that all $c_r(F)$ are integers. Together with the first paragraph this gives both directions of the Newton series for an integer-valued polynomial sequence description.

Let $c_n$ be the Mahler coefficients of $f$. Define its discrete antiderivative byBecause $c_n\to 0$ and the binomial polynomials have field absolute value at most one on ${\mathbb{Z}}_p$, this series converges uniformly to a continuous ${\mathbb{Z}}_p$-valued function. Every summand vanishes at zero. Pascal's identity gives $\left(\genfrac{}{}{0ex}{}{x+1}{n+1}\right)-\left(\genfrac{}{}{0ex}{}{x}{n+1}\right)=\left(\genfrac{}{}{0ex}{}{x}{n}\right)$, so taking differences through the uniformly convergent series proves $g(x+1)-g(x)=f(x)$. For nonnegative integers, induction from $g(0)=0$ gives precisely $g(m)={\sum }_{j=0}^{m-1}f(j)$. Thus this is the required continuous extension, unique because the nonnegative integers are dense in ${\mathbb{Z}}_p$.

The Mahler expansion of $g$ has coefficient zero in degree zero and coefficient $c_{m-1}$ in degree $m\ge 1$. This discrete antidifferentiation on the p-adic integers is the mechanism behind the next part.

The Mahler theorem states that a function $f:{\mathbb{Z}}_p\to {\mathbb{Q}}_p$ is continuous if and only if it has a unique expansionand the expansion converges uniformly. Moreover $\Vert f{\Vert }_{\infty }={\sup }_n|c_n{|}_p$; in particular $f$ takes values in ${\mathbb{Z}}_p$ exactly when all $c_n$ belong to ${\mathbb{Z}}_p$. Here $\left(\genfrac{}{}{0ex}{}{x}{0}\right)=1$ and $\left(\genfrac{}{}{0ex}{}{x}{n}\right)=x(x-1)\cdots (x-n+1)/n!$. These binomial polynomials are continuous and integer-valued on ${\mathbb{Z}}_p$, since they are integer-valued on the dense nonnegative integers. The coefficient condition $c_n\to 0$ therefore makes the series uniformly convergent.

Evaluate at a nonnegative integer $m$. Terms with $n>m$ vanish, giving the finite triangular relation $f(m)={\sum }_{n=0}^m\left(\genfrac{}{}{0ex}{}{m}{n}\right)c_n$. Binomial inversion gives the Mahler coefficientsTo obtain the generating function, multiply the last finite identity by $T^n/n!$ and compare coefficients. Setting $n=j+k$ yieldsThe Mahler coefficient exponential generating function is an identity in ${\mathbb{Q}}_p[[T]]$: each coefficient is a finite sum. It does not assert convergence for every $p$-adic substitution for $T$.

The two cosets of $2{\mathbb{Z}}_2$ are clopen sets in the 2-adic integers. The function is constant on each, so it is a locally constant function and therefore continuous. Its values on the nonnegative integers are $f(j)=(-1{)}^j$.

The Mahler coefficient at index $n$ is the nth forward difference operator applied at zero. HereThus the Mahler expansion of the parity function isTo verify the expansion directly, each binomial polynomial $\left(\genfrac{}{}{0ex}{}{x}{n}\right)$ takes values in ${\mathbb{Z}}_2$: it does so at nonnegative integers, which form a dense subset, and polynomial evaluation is continuous into ${\mathbb{Q}}_2$. Hence the nth summand has 2-adic absolute value at most $2^{-n}$ throughout the domain. The ultrametric inequality proves uniform convergence, and the sum is continuous. At each nonnegative integer $m$, the series becomes the finite binomial theorem identity $(1-2{)}^m=(-1{)}^m$. Equality on a dense subset and continuity prove equality everywhere. Finally, the coefficients in any such expansion are determined successively by its values at $0,1,2,\dots$, since $\left(\genfrac{}{}{0ex}{}{m}{n}\right)=0$ for $n>m$ and $\left(\genfrac{}{}{0ex}{}{m}{m}\right)=1$. This also verifies uniqueness without needing the general Mahler theorem.

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.

Fix a non-Archimedean local field $K$, its valuation ring $R={\mathcal{O}}_K$, a uniformizer $\pi$, and a residue field with cardinality $q$. The construction below applies both in mixed and equal characteristic. A one-dimensional commutative formal group law is $F(X,Y)\in R[[X,Y]]$ with identity, associativity and commutativity, and $F(X,Y)=X+Y+$ higher-degree terms. It has a unique formal inverse. Its morphisms are series $h(T)\in TR[[T]]$ with $h(F(X,Y))=G(h(X),h(Y))$; a unit linear coefficient makes such a morphism invertible by recursive coefficient comparison. Evaluating its integral series on the maximal ideal of a finite extension of $K$ is legitimate because the series converge there.

Choose a Lubin–Tate series $f(T)$ satisfyingA useful choice is $f(T)=\pi T+T^q$. The Lubin–Tate functional equation lemma constructs a unique series with any prescribed linear form $\ell$ and satisfyingHere is its coefficient argument. If terms below degree $d$ have been found, the unknown homogeneous degree-$d$ term appears with multiplier $\pi -{\pi }^d$. The already known error has every coefficient divisible by $\pi$: after reduction, the equation becomes $H(X{)}^q=H(X_1^q,\dots ,X_r^q)$, valid because the residue coefficients lie in ${\mathbb{F}}_q$. Since $\pi -{\pi }^d$ is $\pi$ times a unit for $d\ge 2$, the next term is uniquely determined and integral. Recursion proves existence and uniqueness.

