Proof of A Continuous Lattice-Periodic Function is Uniformly Continuous

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 $d_{E}(a,b)=\lVert a-b\rVert$, by claim 2 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$, and that $|a_{i}-b_{i}|\le\lVert a-b\rVert$ for every $i\in[n]$, by claim 4 of the same lemma. We also use without further comment that a strict inequality $x<y$ between real numbers entails the weak inequality $x\le y$.

Let $w\in C_{\mathrm{per}}$ and let $\theta\in\mathbb{R}$ with $0<\theta$. Let $Q$ denote the half-open unit cell $\{x\in\mathbb{R}^{n}:0\le x_{i}<1\ \text{for every}\ i\in[n]\}$ fixed in that lemma.

Step 1. A compact box on which $w$ is uniformly continuous.

We first record that $\lVert a\rVert\le n$ for every $a\in Q$. Indeed $0\le a_{i}<1$ for every $i\in[n]$, so $a_{i}\,a_{i}\le a_{i}\,1=a_{i}\le1$, the first inequality by claim 5 of Elementary Arithmetic in an Ordered Field with the nonnegative factor $a_{i}$, and hence $a_{i}\,a_{i}\le1$; by claim 1 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ the number $\lVert a\rVert\,\lVert a\rVert$ is the finite sum of the $a_{i}\,a_{i}$, so it is at most the finite sum of $n$ terms equal to $1$, by Comparison and Absolute Value Bounds for Finite Sums of Real Numbers, and that sum is $n$ by The Sum of $n$ Ones is Strictly Increasing in $n$. Thus $\lVert a\rVert\,\lVert a\rVert\le n$. If $n<\lVert a\rVert$ held, then $n\,n<\lVert a\rVert\,n$ by claim 10 of Elementary Order Arithmetic in an Ordered Field, the factor $n$ being positive because $0<1\le n$ by claims 6 and 2 of that lemma, and $\lVert a\rVert\,n\le\lVert a\rVert\,\lVert a\rVert$ by claim 5 of Elementary Arithmetic in an Ordered Field, the factor $\lVert a\rVert$ being nonnegative because $0<n\le\lVert a\rVert$; so $n\,n<\lVert a\rVert\,\lVert a\rVert$ by claim 2 of Elementary Order Arithmetic in an Ordered Field, while $n\le n\,n$ by claim 5 of Elementary Arithmetic in an Ordered Field together with $1\le n$; combining, $n<\lVert a\rVert\,\lVert a\rVert\le n$, a contradiction. Hence $\lVert a\rVert\le n$.

Put $\Lambda=n+1$ and let $E$ be the closed ball of centre $0$ and radius $\Lambda$ in $\mathbb{R}^{n}$ with its Euclidean distance, which is compact by A Closed Euclidean Ball is Convex and Compact. The map $w$ is continuous on $\mathbb{R}^{n}$, hence continuous relative to $E$, so by Heine-Cantor Theorem: Continuity on a Compact Subset Implies Uniform Continuity it is uniformly continuous on $E$: there is a real number $\delta_{0}$ with $0<\delta_{0}$ such that

Put $r=\min\{\delta_{0}/2,\,1\}$, a positive real number by claims 8 and 9 of Elementary Order Arithmetic in an Ordered Field; note $r<\delta_{0}$ and $r\le1$.

Step 2. Transporting the estimate by periodicity.

Let $x,z\in\mathbb{R}^{n}$ with $\lVert x-z\rVert\le r$. By The Half-Open Unit Cell Tiles Euclidean Space §tiling there is exactly one $m\in\mathbb{Z}^{n}$ with $x-m\in Q$. By the bound recorded in step 1, $\lVert x-m\rVert\le n\le\Lambda$, so $x-m\in E$. By the triangle inequality of claim 6 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$,

so $z-m\in E$ as well.

Now $(x-m)-(z-m)=x-z$, so $d_{E}(x-m,z-m)=\lVert x-z\rVert\le r<\delta_{0}$, and step 1 gives

Since $w$ is $\mathbb{Z}^{n}$-periodic and $m\in\mathbb{Z}^{n}$, we have $w(x)=w((x-m)+m)=w(x-m)$ and likewise $w(z)=w(z-m)$. Therefore $|w(x)-w(z)|<\theta$, and in particular $|w(x)-w(z)|\le\theta$. This is the displayed assertion of the claim, with the radius $r$ just constructed.

Step 3. Uniform continuity in the sense of the definition.

Let $\varepsilon\in\mathbb{R}$ with $0<\varepsilon$. Apply steps 1 and 2 with $\theta=\varepsilon/2$, which is positive by claim 8 of Elementary Order Arithmetic in an Ordered Field, obtaining a positive real $r$ such that $|w(x)-w(z)|\le\varepsilon/2$ whenever $\lVert x-z\rVert\le r$. If $d_{E}(x,z)<r$ then $\lVert x-z\rVert\le r$, so $|w(x)-w(z)|\le\varepsilon/2<\varepsilon$, the last inequality because $0<\varepsilon/2$ and claim 3 of Elementary Order Arithmetic in an Ordered Field. So $w$ meets the condition of Uniformly Continuous Map Between Metric Spaces with the radius $r$ for the tolerance $\varepsilon$. $\blacksquare$