Proof of The Trigonometric System is Continuously Differentiable and Orthogonal in the Sobolev Space of the Torus

Each result cited is universally quantified over the data in its own statement, and is applied to the data named here. The field axioms of $\mathbb{R}$ are used freely for associativity, commutativity and distributivity, for $s\cdot1=s$, and for the injectivity of the additive inverse, that is, $-s=-t$ implies $s=t$. By Finite Product Notation in a Field a finite product is determined by the map on the initial segment it is formed over, so two such products agree as soon as the two maps agree; this is used without further comment. Throughout, $\phi_{m}$ for $m\in\mathbb{Z}$ are the one-dimensional maps of The Trigonometric System on the Torus §one-dimensional, so that $e_{k}(x)=\prod_{j=1}^{n}\phi_{k_{j}}(x_{j})$.

Claim 1. (The flipped index.) Let $k\in\mathbb{Z}^{n}$ and $i\in[n]$. Then $k^{(i)}\in\mathbb{Z}^{n}$.

Proof of Claim 1. Every coordinate of $k$ lies in $\mathbb{Z}$ by Lattice-Periodic Functions and the Periodic Function Classes §lattice, and $-k_{i}\in\mathbb{Z}$ by claim 2 of Arithmetic, Order and Discreteness of the Integers; so every coordinate of $k^{(i)}$ lies in $\mathbb{Z}$, and $k^{(i)}\in\mathbb{Z}^{n}$ by the same clause of Lattice-Periodic Functions and the Periodic Function Classes. This proves Claim 1.

Claim 2. (The slice of the product.) Let $k\in\mathbb{Z}^{n}$, let $i\in[n]$ and let $a\in\mathbb{R}^{n}$. For $s\in\mathbb{R}$ let $a[s]$ denote the point of $\mathbb{R}^{n}$ whose $i$th coordinate is $s$ and whose $j$th coordinate is $a_{j}$ for every $j\in[n]$ with $j\ne i$. Let $c\in\mathbb{R}$ be the finite product of the map $[n]\to\mathbb{R}$ taking the value $1$ at $i$ and the value $\phi_{k_{j}}(a_{j})$ at every $j\ne i$. Then

Proof of Claim 2. Write $u:[n]\to\mathbb{R}$ for the map just described, so that $c=\prod_{j=1}^{n}u_{j}$. Fix $s\in\mathbb{R}$ and let $v:[n]\to\mathbb{R}$ take the value $\phi_{k_{i}}(s)$ at $i$ and the value $1$ at every $j\ne i$. Then $u_{j}v_{j}=\phi_{k_{j}}(a[s]_{j})$ for every $j\in[n]$: at $j=i$ both sides are $\phi_{k_{i}}(s)$, and at $j\ne i$ both sides are $\phi_{k_{j}}(a_{j})$, since $a[s]_{j}=a_{j}$ there. Hence by claim 2 of Properties of Finite Products,

the last equality by claim 3 of Properties of Finite Products, applied with the index $i$, since $v_{j}=1$ for every $j\ne i$.

The second identity is the same computation with $w:[n]\to\mathbb{R}$ taking the value $\phi_{-k_{i}}(a_{i})$ at $i$ and $1$ elsewhere: then $u_{j}w_{j}=\phi_{k^{(i)}_{j}}(a_{j})$ for every $j\in[n]$, because the $i$th coordinate of $k^{(i)}$ is $-k_{i}$ and its other coordinates agree with those of $k$; so $e_{k^{(i)}}(a)=c\,\phi_{-k_{i}}(a_{i})$ by the same two clauses. This proves Claim 2.

Proof of claim 1 of the statement. Fix $k\in\mathbb{Z}^{n}$; Claim 1 gives $k^{(i)}\in\mathbb{Z}^{n}$ for every $i\in[n]$.

