Proof of The Radial Compactification of Euclidean Space: Norm Continuity, a Homeomorphism onto the Open Unit Ball, Radial Cutoffs, and Extension of Test Functions

Each result cited below is universally quantified over the data in its own statement, and is applied to the data named where it is used. Throughout, the vector operations of $\mathbb{R}^{m}$ and their identities are those of Euclidean Space $\mathbb{R}^n$ is a Real Vector Space, and the claims of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ are used for the Euclidean norm: claim 2 for $d_{E}(x,y)=\lVert x-y\rVert$, claim 3 for the vanishing of the norm, claim 5 for $\lVert\lambda x\rVert=|\lambda|\,\lVert x\rVert$ and claim 6 for the triangle inequality.

Claim 1. Let $x,y\in\mathbb{R}^{m}$. From $x=(x-y)+y$ and the triangle inequality, $\lVert x\rVert\le\lVert x-y\rVert+\lVert y\rVert$, hence $\lVert x\rVert-\lVert y\rVert\le\lVert x-y\rVert$ by claim 3 of Elementary Arithmetic in an Ordered Field. Exchanging $x$ and $y$ gives $\lVert y\rVert-\lVert x\rVert\le\lVert y-x\rVert$, and $\lVert y-x\rVert=\lVert(-1)(x-y)\rVert=|-1|\,\lVert x-y\rVert=\lVert x-y\rVert$ by homogeneity. By claim 6 of Properties of the Absolute Value in an Ordered Field the two bounds give

that is, $d_{\mathbb{R}}(N(x),N(y))\le 1\cdot d_{E}(x,y)$. So $N$ is Lipschitz with constant $1$ in the sense of Lipschitz Map Between Metric Spaces; it is uniformly continuous on $\mathbb{R}^{m}$ by A Lipschitz Map is Uniformly Continuous, hence continuous on $\mathbb{R}^{m}$ by A Uniformly Continuous Map Between Metric Spaces Is Continuous.

Claim 2. For $y\in\mathbb{R}^{m}$ one has $d_{E}(0,y)=\lVert 0-y\rVert=\lVert y\rVert$, so $\bar{B}$ and $U$ are exactly the closed ball and the open ball of $(\mathbb{R}^{m},d_{E})$ with centre $0$ and radius $1$, in the sense of Closed Ball in a Metric Space and Open Ball in a Metric Space. Since $\lVert 0\rVert=0\le 1$, the set $\bar{B}$ contains $0$ and is nonempty. By claims 2 and 3 of Elementary Properties of the Closed Ball in a Metric Space, $\bar{B}$ is bounded in $(\mathbb{R}^{m},d_{E})$ and closed in $(\mathbb{R}^{m},\mathcal{T})$; therefore $\bar{B}$ is compact in $(\mathbb{R}^{m},\mathcal{T})$ by Heine-Borel Theorem in $\mathbb{R}^n$. By claim 2 of The Restriction of a Metric to a Subset Induces the Subspace Topology the subsets of $\bar{B}$ open in the metric space $(\bar{B},d_{\bar{B}})$ are exactly the members of the subspace topology, so by claim 2 of Compactness in a Subspace Agrees with Compactness in the Ambient Space the metric space $(\bar{B},d_{\bar{B}})$ is compact.

By Open Ball in a Metric Space is Open, $U\in\mathcal{T}$; by claim 1 of Elementary Properties of the Closed Ball in a Metric Space, $U\subseteq\bar{B}$; and $U=U\cap\bar{B}$ is open in $(\bar{B},d_{\bar{B}})$ by claim 2 of The Restriction of a Metric to a Subset Induces the Subspace Topology.

