Proof of The Slice Average of a Continuous Periodic Function

Each result cited is universally quantified over the data in its own statement, and is applied here to the data named below. We use silently that the order of $\mathbb{R}$ is reflexive, transitive and antisymmetric, that equal real numbers satisfy $\le$ in both directions, and that $x<y$ entails $x\le y$.

A member $w$ of $C_{\mathrm{per}}$ is, by Lattice-Periodic Functions and the Periodic Function Classes (iii), a map $\mathbb{R}^{n}\to\mathbb{R}$ that is continuous from $(\mathbb{R}^{n},d_{E})$ to $(\mathbb{R},d_{\mathbb{R}})$ and $\mathbb{Z}^{n}$-periodic; we use this reading throughout without further comment.

(P0) The slice displacement. Let $x\in\mathbb{R}^{n}$, let $i\in[n]$ and let $s,s'\in\mathbb{R}$. Then $\lVert x[i{:}s]-x[i{:}s']\rVert=|s-s'|$.

Indeed, the point $z=x[i{:}s]-x[i{:}s']$ has $k$th coordinate $x_{k}-x_{k}=0$ for every $k\in[n]$ with $k\ne i$, and $i$th coordinate $s-s'$. By claim 1 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square, $\lVert z\rVert^{2}=\sum_{k=1}^{n}z_{k}^{2}$. The map $k\mapsto z_{k}^{2}$ on $[n]$ vanishes at every $k\ne i$, so by claim 1 of Properties of a Sum over a Finite Index Set, claim 4 of Peeling, Splitting, and Interchange for Sums over a Finite Index Set applied with the nonempty subset $\{i\}$ of $[n]$, and claim 1 of that peeling lemma, this sum equals $z_{i}^{2}=(s-s')^{2}$. Now $|s-s'|$ equals $s-s'$ or $-(s-s')$ by claim 1 of Properties of the Absolute Value in an Ordered Field, so in either case $|s-s'|\cdot|s-s'|=(s-s')^{2}$. Thus $\lVert z\rVert$ and $|s-s'|$ are both nonnegative and have the same square, so they are equal by Existence and Uniqueness of the Nonnegative Square Root.

(P1) Slices of continuous maps are continuous. Let $g:\mathbb{R}^{n}\to\mathbb{R}$ be continuous, let $x\in\mathbb{R}^{n}$ and let $i\in[n]$. Then the map $g_{i,x}:\mathbb{R}\to\mathbb{R}$ given by $g_{i,x}(s)=g(x[i{:}s])$ is continuous from $(\mathbb{R},d_{\mathbb{R}})$ to $(\mathbb{R},d_{\mathbb{R}})$.

Indeed, let $s_{0}\in\mathbb{R}$ and let $\varepsilon$ be a real number with $0<\varepsilon$. By Continuous Map Between Metric Spaces, applied to $g$ at the point $x[i{:}s_{0}]$, there is a real $\delta>0$ such that every $y\in\mathbb{R}^{n}$ with $\lVert y-x[i{:}s_{0}]\rVert<\delta$ satisfies $|g(y)-g(x[i{:}s_{0}])|<\varepsilon$. Let $s\in\mathbb{R}$ with $|s-s_{0}|<\delta$. By (P0), $\lVert x[i{:}s]-x[i{:}s_{0}]\rVert=|s-s_{0}|<\delta$, so $|g_{i,x}(s)-g_{i,x}(s_{0})|<\varepsilon$. As $s_{0}$ and $\varepsilon$ were arbitrary, $g_{i,x}$ is continuous by Continuous Map Between Metric Spaces.

(P2) The displacement bound. Let $y,z\in\mathbb{R}^{n}$, let $i\in[n]$ and let $s\in\mathbb{R}$. Then $\lVert y[i{:}s]-z[i{:}s]\rVert\le\lVert y-z\rVert$.

Indeed, put $c_{k}=(y_{k}-z_{k})^{2}$ for $k\in[n]$ with $k\ne i$ and $c_{i}=0$, and $d_{k}=(y_{k}-z_{k})^{2}$ for every $k\in[n]$. By claim 1 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square, $\lVert y[i{:}s]-z[i{:}s]\rVert^{2}=\sum_{k=1}^{n}c_{k}$ and $\lVert y-z\rVert^{2}=\sum_{k=1}^{n}d_{k}$, the $i$th coordinate of $y[i{:}s]-z[i{:}s]$ being $s-s=0$. Each $d_{k}$ is nonnegative: writing $t=y_{k}-z_{k}$ we have $t^{2}=|t|\cdot|t|$ as in (P0), and $0\le|t|$ by claim 1 of Properties of the Absolute Value in an Ordered Field, so $0=|t|\cdot 0\le|t|\cdot|t|$ by claim 1 of Zero Products and Elementary Identities in a Field and claim 5 of Elementary Arithmetic in an Ordered Field. Hence $c_{k}\le d_{k}$ for every $k\in[n]$, and claim 1 of Comparison and Absolute Value Bounds for Finite Sums of Real Numbers gives $\lVert y[i{:}s]-z[i{:}s]\rVert^{2}\le\lVert y-z\rVert^{2}$. Write $A=\lVert y[i{:}s]-z[i{:}s]\rVert$ and $B=\lVert y-z\rVert$, both nonnegative. From $A^{2}\le B^{2}$ and Properties of Real Powers (vi), applied with exponent $2$ in the form $s^{2}\le t\iff s\le t^{1/2}$ with $t=B^{2}$, we get $A\le(B^{2})^{1/2}$; and $(B^{2})^{1/2}=B$ by the identity $(t^{2})^{1/2}=t$ of that same clause. Hence $A\le B$.