The partial derivatives exist and have the stated values. Let $i\in[n]$ and $a\in\mathbb{R}^{n}$, and let $a[s]$ and $c$ be as in Claim 2. Apply Slice Function and the Partial Derivative with the open set $\mathbb{R}^{n}$, the map $e_{k}$, the point $a$ and the index $i$: claim 1 there provides a positive real $\rho$, and claim 2 there introduces the slice function $g$ on the interval $I=\{s\in\mathbb{R}:a_{i}-\rho<s<a_{i}+\rho\}$ given by $g(s)=e_{k}(a[s])$, and states that the partial derivative of $e_{k}$ with respect to the $i$th variable exists at $a$ if and only if $g$ is differentiable at $a_{i}$, in which case $\partial_{i}e_{k}(a)=g'(a_{i})$.

By Claim 2 the map $s\mapsto c\,\phi_{k_{i}}(s)$ on $\mathbb{R}$ agrees with $g$ on $I$. That map is differentiable at every point of $\mathbb{R}$ with derivative $c\,\phi_{k_{i}}'$ there, by The Trigonometric System on the Torus is Orthonormal §derivative and claim 2 of Sum, Constant Multiple, and Product Rules for One-Dimensional Derivatives, applied on the interval $\mathbb{R}$ with the constant $c$ and the companion $\phi_{k_{i}}$; every point of $\mathbb{R}$ is an interior point of $\mathbb{R}$ by Basic Facts about Intervals of the Real Line and Their Interior Points §whole-line, so that claim applies at each of them. The point $a_{i}$ is an interior point of $I$ by Basic Facts about Intervals of the Real Line and Their Interior Points §open-interval, and $I\subseteq\mathbb{R}$, so claim 2 of Restriction Stability of Continuity and of the Derivative gives that $g$ is differentiable at $a_{i}$ with

using The Trigonometric System on the Torus is Orthonormal §derivative for the second equality, commutativity and associativity together with claim 2 of Zero Products and Elementary Identities in a Field for the third, and Claim 2 for the fourth. Hence $\partial_{i}e_{k}(a)$ exists and equals $-2\pi k_{i}\,e_{k^{(i)}}(a)$, which is the displayed identity.

Class $C^{1}$. By The Trigonometric System on the Torus is Orthonormal §classes the map $e_{k}$ lies in $C_{\mathrm{per}}$, hence is continuous on $\mathbb{R}^{n}$ relative to $\mathbb{R}^{n}$ in the sense fixed in Lattice-Periodic Functions and the Periodic Function Classes; by claim 1 of Euclidean Continuity Agrees with Metric Continuity for Real-Valued Functions it is therefore continuous in the Euclidean sense at every point of $\mathbb{R}^{n}$. For $i\in[n]$ the map $\partial_{i}e_{k}$ equals $-2\pi k_{i}\,e_{k^{(i)}}$ by the previous paragraph; the map $e_{k^{(i)}}$ lies in $C_{\mathrm{per}}$ by The Trigonometric System on the Torus is Orthonormal §classes and Claim 1, so $\partial_{i}e_{k}$ is continuous on $\mathbb{R}^{n}$ relative to $\mathbb{R}^{n}$ by claim 5 of Continuity of Sums and Products of Real-Valued Functions on a Metric Space, in its scalar-multiple form, applied on $\mathbb{R}^{n}$ with the real constant $-2\pi k_{i}$ and the companion $e_{k^{(i)}}$, and hence is continuous in the Euclidean sense at every point of $\mathbb{R}^{n}$ by claim 1 of Euclidean Continuity Agrees with Metric Continuity for Real-Valued Functions. So $e_{k}$ is continuous at every point of $\mathbb{R}^{n}$, all of its partial derivatives exist at every point of $\mathbb{R}^{n}$, and each $\partial_{i}e_{k}$ is continuous at every point of $\mathbb{R}^{n}$; by clauses 1 and 3 of C^k Maps on a Euclidean Open Set the map $e_{k}$ is of class $C^{1}$ on $\mathbb{R}^{n}$.

