Proof of Lattice Sums of the Fourier Weights of the Torus, and the Negative Sobolev Norms as Lattice Sums

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

Conventions. Order and arithmetic in $\mathbb{R}$ are handled with Elementary Order Arithmetic in an Ordered Field and Elementary Arithmetic in an Ordered Field, cited by claim number, and the order of $\mathbb{N}$ with Properties of the Order on the Natural Numbers. Natural numbers are read in $\mathbb{R}$ through the canonical map $\iota$, whose properties are those of Properties of the Canonical Map from the Natural Numbers to an Ordered Field; in particular every $m\in\mathbb{N}$ satisfies $1\le m$ and $0<m$ in $\mathbb{R}$ (claims 2 and 3 there). The integers form a subset $\mathbb{Z}\subseteq\mathbb{R}$ by The Integers as a Subset of the Real Numbers, with the properties of Arithmetic, Order, Discreteness and Intervals of the Integers. For $d\in\mathbb{N}$, a point of the lattice $\mathbb{Z}^{d}$ is a map $[d]\to\mathbb{R}$ with integer values (Lattice-Periodic Functions and the Periodic Function Classes §lattice and Euclidean Points as Tuples of Real Numbers), that is, a $d$-tuple in $\mathbb{Z}$ in the sense of Tuples in a Set; points are equal when their components agree, and a point may be defined by prescribing its components (claims 1 and 2 of Euclidean Points as Tuples of Real Numbers). We write $\Gamma_{M}^{(d)}$ for the cube of The Cubes of the Integer Lattice are Finite, Nested, Exhaust the Lattice and Have (2N+1)^n Points in dimension $d$, so $\Gamma_{M}=\Gamma_{M}^{(n)}$; for $t\in\mathbb{Z}$ and $M\in\mathbb{N}$, the condition $-M\le t\le M$ is equivalent to $|t|\le M$ by claim 6 of Properties of the Absolute Value in an Ordered Field. We write $0$ for the point of $\mathbb{Z}^{n}$ all of whose components are $0$. Sums over finite index sets are those of Sum over a Finite Index Set; for every $p\in\mathbb{N}$ the initial segment $[p]$ is a nonempty finite set by claim 1 of Properties of a Sum over a Finite Index Set, which is used below for $[N]$, $[m]$, $[n]$ and $[2m+1]$ without further mention. Numerals abbreviate sums of ones, $2=1+1$, $3=2+1$, $4=2+2$ and $8=4+4$, both in $\mathbb{N}$ and in $\mathbb{R}$; by claims 1 and 4 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field, $\iota$ maps each numeral of $\mathbb{N}$ to the real number of the same name. For $c\in\mathbb{R}$ we have $c^{2}=c^{S(1)}=c^{1}c=cc$ by claim 1 of Properties of Natural Number Powers in a Field, as $2=S(1)$ by clause 1 of Natural Numbers.

For $k\in\mathbb{Z}^{n}$ we have $\lVert k\rVert^{2}=\sum_{i=1}^{n}k_{i}^{2}$, because $\lVert k\rVert$ is by Euclidean Norm on $\mathbb{R}^n$ the nonnegative number whose square is that sum (Existence and Uniqueness of the Nonnegative Square Root). The number $\pi$ is positive, as recorded in the statement of Summability of the Negative Powers of the Fourier Weights of the Torus; hence $\alpha=4\pi^{2}$ is positive by claims 8 and 5 of Elementary Order Arithmetic in an Ordered Field, and $\mu_{k}=1+\alpha\sum_{i=1}^{n}k_{i}^{2}$.

Two facts on natural powers. Let $c\in\mathbb{R}$ and $p,q\in\mathbb{N}$.

(P1) $c^{p}c^{q}=c^{p+q}$. By Principle of Induction for the Natural Numbers, applied (for fixed $p$) to the set of those $q\in\mathbb{N}$ with $c^{p}c^{q}=c^{p+q}$, using claim 1 of Properties of Natural Number Powers in a Field: $c^{p}c^{1}=c^{p}c=c^{S(p)}=c^{p+1}$, and if $c^{p}c^{q}=c^{p+q}$ then $c^{p}c^{S(q)}=(c^{p}c^{q})c=c^{p+q}c=c^{S(p+q)}=c^{p+S(q)}$, by associativity and the recursion of addition in $\mathbb{N}$ (Natural Numbers).

