Proof of The Reproducing Identity for the Fejer Kernels of the Torus

Each result cited is universally quantified over the data in its own statement. Throughout, $\iota$ denotes the canonical map from $\mathbb{N}$ to $\mathbb{R}$, written explicitly where the distinction between a natural number and its image matters; so the integer $a-N-1$ of the statement is $\iota(a)-\iota(N)-1$. Let $2=1+1$, and let $\cos$, $\sin$, $\pi$ and the maps $C_{m},S_{m}$ for $m\in\mathbb{Z}$ be as in Cell Integrals of the Trigonometric Monomials, so that $C_{m}(t)=\cos(2\pi mt)$ and $S_{m}(t)=\sin(2\pi mt)$, and let $\sqrt{2}$ be the nonnegative real number with $\sqrt{2}\sqrt{2}=2$ fixed in The Trigonometric System on the Torus; by The Trigonometric System on the Torus (i), $\phi_{0}(t)=1$, $\phi_{m}=\sqrt{2}\,C_{m}$ for $0<m$ and $\phi_{m}=\sqrt{2}\,S_{-m}$ for $m<0$.

Claim 1. Fix $t,s\in\mathbb{R}$ and put $r=t-s$.

Step 1: the kernel as a cosine sum. By its definition in The One-Dimensional Fejer Kernel: Regularity, Nonnegativity, Mass and Far-Field Decay, $F_{N}(r)=\tfrac{1}{N}\bigl(N+2\sum_{m=1}^{N}(N-m)\cos(2\pi mr)\bigr)$. Distributing $\tfrac{1}{N}$ over the two terms, using $\tfrac{1}{N}N=1$, and applying the homogeneity of finite sums, claim 3 of Properties of Finite Sums, twice to move the factor $\tfrac{2}{N}$ inside the sum, gives

$$F_{N}(r)=1+\sum_{m=1}^{N}c_{m}\cos(2\pi mr),\qquad c_{m}=\frac{2(N-m)}{N}=2\Bigl(1-\frac{m}{N}\Bigr)\quad(m\in[N]).$$

Step 2: the addition formula. Let $m\in\mathbb{N}$, read in $\mathbb{R}$ as $\iota(m)$, so that $0<\iota(m)$ and $-\iota(m)<0$ by claim 1 of Arithmetic, Order, Discreteness and Intervals of the Integers, with $-\iota(m)\in\mathbb{Z}$ by claim 2 of that lemma. Since $2\pi mr=2\pi mt+(-2\pi ms)$, the addition formula for the cosine, Values at Zero, Parity, Addition and Double-Angle Identities for Sine and Cosine (iii), applied with $x=2\pi mt$ and $y=-2\pi ms$, followed by the parity identities Values at Zero, Parity, Addition and Double-Angle Identities for Sine and Cosine (ii), gives

$$\cos(2\pi mr)=\cos(2\pi mt)\cos(2\pi ms)+\sin(2\pi mt)\sin(2\pi ms)=C_{m}(t)C_{m}(s)+S_{m}(t)S_{m}(s).$$

By the sign information just recorded, $\phi_{m}=\sqrt{2}\,C_{m}$ and $\phi_{-m}=\sqrt{2}\,S_{m}$, the latter because $-(-\iota(m))=\iota(m)$. Hence, using $\sqrt{2}\sqrt{2}=2$,

$$\phi_{m}(t)\phi_{m}(s)+\phi_{-m}(t)\phi_{-m}(s)=2\bigl(C_{m}(t)C_{m}(s)+S_{m}(t)S_{m}(s)\bigr)=2\cos(2\pi mr).$$

Step 3: splitting the sum. Let $g:[2N+1]\to\mathbb{R}$ be the map $g(a)=w_{N,a}\,\phi_{a-N-1}(t)\,\phi_{a-N-1}(s)$, so that the right-hand side of claim 1 is $\sum_{a=1}^{2N+1}g(a)$. Since $2N+1=N+(N+1)$, Splitting a Finite Sum at an Index, applied with $m=N$ and $n=N+1$ there, gives

$$\sum_{a=1}^{2N+1}g(a)=\sum_{a=1}^{N}g(a)+\sum_{j=1}^{N+1}g(N+j).$$

Since $N+1=1+N$ by claim 4 of Arithmetic of Addition on the Natural Numbers, the same lemma applied with $m=1$ and $n=N$ there to the map $j\mapsto g(N+j)$ on $[N+1]$, together with $\sum_{k=1}^{1}a_{k}=a_{1}$ from claim 1 of Properties of Finite Sums and the associativity $N+(1+j)=(N+1)+j$ from claim 3 of Arithmetic of Addition on the Natural Numbers, gives

$$\sum_{j=1}^{N+1}g(N+j)=g(N+1)+\sum_{j=1}^{N}g(N+1+j).$$

Step 4: the three pieces. For the middle term, $\iota(N+1)=\iota(N)+1$ by claim 1 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field, so the integer attached to $a=N+1$ is $0$; hence $w_{N,N+1}=1-\tfrac{|0|}{N}=1$, because $|0|=0$ by Absolute Value in an Ordered Field, and $g(N+1)=\phi_{0}(t)\phi_{0}(s)=1$.