Use $\ell =X+Y$ to construct $F_f$, and $\ell =aT$ to construct $[a{]}_f$ for every $a\in R$. Uniqueness gives $F_f(X,0)=X$ and $F_f(0,Y)=Y$. Comparing $[a{]}_f(F_f(X,Y))$ with $F_f([a{]}_f(X),[a{]}_f(Y))$, which satisfy the same functional equation and have the same linear form, proves that $[a{]}_f$ is an endomorphism of the law. The same uniqueness applied to three variables proves associativity: $F_f(F_f(X,Y),Z)$ and $F_f(X,F_f(Y,Z))$ have the same linear form and functional equation. It also proves commutativity and the identitiesThus this Lubin–Tate formal group is equipped with an action of the whole ring $R$, not only its integers. For a unit $a$, the inverse of $[a{]}_f$ is $[a^{-1}{]}_f$. The same recursion, with two series for the fixed $\pi$, gives a unique strict Lubin–Tate change of series. Its integral series and inverse converge on torsion points, so it identifies their torsion fields. We may compute with $f(T)=\pi T+T^q$ without losing the general construction for that uniformizer.

Write $f_0(T)=T$ and $f_n=f^{\circ n}$. The Lubin–Tate torsion of level $n$ isFor $n\ge 1$, its primitive points are the roots ofThis is a monic polynomial of degree $D_n=(q-1)q^{n-1}$. Modulo $\pi$ it is $T^{D_n}$, and its constant coefficient is exactly $\pi$. Thus the Eisenstein layers of Lubin–Tate torsion are Eisenstein polynomials. A root ${\lambda }_n$ generates a totally ramified extension of degree $D_n$, and has $v_K({\lambda }_n)=1/D_n$ in the uniquely extended valuation. Also $f_{n-1}({\lambda }_n{)}^{q-1}=-\pi \ne 0$, so it has exact level $n$.

The mapis well-defined and injective. If $a={\pi }^{r}u$ with $r<n$ and $u$ a unit, then $[a]{\lambda }_n=[u]f_r({\lambda }_n)\ne 0$, because $[u]$ is invertible and ${\lambda }_n$ is primitive. Conversely every multiple of ${\pi }^n$ kills it. The image therefore contains $q^n$ distinct roots of the degree-$q^n$ polynomial $f_n$, so it is all of ${\Lambda }_n$. This proves, without merely assuming the torsion cardinality,Its primitive generators are exactly $[a]{\lambda }_n$ with $a\in (R/{\pi }^{n}R{)}^{\times }$.

Define the Lubin–Tate extension $K_{\pi ,n}=K({\Lambda }_n)$. Every integral endomorphism series evaluated at ${\lambda }_n$ converges in the complete field $K({\lambda }_n)$, so all torsion points already lie there. HenceIt is the splitting field of $f_n$, which has $q^n$ distinct roots by the module calculation. Thus it is Galois. Every K-automorphism preserves the uniquely extended absolute value on a field, hence is continuous and commutes with limits of the integral endomorphism series. Its automorphisms therefore commute with the endomorphism action and send a primitive generator to a primitive generator. There is a unique unit class $a_{\sigma }$ such that $\sigma ({\lambda }_n)=[a_{\sigma }]{\lambda }_n$. The Lubin–Tate Galois action on primitive torsion gives an injective homomorphismBoth sides have order $D_n=(q-1)q^{n-1}$, so it is an isomorphism. In particular every finite layer is abelian and totally ramified.

For an example, take $K={\mathbb{Q}}_p$, $\pi =p$, and the equally admissible series $f(T)=(1+T{)}^p-1$. Its formal group law is the formal multiplicative group $F(X,Y)=X+Y+XY$, with $[a](T)=(1+T{)}^a-1$ for $a\in {\mathbb{Z}}_p$. The binomial polynomials make its coefficients integral. Its level-$n$ torsion points are $\zeta -1$ with ${\zeta }^{p^n}=1$. Thus the Lubin–Tate cyclotomic example gives $K_{p,n}={\mathbb{Q}}_p({\mu }_{p^n})$ and Galois action $\zeta \mapsto {\zeta }^a$. The order formula includes $p=2,n=1$, when the first field and its Galois group are trivial over ${\mathbb{Q}}_2$.

Finally choose primitive points compatibly, $[\pi ]{\lambda }_{n+1}={\lambda }_n$. They exist by taking a root of $f(T)-{\lambda }_n$ in the algebraic closure; its roots have positive valuation and exact level $n+1$. Then $K_{\pi ,n}\subseteq K_{\pi ,n+1}$, and restriction of Galois automorphisms corresponds to reduction of unit classes. The Lubin–Tate tower $K_{\pi }={\bigcup }_n{K}_{\pi ,n}$ consequently has the topological Galois groupThe final isomorphism uses completeness of the discrete valuation ring $R$. This computes the finite-layer and infinite-tower Galois groups directly from the formal action and the primitive Eisenstein polynomials.