Proof of Lattice Sums along Cubes: Linearity, Nonnegative Families, Absolute Summability, Comparison and Finitely Supported Families

Each result cited below is universally quantified over the data in its own statement.

Conventions. For a family $c:\mathbb{Z}^{n}\to\mathbb{R}$ and $N\in\mathbb{N}$, $S_{N}(c)=\sum_{k\in\Gamma_{N}}c(k)$ is the cube sum of Cube Sums of Families on the Integer Lattice §cube-sums; the cube $\Gamma_{N}$ is a nonempty finite set by The Cubes of the Integer Lattice are Finite, Nested, Exhaust the Lattice and Have (2N+1)^n Points §finite, and all sums over finite index sets are those of Sum over a Finite Index Set, a sum over a subset of the domain being the sum of the restriction. By Cube Sums of Families on the Integer Lattice §lattice-sum, $c$ is cube-summable exactly when $(S_{N}(c))_{N\in\mathbb{N}}$ converges in the sense of Limit of a Sequence of Real Numbers, and then $\sum_{k\in\mathbb{Z}^{n}}c(k)$ is its limit, which is unique by claim 1 of Uniqueness of Limits and Boundedness of Convergent Real Sequences. The order of $\mathbb{R}$ is a total order, reflexive, antisymmetric and transitive by clauses 1, 2 and 3 of Total Order on a Set; $x<y$ means $x\le y$ and $x\ne y$. Upper bounds and suprema are as in Upper Bound and Least Upper Bound, and a nonempty set bounded above has a supremum by The Real Numbers: Standing Notation and Background §bounds.

Preliminary (A): the modulus of a finite real sum. Let $F$ be a nonempty finite set and $f:F\to\mathbb{R}$. By claim 3 of Properties of the Absolute Value in an Ordered Field, $-|f(x)|\le f(x)\le|f(x)|$ for every $x\in F$, so by Real Sums over a Finite Index Set: Comparison, Nonnegativity, Monotonicity, Term Bounds, Absolute Values, Counting and Limits §comparison, applied twice,

$$\sum_{x\in F}\bigl(-|f(x)|\bigr)\le\sum_{x\in F}f(x)\le\sum_{x\in F}|f(x)|.$$

Since $-|f(x)|=(-1)|f(x)|$ by the field axioms of Field, homogeneity (claim 4 of Properties of a Sum over a Finite Index Set) identifies the left-hand side with $-\sum_{x\in F}|f(x)|$. The two-sided bound of claim 6 of Properties of the Absolute Value in an Ordered Field then gives

$$\Bigl|\sum_{x\in F}f(x)\Bigr|\le\sum_{x\in F}|f(x)|.$$

Clause 1 (Linearity). Let $A=\sum_{k\in\mathbb{Z}^{n}}a(k)$ and $B=\sum_{k\in\mathbb{Z}^{n}}b(k)$, so that $S_{N}(a)$ converges to $A$ and $S_{N}(b)$ converges to $B$. For every $N$, additivity and homogeneity of sums over the finite set $\Gamma_{N}$ (claims 3 and 4 of Properties of a Sum over a Finite Index Set) give

$$S_{N}(a+b)=S_{N}(a)+S_{N}(b),\qquad S_{N}(\lambda a)=\lambda S_{N}(a).$$

By claim 1 of Arithmetic of Limits of Real Sequences the sequence $(S_{N}(a+b))_{N}$ converges to $A+B$, and by claim 3 of Arithmetic of Limits of Real Sequences the sequence $(S_{N}(\lambda a))_{N}$ converges to $\lambda A$. Hence $a+b$ and $\lambda a$ are cube-summable (Cube Sums of Families on the Integer Lattice §lattice-sum) with lattice sums $A+B$ and $\lambda A$, which are the two displayed identities.

Clause 2 (Nonnegative families). Suppose $0\le a(k)$ for every $k$, and let $T=\{S_{N}(a):N\in\mathbb{N}\}$, a nonempty set.

(a) Monotonicity. Let $N\le M$. Then $\Gamma_{N}\subseteq\Gamma_{M}$ by The Cubes of the Integer Lattice are Finite, Nested, Exhaust the Lattice and Have (2N+1)^n Points §nested, $\Gamma_{N}$ is nonempty, and $a$ is nonnegative on $\Gamma_{M}$, so $S_{N}(a)\le S_{M}(a)$ by Real Sums over a Finite Index Set: Comparison, Nonnegativity, Monotonicity, Term Bounds, Absolute Values, Counting and Limits §monotone.

