Proof of Lattice Sums of Powers of the Fourier Weights in Dimension at Most Two: Power Sums, the Growth of the Wick Constants, and the Quartic Convolution Sums

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

Throughout, $t^{r}$ for $t>0$ and real $r$ is the real power of Real Power of a Positive Real Number. Cube sums are written $S_{N}(a)=\sum_{k\in\Gamma_{N}}a(k)$ as in Cube Sums of Families on the Integer Lattice §cube-sums, and lattice sums are those of Cube Sums of Families on the Integer Lattice §lattice-sum.

Step 0. Preliminaries. By The Number Pi §pi, $\pi=2x_{0}$ with $0<x_{0}$ by The Least Positive Zero of the Cosine §least-zero; hence $0<\pi$ and $c:=\pi^{2}>0$. Every $\mu_{k}$ satisfies $1\le\mu_{k}$, so all real powers of $\mu_{k}$ are defined and positive, $\mu_{k}^{-r}=1/\mu_{k}^{r}$ and $\mu_{k}^{-1}=1/\mu_{k}$, by claim 1 of Real Powers Through the Exponential, and Elementary Asymptotic Tools: Monotonicity, Null Sequences of Negative Powers, Exponential Domination, Integer Rounding, and Square-Root and Exponential Inequalities; the same claim gives $t^{a+b}=t^{a}t^{b}$, $(st)^{a}=s^{a}t^{a}$, $(t^{a})^{b}=t^{ab}$ and the agreement of real powers with natural exponents with natural powers, which are used below without further comment as "claim 1". We also use:

(P) If $0<u\le v$ and $r\ge0$, then $u^{r}\le v^{r}$ and $v^{-r}\le u^{-r}$. For $r>0$ the first inequality is claim 2 of Real Powers Through the Exponential, and Elementary Asymptotic Tools: Monotonicity, Null Sequences of Negative Powers, Exponential Domination, Integer Rounding, and Square-Root and Exponential Inequalities, and the second follows by taking reciprocals of positive numbers; for $r=0$ all four numbers equal $1$ by claim 1 of Real Powers Through the Exponential, and Elementary Asymptotic Tools: Monotonicity, Null Sequences of Negative Powers, Exponential Domination, Integer Rounding, and Square-Root and Exponential Inequalities.

In particular, $1\le\mu_{k}$ and (P) give $1=1^{r}\le\mu_{k}^{r}$ and $0<\mu_{k}^{-r}\le1$ for every $r\ge0$ (that $1^{r}=1$: by claim 2 of Real Powers Through the Exponential, and Elementary Asymptotic Tools: Monotonicity, Null Sequences of Negative Powers, Exponential Domination, Integer Rounding, and Square-Root and Exponential Inequalities applied with $t=1$, once as a base $\ge1$ and once as a base $\le1$, $1^{0}\le1^{r}\le1^{0}=1$). Finally, a point $k\in\mathbb{Z}^{n}$ with components $k_{1},\dots,k_{n}$ satisfies $\lVert k\rVert^{2}=\sum_{i=1}^{n}k_{i}^{2}$ by Euclidean Norm on $\mathbb{R}^n$, the square of the nonnegative square root being its radicand.

Step 1. Claim 1. Let $s\in\mathbb{R}$ with $n<2s$. Since $1\le n$, $s>0$. Put $a=2s-n>0$, $q=2^{-a}$, $f(k)=\mu_{k}^{-s}$, so $0<f(k)\le1$ by Step 0. By claim 2 of Real Powers Through the Exponential, and Elementary Asymptotic Tools: Monotonicity, Null Sequences of Negative Powers, Exponential Domination, Integer Rounding, and Square-Root and Exponential Inequalities, $1=1^{a}<2^{a}$, so $0<q=1/2^{a}<1$.