(P2) If $1\le c$ and $p\le q$, then $c^{p}\le c^{q}$. This is clear if $p=q$. If $p<q$, then $q=p+j$ for some $j\in\mathbb{N}$ by claim 7 of Properties of the Order on the Natural Numbers, so $c^{q}=c^{p}c^{j}$ by (P1); since $0\le1\le c$, claims 2 and 5 of Properties of Natural Number Powers in a Field give $1=1^{j}\le c^{j}$ and $0\le c^{p}$, so $c^{p}=c^{p}\cdot1\le c^{p}c^{j}=c^{q}$ by claim 5 of Elementary Arithmetic in an Ordered Field.

Moreover, if $0<c$ then $0<c^{p}$, by claims 5 and 4 of Properties of Natural Number Powers in a Field.

Three counting facts. (K1) For $p\in\mathbb{N}$, $\sum_{j\in[p]}1=p$. Indeed $\sum_{j\in[p]}1=\sum_{j=1}^{p}1$ by claim 1 of Properties of a Sum over a Finite Index Set, and $\sum_{j=1}^{p}1=p$ by Principle of Induction for the Natural Numbers, applied to the set of those $p\in\mathbb{N}$ for which this identity holds, using the recursion of claim 1 of Properties of Finite Sums and $\iota(p+1)=\iota(p)+1$ (claim 1 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field).

(K2) For $m\in\mathbb{N}$ let $I_{m}=\{t\in\mathbb{Z}:-m\le t\le m\}$ and $2m+1=m+m+1\in\mathbb{N}$, whose image in $\mathbb{R}$ is $2m+1$ by claim 4 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field. The map $[2m+1]\to I_{m}$, $j\mapsto j-(m+1)$, is a bijection: for $j\in[2m+1]$ we have $1\le j\le 2m+1$ in $\mathbb{R}$ (claims 2 and 6 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field), so $-m\le j-(m+1)\le m$, and $j-(m+1)\in\mathbb{Z}$ by claim 2 of Arithmetic, Order, Discreteness and Intervals of the Integers; it is injective by claim 7 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field; and for $t\in I_{m}$ the integer $u=t+m+1$ satisfies $1\le u\le 2m+1$, from $-m\le t\le m$. Since $0<1$ (claim 6 of Elementary Order Arithmetic in an Ordered Field) and $1\le u$, mixed transitivity (claim 2 there) gives $0<u$, so $u=\iota(j)$ for some $j\in\mathbb{N}$ by claim 1 of Arithmetic, Order, Discreteness and Intervals of the Integers. Here $1\le j$ by claim 4 of Properties of the Order on the Natural Numbers, and $j\le 2m+1$: otherwise $2m+1<j$ by trichotomy (claim 3 of Properties of the Order on the Natural Numbers), hence $2m+1<u$ by claim 6 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field; together with $u\le 2m+1$, antisymmetry (clause 2 of Total Order on a Set) would give $u=2m+1$, contradicting $2m+1\ne u$. So $j\in[2m+1]$ and $j$ is mapped to $u-(m+1)=t$. Consequently $I_{m}$ has $2m+1$ elements in the sense of Number of Elements of a Set, so it is finite by Finite Set, and it is nonempty, as the image of the nonempty set $[2m+1]$. By claim 2 of Properties of a Sum over a Finite Index Set and (K1), $\sum_{t\in I_{m}}1=2m+1$.

(K3) For $d,m\in\mathbb{N}$, $\sum_{k\in\Gamma_{m}^{(d)}}1=(2m+1)^{d}$. For fixed $m$ we apply Principle of Induction for the Natural Numbers to the set of those $d\in\mathbb{N}$ for which this identity holds. For $d=1$, the map sending $t\in I_{m}$ to the $1$-tuple with component $t$ is a bijection $I_{m}\to\Gamma_{m}^{(1)}$, so the sum is $2m+1=(2m+1)^{1}$ by claim 2 of Properties of a Sum over a Finite Index Set, (K2) and claim 1 of Properties of Natural Number Powers in a Field. For the step, claim 2 of Finiteness of Cartesian Products, Tuple Sets, and Permutation Sets, applied with $Y=\mathbb{Z}$, gives a bijection $q:\mathbb{Z}^{d}\times\mathbb{Z}\to\mathbb{Z}^{S(d)}$ appending a last component; a tuple lies in $\Gamma_{m}^{(S(d))}$ exactly when all its components lie in $I_{m}$, so $q$ restricts to a bijection from $\Gamma_{m}^{(d)}\times I_{m}$ onto $\Gamma_{m}^{(S(d))}$. By claim 2 of Properties of a Sum over a Finite Index Set, The Product of Two Sums over Finite Index Sets is a Sum over the Cartesian Product (with both factors the constant $1$), the induction hypothesis, (K2) and claim 1 of Properties of Natural Number Powers in a Field,