(P3) Zero extensions of continuous maps. Let $g:\mathbb{R}\to\mathbb{R}$ be continuous from $(\mathbb{R},d_{\mathbb{R}})$ to $(\mathbb{R},d_{\mathbb{R}})$ and let $\tilde{g}:\mathbb{R}\to\mathbb{R}$ take the value $g(s)$ at $s\in[0,1]$ and the value $0$ at every $s\in\mathbb{R}$ with $s\notin[0,1]$. Then $\tilde{g}$ is measurable with respect to $\mathcal{B}(\mathbb{R})$ and integrable with respect to $\lambda$.

Indeed, the restriction $h$ of $g$ to $[0,1]$ is continuous as a map from the subset $[0,1]$ of $(\mathbb{R},d_{\mathbb{R}})$ into $(\mathbb{R},d_{\mathbb{R}})$: for a point of $[0,1]$ and a real $\varepsilon>0$, any $\delta$ witnessing the continuity of $g$ there also witnesses that of $h$, since the requirement is imposed at fewer points. The zero extension of $h$ in the sense of Agreement of the Riemann and Lebesgue Integrals for Continuous Functions on a Closed Interval is exactly $\tilde{g}$, so claim 2 of that lemma, applied with $a=0$ and $b=1$, gives the assertion.

(P4) The indicators of $[0,1]$ and $[0,1)$. Let $[0,1)=\{s\in\mathbb{R}:0\le s\text{ and }s<1\}$. Then $[0,1]$ and $[0,1)$ belong to $\mathcal{B}(\mathbb{R})$, their indicators $\mathbf{1}_{[0,1]}$ and $\mathbf{1}_{[0,1)}$ are measurable with respect to $\mathcal{B}(\mathbb{R})$ and integrable with respect to $\lambda$, and

$$\int_{\mathbb{R}}\mathbf{1}_{[0,1]}\,d\lambda=\lambda([0,1])=1,\qquad\int_{\mathbb{R}}\mathbf{1}_{[0,1)}\,d\lambda=\lambda([0,1))=1 .$$

Moreover $\lambda(\{1\})=0$.

Indeed, $[0,1]$, $[0,1)$ and $\{1\}$ are intervals with endpoints $0\le1$, $0\le1$ and $1\le1$ respectively, the last because $\{1\}$ is the closed interval determined by $1$ and $1$; they lie in $\mathcal{B}(\mathbb{R})$ and have Lebesgue measures $1-0=1$, $1-0=1$ and $1-1=0$ by claim 4 of Lebesgue Measure on the Real Line: Existence, Lengths of Intervals and Uniqueness. Their indicators are measurable by claim 1 of Arithmetic of Measurable Real- and [0,∞]-Valued Maps, and $\int_{\mathbb{R}}\mathbf{1}_{A}\,d\lambda=\lambda(A)$ by The Integral of an Indicator Function is the Measure of the Set. Each indicator takes only the values $0$ and $1$, so it is nonnegative, and by (P6) below it is integrable, its integral $1$ being finite.

(P5) The absolute value of a nonnegative number. Let $t$ be a real number with $0\le t$. Then $|t|=t$.

Indeed, $-t\le0$ by claim 4 of Elementary Order Arithmetic in an Ordered Field together with $-0=0$, recorded in claim 4 of Additive Cancellation and Elementary Additive Identities in a Field; hence $-t\le t$, and $t\le t$, so claim 6 of Properties of the Absolute Value in an Ordered Field, applied with $c=t$, gives $|t|\le t$. Claim 3 of that lemma gives $t\le|t|$. Antisymmetry of the order yields $|t|=t$.

(P6) Nonnegative maps and the two integrals. Let $(X,\mathcal{F},\mu)$ be a measure space and let $h:X\to\mathbb{R}$ be measurable with $0\le h(x)$ for every $x\in X$. Then $h$, regarded as a map into $[0,\infty]$, is measurable in the sense of Measurable Functions and the Lebesgue Integral: Standing Notation (iii); $h$ is integrable if and only if the integral $\int_{X}h\,d\mu\in[0,\infty]$ of Measurable Functions and the Lebesgue Integral: Standing Notation (v) is finite; and in that case that integral is the real number $\int_{X}h\,d\mu$ of Integrable Functions and Their Lebesgue Integral.

Indeed, the two readings of measurability agree for real-valued maps, as recorded in Measurable Functions and the Lebesgue Integral: Standing Notation (iii). With the positive and negative parts of Integrable Functions and Their Lebesgue Integral we have $h^{+}(x)=\max\{h(x),0\}=h(x)$, because $0\le h(x)$, and $h^{-}(x)=\max\{-h(x),0\}=0$, because $-h(x)\le0$ by claim 4 of Elementary Order Arithmetic in an Ordered Field and claim 4 of Additive Cancellation and Elementary Additive Identities in a Field. The map $h^{-}$ is the indicator $\mathbf{1}_{\emptyset}$ of the empty set, which lies in $\mathcal{F}$, so $\int_{X}h^{-}\,d\mu=\mu(\emptyset)=0$ by The Integral of an Indicator Function is the Measure of the Set and the requirement $\mu(\emptyset)=0$ of Measure, Measure Space, and Probability Measure. Hence $h$ is integrable exactly when $\int_{X}h^{+}\,d\mu=\int_{X}h\,d\mu$ is finite, and in that case the integral of Integrable Functions and Their Lebesgue Integral is $\int_{X}h\,d\mu-0=\int_{X}h\,d\mu$.

In particular the map $\mathbf{0}$ on $\mathbb{R}$ with constant value $0$ is the indicator $\mathbf{1}_{\emptyset}$, is measurable by claim 1 of Arithmetic of Measurable Real- and [0,∞]-Valued Maps, and is integrable with $\int_{\mathbb{R}}\mathbf{0}\,d\lambda=\lambda(\emptyset)=0$.