Let $j\in[N]$. By claims 1 and 4 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field, $\iota(N+1+j)=\iota(N)+1+\iota(j)$, so the integer attached to $a=N+1+j$ is $\iota(j)$; it is positive by claim 3 of that lemma, so $|\iota(j)|=\iota(j)$ by Absolute Value in an Ordered Field. Hence $g(N+1+j)=\bigl(1-\tfrac{j}{N}\bigr)\phi_{j}(t)\phi_{j}(s)$.

For the first piece, let $\sigma_{N}:[N]\to[N]$ be the reversing bijection of Reversal of a Finite Sum (i), so that $\iota(\sigma_{N}(a))=\iota(N)+1-\iota(a)$ for $a\in[N]$; the integer attached to $\sigma_{N}(a)$ is therefore $-\iota(a)$, and $|-\iota(a)|=|\iota(a)|=\iota(a)$ by claim 2 of Properties of the Absolute Value in an Ordered Field and Absolute Value in an Ordered Field, the number $\iota(a)$ being positive by claim 3 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field. Hence $g(\sigma_{N}(a))=\bigl(1-\tfrac{a}{N}\bigr)\phi_{-a}(t)\phi_{-a}(s)$, and Reversal of a Finite Sum (ii) gives

$$\sum_{a=1}^{N}g(a)=\sum_{a=1}^{N}g(\sigma_{N}(a))=\sum_{a=1}^{N}\Bigl(1-\frac{a}{N}\Bigr)\phi_{-a}(t)\phi_{-a}(s).$$

Step 5: assembling. Combining Steps 3 and 4 and rearranging the three real summands by the commutativity and associativity of addition,

$$\sum_{a=1}^{2N+1}g(a)=1+\Bigl(\sum_{a=1}^{N}\Bigl(1-\frac{a}{N}\Bigr)\phi_{-a}(t)\phi_{-a}(s)+\sum_{a=1}^{N}\Bigl(1-\frac{a}{N}\Bigr)\phi_{a}(t)\phi_{a}(s)\Bigr).$$

By the additivity of finite sums, claim 2 of Properties of Finite Sums, the two sums in parentheses combine into $\sum_{a=1}^{N}\bigl(1-\tfrac{a}{N}\bigr)\bigl(\phi_{-a}(t)\phi_{-a}(s)+\phi_{a}(t)\phi_{a}(s)\bigr)$, after factoring the common coefficient by distributivity. By Step 2 its $a$th summand equals $\bigl(1-\tfrac{a}{N}\bigr)\,2\cos(2\pi ar)=c_{a}\cos(2\pi ar)$. Two finite sums whose summands agree at every index are equal, so

$$\sum_{a=1}^{2N+1}g(a)=1+\sum_{a=1}^{N}c_{a}\cos(2\pi ar)=F_{N}(r)$$

by Step 1. This proves claim 1.

Claim 2. Fix $x,y\in\mathbb{R}^{n}$. By the definition of $\Phi_{N}$ and the coordinatewise formation of the difference, $(x-y)_{i}=x_{i}-y_{i}$ for every $i\in[n]$ and $\Phi_{N}(x-y)=\prod_{i=1}^{n}F_{N}(x_{i}-y_{i})$. For $i\in[n]$ and $a\in[2N+1]$ put $d_{ia}=w_{N,a}\,\phi_{a-N-1}(x_{i})\,\phi_{a-N-1}(y_{i})$, which defines an $n$-tuple $d$ of $(2N+1)$-tuples of real numbers. By claim 1, $F_{N}(x_{i}-y_{i})=\sum_{a=1}^{2N+1}d_{ia}$ for every $i\in[n]$, so Generalized Distributivity: Expanding a Product of Finite Sums, applied to $d$ with $m=2N+1$ there and with the dimension $n$ as its $n$, gives

$$\Phi_{N}(x-y)=\prod_{i=1}^{n}\Bigl(\sum_{a=1}^{2N+1}d_{ia}\Bigr)=\sum_{a\in[2N+1]^{n}}\ \prod_{i=1}^{n}d_{i\,a_{i}} .$$

Fix $a\in[2N+1]^{n}$. Since $k(a)_{i}=a_{i}-N-1$ for every $i\in[n]$, two applications of the multiplicativity of finite products, claim 2 of Properties of Finite Products, give

$$\prod_{i=1}^{n}d_{i\,a_{i}}=\Bigl(\prod_{i=1}^{n}w_{N,a_{i}}\Bigr)\Bigl(\prod_{i=1}^{n}\phi_{k(a)_{i}}(x_{i})\Bigr)\Bigl(\prod_{i=1}^{n}\phi_{k(a)_{i}}(y_{i})\Bigr)=W_{N,a}\,e_{k(a)}(x)\,e_{k(a)}(y),$$

the last two products being $e_{k(a)}(x)$ and $e_{k(a)}(y)$ by The Trigonometric System on the Torus (ii). Two sums over $[2N+1]^{n}$ whose summands agree at every index are equal, which proves claim 2.

Claim 3. Let $x,y\in\mathbb{R}^{n}$. By claim 2, applied once to the pair $x,y$ and once to the pair $y,x$, the numbers $\Phi_{N}(x-y)$ and $\Phi_{N}(y-x)$ are sums over $[2N+1]^{n}$ whose summands $W_{N,a}\,e_{k(a)}(x)\,e_{k(a)}(y)$ and $W_{N,a}\,e_{k(a)}(y)\,e_{k(a)}(x)$ agree at every $a$ by the commutativity of multiplication in $\mathbb{R}$; hence they are equal.