$$\sum_{k\in\Gamma_{m}^{(S(d))}}1=\sum_{p\in\Gamma_{m}^{(d)}\times I_{m}}1\cdot1=\Bigl(\sum_{k\in\Gamma_{m}^{(d)}}1\Bigr)\Bigl(\sum_{t\in I_{m}}1\Bigr)=(2m+1)^{d}(2m+1)=(2m+1)^{S(d)} .$$

Clause 1. Let $s\in\mathbb{N}$ with $n<2s$, where $2s=s+s$, and put $f(k)=\frac{1}{\mu_{k}^{s}}$. For every $k$, $1\le\mu_{k}$ gives $1=1^{s}\le\mu_{k}^{s}$ (claims 2 and 5 of Properties of Natural Number Powers in a Field), so by Order Reversal under Reciprocals, and Summability of the Reciprocals of the Squares §reciprocal

$$0<f(k)\le1\qquad(k\in\mathbb{Z}^{n}). \tag{1}$$

Case $n\le s$. By The Integer Lattice Admits an Enumeration by the Natural Numbers §enumeration there is a bijection $\kappa:\mathbb{N}\to\mathbb{Z}^{n}$, which is in particular injective, so by Summability of the Negative Powers of the Fourier Weights of the Torus §summable (whose weights $\mu_{k}$ and powers $\mu_{k}^{s}$ are those of the statement) the series $\sum_{j=1}^{\infty}f(\kappa(j))$ converges. Since $f$ is nonnegative by (1), Lattice Sums along Cubes: Linearity, Nonnegative Families, Absolute Summability, Comparison and Finitely Supported Families §enumeration shows that $f$ is cube-summable.

Case $s<n$. Then $2\le n$ (claims 4, 6 and 7 of Properties of the Order on the Natural Numbers), so claim 7 there gives $n=1+(n-1)$ for a natural number $n-1$. Addition in $\mathbb{N}$ commutes with $1$: for $k\in\mathbb{N}$, claims 1 and 4 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field and commutativity of addition in the field $\mathbb{R}$ (Field) give $\iota(1+k)=1+\iota(k)=\iota(k)+1=\iota(k+1)$, so $1+k=k+1$ by claim 7 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field. Hence $n=(n-1)+1$. Also $n<2s$ gives $2s=n+j$ for some $j\in\mathbb{N}$ with $1\le j$, hence $n+1\le 2s$ (claims 7, 4 and 6 there).

Step 1 (pointwise bound). Let $m\in\mathbb{N}$, and let $k\in\mathbb{Z}^{n}$ have a component $k_{i}\in\{m,-m\}$. Then $k_{i}^{2}=m^{2}$ by claim 2 of Zero Products and Elementary Identities in a Field; as $0\le k_{j}^{2}$ for all $j$ (claim 2 of Nonnegativity of Squares in an Ordered Field), claim 6 of Properties of Finite Sums gives $m^{2}\le\sum_{j=1}^{n}k_{j}^{2}$, whence $\alpha m^{2}\le\alpha\sum_{j}k_{j}^{2}\le\mu_{k}$ by claims 5 and 3 of Elementary Arithmetic in an Ordered Field (using $0\le1$). The number $\alpha m^{2}$ is positive (claim 5 of Elementary Order Arithmetic in an Ordered Field), so claims 5 and 3 of Properties of Natural Number Powers in a Field and (P1) give $0<\alpha^{s}m^{s}m^{s}=\alpha^{s}m^{2s}=(\alpha m^{2})^{s}\le\mu_{k}^{s}$, and Order Reversal under Reciprocals, and Summability of the Reciprocals of the Squares §reciprocal yields