Proof of claim 1. Let $i\in[n]$ and $w\in C_{\mathrm{per}}$, and let $x\in\mathbb{R}^{n}$. The map $s\mapsto w(x[i{:}s])$ is continuous by (P1), applied to $g=w$, and $\tilde{w}_{i,x}$ is its zero extension in the sense of (P3), so $\tilde{w}_{i,x}$ is measurable with respect to $\mathcal{B}(\mathbb{R})$ and integrable with respect to $\lambda$ by (P3). Its integral is therefore a real number, so $(P_{i}w)(x)$ is defined for every $x\in\mathbb{R}^{n}$ and $P_{i}w$ is a map from $\mathbb{R}^{n}$ to $\mathbb{R}$.

Nonnegativity. Suppose $0\le w(y)$ for every $y\in\mathbb{R}^{n}$. Then $0\le\tilde{w}_{i,x}(s)$ for every $s\in\mathbb{R}$, the value being $w(x[i{:}s])$ on $[0,1]$ and $0$ elsewhere. By the monotonicity recorded in claim 2 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion, applied to the integrable maps $\mathbf{0}$ and $\tilde{w}_{i,x}$ of (P6), $0=\int_{\mathbb{R}}\mathbf{0}\,d\lambda\le\int_{\mathbb{R}}\tilde{w}_{i,x}\,d\lambda=(P_{i}w)(x)$.

$P_{i}w$ is $\mathbb{Z}^{n}$-periodic. Let $x^{\ast}\in\mathbb{R}^{n}$ and $e\in\mathbb{Z}^{n}$, and let $e'$ be the point of $\mathbb{R}^{n}$ whose $i$th coordinate is $0$ and whose $k$th coordinate is $e_{k}$ for every $k\in[n]$ with $k\ne i$; then $e'\in\mathbb{Z}^{n}$, since $0$ is an integer and each $e_{k}$ is. For $s\in\mathbb{R}$ the points $(x^{\ast}+e)[i{:}s]$ and $x^{\ast}[i{:}s]+e'$ both have $i$th coordinate $s$, the second because $s+0=s$, and for $k\ne i$ both have $k$th coordinate $x^{\ast}_{k}+e_{k}$; so they are equal. By the $\mathbb{Z}^{n}$-periodicity of $w$,

$$w\bigl((x^{\ast}+e)[i{:}s]\bigr)=w\bigl(x^{\ast}[i{:}s]+e'\bigr)=w\bigl(x^{\ast}[i{:}s]\bigr)\qquad(s\in\mathbb{R}),$$

so $\tilde{w}_{i,x^{\ast}+e}=\tilde{w}_{i,x^{\ast}}$ and $(P_{i}w)(x^{\ast}+e)=(P_{i}w)(x^{\ast})$. As $x^{\ast}$ and $e$ were arbitrary, $P_{i}w$ is $\mathbb{Z}^{n}$-periodic in the sense of Lattice-Periodic Functions and the Periodic Function Classes (ii).

$P_{i}w$ is continuous. Let $x_{0}\in\mathbb{R}^{n}$ and let $\varepsilon$ be a real number with $0<\varepsilon$; put $\varepsilon'=\varepsilon/2$, a positive real by claim 8 of Elementary Order Arithmetic in an Ordered Field. Since $w\in C_{\mathrm{per}}$, it is uniformly continuous by A Continuous Lattice-Periodic Function is Uniformly Continuous (i), so there is a real $\delta>0$ such that all $y,z\in\mathbb{R}^{n}$ with $\lVert y-z\rVert<\delta$ satisfy $|w(y)-w(z)|<\varepsilon'$.

Let $y\in\mathbb{R}^{n}$ with $\lVert y-x_{0}\rVert<\delta$. For every $s\in\mathbb{R}$, (P2) gives $\lVert y[i{:}s]-x_{0}[i{:}s]\rVert\le\lVert y-x_{0}\rVert<\delta$, so $|w(y[i{:}s])-w(x_{0}[i{:}s])|<\varepsilon'$. Consequently

$$\bigl|\tilde{w}_{i,y}(s)-\tilde{w}_{i,x_{0}}(s)\bigr|\le\varepsilon'\,\mathbf{1}_{[0,1]}(s)\qquad\text{for every }s\in\mathbb{R},$$

since for $s\in[0,1]$ the left-hand side is $|w(y[i{:}s])-w(x_{0}[i{:}s])|$ and for $s\notin[0,1]$ both sides are $0$. The maps $\tilde{w}_{i,y}$ and $\tilde{w}_{i,x_{0}}$ are integrable by the first paragraph, so their difference is integrable by claim 2 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion, and its absolute value is measurable by claim 4 of Arithmetic of Measurable Real- and [0,∞]-Valued Maps and integrable, being dominated by the map $\varepsilon'\mathbf{1}_{[0,1]}$, which is integrable by (P4) and claim 2 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion; this last step uses claim 1 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion together with the criterion in Integrable Functions and Their Lebesgue Integral. Hence, by claim 2 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion used first for linearity, then for the bound $\bigl|\int f\bigr|\le\int|f|$, and then for monotonicity and homogeneity,

$$\bigl|(P_{i}w)(y)-(P_{i}w)(x_{0})\bigr|=\Bigl|\int_{\mathbb{R}}\bigl(\tilde{w}_{i,y}-\tilde{w}_{i,x_{0}}\bigr)d\lambda\Bigr|\le\int_{\mathbb{R}}\bigl|\tilde{w}_{i,y}-\tilde{w}_{i,x_{0}}\bigr|\,d\lambda\le\varepsilon'\int_{\mathbb{R}}\mathbf{1}_{[0,1]}\,d\lambda=\varepsilon' ,$$

using $\int_{\mathbb{R}}\mathbf{1}_{[0,1]}\,d\lambda=1$ from (P4). Finally $\varepsilon'<\varepsilon$ by claim 8 of Elementary Order Arithmetic in an Ordered Field, so $\bigl|(P_{i}w)(y)-(P_{i}w)(x_{0})\bigr|<\varepsilon$ by claim 2 of that lemma. As $x_{0}$ and $\varepsilon$ were arbitrary, $P_{i}w$ is continuous by Continuous Map Between Metric Spaces, and with the periodicity proved above, $P_{i}w\in C_{\mathrm{per}}$ by Lattice-Periodic Functions and the Periodic Function Classes (iii).

