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 §classes, 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 Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$, $\lVert z\rVert^{2}=\sum_{k=1}^{n}z_{k}^{2}$. 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 Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$, $\lVert y[i{:}s]-z[i{:}s]\rVert^{2}=\sum_{k=1}^{n}c_{k}$ 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 of Nonnegative Real Numbers §inverse, 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

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 Existence of Lebesgue Measure on the Real Line. Their indicators are measurable by claim 1 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions, 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 Measure Spaces and the Lebesgue Integral: Standing Notation §measurable; $h$ is integrable if and only if the integral $\int_{X}h\,d\mu\in[0,\infty]$ of Measure Spaces and the Lebesgue Integral: Standing Notation §integral is finite; and in that case that integral is the real number $\int_{X}h\,d\mu$ of Integrable Function and the Lebesgue Integral.

Indeed, the two readings of measurability agree for real-valued maps, as recorded in Measure Spaces and the Lebesgue Integral: Standing Notation §measurable. With the positive and negative parts of Integrable Function and the 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 Function and the 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, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions, 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, 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 $w$,

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 §periodic.

$P_{i}w$ 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 §uniform, 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

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 its absolute value is measurable by claim 4 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions 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; this last step uses claim 1 of Linearity and Monotonicity of the Lebesgue Integral together with the criterion in Integrable Function and the Lebesgue Integral. Hence, by claim 2 of Linearity and Monotonicity of the Lebesgue Integral used first for linearity, then for the bound $\bigl|\int f\bigr|\le\int|f|$, and then for monotonicity and homogeneity,

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 §classes.

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,

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 §classes.

$w^{a}\in C_{\mathrm{per}}$. 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 of Nonnegative Real Numbers; thus $w^{a}(x)=P_{a}(w(x))$, and $0\le w^{a}(x)$ by Properties of Real Powers of Nonnegative Real Numbers §values. Let $x_{0}\in\mathbb{R}^{n}$ and let $\varepsilon$ be a real number with $0<\varepsilon$. By Properties of Real Powers of Nonnegative Real Numbers §continuity 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 §classes.

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 §classes. 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 §algebra. Let $x\in\mathbb{R}^{n}$. For $s\in[0,1]$,

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, applied to the integrable map $\tilde{w}_{i,x}$ with the real coefficient $v(x)$, gives

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 §algebra, $vw\in C_{\mathrm{per}}$, 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

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 of Nonnegative Real Numbers gives

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 of Nonnegative Real Numbers,

while for $s\notin[0,1]$ one has $\bigl(|f(s)|\bigr)^{2}=0^{2}=0$ by (P5) and Properties of Real Powers of Nonnegative Real Numbers §values. Hence $|f|^{2}=\tilde{v}_{i,x}$ in the notation of Power-Integrable Functions and the p-Seminorm §measurable-power, 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 §space, 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 §conjugate, by the uniqueness asserted there. Applying Hoelder's Inequality, for Two and for Finitely Many Factors §holder to the measure space $(\mathbb{R},\mathcal{B}(\mathbb{R}),\lambda)$ with $p=q=2$ and the functions $f$ and $g$ gives

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 §bounded 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 §member, 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

maps from $\mathbb{R}^{n}$ to $\mathbb{R}$. By Continuous Periodic Functions are Power-Integrable and Dense on the Torus §integral, 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