Proof of Existence of Mollifier Kernels of Every Radius

Throughout, $\mathbb{R}^{n}$ and $\mathbb{R}^{1}$ are open subsets of themselves by claim 1 of Euclidean Space is Open in Itself, and $C^k$ Maps are Continuous; the order arithmetic used below is that of Elementary Order Arithmetic in an Ordered Field and Elementary Arithmetic in an Ordered Field. Smoothness is that of Smooth Map on a Euclidean Open Set.

Step 1 (smoothness of $\rho_{0}$). Let $q:\mathbb{R}^{n}\to\mathbb{R}$ be given by $q(y)=\lVert y\rVert^{2}$; it is smooth on $\mathbb{R}^{n}$ by The Squared Euclidean Norm is Smooth. By claim 2 of Constants, Coordinate Functions, Sums and Products of $C^k$ Functions on a Euclidean Open Set the function on $\mathbb{R}^{n}$ with constant value $\delta^{2}$ is smooth on $\mathbb{R}^{n}$, and by claim 3 there so is the function $N:\mathbb{R}^{n}\to\mathbb{R}$ given by

Regard $N$ as a map from $\mathbb{R}^{n}$ into $\mathbb{R}^{1}$ with single coordinate function $N$, in accordance with the scalar convention of clause 3 of C^k Maps on a Euclidean Open Set; its values lie in $\mathbb{R}^{1}$. By claim 2 of The Exponential Bump Building Block is Smooth on the Real Line, $\varphi$ is smooth on $\mathbb{R}^{1}$. Claim 3 of A Composition of $C^k$ Maps Between Euclidean Open Sets is of Class $C^k$, applied with $U=\mathbb{R}^{n}$, $V=\mathbb{R}^{1}$, $F=N$ and $G=\varphi$, therefore shows that $\varphi\circ N$ is smooth on $\mathbb{R}^{n}$. Since $(\varphi\circ N)(y)=\rho_{0}(y)$ for every $y\in\mathbb{R}^{n}$, the function $\rho_{0}$ is smooth on $\mathbb{R}^{n}$.

Step 2 (sign and support). By claim 1 of The Exponential Bump Building Block is Smooth on the Real Line we have $0\le\varphi(s)$ for every real $s$; hence $0\le\rho_{0}(y)$ for every $y\in\mathbb{R}^{n}$.

Let $y\in\mathbb{R}^{n}$ satisfy $\delta<\lVert y\rVert$. By claim 1 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ we have $0\le\lVert y\rVert$, and $0\le\delta$ since $0<\delta$. Claim 1 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field therefore gives $\delta^{2}<\lVert y\rVert^{2}$, and adding $-\lVert y\rVert^{2}$ to both sides (claim 1 of Elementary Order Arithmetic in an Ordered Field) gives $N(y)<0$. If $0<\varphi(N(y))$ held, then claim 1 of The Exponential Bump Building Block is Smooth on the Real Line would force $0<N(y)$; together with $N(y)<0$ this would give $0<0$ by claim 2 of Elementary Order Arithmetic in an Ordered Field, which is impossible because $0<0$ includes $0\ne 0$. Hence $0<\varphi(N(y))$ fails. Since $0\le\varphi(N(y))$, and since $0<t$ means $0\le t$ together with $0\ne t$, we conclude $\rho_{0}(y)=\varphi(N(y))=0$.

Step 3 (compact support, continuity and integrability). Since $0<\delta$ and $\rho_{0}(y)=0$ for every $y$ with $\delta<\lVert y\rVert$, claim 2 of Compact Support on $\mathbb{R}^n$ Means Vanishing Outside a Bounded Set shows that $\rho_{0}$ is compactly supported. By Step 1 and claim 3 of Euclidean Space is Open in Itself, and $C^k$ Maps are Continuous, applied with $U=\mathbb{R}^{n}$, with $m=1$ and with $\rho_{0}$ as the map, $\rho_{0}$ is continuous at every point of $\mathbb{R}^{n}$ as a map from $(\mathbb{R}^{n},d_{E})$ into $(\mathbb{R},d_{\mathbb{R}})$, where $d_{\mathbb{R}}$ is the metric of The Absolute Value Metric on the Real Line. Claim 2 of A Continuous Compactly Supported Function on $\mathbb{R}^n$ is Bounded and Integrable therefore shows that $\rho_{0}$ is integrable with respect to $\lambda_{n}$.