Claim 3. Write $a_{x}=(1+\lVert x\rVert)^{-1}$ for $x\in\mathbb{R}^{m}$; since $0\le\lVert x\rVert$ we have $1\le 1+\lVert x\rVert$ by claim 3 of Elementary Arithmetic in an Ordered Field, and $0<1$ by claim 6 of Elementary Order Arithmetic in an Ordered Field, so $0<1+\lVert x\rVert$ by claim 2 of that lemma. Hence $a_{x}$ is defined and $0<a_{x}$ by claim 7 of Elementary Order Arithmetic in an Ordered Field; and claim 5 of Elementary Arithmetic in an Ordered Field, applied to $1\le 1+\lVert x\rVert$ with the nonnegative factor $a_{x}$, gives $a_{x}=a_{x}\cdot 1\le a_{x}(1+\lVert x\rVert)=1$. By homogeneity, $\lVert h(x)\rVert=a_{x}\lVert x\rVert$, and from $\lVert x\rVert<1+\lVert x\rVert$ we get $a_{x}\lVert x\rVert<1$ on multiplying by the positive number $a_{x}$, using claim 10 of Elementary Order Arithmetic in an Ordered Field. Hence $h(x)\in U$.

Inverse. Let $x\in\mathbb{R}^{m}$, put $y=h(x)$ and $s=\lVert y\rVert=a_{x}\lVert x\rVert$. Then

so $(1-s)^{-1}=1+\lVert x\rVert$ and $g(y)=(1+\lVert x\rVert)a_{x}x=x$. Thus $g(h(x))=x$ for every $x\in\mathbb{R}^{m}$, so $h$ is injective. Conversely let $y\in U$, put $x=g(y)$ and $t=\lVert x\rVert=(1-\lVert y\rVert)^{-1}\lVert y\rVert$, the factor $(1-\lVert y\rVert)^{-1}$ being positive. Then

so $a_{x}=1-\lVert y\rVert$ and $h(x)=(1-\lVert y\rVert)(1-\lVert y\rVert)^{-1}y=y$. Hence $h$ maps $\mathbb{R}^{m}$ onto $U$ and $g$ is its inverse.

$h$ is Lipschitz with constant $2$. Let $x,y\in\mathbb{R}^{m}$ and write $a=a_{x}$, $b=a_{y}$. Then $h(x)-h(y)=a(x-y)+(a-b)y$, and

Since $a$ and $b$ are positive, $|a-b|=ab\,\bigl|\lVert y\rVert-\lVert x\rVert\bigr|$. Now $0<a\le 1$, and $b\lVert y\rVert=\lVert h(y)\rVert\le 1$ by the first paragraph of the present claim; hence, using claim 1,

By the triangle inequality and homogeneity, $\lVert h(x)-h(y)\rVert\le a\lVert x-y\rVert+|a-b|\,\lVert y\rVert\le 2\lVert x-y\rVert$, that is, $d_{E}(h(x),h(y))\le 2\,d_{E}(x,y)$. So $h$ is Lipschitz with constant $2$, hence continuous on $\mathbb{R}^{m}$ by A Lipschitz Map is Uniformly Continuous and A Uniformly Continuous Map Between Metric Spaces Is Continuous.

$g$ is continuous on $U$. Let $y_{0}\in U$, put $s_{0}=\lVert y_{0}\rVert<1$, $c=2^{-1}(1-s_{0})>0$ and $\delta_{0}=c$. Let $y\in U$ satisfy $d_{E}(y,y_{0})<\delta_{0}$. By claim 1, $\lVert y\rVert\le s_{0}+\lVert y-y_{0}\rVert<s_{0}+c$, so $1-\lVert y\rVert>1-s_{0}-c=c$ and therefore $\beta:=(1-\lVert y\rVert)^{-1}<c^{-1}$. Put $\beta_{0}=(1-s_{0})^{-1}$. As above,

so $|\beta-\beta_{0}|\le c^{-1}\beta_{0}\,\lVert y-y_{0}\rVert$ by claim 1. Since $g(y)-g(y_{0})=\beta(y-y_{0})+(\beta-\beta_{0})y_{0}$ and $\lVert y_{0}\rVert\le 1$, the triangle inequality and homogeneity give

