Proof of The Periodised Kernel of a Periodic Convolution

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. We use throughout that $d_{E}(a,b)=\lVert a-b\rVert$ and that $|a_{i}|\le\lVert a\rVert$ for every $i\in[n]$, by claims 2 and 4 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square, and the triangle inequality of claim 6 of that lemma. We also use 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, and the Integrability Criterion, so $f$ is integrable by Integrable Functions and Their Lebesgue Integral, a measurable map being integrable exactly when the integral of its absolute value is finite.

Step 0. Bounded sets of lattice vectors are finite.

For a natural number $N$ put $G_{N}=\{m\in\mathbb{Z}^{n}:-N\le m_{i}\le N\ \text{for every}\ i\in[n]\}$. 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 $l\in[2N+1]$ to $l-N-1$ is onto $J$: its values are integers by claim 2 of Arithmetic, Order, Discreteness and Intervals of the Integers and satisfy $-N\le l-N-1\le N$ because $1\le l\le2N+1$, while a given $j\in J$ is its value at $l=j+N+1$, which lies in $[2N+1]$ and is a natural number by claim 1 of Arithmetic, Order, Discreteness and Intervals of the Integers. So $J$ is finite by claim 4 of Basic Properties of Finite Sets, and $G_{N}$, 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.

Consequently: (F) if $\Lambda$ is a nonnegative real number and $T\subseteq\mathbb{Z}^{n}$ is such that $|m_{i}|\le\Lambda$ for every $m\in T$ and every $i\in[n]$, then $T$ is finite. Indeed by The Archimedean Property of the Real Numbers there is a natural number $N$ with $\Lambda\le N$, so $T\subseteq G_{N}$, and subsets of finite sets are finite by claim 3 of Basic Properties of Finite Sets.

Step 1. Proof of claim 1.

Put $S=\{y\in\mathbb{R}^{n}:\psi(y)\ne0\}$. If $y$ lies outside the closed ball $\bar{B}$ of centre $0$ and radius $R$ then $R<\lVert y\rVert$, so $\psi(y)=0$; hence $S\subseteq\bar{B}$. The support of $\psi$ is the closure of $S$, the intersection of all closed sets containing $S$; since $\bar{B}$ is closed by claim 3 of Elementary Properties of the Closed Ball in a Metric Space and bounded by claim 2 of the same lemma, that support is closed and bounded, hence compact by The Heine-Borel Theorem in $\mathbb{R}^n$, and Compactness of Closed Balls and the Unit Sphere, so $\psi$ is compactly supported. Claims 1 and 2 of A Continuous Compactly Supported Function on $\mathbb{R}^n$ is Bounded and Integrable now give that $\psi$ is bounded and integrable with respect to $\lambda_{n}$; fix a real $M$ with $0\le M$ and $|\psi(z)|\le M$ for every $z$. Since $\psi$ is integrable, $|\psi|$ is measurable with finite integral, so $K=\int_{\mathbb{R}^{n}}|\psi|\,d\lambda_{n}$ is a real number, and $0\le K$ by claim 1 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion.

Step 2. Proof of claim 2.

Fix $x,y\in\mathbb{R}^{n}$ and let $m\in S(x,y)$, so $\psi(x-y-m)\ne0$ and therefore $\lVert x-y-m\rVert\le R$. For each $i\in[n]$,

$$|m_{i}|=\bigl|(x_{i}-y_{i})-(x_{i}-y_{i}-m_{i})\bigr|\le|x_{i}-y_{i}|+|x_{i}-y_{i}-m_{i}|\le\lVert x-y\rVert+R,$$

using claim 5 of Properties of the Absolute Value in an Ordered Field. By (F) with $\Lambda=\lVert x-y\rVert+R$, the set $S(x,y)$ is finite.