Proof of claim 2. Let $i\in[n]$ and $w\in C_{\mathrm{per}}$.

$P_{i}w$ is free of the $i$th coordinate. Let $x\in\mathbb{R}^{n}$ and $s\in\mathbb{R}$, and put $y=x[i{:}s]$. For every $\sigma\in\mathbb{R}$ the points $y[i{:}\sigma]$ and $x[i{:}\sigma]$ have $i$th coordinate $\sigma$ and, for $k\ne i$, the common $k$th coordinate $x_{k}$, because $y_{k}=x_{k}$ for $k\ne i$; so $y[i{:}\sigma]=x[i{:}\sigma]$. Hence $\tilde{w}_{i,y}=\tilde{w}_{i,x}$ and $(P_{i}w)(x[i{:}s])=(P_{i}w)(x)$.

Other free coordinates are inherited. Let $j\in[n]$ and suppose $w$ is free of the $j$th coordinate. If $j=i$ the conclusion is the previous paragraph, so assume $j\ne i$. Let $x\in\mathbb{R}^{n}$ and $s\in\mathbb{R}$, and put $y=x[j{:}s]$. For $\sigma\in\mathbb{R}$ the points $y[i{:}\sigma]$ and $\bigl(x[i{:}\sigma]\bigr)[j{:}s]$ have $i$th coordinate $\sigma$, $j$th coordinate $s$, and $k$th coordinate $x_{k}$ for every $k\in[n]$ with $k\ne i$ and $k\ne j$; so they are equal. Since $w$ is free of the $j$th coordinate,

$$w\bigl(y[i{:}\sigma]\bigr)=w\Bigl(\bigl(x[i{:}\sigma]\bigr)[j{:}s]\Bigr)=w\bigl(x[i{:}\sigma]\bigr)\qquad(\sigma\in\mathbb{R}),$$

so $\tilde{w}_{i,y}=\tilde{w}_{i,x}$ and $(P_{i}w)(x[j{:}s])=(P_{i}w)(x)$.

Proof of claim 5. Let $w\in C_{\mathrm{per}}$.

$|w|\in C_{\mathrm{per}}$. For all real $c,d$ one has $\bigl||c|-|d|\bigr|\le|c-d|$ by claim 7 of Properties of the Absolute Value in an Ordered Field, so any $\delta$ witnessing the continuity of $w$ at a point of $\mathbb{R}^{n}$ witnesses that of $|w|$ there; thus $|w|$ is continuous. If $e\in\mathbb{Z}^{n}$ then $|w|(x+e)=|w(x+e)|=|w(x)|=|w|(x)$, so $|w|$ is $\mathbb{Z}^{n}$-periodic. Hence $|w|\in C_{\mathrm{per}}$ by Lattice-Periodic Functions and the Periodic Function Classes (iii).

$w^{a}\in C_{\mathrm{per}}$. Suppose $0\le w(x)$ for every $x\in\mathbb{R}^{n}$ and let $a$ be a positive real number. Write $\mathbb{R}_{+}$ for the nonnegative reals with the metric $d_{+}$, the restriction of $d_{\mathbb{R}}$, and $P_{a}:\mathbb{R}_{+}\to\mathbb{R}_{+}$ for the map $t\mapsto t^{a}$, as in Properties of Real Powers; thus $w^{a}(x)=P_{a}(w(x))$, and $0\le w^{a}(x)$ by Properties of Real Powers (i). Let $x_{0}\in\mathbb{R}^{n}$ and let $\varepsilon$ be a real number with $0<\varepsilon$. By Properties of Real Powers (vii) and Continuous Map Between Metric Spaces, applied to $P_{a}$ at the point $w(x_{0})$ of $\mathbb{R}_{+}$, there is a real $\eta>0$ such that every $t\in\mathbb{R}_{+}$ with $|t-w(x_{0})|<\eta$ satisfies $|t^{a}-(w(x_{0}))^{a}|<\varepsilon$. By the continuity of $w$ at $x_{0}$ there is a real $\delta>0$ such that every $y\in\mathbb{R}^{n}$ with $\lVert y-x_{0}\rVert<\delta$ satisfies $|w(y)-w(x_{0})|<\eta$; for such $y$ the value $w(y)$ lies in $\mathbb{R}_{+}$, so $|w^{a}(y)-w^{a}(x_{0})|<\varepsilon$. As $x_{0}$ and $\varepsilon$ were arbitrary, $w^{a}$ is continuous by Continuous Map Between Metric Spaces. If $e\in\mathbb{Z}^{n}$ then $w^{a}(x+e)=(w(x+e))^{a}=(w(x))^{a}=w^{a}(x)$, so $w^{a}$ is $\mathbb{Z}^{n}$-periodic and $w^{a}\in C_{\mathrm{per}}$ by Lattice-Periodic Functions and the Periodic Function Classes (iii).

Free coordinates. Let $j\in[n]$ and suppose $w$ is free of the $j$th coordinate. Then $|w|(x[j{:}s])=|w(x[j{:}s])|=|w(x)|=|w|(x)$ for all $x$ and $s$, and likewise $w^{a}(x[j{:}s])=(w(x[j{:}s]))^{a}=(w(x))^{a}=w^{a}(x)$ in the case just described.