(b) If $T$ is bounded above, then $(S_{N}(a))$ converges to $\sigma=\sup T$. The supremum exists by The Real Numbers: Standing Notation and Background §bounds. Let $0<\varepsilon$. By claim 3 of Approximation Property of the Supremum and the Infimum in $\mathbb{R}$ there is $N_{0}\in\mathbb{N}$ with $\sigma-\varepsilon<S_{N_{0}}(a)$. Let $N\ge N_{0}$. By (a), $S_{N_{0}}(a)\le S_{N}(a)$, and $S_{N}(a)\le\sigma$ because $\sigma$ is an upper bound of $T$. By mixed transitivity (claim 2 of Elementary Order Arithmetic in an Ordered Field), $\sigma-\varepsilon<S_{N}(a)$, and adding $-\sigma$ (claim 1 of Elementary Order Arithmetic in an Ordered Field) gives $-\varepsilon<S_{N}(a)-\sigma$. Adding $-\sigma$ to $S_{N}(a)\le\sigma$ (clause 1 of Ordered Field) gives $S_{N}(a)-\sigma\le0$, and with $0<\varepsilon$ mixed transitivity (claim 2 of Elementary Order Arithmetic in an Ordered Field) gives $S_{N}(a)-\sigma<\varepsilon$. By claim 9 of Properties of the Absolute Value in an Ordered Field, $|S_{N}(a)-\sigma|<\varepsilon$. As $\varepsilon$ was arbitrary, $(S_{N}(a))$ converges to $\sigma$ by Limit of a Sequence of Real Numbers.

(c) The equivalence and the value of the sum. If $T$ is bounded above, $a$ is cube-summable by (b). Conversely, if $a$ is cube-summable, the convergent sequence $(S_{N}(a))$ is bounded by claim 2 of Uniqueness of Limits and Boundedness of Convergent Real Sequences, that is, by Bounded Sequence of Real Numbers there is a real $M>0$ with $|S_{N}(a)|\le M$ for every $N$; since $S_{N}(a)\le|S_{N}(a)|$ by claim 3 of Properties of the Absolute Value in an Ordered Field, transitivity gives $S_{N}(a)\le M$, so $T$ is bounded above. In either case, by (b) and uniqueness of limits (claim 1 of Uniqueness of Limits and Boundedness of Convergent Real Sequences), $\sum_{k\in\mathbb{Z}^{n}}a(k)=\sup T$.

(d) The two inequalities. Assume $a$ is cube-summable, so the lattice sum is $\sup T$ by (c). By Real Sums over a Finite Index Set: Comparison, Nonnegativity, Monotonicity, Term Bounds, Absolute Values, Counting and Limits §nonnegative, $0\le S_{1}(a)$, and $S_{1}(a)\le\sup T$, so $0\le\sum_{k\in\mathbb{Z}^{n}}a(k)$ by transitivity. Let $E\subseteq\mathbb{Z}^{n}$ be nonempty and finite. By The Cubes of the Integer Lattice are Finite, Nested, Exhaust the Lattice and Have (2N+1)^n Points §exhaust there is $N$ with $E\subseteq\Gamma_{N}$, and then Real Sums over a Finite Index Set: Comparison, Nonnegativity, Monotonicity, Term Bounds, Absolute Values, Counting and Limits §monotone gives $\sum_{k\in E}a(k)\le S_{N}(a)\le\sup T$; transitivity finishes the proof of clause 2.

Clause 3 (Enumerations). Suppose $0\le a(k)$ for every $k$ and let $\kappa:\mathbb{N}\to\mathbb{Z}^{n}$ be a bijection. The series $\sum_{j=1}^{\infty}a(\kappa(j))$ has the nonnegative terms $a(\kappa(j))$ and partial sums $s_{J}=\sum_{j=1}^{J}a(\kappa(j))$ (Series of Real Numbers §partial-sums). Put $U=\{s_{J}:J\in\mathbb{N}\}$ and $T=\{S_{N}(a):N\in\mathbb{N}\}$.

