Proof of The Periodic Extension of a Function on the Unit Cell

Each result cited below is universally quantified over the data appearing in its own statement, and is applied here to the data named in the statement of this lemma. Throughout we use that $\mathcal{B}_{Q}=\{A\in\mathcal{B}(\mathbb{R}^{n}):A\subseteq Q\}$ and that $\lambda_{Q}$ is the restriction of $\lambda_{n}$ to $\mathcal{B}_{Q}$, as fixed in The Flat Torus: Standing Notation §measure; in particular $\mathcal{B}_{Q}\subseteq\mathcal{B}(\mathbb{R}^{n})$ and $\lambda_{Q}(A)=\lambda_{n}(A)$ for $A\in\mathcal{B}_{Q}$, and $\lambda_{Q}(Q)=1$. We also use repeatedly the following domination criterion: if $f$ is measurable and $|f|\le g$ pointwise with $g$ measurable, nonnegative and of finite integral, then $\int|f|\le\int g<\infty$ by claim 1 of Linearity and Monotonicity of the Lebesgue Integral, so $f$ is integrable by Integrable Function and the Lebesgue Integral, a measurable map being integrable exactly when the integral of its absolute value is finite.

Step 1. Proof of claim 1.

Let $x\in\mathbb{R}^{n}$ and $m\in\mathbb{Z}^{n}$. By The Half-Open Unit Cell Tiles Euclidean Space §wrap we have $\pi(x+m)=\pi(x)$, hence $\tilde{v}(x+m)=v(\pi(x+m))=v(\pi(x))=\tilde{v}(x)$; so $\tilde{v}$ is $\mathbb{Z}^{n}$-periodic. By the same clause $\pi(x)=x$ exactly when $x\in Q$, so $\tilde{v}(x)=v(x)$ for $x\in Q$.

Suppose $v$ is measurable with respect to $\mathcal{B}_{Q}$, and let $E$ be a Borel subset of $\mathbb{R}$. The set $A=\{y\in Q:v(y)\in E\}$ lies in $\mathcal{B}_{Q}$, hence in $\mathcal{B}(\mathbb{R}^{n})$. Since $\pi(x)\in Q$ for every $x$, we have $\{x\in\mathbb{R}^{n}:\tilde{v}(x)\in E\}=\{x\in\mathbb{R}^{n}:\pi(x)\in A\}$, which belongs to $\mathcal{B}(\mathbb{R}^{n})$ because $\pi$ is measurable with respect to $\mathcal{B}(\mathbb{R}^{n})$ and $\mathcal{B}(\mathbb{R}^{n})$ by The Half-Open Unit Cell Tiles Euclidean Space §wrap. So $\tilde{v}$ is measurable.

Finally let $u:\mathbb{R}^{n}\to\mathbb{R}$ be $\mathbb{Z}^{n}$-periodic and let $x\in\mathbb{R}^{n}$. By The Half-Open Unit Cell Tiles Euclidean Space §tiling there is exactly one $m\in\mathbb{Z}^{n}$ with $x-m\in Q$, and $\pi(x)=x-m$ by The Half-Open Unit Cell Tiles Euclidean Space §wrap. Hence the periodic extension of $u|_{Q}$ has value $(u|_{Q})(x-m)=u(x-m)=u(x)$ at $x$, the last equality by periodicity of $u$ applied with the lattice vector $-m$, which lies in $\mathbb{Z}^{n}$. So that extension is $u$.

Step 2. Proof of claim 2.

For $x\in\mathbb{R}^{n}$ the value of the periodic extension of $v+c\,v'$ at $x$ is $(v+c\,v')(\pi(x))=v(\pi(x))+c\,v'(\pi(x))=\tilde{v}(x)+c\,\tilde{v}'(x)$, the middle equality being the definition of the pointwise sum and scalar multiple.

Step 3. Proof of claim 3.

Put $N=\{y\in Q:v(y)\ne v'(y)\}$. The map $v-v'$ is measurable with respect to $\mathcal{B}_{Q}$ by Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions, so $N=\{y\in Q:0<|v(y)-v'(y)|\}$ lies in $\mathcal{B}_{Q}$ by claim 4 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions and claim 3 of Rational Intervals and Rays Generate the Borel Sigma-Algebra of the Real Line; and $\lambda_{Q}(N)=0$ by hypothesis, hence $\lambda_{n}(N)=0$.