Proof of claim 3. Let $i\in[n]$.

The constant map. The map $\mathbf{1}$ is continuous, since for a point of $\mathbb{R}^{n}$ and a real $\varepsilon>0$ any positive $\delta$ witnesses the condition of Continuous Map Between Metric Spaces, the difference of values being $|1-1|=0<\varepsilon$; and $\mathbf{1}(x+e)=1=\mathbf{1}(x)$ for $e\in\mathbb{Z}^{n}$, so $\mathbf{1}$ is $\mathbb{Z}^{n}$-periodic. Hence $\mathbf{1}\in C_{\mathrm{per}}$ by Lattice-Periodic Functions and the Periodic Function Classes (iii). For $x\in\mathbb{R}^{n}$ the map $\tilde{\mathbf{1}}_{i,x}$ takes the value $1$ at every $s\in[0,1]$ and $0$ elsewhere, so it is the indicator $\mathbf{1}_{[0,1]}$, and $(P_{i}\mathbf{1})(x)=\int_{\mathbb{R}}\mathbf{1}_{[0,1]}\,d\lambda=1$ by (P4). Thus $P_{i}\mathbf{1}=\mathbf{1}$.

Factoring. Let $v,w\in C_{\mathrm{per}}$ with $v$ free of the $i$th coordinate. Then $vw\in C_{\mathrm{per}}$ by Elementary Properties of Lattice-Periodic Functions (i). Let $x\in\mathbb{R}^{n}$. For $s\in[0,1]$,

$$\widetilde{(vw)}_{i,x}(s)=v(x[i{:}s])\,w(x[i{:}s])=v(x)\,w(x[i{:}s])=v(x)\,\tilde{w}_{i,x}(s),$$

the middle equality because $v$ is free of the $i$th coordinate; and for $s\in\mathbb{R}$ with $s\notin[0,1]$ both ends are $0$, the right-hand one by claim 1 of Zero Products and Elementary Identities in a Field. So $\widetilde{(vw)}_{i,x}=v(x)\,\tilde{w}_{i,x}$, and the linearity in claim 2 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion, applied to the integrable map $\tilde{w}_{i,x}$ with the real coefficient $v(x)$, gives

$$\bigl(P_{i}(vw)\bigr)(x)=v(x)\int_{\mathbb{R}}\tilde{w}_{i,x}\,d\lambda=v(x)\,(P_{i}w)(x)=\bigl(v\,P_{i}w\bigr)(x).$$

As $x$ was arbitrary, $P_{i}(vw)=v\,P_{i}w$. Taking $w=\mathbf{1}$ and using $v(x)\cdot1=v(x)$ gives $v\mathbf{1}=v$ and $P_{i}v=P_{i}(v\mathbf{1})=v\,P_{i}\mathbf{1}=v\mathbf{1}=v$.

Proof of claim 4. Let $i\in[n]$ and let $v,w\in C_{\mathrm{per}}$ satisfy $0\le v(y)$ and $0\le w(y)$ for every $y\in\mathbb{R}^{n}$. By Elementary Properties of Lattice-Periodic Functions (i), $vw\in C_{\mathrm{per}}$, and $0=0\cdot w(y)\le v(y)w(y)$ for every $y$, by claim 1 of Zero Products and Elementary Identities in a Field and claim 5 of Elementary Arithmetic in an Ordered Field. Hence, by claim 5, the maps $v^{1/2}$, $w^{1/2}$ and $(vw)^{1/2}$ lie in $C_{\mathrm{per}}$ and take nonnegative values.

Fix $x\in\mathbb{R}^{n}$ and put

$$f=\widetilde{\bigl(v^{1/2}\bigr)}_{i,x},\qquad g=\widetilde{\bigl(w^{1/2}\bigr)}_{i,x},\qquad F=\widetilde{\bigl((vw)^{1/2}\bigr)}_{i,x},$$

which by claim 1 are measurable with respect to $\mathcal{B}(\mathbb{R})$ and integrable with respect to $\lambda$, and take nonnegative values.

The pointwise product. For $s\in[0,1]$, claim 3 of Properties of Real Powers gives

$$f(s)g(s)=\bigl(v(x[i{:}s])\bigr)^{1/2}\bigl(w(x[i{:}s])\bigr)^{1/2}=\bigl(v(x[i{:}s])\,w(x[i{:}s])\bigr)^{1/2}=F(s),$$

and for $s\notin[0,1]$ both sides are $0$. So $fg=F$, and $|fg|=F$ by (P5), the values of $F$ being nonnegative.

The second powers. For $s\in[0,1]$, using (P5) and then claims 4 and 2 of Properties of Real Powers,

$$\bigl(|f(s)|\bigr)^{2}=\bigl(f(s)\bigr)^{2}=\Bigl(\bigl(v(x[i{:}s])\bigr)^{1/2}\Bigr)^{2}=\bigl(v(x[i{:}s])\bigr)^{1}=v(x[i{:}s]) ,$$

while for $s\notin[0,1]$ one has $\bigl(|f(s)|\bigr)^{2}=0^{2}=0$ by (P5) and Properties of Real Powers (i). Hence $|f|^{2}=\tilde{v}_{i,x}$ in the notation of Power-Integrable Functions and the p-Seminorm (i), and likewise $|g|^{2}=\tilde{w}_{i,x}$.

Since $\tilde{v}_{i,x}$ is nonnegative and integrable, (P6) shows that its integral in $[0,\infty]$ is the real number $(P_{i}v)(x)$, in particular finite. So $f$ is a $2$-integrable function on $(\mathbb{R},\mathcal{B}(\mathbb{R}),\lambda)$ in the sense of Power-Integrable Functions and the p-Seminorm (ii), with $2$-seminorm $\lVert f\rVert_{2}=\bigl((P_{i}v)(x)\bigr)^{1/2}$; and likewise $\lVert g\rVert_{2}=\bigl((P_{i}w)(x)\bigr)^{1/2}$.