Given a positive real $\varepsilon$, choose $\delta$ to be the least of $\delta_{0}$ and $\varepsilon L^{-1}$, which is positive. Every $y\in U$ with $d_{U}(y,y_{0})=d_{E}(y,y_{0})<\delta$ then satisfies $d_{E}(g(y),g(y_{0}))\le L\lVert y-y_{0}\rVert<L\,\varepsilon L^{-1}=\varepsilon$. By Continuous Map Between Metric Spaces, $g$ is continuous at $y_{0}$ relative to $U$, and as $y_{0}\in U$ was arbitrary, $g$ is continuous on $U$ as a map from $(U,d_{U})$ into $(\mathbb{R}^{m},d_{E})$.

Claim 4. The map $N$ is continuous on $\mathbb{R}^{m}$ by claim 1, so its restriction to $K$ is continuous on $K$ by claim 4 of Semicontinuity and Continuity Under Composition with a Continuous Map. Since $K$ is nonempty and compact in $(\mathbb{R}^{m},\mathcal{T})$, Extreme Value Theorem on a Compact Subset of a Metric Space provides $x_{0}\in K$ with $\lVert x\rVert\le\lVert x_{0}\rVert$ for every $x\in K$. Put $\rho=\lVert h(x_{0})\rVert$, so that $0\le\rho<1$ by claim 3.

Let $x\in K$ and write $s=\lVert x\rVert$, $t=\lVert x_{0}\rVert$, so that $0\le s\le t$. Then

because $t-s\ge 0$ and the two factors are positive. Hence $\lVert h(x)\rVert=(1+s)^{-1}s\le(1+t)^{-1}t=\rho$.

Claim 5. Since $0\le\rho<1$, we have $\sigma-\rho=2^{-1}(1-\rho)>0$ and $1-\sigma=2^{-1}(1-\rho)>0$, so $\rho<\sigma<1$. Put $\psi(y)=(\sigma-\rho)^{-1}(\sigma-N(y))$ for $y\in\mathbb{R}^{m}$. By claim 1 the map $N$ is continuous on $\mathbb{R}^{m}$, and by claims 1, 2, 4 and 5 of Continuity of Sums and Products of Real-Valued Functions on a Metric Space the constant maps are continuous and sums and scalar multiples of continuous real-valued maps are continuous; hence $\psi$ is continuous on $\mathbb{R}^{m}$. By claim 4 of Continuity of the Absolute Value, Maximum and Minimum of Real-Valued Functions on a Metric Space, first $\max\{0,\psi\}$ and then $\chi=\min\{1,\max\{0,\psi\}\}$ are continuous on $\mathbb{R}^{m}$, the constant maps $0$ and $1$ being continuous.

By Minimum of Two Elements of a Totally Ordered Set and Elementary Properties of the Minimum of Two Elements, $\max\{0,\psi(y)\}\ge 0$ and $\chi(y)\le 1$, while $\chi(y)$ is one of the two numbers $1$ and $\max\{0,\psi(y)\}$, both nonnegative; so $0\le\chi(y)\le 1$. If $\lVert y\rVert\le\rho$ then $\sigma-\lVert y\rVert\ge\sigma-\rho>0$, so $\psi(y)\ge 1$ on multiplying by the positive number $(\sigma-\rho)^{-1}$, whence $\max\{0,\psi(y)\}=\psi(y)\ge 1$ and $\chi(y)=1$. If $\sigma\le\lVert y\rVert$ then $\sigma-\lVert y\rVert\le 0$, so $\psi(y)\le 0$, whence $\max\{0,\psi(y)\}=0$ and $\chi(y)=\min\{1,0\}=0$.

Claim 6. Vanishing near the unit sphere. Let $y\in\bar{B}$ satisfy $\sigma\le\lVert y\rVert$. If $y\in U$ then $F(y)=\chi(y)f(g(y))=0$ because $\chi(y)=0$ by claim 5; if $y\notin U$ then $F(y)=0$ by definition.