Step 1: partial sums as sums over finite sets. Fix $J\in\mathbb{N}$ and let $K_{J}=\{\kappa(j):j\in[J]\}$. The map $\kappa_{J}:[J]\to K_{J}$, $j\mapsto\kappa(j)$, is a bijection in the sense of Bijection of Sets: every element of $K_{J}$ is $\kappa(j)$ for some $j\in[J]$, and it is not $\kappa(j')$ for any other $j'\in[J]$, since each point of $\mathbb{Z}^{n}$ has exactly one preimage under the bijection $\kappa$. Hence $K_{J}$ has $J$ elements (Number of Elements of a Set), so it is a nonempty finite subset of $\mathbb{Z}^{n}$, and by Sum over a Finite Index Set, computed with the bijection $\kappa_{J}$,

$$\sum_{k\in K_{J}}a(k)=\sum_{j=1}^{J}a(\kappa(j))=s_{J};$$

here the finite sum is that of the restriction to $[J]$ of $j\mapsto a(\kappa(j))$, as in The Real Numbers: Standing Notation and Background §naturals.

Step 2: each $s_{J}$ is at most some $S_{N}(a)$. By The Cubes of the Integer Lattice are Finite, Nested, Exhaust the Lattice and Have (2N+1)^n Points §exhaust there is $N$ with $K_{J}\subseteq\Gamma_{N}$, and by Step 1 and Real Sums over a Finite Index Set: Comparison, Nonnegativity, Monotonicity, Term Bounds, Absolute Values, Counting and Limits §monotone, $s_{J}\le S_{N}(a)$.

Step 3: each $S_{N}(a)$ is at most some $s_{J}$. Fix $N$. For $k\in\Gamma_{N}$ let $\mu(k)\in\mathbb{N}$ be the unique natural number with $\kappa(\mu(k))=k$ (Bijection of Sets), and let $\iota$ be the canonical map of The Real Numbers: Standing Notation and Background §numbers. Each $\iota(\mu(k))$ is positive by claim 3 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field, in particular nonnegative. Put $C=\sum_{k\in\Gamma_{N}}\iota(\mu(k))$. For $k\in\Gamma_{N}$ the sum over the singleton $\{k\}$ is $\iota(\mu(k))$ by claim 1 of Peeling, Splitting, and Interchange for Sums over a Finite Index Set, so $\iota(\mu(k))\le C$ by Real Sums over a Finite Index Set: Comparison, Nonnegativity, Monotonicity, Term Bounds, Absolute Values, Counting and Limits §monotone. By claim 1 of The Archimedean Property of the Real Numbers there is $J\in\mathbb{N}$ with $C<\iota(J)$, whence $\iota(\mu(k))<\iota(J)$ by mixed transitivity (claim 2 of Elementary Order Arithmetic in an Ordered Field). If $J\le\mu(k)$ held, then $\iota(J)\le\iota(\mu(k))$ (equality if $J=\mu(k)$, claim 6 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field if $J<\mu(k)$), and antisymmetry would force $\iota(\mu(k))=\iota(J)$, which is false. So by the trichotomy of claim 3 of Properties of the Order on the Natural Numbers, $\mu(k)<J$, hence $\mu(k)\in[J]$ and $k=\kappa(\mu(k))\in K_{J}$. Thus $\Gamma_{N}\subseteq K_{J}$, and Real Sums over a Finite Index Set: Comparison, Nonnegativity, Monotonicity, Term Bounds, Absolute Values, Counting and Limits §monotone with Step 1 gives $S_{N}(a)\le s_{J}$.

Step 4: conclusion. By Steps 2 and 3 and transitivity, every upper bound of $T$ is an upper bound of $U$ and every upper bound of $U$ is an upper bound of $T$; in particular $U$ is bounded above if and only if $T$ is. By Series of Nonnegative Real Numbers, Comparison, and the Geometric Series §criterion the series $\sum_{j=1}^{\infty}a(\kappa(j))$ converges if and only if $U$ is bounded above, and by clause 2 (proved above) $a$ is cube-summable if and only if $T$ is bounded above; so the two conditions are equivalent. When they hold, the series has sum $\sup U$ by Series of Nonnegative Real Numbers, Comparison, and the Geometric Series §criterion, and the lattice sum is $\sup T$ by clause 2. Since $\sup T$ is an upper bound of $U$, $\sup U\le\sup T$ by Upper Bound and Least Upper Bound; symmetrically $\sup T\le\sup U$; by antisymmetry $\sup U=\sup T$, which is the displayed identity.