Since $\mathbb{Z}^{n}$ contains $0$, the set $S(x,y)\cup\{0\}$ is a nonempty finite superset of $S(x,y)$, finite by Peeling an Element off a Finite Set, and Unions of Finite Sets; so at least one admissible index set exists. Let $F_{1},F_{2}\subseteq\mathbb{Z}^{n}$ be nonempty and finite with $S(x,y)\subseteq F_{j}$ for $j\in\{1,2\}$, and put $F=F_{1}\cup F_{2}$, again nonempty and finite by Peeling an Element off a Finite Set, and Unions of Finite Sets. Fix $j$. For $m\in F\setminus F_{j}$ we have $m\notin S(x,y)$, so $\psi(x-y-m)=0$; hence claim 4 of Peeling, Splitting, and Interchange for Sums over a Finite Index Set, on a sum whose terms vanish outside a nonempty subset, applied to the subset $F_{j}$ of $F$, gives

$$\sum_{m\in F}\psi(x-y-m)=\sum_{m\in F_{j}}\psi(x-y-m).$$

As this holds for $j=1$ and for $j=2$, the sums over $F_{1}$ and over $F_{2}$ agree. Thus $\Psi(x,y)$ is well defined.

Step 3. Proof of claim 3.

Periodicity in the second variable. Fix $x$ and let $k\in\mathbb{Z}^{n}$. For every $m$, $\psi\bigl(x-(y+k)-m\bigr)=\psi\bigl(x-y-(k+m)\bigr)$, so $m\in S(x,y+k)$ exactly when $k+m\in S(x,y)$. Let $F$ be finite with $S(x,y)\subseteq F$ and put $F'=\{m\in\mathbb{Z}^{n}:k+m\in F\}$, the image of $F$ under $m'\mapsto m'-k$ and hence finite by claim 4 of Basic Properties of Finite Sets; then $S(x,y+k)\subseteq F'$. The map $m\mapsto k+m$ is a bijection of $F'$ onto $F$, so composing an enumeration of $F'$ with it gives an enumeration of $F$, and the corresponding terms coincide; by A Sum over a Finite Index Set Does Not Depend on the Enumeration, which says a sum over a finite index set may be computed from any enumeration, the two sums are equal. Hence $\Psi(x,y+k)=\Psi(x,y)$, so $y\mapsto\Psi(x,y)$ is $\mathbb{Z}^{n}$-periodic.

Local representation. Fix $x$ and $y_{0}$, and put

$$T_{0}=\{m\in\mathbb{Z}^{n}:\ \psi(x-y-m)\ne0\ \text{for some}\ y\ \text{with}\ \lVert y-y_{0}\rVert\le1\},\qquad T=T_{0}\cup\{0\}.$$

For $m\in T_{0}$, with a witness $y$, we have $\lVert x-y_{0}-m\rVert\le\lVert x-y-m\rVert+\lVert y-y_{0}\rVert\le R+1$, so $|m_{i}|\le\lVert x-y_{0}\rVert+R+1$ for every $i$, exactly as in step 2; by (F) the set $T_{0}$ is finite, hence $T$ is finite by Peeling an Element off a Finite Set, and Unions of Finite Sets, and $T$ is nonempty, so sums indexed by $T$ are defined. For every $y$ with $\lVert y-y_{0}\rVert\le1$ we have $S(x,y)\subseteq T$, so by claim 2,

$$\Psi(x,y)=\sum_{m\in T}\psi(x-y-m).$$

Continuity and smoothness in the second variable. For $m\in\mathbb{Z}^{n}$ let $\alpha_{m}(y)=x-y-m$. Then $\alpha_{m}(y)-\alpha_{m}(y')=y'-y$, so $d_{E}(\alpha_{m}(y),\alpha_{m}(y'))=\lVert y'-y\rVert=\lVert y-y'\rVert$ by claim 5 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square with the scalar $-1$ and claim 2 of Properties of the Absolute Value in an Ordered Field; hence $\alpha_{m}$ is continuous, with $\delta=\varepsilon$ at each point, and $\psi\circ\alpha_{m}$ is continuous by Composition of Continuous Euclidean Maps. If $\psi$ is smooth, then each component $y\mapsto x_{i}-y_{i}-m_{i}$ of $\alpha_{m}$ is of class $C^{k}$ for every natural $k$ by claims 2 and 3 of Constants, Affine Functions, Sums, Products and Quotients of Cᵏ Functions, so $\alpha_{m}$ is smooth and $\psi\circ\alpha_{m}$ is smooth by Composites of Cᵏ Functions Are of Class Cᵏ.

