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The scaled backward finite difference uses the current and preceding nodal values to approximate a first derivative. If a function has two continuous derivatives, Taylor's theorem gives $D_h^{-}u(x)=u^{\prime }(x)-h{u}^{\prime \prime }(x)/2+o(h)$ as $h\to 0$. Composition with the scaled forward difference operator gives $D_h^{+}{D}_h^{-}u(x)=(u(x+h)-2u(x)+u(x-h))/h^2$, the second-order central difference for the second derivative.

For $n\ge 0$, define $B_0(X)=1$ and $B_n(X)=X(X-1)\cdots (X-n+1)/n!$. The binomial coefficient formula shows $B_n(m)\in \mathbb{Z}$ for every nonnegative integer $m$. Since these integers are dense in the p-adic integers, continuity of the polynomial implies $B_n({\mathbb{Z}}_p)\subseteq {\mathbb{Z}}_p$, so it is an integer-valued polynomial there and $\Vert B_n{\Vert }_{\infty }=1$. The Pascal's identity gives $\Delta B_{n+1}=B_n$ for the forward difference operator.

The continuous functions from the p-adic integers to the P-adic numbers form a Banach space over ${\mathbb{Q}}_p$ with supremum norm $\Vert f{\Vert }_{\infty }={\sup }_{x\in {\mathbb{Z}}_p}|f(x){|}_p$. Compactness of the domain bounds the norm, and uniform convergence in the complete codomain proves completeness. Translation and the forward difference operator $\Delta f(x)=f(x+1)-f(x)$ are bounded operators, with $\Vert \Delta f{\Vert }_{\infty }\le \Vert f{\Vert }_{\infty }$.

The Mahler coefficients tend to zero, so the displayed series defines a continuous function and has value zero at zero. The Pascal's identity and uniform convergence give $\Delta Jf=f$. Thus the forward difference operator is surjective on continuous functions on the p-adic integers. Its kernel consists of constants: period one implies agreement on the dense nonnegative integers, and continuity then implies constancy. This selects the unique discrete antiderivative vanishing at zero.

There is also a direct construction on locally constant functions. If $h$ has period $M=p^r$, put $a_j=h(j)$, $S={\sum }_{i=0}^{M-1}{a}_i$, and $P_j={\sum }_{i<j}{a}_i$. For $x=j+My$ with $0\le j<M$ and $y\in {\mathbb{Z}}_p$, setThis is a continuous function on each residue class. Increasing $j$ gives $Jh(x+1)-Jh(x)=a_j$; at $j=M-1$, the next point has residue zero and quotient $y+1$, giving the same identity. The ultrametric inequality gives $\Vert Jh{\Vert }_{\infty }\le \Vert h{\Vert }_{\infty }$, and $Jh(0)=0$. The construction is independent of the chosen period, since two normalized discrete antiderivatives agree on the nonnegative integers and then on the p-adic integers by continuity. It is linear on the locally constant functions, which form a dense subset in the supremum norm. Completeness therefore extends it to every continuous function, retaining $\Delta J=I$ and the norm bound. This proves surjectivity without first using the Mahler theorem.

For a continuous function $f:{\mathbb{Z}}_p\to {\mathbb{Q}}_p$, its nth Mahler coefficient isThis is the explicit iterate of the forward difference operator. These coefficients are exactly those of its Mahler expansion.

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.

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.

Take an arbitrary $f\in C({\mathbb{Z}}_p,{\mathbb{Q}}_p)$. Compactness of ${\mathbb{Z}}_p$ bounds its field absolute value, so some $p^{m}f$ takes values in ${\mathbb{Z}}_p$. Apply part (ii) to obtain a continuous $h$ with $h(x+1)-h(x)=p^{m}f(x)$. Then $g=p^{-m}h$ belongs to the same continuous-function space and satisfies $f=g(\cdot +1)-g$.

Translation invariance by one alone now gives $L(f)=L(g(\cdot +1))-L(g)=0$. Since $f$ was arbitrary,This proves translation-invariant linear forms on p-adic continuous functions vanish without assuming that the linear form itself is continuous. A proof using only density of polynomials would not suffice for an arbitrary, potentially discontinuous linear form; surjectivity of the forward difference operator on the entire function space is what is needed.

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.

Substitution of $y_n={\lambda }^n$ into the linear recurrence relation gives ${\lambda }^{n}p(\lambda )=0$. Since $\lambda \ne 0$, division by ${\lambda }^n$ proves the sequence is a solution exactly when $p(\lambda )=0$.

For a nonzero triple root of a polynomial $\mu$, the repeated characteristic root of a linear recurrence rule gives the three independent solutions ${\mu }^n,n{\mu }^n,n^2{\mu }^n$. One way to see the polynomial factors is to write $y_n={\mu }^n{z}_n$: the recurrence becomes the third forward difference operator applied to $z_n$ equal to zero, so $z_n$ is a quadratic polynomial. ThusThese three constants give arbitrary initial values $y_0,y_1,y_2$, hence the full third-order solution. If $\mu =0$, the recurrence instead reduces to $y_{n+3}=0$: the first three values are arbitrary and every subsequent value is zero. The nonzero-root formula must not be used to discard those initial values.

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.

Define the binomial coefficient $\left(\genfrac{}{}{0ex}{}{n}{r}\right)$ as the number of $r$-element subsets of an $n$-element set, equivalently as $n!/[r!(n-r)!]$. Partitioning the $r$-element subsets of $\{1,\dots ,n+1\}$ according to whether they contain $n+1$ proves Pascal's identityThus the forward difference operator satisfies

Let $f_i=f(ih)$ and $g_i=g(ih)$, and define the scaled forward difference operator, backward difference operator and second-order central difference byThe two finite difference methods are the nonlinear systemswith the prescribed Dirichlet boundary conditions $U_0=V_0=A$ and $U_N=V_N=B$. These equations determine the interior nodal values.

For clarity distinguish the interior and edge inner products and their norms:The bracketed norm includes the last edge $i=N$. Let $W=U-V$, so $W_0=W_N=0$, and put $q_i=D_x^{-}{W}_i$ for $1\le i\le N$. Direct telescoping gives discrete summation by parts:Subtract the two nonlinear systems, multiply each equation by $h{W}_i$ and sum. The resulting discrete energy identity isThe nonlinear summands are nonnegative by the factorization in (a). Using Young's inequality for products at every node givesThis also proves uniqueness of each discrete solution. Existence can be seen by minimizing, with the endpoints fixed, the strictly convex energy functionalIt is a coercive function of its finitely many interior values, so it attains a minimum. Its first derivatives are $h(-D_x^{+}{D}_x^{-}{W}_i+W_i+W_i^3-f_i)$; thus its minimizer solves the scheme. The same argument applies to $g$.

A nonzero root $\mu$ of multiplicity $m$ supplies $m$ independent solutions $n^j{\mu }^n$ to a constant-coefficient linear recurrence relation. Substitution $y_n={\mu }^n{z}_n$ reduces its repeated factor to the $m$th forward difference operator. A zero root needs separate initial-data treatment: for characteristic polynomial ${\lambda }^m$, the recurrence is $y_{n+m}=0$, leaving the first $m$ values arbitrary.