Behaviour along $h$, and the bound. For $x\in\mathbb{R}^{m}$ we have $h(x)\in U$ by claim 3, so $F(h(x))=\chi(h(x))f(g(h(x)))=\chi(h(x))f(x)$, again by claim 3. If $|f(x)|\le M$ for every $x\in\mathbb{R}^{m}$, then $M\ge 0$; for $y\in U$, claim 4 of Properties of the Absolute Value in an Ordered Field gives $|F(y)|=|\chi(y)|\,|f(g(y))|$; from $0\le\chi(y)\le 1$ by claim 5 of the present lemma, together with $-1\le 0$, which follows from $0<1$ (claim 6 of Elementary Order Arithmetic in an Ordered Field) by the sign reversal of claim 4 there, claim 6 of Properties of the Absolute Value in an Ordered Field gives $|\chi(y)|\le 1$; so claim 5 of Elementary Arithmetic in an Ordered Field, applied to $|\chi(y)|\le 1$ with the nonnegative factor $|f(g(y))|$, gives $|F(y)|\le|f(g(y))|\le M$. For $y\in\bar{B}\setminus U$, $|F(y)|=0\le M$.

Continuity. Let $y_{0}\in\bar{B}$ and let $\varepsilon$ be a positive real number.

Suppose first that $y_{0}\in U$, and put $\delta_{1}=1-\lVert y_{0}\rVert>0$. If $y\in\bar{B}$ satisfies $d_{E}(y,y_{0})<\delta_{1}$, then $\lVert y\rVert\le\lVert y_{0}\rVert+\lVert y-y_{0}\rVert<1$ by claim 1, so $y\in U$. On $U$ the map $y\mapsto\chi(y)f(g(y))$ is continuous relative to $U$: the map $g$ is continuous on $U$ as a map into $(\mathbb{R}^{m},d_{E})$ by claim 3 and $f$ is continuous on $\mathbb{R}^{m}$, so $f\circ g$ is continuous on $U$ by claim 3 of Semicontinuity and Continuity Under Composition with a Continuous Map; the restriction of $\chi$ to $U$ is continuous on $U$ by claim 4 of that lemma; and the pointwise product of two maps continuous on $U$ is continuous on $U$ by claim 5 of Continuity of Sums and Products of Real-Valued Functions on a Metric Space. Hence there is a positive real $\delta_{2}$ such that every $y\in U$ with $d_{E}(y,y_{0})<\delta_{2}$ satisfies $|\chi(y)f(g(y))-\chi(y_{0})f(g(y_{0}))|<\varepsilon$. Let $\delta$ be the least of $\delta_{1}$ and $\delta_{2}$, which is positive. Every $y\in\bar{B}$ with $d_{\bar{B}}(y,y_{0})=d_{E}(y,y_{0})<\delta$ then lies in $U$, so $F(y)=\chi(y)f(g(y))$ and $F(y_{0})=\chi(y_{0})f(g(y_{0}))$, and $|F(y)-F(y_{0})|<\varepsilon$.

Suppose next that $y_{0}\in\bar{B}\setminus U$, so that $\lVert y_{0}\rVert=1$, since $\lVert y_{0}\rVert\le 1$ and not $\lVert y_{0}\rVert<1$. Then $F(y_{0})=0$. Put $\delta=1-\sigma>0$. If $y\in\bar{B}$ satisfies $d_{\bar{B}}(y,y_{0})=d_{E}(y,y_{0})<\delta$, then by claim 1

so $F(y)=0$ by the vanishing established above, and $|F(y)-F(y_{0})|=0<\varepsilon$.

In both cases $F$ is continuous at $y_{0}$ relative to $\bar{B}$ in the sense of Continuous Map Between Metric Spaces, read for the metric space $(\bar{B},d_{\bar{B}})$; as $y_{0}\in\bar{B}$ was arbitrary, $F$ is continuous on $\bar{B}$.