Let $g$ be the map $y\mapsto\sum_{m\in T}\psi(\alpha_{m}(y))$. By induction over the finitely many members of $T$, using claim 5 of Continuity of Sums and Products of Real-Valued Functions on a Metric Space for continuity and claim 3 of Constants, Affine Functions, Sums, Products and Quotients of Cᵏ Functions for class $C^{k}$, the map $g$ is continuous, and is smooth if $\psi$ is smooth. By the local representation, $\Psi(x,\cdot)$ and $g$ agree at every $y$ with $\lVert y-y_{0}\rVert\le1$, hence on the open ball $B$ of centre $y_{0}$ and radius $1$, which is open by Open Ball in a Metric Space is Open.

Continuity of $\Psi(x,\cdot)$ at $y_{0}$ follows, since the two maps agree on the open set $B$ containing $y_{0}$ and $g$ is continuous at $y_{0}$. As $y_{0}$ was arbitrary, $\Psi(x,\cdot)$ is continuous on $\mathbb{R}^{n}$.

For smoothness, suppose $\psi$ is smooth, so that $g$ is smooth. We first record a locality principle. Let $U\subseteq\mathbb{R}^{n}$ be open and let $\phi_{1},\phi_{2}$ agree at every point of $U$. For $y_{1}\in U$ and $i\in[n]$ there is, by Open Subset of a Metric Space, a positive real $s$ with $y_{1}+t\,e_{i}\in U$ whenever $|t|<s$; by Partial Derivative of a Real-Valued Function on an Open Subset of $\mathbb{R}^n$ the partial derivative at $y_{1}$ is the limit as $t$ tends to $0$ of the quotients $(\phi(y_{1}+t\,e_{i})-\phi(y_{1}))/t$, and for $|t|<s$ these quotients agree for $\phi_{1}$ and $\phi_{2}$. Hence $\partial_{i}\phi_{1}(y_{1})$ exists exactly when $\partial_{i}\phi_{2}(y_{1})$ does, and then they are equal. So: if $\phi_{1}$ and $\phi_{2}$ agree on $U$ then their first partial derivatives exist and agree at every point of $U$.

Now argue by induction on the natural number $k$ for the following assertion $P(k)$, quantified over all open sets $U\subseteq\mathbb{R}^{n}$ and all pairs of maps: if $\phi_{1},\phi_{2}:\mathbb{R}^{n}\to\mathbb{R}$ agree at every point of $U$ and $\phi_{2}$ is of class $C^{k}$ on $U$, then $\phi_{1}$ is of class $C^{k}$ on $U$, with the same iterated partial derivatives as $\phi_{2}$ there. For $k=1$: the locality principle applied on $U$ gives that each $\partial_{i}\phi_{1}$ exists on $U$ and equals $\partial_{i}\phi_{2}$ there, and $\partial_{i}\phi_{2}$ is continuous on $U$ since $\phi_{2}$ is of class $C^{1}$; so $\phi_{1}$ is of class $C^{1}$ on $U$ by Functions of Class Cᵏ on an Open Subset of ℝⁿ. For the step, suppose $P(k)$, let $U$ be open, and let $\phi_{1},\phi_{2}$ agree at every point of $U$ with $\phi_{2}$ of class $C^{k+1}$ on $U$. For each $i$ the maps $\partial_{i}\phi_{1}$ and $\partial_{i}\phi_{2}$ exist and agree at every point of $U$, by the locality principle, and $\partial_{i}\phi_{2}$ is of class $C^{k}$ on $U$ because $\phi_{2}$ is of class $C^{k+1}$; so $P(k)$, applied to $U$ and to the pair $\partial_{i}\phi_{1}$, $\partial_{i}\phi_{2}$, gives that $\partial_{i}\phi_{1}$ is of class $C^{k}$ on $U$ with the same iterated partial derivatives as $\partial_{i}\phi_{2}$. With the case $k=1$ this makes $\phi_{1}$ of class $C^{k+1}$ on $U$, with the same iterated partial derivatives as $\phi_{2}$, by Functions of Class Cᵏ on an Open Subset of ℝⁿ. This proves $P(k)$ for every natural number $k$. Since $\Psi(x,\cdot)$ and $g$ agree at every point of the open set $B$ and $g$ is smooth, hence of class $C^{k}$ on $B$ for every $k$, $P(k)$ applied with $U=B$ gives that $\Psi(x,\cdot)$ is of class $C^{k}$ on $B$ for every natural number $k$, with the same iterated partial derivatives as $g$ there.