$$f(k)\le\frac{1}{\alpha^{s}m^{2s}} . \tag{2}$$

Step 2 (faces). For $m\in\mathbb{N}$, $i\in[n]$ and $\sigma\in\{1,-1\}$ let $B(m,i,\sigma)=\{k\in\Gamma_{m}:k_{i}=\sigma m\}$. Sending $u\in\Gamma_{m}^{(n-1)}$ to the point $k\in\mathbb{Z}^{n}$ with $k_{j}=u_{j}$ for $j<i$, $k_{i}=\sigma m$ and $k_{j}=u_{j-1}$ for $j>i$ defines a bijection $\Gamma_{m}^{(n-1)}\to B(m,i,\sigma)$, whose inverse deletes the $i$th component (both preserve the cube conditions, since $|\sigma m|=m$). Both maps are well defined, since a point is defined by prescribing its components (claim 2 of Euclidean Points as Tuples of Real Numbers), and they are mutually inverse, since in each composite every component is returned unchanged and points with the same components are equal (claim 1 there). As $\Gamma_{m}^{(n-1)}$ is nonempty and finite (The Cubes of the Integer Lattice are Finite, Nested, Exhaust the Lattice and Have (2N+1)^n Points §finite in dimension $n-1$), it has $p$ elements for some $p\in\mathbb{N}$ (Finite Set); composing a bijection $[p]\to\Gamma_{m}^{(n-1)}$ with the bijection above gives a bijection $[p]\to B(m,i,\sigma)$, so $B(m,i,\sigma)$ has $p$ elements (Number of Elements of a Set) and is therefore nonempty and finite (Finite Set). By claim 2 of Properties of a Sum over a Finite Index Set and (K3), $\sum_{k\in B(m,i,\sigma)}1=(2m+1)^{n-1}$. By (2), Real Sums over a Finite Index Set: Comparison, Nonnegativity, Monotonicity, Term Bounds, Absolute Values, Counting and Limits §comparison and claim 4 of Properties of a Sum over a Finite Index Set,

$$\sum_{k\in B(m,i,\sigma)}f(k)\le\frac{(2m+1)^{n-1}}{\alpha^{s}m^{2s}} .$$

Since $1\le m$ we have $0\le 2m+1\le 3m$ (claim 3 of Elementary Arithmetic in an Ordered Field), so $(2m+1)^{n-1}\le(3m)^{n-1}=3^{n-1}m^{n-1}$ by claims 5 and 3 of Properties of Natural Number Powers in a Field. By (P1) and (P2), $m^{n-1}m^{2}=m^{n+1}\le m^{2s}$; multiplying by the positive number $\frac{1}{m^{2}m^{2s}}$ (claim 5 of Elementary Arithmetic in an Ordered Field) gives $\frac{m^{n-1}}{m^{2s}}\le\frac{1}{m^{2}}$. Hence, with the constant $C_{0}=\frac{3^{n-1}}{\alpha^{s}}$, which is positive and does not depend on $m$, $i$, $\sigma$,

$$\sum_{k\in B(m,i,\sigma)}f(k)\le\frac{C_{0}}{m^{2}} . \tag{3}$$

Step 3 (covering a cube by faces). Fix $N\in\mathbb{N}$. The set $\{1,-1\}$ is nonempty and finite by claim 2 of Peeling, Splitting, and Interchange for Sums over a Finite Index Set (note $-1<0<1$ by claims 6 and 4 of Elementary Order Arithmetic in an Ordered Field), so $P=[n]\times\{1,-1\}$ and $A=[N]\times P$ are nonempty finite sets by claim 1 of Finiteness of Cartesian Products, Tuple Sets, and Permutation Sets, as $[n]$ and $[N]$ are nonempty and finite by claim 1 of Properties of a Sum over a Finite Index Set. For $a=(m,(i,\sigma))\in A$ write $B(a)=B(m,i,\sigma)$, which is nonempty (it contains the point with $i$th component $\sigma m$ and all other components $0$) and finite (Step 2). Let $T=\{(a,k):a\in A,\ k\in B(a)\}$, which is nonempty and finite by Sums over Finite Index Sets: Finite Unions, Disjoint Unions, Vanishing Terms, Dependent Pairs, Conjugation and the Modulus §pairs.