Finally $e_{k}$ is $\mathbb{Z}^{n}$-periodic, being a member of $C_{\mathrm{per}}$, so $e_{k}\in C^{1}_{\mathrm{per}}$ by Lattice-Periodic Functions and the Periodic Function Classes §classes.

Proof of claim 2 of the statement. Let $k\in\mathbb{Z}^{n}$. By claim 1 of the statement $e_{k}\in C^{1}_{\mathrm{per}}$, so $E_{k}=[\,e_{k}|_{Q}\,]$ lies in $H^{1}(\mathbb{T}^{n})$ by The Sobolev Space of Once Weakly Differentiable Square-Integrable Classes on the Torus §space.

Let $i\in[n]$. By Elementary Properties of the Weak Partial Derivative on the Torus §classical, applied to $e_{k}\in C^{1}_{\mathrm{per}}$ with $r=2$, the class $[\,(\partial_{i}e_{k})|_{Q}\,]$ is the $i$-th weak partial derivative of $E_{k}$ in $L^{2}(\mathbb{T}^{n})$. By claim 1 of the statement $\partial_{i}e_{k}=-2\pi k_{i}\,e_{k^{(i)}}$ as maps on $\mathbb{R}^{n}$, and $e_{k^{(i)}}\in C_{\mathrm{per}}$ by The Trigonometric System on the Torus is Orthonormal §classes, so Continuous Periodic Functions are Power-Integrable and Dense on the Torus §linear, used with $p=2$ and the scalar $-2\pi k_{i}$, gives

Hence $\partial_{i}E_{k}=-2\pi k_{i}E_{k^{(i)}}$.

Proof of claim 3 of the statement. Let $k,m\in\mathbb{Z}^{n}$. By The Sobolev Space of Once Weakly Differentiable Square-Integrable Classes on the Torus §inner-product and claim 2 of the statement,

For each $j$, the linearity of the inner product in its first argument and its symmetry, both recorded in Real Inner Product Space §inner-product, give

the second equality by claim 2 of Zero Products and Elementary Identities in a Field and $4=2+2=2\cdot2$.

Suppose first $k=m$. Then $k^{(j)}=m^{(j)}$ for every $j$, so $\langle E_{k},E_{m}\rangle_{L^{2}}=1$ and $\langle E_{k^{(j)}},E_{m^{(j)}}\rangle_{L^{2}}=1$ for every $j$, by The Trigonometric System on the Torus is Orthonormal §orthonormal. Hence the $j$th summand is $4\pi^{2}k_{j}k_{j}=4\pi^{2}k_{j}^{2}$, and by claim 3 of Properties of Finite Sums, applied with the constant $4\pi^{2}$, together with claim 1 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$,

so $\langle E_{k},E_{m}\rangle_{H^{1}}=1+4\pi^{2}\lVert k\rVert^{2}$.

Suppose now $k\ne m$. Then $\langle E_{k},E_{m}\rangle_{L^{2}}=0$ by The Trigonometric System on the Torus is Orthonormal §orthonormal. Moreover $k^{(j)}\ne m^{(j)}$ for every $j\in[n]$: points of $\mathbb{R}^{n}$ are $n$-tuples of real numbers, and $k^{(j)}=m^{(j)}$ would give $-k_{j}=-m_{j}$, hence $k_{j}=m_{j}$, together with $k_{l}=m_{l}$ for every $l\ne j$, that is $k=m$. So each $\langle E_{k^{(j)}},E_{m^{(j)}}\rangle_{L^{2}}$ vanishes by the same clause, each summand is $0$ by claim 1 of Zero Products and Elementary Identities in a Field, and the sum is $0$ by claim 7 of Properties of Finite Sums, applied with the index $1$, every other summand being $0$. Hence $\langle E_{k},E_{m}\rangle_{H^{1}}=0+0=0$.