Finally, Functions of Class Cᵏ on an Open Subset of ℝⁿ requires, for class $C^{k}$ on $\mathbb{R}^{n}$, the existence of the relevant iterated partial derivatives at every point of $\mathbb{R}^{n}$ and their continuity there, and continuity on $\mathbb{R}^{n}$ is continuity at each point. Each point $y_{0}$ has an open ball $B$ about it on which, by the induction just performed, those derivatives exist and are continuous; continuity at $y_{0}$ relative to $\mathbb{R}^{n}$ follows from continuity relative to the open set $B$. Hence $\Psi(x,\cdot)$ is of class $C^{k}$ on $\mathbb{R}^{n}$ for every natural $k$, so smooth by Smooth Functions on an Open Subset of ℝⁿ.

Therefore $\Psi(x,\cdot)\in C_{\mathrm{per}}$, and $\Psi(x,\cdot)\in C^{\infty}_{\mathrm{per}}$ when $\psi$ is smooth, by Lattice-Periodic Functions and the Periodic Function Classes (iii).

The symmetric statement. Fix $y$ and put $\beta_{m}(x)=x-y-m$, which satisfies $\beta_{m}(x)-\beta_{m}(x')=x-x'$ and is therefore continuous, and smooth by the same argument. For $k\in\mathbb{Z}^{n}$ we have $\psi(x+k-y-m)=\psi\bigl(x-y-(m-k)\bigr)$, so $m\in S(x+k,y)$ exactly when $m-k\in S(x,y)$, and the reindexing argument above applies with the bijection $m\mapsto m-k$. The local representation and the continuity and smoothness arguments are unchanged with $x$ in place of $y$. Hence $x\mapsto\Psi(x,y)$ lies in $C_{\mathrm{per}}$, and in $C^{\infty}_{\mathrm{per}}$ when $\psi$ is smooth.

Step 4. Proof of claim 4.

Fix $x$. By claim 3 the map $\Psi(x,\cdot)$ lies in $C_{\mathrm{per}}$, so $\Psi_{x}\in\mathcal{L}^{1}(\mathbb{T}^{n})$ by Continuous Periodic Functions are Power-Integrable and Dense on the Torus (i), used with the exponent $1$.

By The Archimedean Property of the Real Numbers choose a natural number $N$ with $\lVert x\rVert+R+1\le N$. Let $y\in Q$ and $m\in S(x,y)$. Then $0\le y_{i}<1$ for every $i$, so $|y_{i}|\le1$, and as in step 2,

$$|m_{i}|\le|x_{i}|+|y_{i}|+|x_{i}-y_{i}-m_{i}|\le\lVert x\rVert+1+R\le N .$$

Hence $S(x,y)\subseteq G_{N}$ for every $y\in Q$, and by claim 2, $\Psi(x,y)=\sum_{m\in G_{N}}\psi(x-y-m)$ for $y\in Q$. By the triangle inequality for a sum over a finite index set, which follows by induction over the members of $G_{N}$ from the peeling identity of Peeling, Splitting, and Interchange for Sums over a Finite Index Set and claim 5 of Properties of the Absolute Value in an Ordered Field,

$$|\Psi(x,y)|\le\sum_{m\in G_{N}}|\psi(x-y-m)|\qquad(y\in Q).$$

Integrating over $Q$ and using claim 1 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion for monotonicity and for the finitely many summands,

$$\int_{\mathbb{T}^{n}}|\Psi_{x}|\,dy\le\sum_{m\in G_{N}}\int_{Q}|\psi(x-y-m)|\,d\lambda_{Q}(y).$$