Let $k\in\Gamma_{N}\setminus\{0\}$. The set of $M\in\mathbb{N}$ with $k\in\Gamma_{M}$ contains $N$, so it has a least element $m(k)$ by The Natural Numbers Are Well Ordered, and $m(k)\in[N]$. We claim that $|k_{i}|=m(k)$ for some $i\in[n]$. All components satisfy $|k_{j}|\le m(k)$, and each $|k_{j}|$ is a nonnegative integer (claim 1 of Properties of the Absolute Value in an Ordered Field, claim 2 of Arithmetic, Order, Discreteness and Intervals of the Integers). If $m(k)=1$: as $k\ne0$, some $k_{i}\ne0$ (claim 1 of Euclidean Points as Tuples of Real Numbers), so $0<|k_{i}|$ and hence $1\le|k_{i}|\le1$ by claim 3 of Arithmetic, Order, Discreteness and Intervals of the Integers. If $m(k)\ne1$: then $m(k)=1+M'=M'+1$ for some $M'\in\mathbb{N}$ (claims 4 and 7 of Properties of the Order on the Natural Numbers, and $1+M'=M'+1$ as shown in the case $s<n$ above), and $k\notin\Gamma_{M'}$ by minimality, so some $i$ has $M'<|k_{i}|$ (totality of the order), and claim 3 of Arithmetic, Order, Discreteness and Intervals of the Integers gives $m(k)=M'+1\le|k_{i}|\le m(k)$. Let $i(k)$ be the least such $i$ (The Natural Numbers Are Well Ordered), and let $\sigma(k)=1$ if $k_{i(k)}=m(k)$ and $\sigma(k)=-1$ otherwise, in which case $k_{i(k)}=-m(k)$ by claim 1 of Properties of the Absolute Value in an Ordered Field. Since $k\in\Gamma_{m(k)}$, we get $k\in B(m(k),i(k),\sigma(k))$, so

$$\theta(k)=\bigl((m(k),(i(k),\sigma(k))),\,k\bigr)\in T .$$

The map $\theta$ is injective (its second component is $k$), hence a bijection from $\Gamma_{N}\setminus\{0\}$ onto its image $E\subseteq T$. The set $\Gamma_{N}\setminus\{0\}$ is nonempty: it contains the point with first component $1$ and all others $0$. It is finite by claim 3 of Basic Properties of Finite Sets, being a subset of the finite set $\Gamma_{N}$ (The Cubes of the Integer Lattice are Finite, Nested, Exhaust the Lattice and Have (2N+1)^n Points §finite); so it has $p$ elements for some $p\in\mathbb{N}$ (Finite Set), and composing a bijection $[p]\to\Gamma_{N}\setminus\{0\}$ with $\theta$ shows that $E$ has $p$ elements (Number of Elements of a Set), hence is nonempty and finite (Finite Set). Define $h:T\to\mathbb{R}$ by $h((a,k))=f(k)$, which is nonnegative by (1). By claim 2 of Properties of a Sum over a Finite Index Set, Real Sums over a Finite Index Set: Comparison, Nonnegativity, Monotonicity, Term Bounds, Absolute Values, Counting and Limits §monotone, Sums over Finite Index Sets: Finite Unions, Disjoint Unions, Vanishing Terms, Dependent Pairs, Conjugation and the Modulus §pairs, (3) with Real Sums over a Finite Index Set: Comparison, Nonnegativity, Monotonicity, Term Bounds, Absolute Values, Counting and Limits §comparison, and The Product of Two Sums over Finite Index Sets is a Sum over the Cartesian Product (with the factors $m\mapsto\frac{C_{0}}{m^{2}}$ on $[N]$ and the constant $1$ on $P$),

$$\sum_{k\in\Gamma_{N}\setminus\{0\}}f(k)=\sum_{t\in E}h(t)\le\sum_{t\in T}h(t)=\sum_{a\in A}\Bigl(\sum_{k\in B(a)}f(k)\Bigr)\le\sum_{(m,p)\in A}\frac{C_{0}}{m^{2}}\cdot1=\Bigl(\sum_{m\in[N]}\frac{C_{0}}{m^{2}}\Bigr)c_{P},$$

where $c_{P}=\sum_{p\in P}1$ is a nonnegative real number (Real Sums over a Finite Index Set: Comparison, Nonnegativity, Monotonicity, Term Bounds, Absolute Values, Counting and Limits §nonnegative) not depending on $N$. By claims 1 and 4 of Properties of a Sum over a Finite Index Set and Order Reversal under Reciprocals, and Summability of the Reciprocals of the Squares §partial, $\sum_{m\in[N]}\frac{C_{0}}{m^{2}}=C_{0}\sum_{m=1}^{N}\frac{1}{m^{2}}\le C_{0}\bigl(2-\frac{1}{N}\bigr)\le 2C_{0}$, using claim 5 of Elementary Arithmetic in an Ordered Field and $0<\frac{1}{N}$; multiplying by $0\le c_{P}$ (the same claim) gives