H"older. Since $1<2$ and $\tfrac12+\tfrac12=1$, the number $2$ is the exponent conjugate to $2$ in the sense of Conjugate Exponents and Young's Inequality (i), by the uniqueness asserted there. Applying Hölder's Inequality (i) to the measure space $(\mathbb{R},\mathcal{B}(\mathbb{R}),\lambda)$ with $p=q=2$ and the functions $f$ and $g$ gives

$$\int_{\mathbb{R}}|fg|\,d\lambda\le\lVert f\rVert_{2}\,\lVert g\rVert_{2}=\bigl((P_{i}v)(x)\bigr)^{1/2}\bigl((P_{i}w)(x)\bigr)^{1/2}.$$

Finally $|fg|=F$, and $F$ is nonnegative and integrable, so by (P6) the left-hand side is the real number $\int_{\mathbb{R}}F\,d\lambda=\bigl(P_{i}\bigl[(vw)^{1/2}\bigr]\bigr)(x)$. As $x$ was arbitrary, claim 4 follows.

Proof of claim 6. Let $v:\mathbb{R}^{n}\to\mathbb{R}$ be free of the $j$th coordinate for every $j\in[n]$, and let $x\in\mathbb{R}^{n}$. For $k\in\mathbb{N}$ let $y^{(k)}$ be the point of $\mathbb{R}^{n}$ whose $j$th coordinate is $0$ for every $j\in[n]$ with $j\le k$ and is $x_{j}$ for every $j\in[n]$ with $k<j$; this is well defined because the order of $\mathbb{N}$ is total by claim 3 of Properties of the Order on the Natural Numbers. Put $A=\{k\in\mathbb{N}:v(y^{(k)})=v(x)\}$.

$1\in A$. Every $j\in[n]$ satisfies $1\le j$ by claim 4 of Properties of the Order on the Natural Numbers, so the $j$th coordinate of $y^{(1)}$ is $0$ when $j=1$ and $x_{j}$ when $1<j$; that is, $y^{(1)}=x[1{:}0]$. Since $1\le n$ we have $1\in[n]$, so freeness of the first coordinate gives $v(y^{(1)})=v(x)$.

$A$ is closed under the successor map $S$. Let $k\in A$. By claim 3 of Properties of the Order on the Natural Numbers either $S(k)\le n$ or $n<S(k)$.

Suppose first $S(k)\le n$, so $S(k)\in[n]$. We claim $y^{(S(k))}=y^{(k)}[S(k){:}0]$. The $S(k)$th coordinates are both $0$. Let $j\in[n]$ with $j\ne S(k)$. If $j\le S(k)$ then $j\le k$ by claim 5 of Properties of the Order on the Natural Numbers, so both points have $j$th coordinate $0$. Otherwise $S(k)<j$, and $k<S(k)$ by claim 5 of that lemma, so $k<j$ and both points have $j$th coordinate $x_{j}$. The claim follows, and freeness of the $S(k)$th coordinate gives $v(y^{(S(k))})=v(y^{(k)})=v(x)$.

Suppose instead $n<S(k)$. Then $n\le S(k)$ and $n\ne S(k)$, so $n\le k$ by claim 5 of Properties of the Order on the Natural Numbers, and $k\le S(k)$ by that same claim. Every $j\in[n]$ then satisfies $j\le n\le k\le S(k)$, so $y^{(k)}$ and $y^{(S(k))}$ have $j$th coordinate $0$ for every $j\in[n]$ and are equal; hence $v(y^{(S(k))})=v(x)$.

In both cases $S(k)\in A$. By Principle of Induction for the Natural Numbers, $A=\mathbb{N}$; in particular $n\in A$. Every $j\in[n]$ satisfies $j\le n$, so $y^{(n)}$ is the point $\mathbf{0}$ all of whose coordinates are $0$, and $v(x)=v(y^{(n)})=v(\mathbf{0})$. Since $x$ was arbitrary, $c=v(\mathbf{0})$ has the required property.

Proof of claim 7. Suppose $n\le3$, let $i\in[n]$ and let $w\in C_{\mathrm{per}}$ satisfy $0\le w(y)$ for every $y\in\mathbb{R}^{n}$. By claim 1, $P_{i}w\in C_{\mathrm{per}}$ and $0\le(P_{i}w)(y)$ for every $y\in\mathbb{R}^{n}$.

Membership, and reduction to an identity on $\mathbb{R}^{n}$. By Elementary Properties of Lattice-Periodic Functions (iii) there are nonnegative real numbers bounding $|w|$ and $|P_{i}w|$ on $\mathbb{R}^{n}$, so Continuous Periodic Functions are Power-Integrable and Dense on the Torus (i), applied with $p=1$ to $w$ and to $P_{i}w$, shows that $w|_{Q}$ and $(P_{i}w)|_{Q}$ are measurable with respect to $\mathcal{B}_{Q}$ and belong to $\mathcal{L}^{1}(\mathbb{T}^{n})$. Put

$$f=\mathbf{1}_{Q}\,w,\qquad h=\mathbf{1}_{Q}\,(P_{i}w),$$

maps from $\mathbb{R}^{n}$ to $\mathbb{R}$. By Continuous Periodic Functions are Power-Integrable and Dense on the Torus (ii), applied to $w$ and to $P_{i}w$, both are measurable with respect to $\mathcal{B}(\mathbb{R}^{n})$ and integrable with respect to $\lambda_{n}$, and

$$\int_{\mathbb{T}^{n}}w|_{Q}\,dx=\int_{\mathbb{R}^{n}}f\,d\lambda_{n},\qquad\int_{\mathbb{T}^{n}}(P_{i}w)|_{Q}\,dx=\int_{\mathbb{R}^{n}}h\,d\lambda_{n}.$$