Shell estimate. Let $j\in\mathbb{N}$ and let $D_{j}$ be the set of $k\in\Gamma_{2^{j}}$ with $k\notin\Gamma_{2^{j-1}}$ (where $2^{0}=1$). By The Cubes of the Integer Lattice are Finite, Nested, Exhaust the Lattice and Have (2N+1)^n Points §nested, $\Gamma_{2^{j-1}}\subseteq\Gamma_{2^{j}}$, so $\Gamma_{2^{j}}$ is the disjoint union of $\Gamma_{2^{j-1}}$ and $D_{j}$; $D_{j}$ is nonempty, since it contains the point with first component $2^{j}$ and all other components $0$. Let $k\in D_{j}$. As $k\notin\Gamma_{2^{j-1}}$, some component satisfies $|k_{i}|>2^{j-1}$, so $k_{i}^{2}\ge2^{2j-2}$, and since the other squares are nonnegative, $\lVert k\rVert^{2}\ge2^{2j-2}$. Hence $\mu_{k}\ge4\pi^{2}2^{2j-2}=c\,2^{2j}>0$, and by (P) and claim 1,

$$f(k)\le(c\,2^{2j})^{-s}=c^{-s}\,2^{-2js}.$$

By The Cubes of the Integer Lattice are Finite, Nested, Exhaust the Lattice and Have (2N+1)^n Points §count, $\sum_{k\in\Gamma_{2^{j}}}1=(2^{j+1}+1)^{n}\le(2^{j+2})^{n}=2^{2n}2^{jn}$, since $1\le2^{j+1}$ and natural powers preserve the order of nonnegative reals. Using Real Sums over a Finite Index Set: Comparison, Nonnegativity, Monotonicity, Term Bounds, Absolute Values, Counting and Limits §comparison, then Real Sums over a Finite Index Set: Comparison, Nonnegativity, Monotonicity, Term Bounds, Absolute Values, Counting and Limits §monotone (the constant being nonnegative) and the homogeneity claim 4 of Properties of a Sum over a Finite Index Set,

$$\sum_{k\in D_{j}}f(k)\le\sum_{k\in D_{j}}c^{-s}2^{-2js}\le\sum_{k\in\Gamma_{2^{j}}}c^{-s}2^{-2js}=c^{-s}2^{-2js}(2^{j+1}+1)^{n}\le K\,2^{jn}2^{-2js}=K\,q^{j},$$

where $K=2^{2n}c^{-s}$ and $2^{jn}2^{-2js}=2^{j(n-2s)}=(2^{-a})^{j}$ by claim 1.

Summation. By Sums over Finite Index Sets: Finite Unions, Disjoint Unions, Vanishing Terms, Dependent Pairs, Conjugation and the Modulus §disjoint-union, $S_{2^{j}}(f)=S_{2^{j-1}}(f)+\sum_{k\in D_{j}}f(k)$ for every $j\in\mathbb{N}$, so by induction on $J\in\mathbb{N}$,

$$S_{2^{J}}(f)=S_{1}(f)+\sum_{j=1}^{J}\sum_{k\in D_{j}}f(k)\le S_{1}(f)+K\sum_{j=1}^{J}q^{j}.$$

Since $f\le1$, Real Sums over a Finite Index Set: Comparison, Nonnegativity, Monotonicity, Term Bounds, Absolute Values, Counting and Limits §comparison and The Cubes of the Integer Lattice are Finite, Nested, Exhaust the Lattice and Have (2N+1)^n Points §count give $S_{1}(f)\le3^{n}$. By Series of Nonnegative Real Numbers, Comparison, and the Geometric Series §geometric, $\sum_{j=1}^{J}q^{j}=(q-q^{J+1})/(1-q)\le q/(1-q)$. Now let $N\in\mathbb{N}$; then $N\le2^{N}$ (induction on $N$), and the family $f$ is nonnegative, so by Lattice Sums along Cubes: Linearity, Nonnegative Families, Absolute Summability, Comparison and Finitely Supported Families §nonnegative

$$S_{N}(f)\le S_{2^{N}}(f)\le3^{n}+\frac{Kq}{1-q}.$$

Thus $\{S_{N}(f):N\in\mathbb{N}\}$ is bounded above, and $f$ is cube-summable by Lattice Sums along Cubes: Linearity, Nonnegative Families, Absolute Summability, Comparison and Finitely Supported Families §nonnegative. This proves claim 1.