$$\sum_{k\in\Gamma_{N}\setminus\{0\}}f(k)\le 2C_{0}c_{P} .$$

Step 4 (conclusion). The point $0$ lies in $\Gamma_{N}$ (The Cubes of the Integer Lattice are Finite, Nested, Exhaust the Lattice and Have (2N+1)^n Points §finite); by claims 3 and 1 of Peeling, Splitting, and Interchange for Sums over a Finite Index Set (splitting $\Gamma_{N}$ into $\{0\}$ and $\Gamma_{N}\setminus\{0\}$), (1) and Step 3,

$$\sum_{k\in\Gamma_{N}}f(k)=f(0)+\sum_{k\in\Gamma_{N}\setminus\{0\}}f(k)\le1+2C_{0}c_{P}$$

for every $N\in\mathbb{N}$. So the set of cube sums of the nonnegative family $f$ is bounded above, and $f$ is cube-summable by Lattice Sums along Cubes: Linearity, Nonnegative Families, Absolute Summability, Comparison and Finitely Supported Families §nonnegative.

Clause 2. Suppose $2\le n$ and fix $N\in\mathbb{N}$. Let $D_{N}=\{(m,j):m\in[N],\ j\in[m]\}$, a nonempty finite set by Sums over Finite Index Sets: Finite Unions, Disjoint Unions, Vanishing Terms, Dependent Pairs, Conjugation and the Modulus §pairs, applied with the index set $[N]$ and the sets $[m]$, which are nonempty and finite by claim 1 of Properties of a Sum over a Finite Index Set. For $(m,j)\in D_{N}$ let $\psi(m,j)\in\mathbb{Z}^{n}$ be the point with first component $m$, second component $j$ and all other components $0$. Since $1\le j\le m\le N$ in $\mathbb{R}$ (claims 2 and 6 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field), $\psi(m,j)\in\Gamma_{N}$; and $\psi$ is injective by claim 7 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field. By claim 1 of Properties of a Sum over a Finite Index Set, Sums over Finite Index Sets: Finite Unions, Disjoint Unions, Vanishing Terms, Dependent Pairs, Conjugation and the Modulus §vanishing (the terms with index outside $[2]\subseteq[n]$ vanish, as $0\cdot0=0$) and claim 1 of Properties of Finite Sums, $\lVert\psi(m,j)\rVert^{2}=m^{2}+j^{2}$. Now $j^{2}\le m^{2}$ by claim 2 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field, and $1\le m^{2}$ by the same claim applied to $0\le1\le m$. Hence, with claims 3 and 5 of Elementary Arithmetic in an Ordered Field,

$$\mu_{\psi(m,j)}=1+\alpha(m^{2}+j^{2})\le1+2\alpha m^{2}\le m^{2}+2\alpha m^{2}=c_{1}m^{2},\qquad c_{1}=1+2\alpha=1+8\pi^{2},$$

and $c_{1}$ is positive; so $\frac{1}{c_{1}m^{2}}\le\frac{1}{\mu_{\psi(m,j)}}$ by Order Reversal under Reciprocals, and Summability of the Reciprocals of the Squares §reciprocal. The family $k\mapsto\frac{1}{\mu_{k}}$ is positive, by the same reference. By Real Sums over a Finite Index Set: Comparison, Nonnegativity, Monotonicity, Term Bounds, Absolute Values, Counting and Limits §monotone, claim 2 of Properties of a Sum over a Finite Index Set, Real Sums over a Finite Index Set: Comparison, Nonnegativity, Monotonicity, Term Bounds, Absolute Values, Counting and Limits §comparison, Sums over Finite Index Sets: Finite Unions, Disjoint Unions, Vanishing Terms, Dependent Pairs, Conjugation and the Modulus §pairs, claim 4 of Properties of a Sum over a Finite Index Set with (K1), and claims 1 and 4 of Properties of a Sum over a Finite Index Set,

$$\sum_{k\in\Gamma_{N}}\frac{1}{\mu_{k}}\ge\sum_{k\in\psi(D_{N})}\frac{1}{\mu_{k}}=\sum_{(m,j)\in D_{N}}\frac{1}{\mu_{\psi(m,j)}}\ge\sum_{m\in[N]}\Bigl(\sum_{j\in[m]}\frac{1}{c_{1}m^{2}}\Bigr)=\sum_{m\in[N]}\frac{m}{c_{1}m^{2}}=\frac{1}{c_{1}}\sum_{m=1}^{N}\frac{1}{m} .$$