Since $\tilde{v}(x)=v(\pi(x))$ and $\tilde{v}'(x)=v'(\pi(x))$, we have $\tilde{v}(x)\ne\tilde{v}'(x)$ exactly when $\pi(x)\in N$. By The Half-Open Unit Cell Tiles Euclidean Space §tiling and The Half-Open Unit Cell Tiles Euclidean Space §wrap, $\pi(x)\in N$ holds exactly when $x\in N+m$ for some $m\in\mathbb{Z}^{n}$; so

For each $m$, the set $N+m$ lies in $\mathcal{B}(\mathbb{R}^{n})$ with $\lambda_{n}(N+m)=\lambda_{n}(N)=0$, by claim 1 of Translation and Reflection Invariance of Lebesgue Measure on $\mathbb{R}^n$. The lattice $\mathbb{Z}^{n}$ is the set of $n$-tuples of integers by Lattice-Periodic Functions and the Periodic Function Classes §lattice, hence is countable by claim 1 of The Integers and the Rational Numbers are Countable together with claim 2 of Products and Powers of Countable Sets. So the union above is a countable union of null sets, hence null by The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere §null-union; that is, $\tilde{v}=\tilde{v}'$ almost everywhere.

Step 4. Proof of claim 4.

Let $v\in\mathcal{L}^{p}(\mathbb{T}^{n})$. If $p=1$ there is nothing to prove: $v$ is measurable with $\int_{Q}|v|\,d\lambda_{Q}<\infty$ by Power-Integrable Functions and the p-Seminorm §space, hence integrable by Integrable Function and the Lebesgue Integral, and the asserted inequality reads $\lVert v\rVert_{1}\le\lVert v\rVert_{1}$.

Suppose $1<p$ and let $q$ be the conjugate exponent of $p$, which exists by that clause. We first record that $1^{a}=1$ for every real number $a$ with $0<a$. Indeed $0<1$ by claim 6 of Elementary Order Arithmetic in an Ordered Field, so $0<1^{a}$ by Properties of Real Powers of Nonnegative Real Numbers §values; and $1^{a}=(1\cdot1)^{a}=1^{a}\cdot1^{a}$ by Properties of Real Powers of Nonnegative Real Numbers §product, which gives the power of a product as the product of the powers. Multiplying by the inverse of the nonzero number $1^{a}$ yields $1=1^{a}$.

Let $\mathbf{e}:Q\to\mathbb{R}$ be the map with constant value $1$, which is measurable with respect to $\mathcal{B}_{Q}$ and is the indicator of $Q$ read as a map on $Q$. By the previous paragraph and claim 1 of Properties of the Absolute Value in an Ordered Field we have $|\mathbf{e}|^{q}=\mathbf{e}$, so

by The Integral of an Indicator Function is the Measure of the Set. Hence $\mathbf{e}\in\mathcal{L}^{q}(\mathbb{T}^{n})$ by Power-Integrable Functions and the p-Seminorm §space, and $\lVert\mathbf{e}\rVert_{q}=1^{1/q}=1$ by Power-Integrable Functions and the p-Seminorm §seminorm and the previous paragraph. The pointwise product $v\,\mathbf{e}$ equals $v$. By Hoelder's Inequality, for Two and for Finitely Many Factors §holder applied to $v\in\mathcal{L}^{p}$ and $\mathbf{e}\in\mathcal{L}^{q}$, that product lies in $\mathcal{L}^{1}(\mathbb{T}^{n})$ and

Membership in $\mathcal{L}^{1}(\mathbb{T}^{n})$ is exactly integrability with respect to $\lambda_{Q}$, by Power-Integrable Functions and the p-Seminorm §space and Integrable Function and the Lebesgue Integral.

Step 5. Proof of claim 5.

Let $v$ be measurable with respect to $\mathcal{B}_{Q}$ and integrable with respect to $\lambda_{Q}$. By claim 1 the extension $\tilde{v}$ is measurable and $\mathbb{Z}^{n}$-periodic, and $\mathbf{1}_{Q}\tilde{v}$, the map on $\mathbb{R}^{n}$ agreeing with $v$ on $Q$ and vanishing off $Q$, is integrable with respect to $\lambda_{n}$ with

by claim 1 of Assembly of Measure Spaces: Restriction, Transport, One-Point Spaces, and Countable Disjoint Unions, which identifies integration against the restricted measure with integration of the zero extension, and by The Flat Torus: Standing Notation §measure. Applying The Half-Open Unit Cell Tiles Euclidean Space §translate-integrable to the measurable periodic map $\tilde{v}$ and to $h$ gives that $\mathbf{1}_{Q+h}\tilde{v}$ is integrable with the same integral, which is the displayed identity of claim 5.