The indicator $\mathbf{1}_{Q}$ takes only the values $0$ and $1$, so $f$ and $h$ are nonnegative: at $y\in Q$ their values are $w(y)$ and $(P_{i}w)(y)$, and at $y\notin Q$ both are $0$ by claim 1 of Zero Products and Elementary Identities in a Field. By (P6) they are therefore measurable as maps into $[0,\infty]$, and the two displayed real numbers are their integrals in $[0,\infty]$. It suffices to prove

$$\int_{\mathbb{R}^{n}}f\,d\lambda_{n}=\int_{\mathbb{R}^{n}}h\,d\lambda_{n}\qquad\text{in }[0,\infty].$$

Step A: the slices in the $i$th coordinate. We claim that for every $x\in\mathbb{R}^{n}$ the maps $s\mapsto f(x[i{:}s])$ and $s\mapsto h(x[i{:}s])$ are measurable with respect to $\mathcal{B}(\mathbb{R})$, take nonnegative values, and have the same integral in $[0,\infty]$.

Fix $x\in\mathbb{R}^{n}$ and let $c$ be the real number $1$ if $0\le x_{k}$ and $x_{k}<1$ for every $k\in[n]$ with $k\ne i$, and $0$ otherwise. By The Flat Torus: Standing Notation (ii) and The Half-Open Unit Cell Tiles Euclidean Space (i), $Q$ is the set of $y\in\mathbb{R}^{n}$ with $0\le y_{k}$ and $y_{k}<1$ for every $k\in[n]$. The point $x[i{:}s]$ has $i$th coordinate $s$ and $k$th coordinate $x_{k}$ for every $k\ne i$, so $x[i{:}s]\in Q$ holds exactly when $c=1$ and $s\in[0,1)$, with $[0,1)$ as in (P4). Hence

$$\mathbf{1}_{Q}(x[i{:}s])=c\,\mathbf{1}_{[0,1)}(s)\qquad(s\in\mathbb{R}).$$

By claim 2 the map $P_{i}w$ is free of the $i$th coordinate, so $(P_{i}w)(x[i{:}s])=(P_{i}w)(x)$ for every $s\in\mathbb{R}$ and

$$h(x[i{:}s])=\alpha\,\mathbf{1}_{[0,1)}(s),\qquad\text{where }\alpha=c\,(P_{i}w)(x).$$

The number $\alpha$ is nonnegative, by claim 1 and claim 5 of Elementary Arithmetic in an Ordered Field together with claim 1 of Zero Products and Elementary Identities in a Field. The map $s\mapsto h(x[i{:}s])$ is measurable by claims 1 and 2 of Arithmetic of Measurable Real- and [0,∞]-Valued Maps and nonnegative, so by (P6) and claim 1 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion, together with $\int_{\mathbb{R}}\mathbf{1}_{[0,1)}\,d\lambda=1$ from (P4),

$$\int_{\mathbb{R}}h(x[i{:}s])\,d\lambda(s)=\alpha .$$

For $f$ we have $f(x[i{:}s])=c\,\mathbf{1}_{[0,1)}(s)\,w(x[i{:}s])$. The map $s\mapsto w(x[i{:}s])$ is continuous by (P1), hence measurable with respect to $\mathcal{B}(\mathbb{R})$ by claim 3(a) of The Borel Sigma-Algebra of a Euclidean Space as a Product, and Measurability of Projections, Sequentially Continuous Maps, and Open and Closed Sets, applied in dimension $1$, together with claim 5 of that lemma applied with $m=1$; so $s\mapsto f(x[i{:}s])$ is measurable by claims 1, 2 and 3 of Arithmetic of Measurable Real- and [0,∞]-Valued Maps, and it is nonnegative.

If $c=0$ then $f(x[i{:}s])=0$ for every $s$, by claim 1 of Zero Products and Elementary Identities in a Field, and also $\alpha=0$; the map is the indicator of the empty set, so its integral is $\lambda(\emptyset)=0=\alpha$ by The Integral of an Indicator Function is the Measure of the Set and Measure, Measure Space, and Probability Measure. If $c=1$ then $f(x[i{:}s])=\mathbf{1}_{[0,1)}(s)\,w(x[i{:}s])$, and this agrees with $\tilde{w}_{i,x}(s)$ at every real $s$ with $s\ne1$: for $s\in[0,1)$ both equal $w(x[i{:}s])$, and for $s\in\mathbb{R}$ with $s\notin[0,1]$ both are $0$, while $[0,1]$ and $[0,1)$ have the same members apart from $1$. The set $\{1\}$ is null by (P4), so the two maps agree almost everywhere, and The Lebesgue Integral and Null Sets (iii) gives that their integrals in $[0,\infty]$ are equal. Since $\tilde{w}_{i,x}$ is nonnegative and integrable, (P6) identifies its integral in $[0,\infty]$ with the real number $(P_{i}w)(x)=\alpha$. In both cases

$$\int_{\mathbb{R}}f(x[i{:}s])\,d\lambda(s)=\alpha=\int_{\mathbb{R}}h(x[i{:}s])\,d\lambda(s),$$

which proves the claim of Step A.