Step 2. Claim 2. Let $\varepsilon>0$, $N\in\mathbb{N}$ and $A=1+4\pi^{2}nN^{2}$. For $k\in\Gamma_{N}$ every component satisfies $|k_{i}|\le N$, so $k_{i}^{2}\le N^{2}$, $\lVert k\rVert^{2}=\sum_{i=1}^{n}k_{i}^{2}\le nN^{2}$ and $1\le\mu_{k}\le A$. By claim 1, $\mu_{k}^{-1}=\mu_{k}^{\varepsilon}\mu_{k}^{-(1+\varepsilon)}$, and $\mu_{k}^{\varepsilon}\le A^{\varepsilon}$ by (P). Let $g(k)=\mu_{k}^{-(1+\varepsilon)}>0$; as $n\le2<2(1+\varepsilon)$, $g$ is cube-summable by Step 1. Hence, by Real Sums over a Finite Index Set: Comparison, Nonnegativity, Monotonicity, Term Bounds, Absolute Values, Counting and Limits §comparison, claim 4 of Properties of a Sum over a Finite Index Set, and the last assertion of Lattice Sums along Cubes: Linearity, Nonnegative Families, Absolute Summability, Comparison and Finitely Supported Families §nonnegative with $E=\Gamma_{N}$ (together with $A^{\varepsilon}>0$),

$$\sum_{k\in\Gamma_{N}}\mu_{k}^{-1}\le\sum_{k\in\Gamma_{N}}A^{\varepsilon}g(k)=A^{\varepsilon}S_{N}(g)\le A^{\varepsilon}\sum_{k\in\mathbb{Z}^{n}}\mu_{k}^{-(1+\varepsilon)}.$$

This proves claim 2.

Step 3. Claim 3: a pointwise bound. Fix $\delta$ with $0\le\delta<1/4$ and $i\in\{1,2,3,4\}$. Put $b=(1-\delta)/3>0$ and $\sigma=1+b$. Since $2\sigma>2\ge n$, the family $f_{\sigma}(k)=\mu_{k}^{-\sigma}$ is cube-summable by Step 1; let $L=\sum_{k\in\mathbb{Z}^{n}}\mu_{k}^{-\sigma}$. By Lattice Sums along Cubes: Linearity, Nonnegative Families, Absolute Summability, Comparison and Finitely Supported Families §nonnegative, $0\le S_{M}(f_{\sigma})\le L$ for every $M\in\mathbb{N}$.

Let $T$ denote the family of claim 3 for these $\delta,i$. For $(k_{1},k_{2},k_{3})\in\mathbb{Z}^{3n}$, with $k_{4}=-(k_{1}+k_{2}+k_{3})$, and for $j'\in\{1,2,3,4\}$ put

$$P_{j'}(k_{1},k_{2},k_{3})=\prod_{l\in\{1,2,3,4\},\,l\ne j'}\mu_{k_{l}}^{-\sigma},$$

a product of three positive factors. We show $0<T(k_{1},k_{2},k_{3})\le P_{1}+P_{2}+P_{3}+P_{4}$ (values at $(k_{1},k_{2},k_{3})$). Choose $j\in\{1,2,3,4\}$ with $\mu_{k_{l}}\le\mu_{k_{j}}$ for all $l$, and put $m=\mu_{k_{j}}$. Then $\mu_{k_{i}}^{\delta}\le m^{\delta}$ by (P), and since $\mu_{k_{l}}^{-1}=1/\mu_{k_{l}}$,

$$T(k_{1},k_{2},k_{3})\le m^{\delta}m^{-1}\prod_{l\ne j}\mu_{k_{l}}^{-1}=(m^{3})^{-b}\prod_{l\ne j}\mu_{k_{l}}^{-1},$$

because $m^{\delta}m^{-1}=m^{-(1-\delta)}=m^{-3b}=(m^{3})^{-b}$ by claim 1. The product $Q=\prod_{l\ne j}\mu_{k_{l}}$ of three numbers in $[1,m]$ satisfies $0<Q\le m^{3}$, so by (P) and claim 1, $(m^{3})^{-b}\le Q^{-b}=\prod_{l\ne j}\mu_{k_{l}}^{-b}$. Combining with $\mu_{k_{l}}^{-1}\mu_{k_{l}}^{-b}=\mu_{k_{l}}^{-\sigma}$ (claim 1) gives $T(k_{1},k_{2},k_{3})\le P_{j}(k_{1},k_{2},k_{3})$, and the remaining three $P_{j'}$ are positive. Positivity of $T$ is clear from Step 0.