If some $U\in\mathbb{R}$ were an upper bound of the set of cube sums, then multiplying by $c_{1}>0$ (claim 5 of Elementary Arithmetic in an Ordered Field) would give $\sum_{m=1}^{N}\frac1m\le c_{1}U$ for every $N\in\mathbb{N}$, contradicting The Harmonic Series Diverges §harmonic. Hence the set of cube sums $\bigl\{\sum_{k\in\Gamma_{N}}\frac{1}{\mu_{k}}:N\in\mathbb{N}\bigr\}$ is not bounded above.

Clause 3. Let $m\in\mathbb{N}$ and $c\in\mathrm{Map}(\mathbb{Z}^{n},\mathbb{R})$, and put $a(k)=\frac{c(k)^{2}}{\mu_{k}^{m}}$. Each $a(k)$ is nonnegative, since $0\le c(k)^{2}$ by claim 2 of Nonnegativity of Squares in an Ordered Field and $0<\frac{1}{\mu_{k}^{m}}$ by (1) applied with $m$ in place of $s$ (the derivation of (1) used nothing about $s$); as $0<\frac{1}{\mu_{k}^{m}}$ includes $0\le\frac{1}{\mu_{k}^{m}}$, the product $a(k)=c(k)^{2}\cdot\frac{1}{\mu_{k}^{m}}$ is nonnegative by clause 2 of Ordered Field. By The Integer Lattice Admits an Enumeration by the Natural Numbers §enumeration there is a bijection $\kappa:\mathbb{N}\to\mathbb{Z}^{n}$, an enumeration in the sense of Properties of the Fourier Coefficients on the Torus, and the Realisation of Weighted Coefficient Families. By The Negative-Order Sobolev Spaces of the Torus are Hilbert Spaces: the Embedding of the Square-Integrable Classes, the Rescaled Trigonometric Basis, the Series Form of the Inner Product and the Inclusion of the Scale §weights, $(\rho_{k}^{m})^{2}=\frac{1}{\mu_{k}^{m}}$, so the terms of the series in The Negative-Order Sobolev Spaces of the Torus are Hilbert Spaces: the Embedding of the Square-Integrable Classes, the Rescaled Trigonometric Basis, the Series Form of the Inner Product and the Inclusion of the Scale §series are $(\rho_{\kappa(j)}^{m})^{2}c(\kappa(j))^{2}=a(\kappa(j))$. That clause therefore says: $c\in H^{-m}(\mathbb{T}^{n})$ if and only if $\sum_{j=1}^{\infty}a(\kappa(j))$ converges, and then $|c|_{H^{-m}}^{2}=\sum_{j=1}^{\infty}a(\kappa(j))$. By Lattice Sums along Cubes: Linearity, Nonnegative Families, Absolute Summability, Comparison and Finitely Supported Families §enumeration, applied to the nonnegative family $a$, the series converges if and only if $a$ is cube-summable, and then its sum is $\sum_{k\in\mathbb{Z}^{n}}a(k)$. This is clause 3.

Clause 4. Let $m\in\mathbb{N}$ and $k\in\mathbb{Z}^{n}$, and let $a(k')=\frac{e_{k}(k')^{2}}{\mu_{k'}^{m}}$. Since $1\cdot1=1$ and $0\cdot0=0$, $a(k)=\frac{1}{\mu_{k}^{m}}$ and $a(k')=0$ for $k'\ne k$. By Lattice Sums along Cubes: Linearity, Nonnegative Families, Absolute Summability, Comparison and Finitely Supported Families §finite-support with $E=\{k\}$ and claim 1 of Peeling, Splitting, and Interchange for Sums over a Finite Index Set, $a$ is cube-summable with $\sum_{k'\in\mathbb{Z}^{n}}a(k')=a(k)=\frac{1}{\mu_{k}^{m}}$. By clause 3, $e_{k}\in H^{-m}(\mathbb{T}^{n})$ and $|e_{k}|_{H^{-m}}^{2}=\frac{1}{\mu_{k}^{m}}$.