Step B: the iterated integrals. By claim 1 of Finite Products of Lebesgue Measure and Coordinate Integration on $\mathbb{R}^l$ and claim 5 of The Borel Sigma-Algebra of a Euclidean Space as a Product, and Measurability of Projections, Sequentially Continuous Maps, and Open and Closed Sets, for every natural number $m$ the $\sigma$-algebra $\mathcal{B}_{m}$ of the first of those lemmas is $\mathcal{B}(\mathbb{R}^{m})$, with $\mathcal{B}_{1}=\mathcal{B}(\mathbb{R})$ and $\mathcal{B}_{m}=\mathcal{B}_{m-1}\otimes\mathcal{B}(\mathbb{R})$ for $m\ge2$, under the identification of $\mathbb{R}^{m}$ with $\mathbb{R}^{m-1}\times\mathbb{R}$ recorded there; and by Lebesgue Measure on $\mathbb{R}^n$ Lebesgue measure $\lambda_{m}$ on $\mathcal{B}(\mathbb{R}^{m})$ is the measure $\lambda_{m}$ of that lemma, so that $\lambda_{1}=\lambda$ and $\lambda_{m}=\lambda_{m-1}\otimes\lambda$ for $m\ge2$. Every $\lambda_{m}$ is $\sigma$-finite by claim 1 of that lemma. We may therefore apply Tonelli's and Fubini's Theorems to the $\sigma$-finite measure spaces $(\mathbb{R}^{m-1},\mathcal{B}(\mathbb{R}^{m-1}),\lambda_{m-1})$ and $(\mathbb{R},\mathcal{B}(\mathbb{R}),\lambda)$, for $m=2$ and, when $n=3$, for $m=3$. Points of $\mathbb{R}^{2}$ are written $(r,s)$ and points of $\mathbb{R}^{3}$ are written $((r,s),t)$, in accordance with these identifications; thus in $\mathbb{R}^{3}$ the coordinates of $((r,s),t)$ are $r$, $s$ and $t$ in this order.

Let $g$ denote either $f$ or $h$. In each case below the integral $\int_{\mathbb{R}^{n}}g\,d\lambda_{n}$ is expressed as an iterated integral in which, for each value of the outer variables, the innermost integral is one of the integrals of Step A. Those integrals have the same value for $g=f$ and for $g=h$, so at each stage the measurable functions being integrated coincide for $f$ and for $h$, and therefore so do the iterated integrals.

The case $n=1$. Here $i=1$. Under the identification of $\mathbb{R}^{1}$ with $\mathbb{R}$ recorded above we have $\lambda_{1}=\lambda$, and for any $x\in\mathbb{R}^{1}$ the map $s\mapsto x[1{:}s]$ is that identification itself, the point $x[1{:}s]$ having single coordinate $s$. Hence $\int_{\mathbb{R}^{1}}g\,d\lambda_{1}=\int_{\mathbb{R}}g(x[1{:}s])\,d\lambda(s)$.

The case $n=2$. By Tonelli's and Fubini's Theorems applied with $X=Y=\mathbb{R}$,

$$\int_{\mathbb{R}^{2}}g\,d\lambda_{2}=\int_{\mathbb{R}}\Bigl(\int_{\mathbb{R}}g((r,s))\,d\lambda(s)\Bigr)d\lambda(r)=\int_{\mathbb{R}}\Bigl(\int_{\mathbb{R}}g((r,s))\,d\lambda(r)\Bigr)d\lambda(s),$$

the sections and the functions defined by the inner integrals being measurable by that theorem. If $i=2$ we use the first form: for fixed $r$ and every $s$, the point $(r,s)$ equals $x[2{:}s]$ for $x=(r,0)$, so the inner integral is $\int_{\mathbb{R}}g(x[2{:}s])\,d\lambda(s)$ with that $x$. If $i=1$ we use the second form: for fixed $s$ and every $r$, the point $(r,s)$ equals $x[1{:}r]$ for $x=(0,s)$.

The case $n=3$. By Tonelli's and Fubini's Theorems applied with $X=\mathbb{R}^{2}$ and $Y=\mathbb{R}$,

$$\int_{\mathbb{R}^{3}}g\,d\lambda_{3}=\int_{\mathbb{R}^{2}}\Bigl(\int_{\mathbb{R}}g((\theta,t))\,d\lambda(t)\Bigr)d\lambda_{2}(\theta)=\int_{\mathbb{R}}\Bigl(\int_{\mathbb{R}^{2}}g((\theta,t))\,d\lambda_{2}(\theta)\Bigr)d\lambda(t),$$

and for every $t\in\mathbb{R}$ the section $\theta\mapsto g((\theta,t))$ is measurable with respect to $\mathcal{B}(\mathbb{R}^{2})$ by the part of that theorem concerning sections.

If $i=3$ we use the first form: for fixed $\theta=(r,s)$ and every $t$, the point $((r,s),t)$ equals $x[3{:}t]$ for $x=((r,s),0)$, so the inner integral is one of the integrals of Step A.

If $i\in\{1,2\}$ we use the second form and apply Tonelli's and Fubini's Theorems once more, with $X=Y=\mathbb{R}$, to the section $\theta\mapsto g((\theta,t))$ for each fixed $t$:

$$\int_{\mathbb{R}^{2}}g((\theta,t))\,d\lambda_{2}(\theta)=\int_{\mathbb{R}}\Bigl(\int_{\mathbb{R}}g(((r,s),t))\,d\lambda(r)\Bigr)d\lambda(s)=\int_{\mathbb{R}}\Bigl(\int_{\mathbb{R}}g(((r,s),t))\,d\lambda(s)\Bigr)d\lambda(r).$$

For $i=1$ we take the first form: for fixed $s$ and $t$ and every $r$, the point $((r,s),t)$ equals $x[1{:}r]$ for $x=((0,s),t)$. For $i=2$ we take the second: for fixed $r$ and $t$ and every $s$, the point $((r,s),t)$ equals $x[2{:}s]$ for $x=((r,0),t)$.

In every case, by Step A the innermost integrals agree for $g=f$ and $g=h$ at every value of the outer variables. Hence the functions integrated at the next level agree, and, iterating, $\int_{\mathbb{R}^{n}}f\,d\lambda_{n}=\int_{\mathbb{R}^{n}}h\,d\lambda_{n}$. With the reduction at the start of this proof, claim 7 follows.