Fix $m\in G_{N}$ and put $h(z)=\mathbf{1}_{Q+m}(z)\,|\psi(x-z)|$, a measurable map into $[0,\infty)$. Then $h(y+m)=\mathbf{1}_{Q}(y)\,|\psi(x-y-m)|$, since $y+m\in Q+m$ exactly when $y\in Q$. By claim 2 of Translation and Reflection Invariance of Lebesgue Measure on ℝⁿ, applied with $a=m$, and 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,

$$\int_{Q}|\psi(x-y-m)|\,d\lambda_{Q}(y)=\int_{\mathbb{R}^{n}}h(y+m)\,d\lambda_{n}(y)=\int_{\mathbb{R}^{n}}h\,d\lambda_{n}=\int_{\mathbb{R}^{n}}\mathbf{1}_{Q+m}(z)\,|\psi(x-z)|\,d\lambda_{n}(z).$$

The sets $Q+m$, for $m\in\mathbb{Z}^{n}$, are pairwise disjoint by The Half-Open Unit Cell Tiles Euclidean Space (ii), so $\sum_{m\in G_{N}}\mathbf{1}_{Q+m}(z)\le1$ for every $z$. Summing the previous display over $m\in G_{N}$ and applying claim 1 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion again,

$$\sum_{m\in G_{N}}\int_{Q}|\psi(x-y-m)|\,d\lambda_{Q}(y)=\int_{\mathbb{R}^{n}}\Bigl(\sum_{m\in G_{N}}\mathbf{1}_{Q+m}\Bigr)|\psi(x-\,\cdot\,)|\,d\lambda_{n}\le\int_{\mathbb{R}^{n}}|\psi(x-z)|\,d\lambda_{n}(z)=K,$$

the last equality by claim 2 of Translation and Reflection Invariance of Lebesgue Measure on ℝⁿ, applied with $a=x$ to the measurable nonnegative map $|\psi|$, whose reflection identity gives $\int|\psi(x-z)|\,d\lambda_{n}(z)=\int|\psi|\,d\lambda_{n}$. This proves the bound for $\Psi_{x}$.

For the symmetric statement, fix $y$ and choose $N$ with $\lVert y\rVert+R+1\le N$; the same computation applies with the roles of $x$ and $y$ exchanged, the relevant cells being $Q-m$ for $m\in G_{N}$, which are again pairwise disjoint since $Q-m=Q+(-m)$ and $-m$ runs over lattice vectors.

Step 5. Proof of claim 5.

Let $u\in\mathcal{L}^{1}(\mathbb{T}^{n})$ and fix $x$. The map $\Psi(x,\cdot)$ lies in $C_{\mathrm{per}}$, hence is bounded by Elementary Properties of Lattice-Periodic Functions (iii); say $|\Psi(x,y)|\le M'$ for every $y$. Both $\Psi_{x}$ and $u$ are measurable with respect to $\mathcal{B}_{Q}$, so their product is measurable by claim 3 of Arithmetic of Measurable Real- and [0,∞]-Valued Maps, and $|\Psi_{x}u|\le M'|u|$ pointwise with $M'|u|$ integrable by claim 2 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion; the domination criterion gives that $\Psi_{x}u$ is integrable with respect to $\lambda_{Q}$.

Let $N$ be as in step 4, so that $S(x,y)\subseteq G_{N}$ and $\Psi(x,y)=\sum_{m\in G_{N}}\psi(x-y-m)$ for every $y\in Q$. By Periodic Convolution and Mollification on the Torus (i),

$$(\psi\star u)(x)=\int_{\mathbb{R}^{n}}\psi(x-z)\,\tilde{u}(z)\,d\lambda_{n}(z),$$