Clause 4 (Absolute summability). Suppose $|a|$ is cube-summable and let $L=\sum_{k\in\mathbb{Z}^{n}}|a(k)|$. Since $0\le|a(k)|$ by claim 1 of Properties of the Absolute Value in an Ordered Field, clause 2 applies to $|a|$ and gives $S_{N}(|a|)\le L$ for every $N$. Define $p:\mathbb{Z}^{n}\to\mathbb{R}$ by $p(k)=|a(k)|+a(k)$. By claim 3 of Properties of the Absolute Value in an Ordered Field, $-|a(k)|\le a(k)$ and $a(k)\le|a(k)|$; adding $|a(k)|$ (clause 1 of Ordered Field) gives $0\le p(k)\le|a(k)|+|a(k)|$. By Real Sums over a Finite Index Set: Comparison, Nonnegativity, Monotonicity, Term Bounds, Absolute Values, Counting and Limits §comparison and additivity (claim 3 of Properties of a Sum over a Finite Index Set),

$$S_{N}(p)\le S_{N}(|a|)+S_{N}(|a|)\le L+L,$$

the second inequality by adding $S_{N}(|a|)\le L$ twice (clause 1 of Ordered Field) and transitivity. So $\{S_{N}(p):N\in\mathbb{N}\}$ is bounded above, and $p$ is cube-summable by clause 2. Since $a(k)=p(k)+(-1)|a(k)|$ for every $k$ by the field axioms of Field, clause 1 (with $\lambda=-1$, then for the sum) shows that $a$ is cube-summable.

For the inequality, Preliminary (A) with $F=\Gamma_{N}$ gives $|S_{N}(a)|\le S_{N}(|a|)$ for every $N$. The sequence $(S_{N}(a))$ converges to $\sum_{k\in\mathbb{Z}^{n}}a(k)$, so $(|S_{N}(a)|)$ converges to $\bigl|\sum_{k\in\mathbb{Z}^{n}}a(k)\bigr|$ by claim 4 of Order Properties of Limits of Real Sequences, while $(S_{N}(|a|))$ converges to $L$. The comparison of limits in claim 1 of Order Properties of Limits of Real Sequences gives $\bigl|\sum_{k\in\mathbb{Z}^{n}}a(k)\bigr|\le L$.

Clause 5 (Comparison). Suppose $b$ is cube-summable and $|a(k)|\le b(k)$ for all $k$. Since $0\le|a(k)|$ (claim 1 of Properties of the Absolute Value in an Ordered Field), transitivity gives $0\le b(k)$, so by clause 2 $S_{N}(b)\le\sum_{k\in\mathbb{Z}^{n}}b(k)$ for every $N$. By Real Sums over a Finite Index Set: Comparison, Nonnegativity, Monotonicity, Term Bounds, Absolute Values, Counting and Limits §comparison, $S_{N}(|a|)\le S_{N}(b)$, so by transitivity $\sum_{k\in\mathbb{Z}^{n}}b(k)$ is an upper bound of $\{S_{N}(|a|):N\in\mathbb{N}\}$. The family $|a|$ is nonnegative, so by clause 2 it is cube-summable and its lattice sum is the supremum of that set, which is at most the upper bound $\sum_{k\in\mathbb{Z}^{n}}b(k)$ by Upper Bound and Least Upper Bound. Finally $a$ is cube-summable by clause 4.

Clause 6 (Finitely supported families). Let $E$ be nonempty and finite with $a(k)=0$ for $k\notin E$, and put $c=\sum_{k\in E}a(k)$. If $E\subseteq\Gamma_{N}$, then $E$ is a nonempty subset of the nonempty finite set $\Gamma_{N}$ and $a$ vanishes on $\Gamma_{N}\setminus E$, so $S_{N}(a)=c$ by Sums over Finite Index Sets: Finite Unions, Disjoint Unions, Vanishing Terms, Dependent Pairs, Conjugation and the Modulus §vanishing. By The Cubes of the Integer Lattice are Finite, Nested, Exhaust the Lattice and Have (2N+1)^n Points §exhaust there is $N_{0}$ with $E\subseteq\Gamma_{N_{0}}$, and for $N\ge N_{0}$ we have $\Gamma_{N_{0}}\subseteq\Gamma_{N}$ by The Cubes of the Integer Lattice are Finite, Nested, Exhaust the Lattice and Have (2N+1)^n Points §nested, so $E\subseteq\Gamma_{N}$ and $S_{N}(a)=c$. Given $0<\varepsilon$, for every $N\ge N_{0}$ we get $|S_{N}(a)-c|=|0|=0<\varepsilon$, using claim 1 of Properties of the Absolute Value in an Ordered Field. Hence $(S_{N}(a))$ converges to $c$ (Limit of a Sequence of Real Numbers), that is, $a$ is cube-summable with $\sum_{k\in\mathbb{Z}^{n}}a(k)=c$.