Step 4 (positivity of the integral). Let $0_{\mathbb{R}^{n}}$ be the origin of $\mathbb{R}^{n}$. By claim 3 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$, $\lVert 0_{\mathbb{R}^{n}}\rVert=0$, hence $\lVert 0_{\mathbb{R}^{n}}\rVert^{2}=0$ by claim 1 of Zero Products and Elementary Identities in a Field, and therefore $N(0_{\mathbb{R}^{n}})=\delta^{2}$. Since $0<\delta$, claim 5 of Elementary Order Arithmetic in an Ordered Field gives $0<\delta^{2}$, so claim 1 of The Exponential Bump Building Block is Smooth on the Real Line gives $0<\varphi(\delta^{2})=\rho_{0}(0_{\mathbb{R}^{n}})$. Together with Steps 2 and 3, claim 3 of A Continuous Compactly Supported Function on $\mathbb{R}^n$ is Bounded and Integrable applies to $\rho_{0}$ and yields

This completes the proof of claim 1.

Step 5 (proof of claim 2). Write $J=\int_{\mathbb{R}^{n}}\rho_{0}\,d\lambda_{n}$. By Step 4, $0<J$, so the multiplicative inverse $c=J^{-1}$ exists and $0<c$ by claim 7 of Elementary Order Arithmetic in an Ordered Field. We check the four conditions of Mollifier Kernel of Radius $\delta$ on $\mathbb{R}^n$ for $\rho$.

Condition 1. $\rho$ is the pointwise scalar multiple $c\rho_{0}$, hence smooth on $\mathbb{R}^{n}$ by Step 1 and claim 3 of Constants, Coordinate Functions, Sums and Products of $C^k$ Functions on a Euclidean Open Set.

Condition 2. Let $y\in\mathbb{R}^{n}$. By Step 2, $0\le\rho_{0}(y)$, and $0\le c$; claim 5 of Elementary Arithmetic in an Ordered Field gives $c\cdot 0\le c\,\rho_{0}(y)$, and $c\cdot 0=0$ by claim 1 of Zero Products and Elementary Identities in a Field. Hence $0\le\rho(y)$.

Condition 3. If $\delta<\lVert y\rVert$ then $\rho_{0}(y)=0$ by Step 2, so $\rho(y)=c\cdot 0=0$ by claim 1 of Zero Products and Elementary Identities in a Field.

Condition 4. By Lebesgue Measure on $\mathbb{R}^n$, $\lambda_{n}$ is a measure on the Borel $\sigma$-algebra of $\mathbb{R}^{n}$, so that triple is a measure space. Apply claim 2 of Linearity and Monotonicity of the Lebesgue Integral with $f=g=\rho_{0}$, which is integrable by Step 3, and with $a=c$ and $b=0$: the function $c\rho_{0}+0\rho_{0}$ is integrable with respect to $\lambda_{n}$ and

By claim 1 of Zero Products and Elementary Identities in a Field we have $0\,\rho_{0}(y)=0$ for every $y$ and $0\cdot J=0$; since $0$ is the additive identity of $\mathbb{R}$, the function $c\rho_{0}+0\rho_{0}$ is $\rho$ and the right-hand side is $cJ=J^{-1}J=1$. Hence $\rho$ is integrable with respect to $\lambda_{n}$ and $\int_{\mathbb{R}^{n}}\rho\,d\lambda_{n}=1$.

All four conditions of Mollifier Kernel of Radius $\delta$ on $\mathbb{R}^n$ hold, so $\rho$ is a mollifier kernel of radius $\delta$ on $\mathbb{R}^{n}$. As $n$ with $1\le n$ and $\delta$ with $0<\delta$ were arbitrary, this also proves the final assertion. $\blacksquare$