Now let $B\in\mathcal{B}(\mathbb{R}^{n})$ be bounded. Choose a real $\Lambda$ with $0<\Lambda$ such that $\lVert x\rVert\le\Lambda$ for every $x\in B$, and, by The Archimedean Property of the Real Numbers, a natural number $N$ with $\Lambda+1\le N$. Put

$G$ is finite. Let $J=\{j\in\mathbb{Z}:-N\le j\le N\}$. The initial segment $[2N+1]$ is finite by claim 1 of Basic Properties of Finite Sets, and the map sending $k\in[2N+1]$ to $k-N-1$ takes values in $J$ and is onto $J$: it takes values in $\mathbb{Z}$ by claim 2 of Arithmetic, Order and Discreteness of the Integers, and $-N\le k-N-1\le N$ because $1\le k\le 2N+1$; conversely a given $j\in J$ is the value at $k=j+N+1$, which satisfies $1\le k\le 2N+1$ and is a natural number by claim 1 of Arithmetic, Order and Discreteness of the Integers. Hence $J$ is finite by claim 4 of Basic Properties of Finite Sets, and $G$, being the set of $n$-tuples with all entries in $J$, is finite by claim 3 of Finiteness of Cartesian Products, Tuple Sets, and Permutation Sets.

$B$ is covered by the cells indexed by $G$. Let $x\in B$ and let $m\in\mathbb{Z}^{n}$ be the unique lattice vector with $x-m\in Q$, given by The Half-Open Unit Cell Tiles Euclidean Space §tiling. For each $i\in[n]$ we have $0\le x_{i}-m_{i}<1$, so $x_{i}-1<m_{i}\le x_{i}$; and $|x_{i}|\le\lVert x\rVert\le\Lambda$ by claim 4 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$. Hence $-N\le-\Lambda-1<m_{i}\le\Lambda\le N$, so $m\in G$ and $x\in Q+m$.

Consequently, for every $x\in\mathbb{R}^{n}$,

since for $x\in B$ the right-hand side contains the term with the $m$ just produced and all terms are nonnegative, while for $x\notin B$ the left-hand side is $0$. The map $|\tilde{v}|$ is measurable by claim 4 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions and is $\mathbb{Z}^{n}$-periodic, being the periodic extension of $|v|$; and $\mathbf{1}_{Q}|\tilde{v}|$ is integrable, because $|v|$ is integrable by Integrable Function and the Lebesgue Integral and claim 1 of Assembly of Measure Spaces: Restriction, Transport, One-Point Spaces, and Countable Disjoint Unions applies as above. So The Half-Open Unit Cell Tiles Euclidean Space §translate-integrable gives that each $\mathbf{1}_{Q+m}|\tilde{v}|$ is integrable, and the finite sum over $m\in G$ is integrable by claim 2 of Linearity and Monotonicity of the Lebesgue Integral, applied repeatedly over the finitely many summands. The map $\mathbf{1}_{B}\tilde{v}$ is measurable by Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions, since $B\in\mathcal{B}(\mathbb{R}^{n})$, and $|\mathbf{1}_{B}\tilde{v}|=\mathbf{1}_{B}|\tilde{v}|$ by claim 4 of Properties of the Absolute Value in an Ordered Field; the domination criterion now shows that $\mathbf{1}_{B}\tilde{v}$ is integrable.

Step 6. Proof of claim 6.

Let $v\in\mathcal{L}^{p}(\mathbb{T}^{n})$. The map $|v|^{p}$ is measurable with respect to $\mathcal{B}_{Q}$ by Power-Integrable Functions and the p-Seminorm §measurable-power, and $\int_{Q}|v|^{p}\,d\lambda_{Q}<\infty$ by Power-Integrable Functions and the p-Seminorm §space; being nonnegative, it is therefore integrable with respect to $\lambda_{Q}$ by Integrable Function and the Lebesgue Integral. For $x\in\mathbb{R}^{n}$ the periodic extension of $|v|^{p}$ has value $\bigl(|v(\pi(x))|\bigr)^{p}=\bigl(|\tilde{v}(x)|\bigr)^{p}$, which is the value of $|\tilde{v}|^{p}$ at $x$; so that extension is $|\tilde{v}|^{p}$.

Claim 5 applied to $|v|^{p}$ in place of $v$ therefore gives that $\mathbf{1}_{B}|\tilde{v}|^{p}$ is integrable for every bounded $B\in\mathcal{B}(\mathbb{R}^{n})$ and that, for every $h\in\mathbb{R}^{n}$,

Finally, $\lVert v\rVert_{p}$ is the power of $\int_{Q}|v|^{p}\,d\lambda_{Q}$ with exponent $1/p$ by Power-Integrable Functions and the p-Seminorm §seminorm, so raising it to the power $p$ returns $\int_{Q}|v|^{p}\,d\lambda_{Q}$ by Properties of Real Powers of Nonnegative Real Numbers §inverse. This is the displayed identity of claim 6. $\blacksquare$