where $\tilde{u}$ is the periodic extension of $u$. The integrand vanishes off the union of the sets $Q+m$ with $m\in G_{N}$: if $z$ lies outside that union, let $m$ be the unique lattice vector with $z-m\in Q$, given by The Half-Open Unit Cell Tiles Euclidean Space (ii); then $m\notin G_{N}$, and if $\psi(x-z)\ne0$ we would have $m\in S(x,z-m)\subseteq G_{N}$, a contradiction. Since the cells are pairwise disjoint, $\sum_{m\in G_{N}}\mathbf{1}_{Q+m}$ is the indicator of their union, so

$$\psi(x-z)\,\tilde{u}(z)=\sum_{m\in G_{N}}\mathbf{1}_{Q+m}(z)\,\psi(x-z)\,\tilde{u}(z)\qquad(z\in\mathbb{R}^{n}).$$

Each summand is measurable and dominated in absolute value by $M\,\mathbf{1}_{Q+m}|\tilde{u}|$. Here $|u|$ is measurable with respect to $\mathcal{B}_{Q}$ by claim 4 of Arithmetic of Measurable Real- and [0,∞]-Valued Maps and integrable with respect to $\lambda_{Q}$, since $\int_{Q}|u|\,d\lambda_{Q}$ is finite by Integrable Functions and Their Lebesgue Integral, and the periodic extension of $|u|$ is $|\tilde{u}|$, because $|u|(\pi(z))=|u(\pi(z))|=|\tilde{u}(z)|$ for every $z\in\mathbb{R}^{n}$. So The Periodic Extension of a Function on the Unit Cell (v), applied to $|u|$ with $h=m$, gives that $\mathbf{1}_{Q+m}|\tilde{u}|$ has integral $\int_{\mathbb{T}^{n}}|u|\,dx$, a real number; being measurable and nonnegative, it is therefore integrable by Integrable Functions and Their Lebesgue Integral, and so is $M\,\mathbf{1}_{Q+m}|\tilde{u}|$ by claim 2 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion. Each summand is thus integrable by the domination criterion, and claim 2 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion gives

$$(\psi\star u)(x)=\sum_{m\in G_{N}}\int_{\mathbb{R}^{n}}\mathbf{1}_{Q+m}(z)\,\psi(x-z)\,\tilde{u}(z)\,d\lambda_{n}(z).$$

Fix $m\in G_{N}$ and let $h(z)=\mathbf{1}_{Q+m}(z)\psi(x-z)\tilde{u}(z)$. Then $h(y+m)=\mathbf{1}_{Q}(y)\,\psi(x-y-m)\,\tilde{u}(y+m)=\mathbf{1}_{Q}(y)\,\psi(x-y-m)\,\tilde{u}(y)$, the last step by $\mathbb{Z}^{n}$-periodicity of $\tilde{u}$, which holds by The Periodic Extension of a Function on the Unit Cell (i). By claim 3 of Translation and Reflection Invariance of Lebesgue Measure on ℝⁿ with $a=m$, the map $y\mapsto h(y+m)$ is integrable and $\int_{\mathbb{R}^{n}}h\,d\lambda_{n}=\int_{\mathbb{R}^{n}}h(y+m)\,d\lambda_{n}(y)$.

Let $g_{m}:Q\to\mathbb{R}$ be given by $g_{m}(y)=\psi(x-y-m)\,u(y)$, so that $y\mapsto h(y+m)$ is the extension of $g_{m}$ by the value $0$ off $Q$, using $\tilde{u}=u$ on $Q$. Claim 1 of Assembly of Measure Spaces: Restriction, Transport, One-Point Spaces, and Countable Disjoint Unions identifies the integral of such a zero extension with the integral over $Q$ against $\lambda_{Q}$, but is stated for maps into $[0,\infty]$; we therefore split $g_{m}$ into nonnegative parts. Put

$$g_{m}^{+}=\tfrac{1}{2}\bigl(|g_{m}|+g_{m}\bigr),\qquad g_{m}^{-}=\tfrac{1}{2}\bigl(|g_{m}|-g_{m}\bigr),$$