Step 4. Claim 3: bounded cube sums. Let $\Gamma_{M}^{(3)}$ denote the cube of $\mathbb{Z}^{3n}$; as its points are those all of whose $3n$ components lie between $-M$ and $M$, and the $3n$ components of a point $(k_{1},k_{2},k_{3})$ of $\mathbb{Z}^{3n}$ are read as the concatenation of the components of $k_{1}$, $k_{2}$ and $k_{3}$, under the convention of the statement $(k_{1},k_{2},k_{3})\in\Gamma_{M}^{(3)}$ if and only if $k_{1},k_{2},k_{3}\in\Gamma_{M}$. Define $F$ on $\mathbb{Z}^{3n}$ by $F(x_{1},x_{2},x_{3})=f_{\sigma}(x_{1})f_{\sigma}(x_{2})f_{\sigma}(x_{3})$. Applying The Product of Two Sums over Finite Index Sets is a Sum over the Cartesian Product twice (to $\Gamma_{M}\times\Gamma_{M}$, then to $(\Gamma_{M}\times\Gamma_{M})\times\Gamma_{M}$) and reindexing along the bijection $((x_{1},x_{2}),x_{3})\mapsto(x_{1},x_{2},x_{3})$ onto $\Gamma_{M}^{(3)}$ (claim 2 of Properties of a Sum over a Finite Index Set), for every $M\in\mathbb{N}$

$$\sum_{x\in\Gamma_{M}^{(3)}}F(x)=S_{M}(f_{\sigma})^{3}\le L^{3}.\qquad(*)$$

Fix $j'\in\{1,2,3,4\}$ and let $\theta_{j'}(k_{1},k_{2},k_{3})$ be the triple formed by the $k_{l}$, $l\in\{1,2,3,4\}\setminus\{j'\}$, in increasing order of $l$; thus $P_{j'}=F\circ\theta_{j'}$. The map $\theta_{j'}$ is injective on $\mathbb{Z}^{3n}$: since $k_{1}+k_{2}+k_{3}+k_{4}=0$, the omitted $k_{j'}$ is minus the sum of the three entries of $\theta_{j'}(k_{1},k_{2},k_{3})$, so this triple determines $k_{1},\dots,k_{4}$. Let $N\in\mathbb{N}$ and $(k_{1},k_{2},k_{3})\in\Gamma_{N}^{(3)}$. Each component of $k_{4}$ is minus a sum of three integers between $-N$ and $N$, hence lies between $-3N$ and $3N$, so $k_{4}\in\Gamma_{3N}$; and $k_{1},k_{2},k_{3}\in\Gamma_{N}\subseteq\Gamma_{3N}$ by The Cubes of the Integer Lattice are Finite, Nested, Exhaust the Lattice and Have (2N+1)^n Points §nested. So $E=\theta_{j'}(\Gamma_{N}^{(3)})$ is a nonempty subset of $\Gamma_{3N}^{(3)}$ onto which $\theta_{j'}$ maps $\Gamma_{N}^{(3)}$ bijectively. By claim 2 of Properties of a Sum over a Finite Index Set, then Real Sums over a Finite Index Set: Comparison, Nonnegativity, Monotonicity, Term Bounds, Absolute Values, Counting and Limits §monotone ($F>0$), then $(*)$ with $M=3N$,

$$S_{N}(P_{j'})=\sum_{x\in\Gamma_{N}^{(3)}}F(\theta_{j'}(x))=\sum_{y\in E}F(y)\le\sum_{y\in\Gamma_{3N}^{(3)}}F(y)\le L^{3}.$$

By Step 3, Real Sums over a Finite Index Set: Comparison, Nonnegativity, Monotonicity, Term Bounds, Absolute Values, Counting and Limits §comparison and iterated additivity (claim 3 of Properties of a Sum over a Finite Index Set, applied three times to the four summands),

$$S_{N}(T)\le\sum_{j'=1}^{4}S_{N}(P_{j'})\le4L^{3}\qquad\text{for every }N\in\mathbb{N}.$$

Since $T$ is nonnegative, its cube sums being bounded above makes it cube-summable by Lattice Sums along Cubes: Linearity, Nonnegative Families, Absolute Summability, Comparison and Finitely Supported Families §nonnegative. As $i\in\{1,2,3,4\}$ was arbitrary, claim 3 is proved. (Only $\delta<1$ was used.)