which are measurable with respect to $\mathcal{B}_{Q}$ by claims 2 and 4 of Arithmetic of Measurable Real- and [0,∞]-Valued Maps, take values in $[0,\infty)$ by claim 3 of Properties of the Absolute Value in an Ordered Field, satisfy $g_{m}=g_{m}^{+}-g_{m}^{-}$ and $g_{m}^{\pm}\le|g_{m}|$, and are integrable by the domination criterion. The zero extension of $g_{m}$ is likewise the difference of the zero extensions of $g_{m}^{+}$ and $g_{m}^{-}$. Applying claim 1 of Assembly of Measure Spaces: Restriction, Transport, One-Point Spaces, and Countable Disjoint Unions to $g_{m}^{+}$ and to $g_{m}^{-}$ separately and subtracting, which is legitimate by claim 2 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion since all four integrals are finite,

$$\int_{\mathbb{R}^{n}}h(y+m)\,d\lambda_{n}(y)=\int_{Q}\psi(x-y-m)\,u(y)\,d\lambda_{Q}(y).$$

Summing over $m\in G_{N}$ and using claim 2 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion once more, together with the representation of $\Psi(x,y)$ on $Q$,

$$(\psi\star u)(x)=\int_{Q}\Bigl(\sum_{m\in G_{N}}\psi(x-y-m)\Bigr)u(y)\,d\lambda_{Q}(y)=\int_{\mathbb{T}^{n}}\Psi_{x}\,u\,dy .$$

Step 6. Proof of claim 6.

By The Archimedean Property of the Real Numbers choose a natural number $N$ with $R+2\le N$. For $x,y\in Q$ and $m\in S(x,y)$ we have $|x_{i}|\le1$ and $|y_{i}|\le1$, so as in step 4, $|m_{i}|\le|x_{i}|+|y_{i}|+R\le R+2\le N$; hence $S(x,y)\subseteq G_{N}$ and

$$\Psi(x,y)=\sum_{m\in G_{N}}\psi(x-y-m)\qquad(x,y\in Q).$$

The projections $(x,y)\mapsto x$ and $(x,y)\mapsto y$ from $Q\times Q$ to $Q$ are measurable with respect to $\mathcal{B}_{Q}\otimes\mathcal{B}_{Q}$ and $\mathcal{B}_{Q}$, since the preimage of $A\in\mathcal{B}_{Q}$ is $A\times Q$, respectively $Q\times A$, a measurable rectangle and hence a member of the product $\sigma$-algebra by Measurable Rectangles and the Product σ-Algebra. For $i\in[n]$ the coordinate map $z\mapsto z_{i}$ on $\mathbb{R}^{n}$ is measurable by claims 1 and 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, and its restriction to $Q$ is measurable with respect to $\mathcal{B}_{Q}$, because for a Borel $A\subseteq\mathbb{R}$ the set $\{z\in Q:z_{i}\in A\}$ is the intersection with $Q$ of a member of $\mathcal{B}(\mathbb{R}^{n})$ and so lies in $\mathcal{B}_{Q}$. Hence for each $i$ the map $(x,y)\mapsto x_{i}-y_{i}-m_{i}$ is measurable with respect to $\mathcal{B}_{Q}\otimes\mathcal{B}_{Q}$, by claims 1 and 2 of Arithmetic of Measurable Real- and [0,∞]-Valued Maps.

By claims 2 and 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, the componentwise criterion, the map $(x,y)\mapsto x-y-m$ from $Q\times Q$ to $\mathbb{R}^{n}$ is measurable with respect to $\mathcal{B}_{Q}\otimes\mathcal{B}_{Q}$ and $\mathcal{B}(\mathbb{R}^{n})$. Since $\psi$ is continuous, it is measurable by claims 3(a) and 5 of the same lemma, so the composite $(x,y)\mapsto\psi(x-y-m)$ is measurable. Finally, by induction over the finitely many members of $G_{N}$ and claim 2 of Arithmetic of Measurable Real- and [0,∞]-Valued Maps, the finite sum displayed above is measurable with respect to $\mathcal{B}_{Q}\otimes\mathcal{B}_{Q}$; that sum is the restriction of $\Psi$ to $Q\times Q$. $\blacksquare$