Proof of Rescaling a Mollifier Kernel

Throughout, order arithmetic is that of Elementary Order Arithmetic in an Ordered Field and Elementary Arithmetic in an Ordered Field, absolute values are those of that definition with the properties of Properties of the Absolute Value in an Ordered Field, and $\mathbb{R}^{n}$ is an open subset of itself by claim 1 of Euclidean Space is Open in Itself, and $C^k$ Maps are Continuous. Let $\lambda_{n}$ be Lebesgue measure on the Borel $\sigma$-algebra $\mathcal{B}(\mathbb{R}^{n})$, and let $d_{E}$ be the Euclidean distance, a metric on $\mathbb{R}^{n}$, and $d_{\mathbb{R}}$ the metric of The Absolute Value Metric on the Real Numbers on $\mathbb{R}$.

Step 0 (notation and elementary facts). Put $\alpha=\varepsilon^{-1}$ and $\mu=\alpha^{n}$, so that $\rho_{\varepsilon}(y)=\mu\,\rho(\alpha y)$ for every $y\in\mathbb{R}^{n}$.

Since $0<\varepsilon$, claim 7 of Elementary Order Arithmetic in an Ordered Field gives $0<\alpha$; in particular $\alpha\ne 0$, so $\mu\ne 0$ by claim 4 of Properties of Natural Number Powers in a Field, and $0\le\mu$ by claim 5 of that lemma, whence $0<\mu$. By claims 3 and 2 of Properties of Natural Number Powers in a Field,

$$\mu\,\varepsilon^{n}=\alpha^{n}\varepsilon^{n}=(\alpha\varepsilon)^{n}=1^{n}=1,$$

so $\varepsilon^{n}$ is the multiplicative inverse of $\mu$. Also $|\alpha|=\alpha$: by claim 1 of Properties of the Absolute Value in an Ordered Field the value $|\alpha|$ is $\alpha$ or $-\alpha$, and $|\alpha|=-\alpha$ together with $0\le|\alpha|$ would give $\alpha\le 0$ by claim 4 of Elementary Order Arithmetic in an Ordered Field, contradicting $0<\alpha$. Hence $|\alpha|^{n}=\mu$.

Finally $0<\varepsilon\delta$ by claim 5 of Elementary Order Arithmetic in an Ordered Field, so "mollifier kernel of radius $\varepsilon\delta$" is meaningful. We verify the four conditions of Mollifier Kernel of Radius $\delta$ on $\mathbb{R}^n$ for $\rho_{\varepsilon}$ with $\varepsilon\delta$ in place of $\delta$; conditions 1 to 4 for $\rho$ with radius $\delta$ are available by hypothesis.

Step 1 (smoothness). Apply Partial Derivatives, Continuity and $C^k$ Regularity under a Scaling Substitution with $m=1$, with $U=\mathbb{R}^{n}$, with $f=\rho$, with the point $c$ of that lemma taken to be the origin $0_{\mathbb{R}^{n}}$, with $\lambda=\alpha$ and with the scalar $\mu$ of that lemma taken to be $\mu=\alpha^{n}$; both $\alpha$ and $\mu$ are nonzero by Step 0. Since $0_{\mathbb{R}^{n}}+v=v$ for every $v\in\mathbb{R}^{n}$ by the vector space structure of $\mathbb{R}^{n}$, the set $V$ of that lemma is $\{x\in\mathbb{R}^{n}:\alpha x\in\mathbb{R}^{n}\}=\mathbb{R}^{n}$, and the map $g$ of that lemma is given by $g(x)=\mu\,\rho(\alpha x)=\rho_{\varepsilon}(x)$.

By condition 1 of Mollifier Kernel of Radius $\delta$ on $\mathbb{R}^n$ for $\rho$, the map $\rho$ is smooth on $\mathbb{R}^{n}$. Claim 5 of Partial Derivatives, Continuity and $C^k$ Regularity under a Scaling Substitution therefore gives that $\rho_{\varepsilon}$ is smooth on $\mathbb{R}^{n}$.

Step 2 (nonnegativity). Let $y\in\mathbb{R}^{n}$. By condition 2 for $\rho$ we have $0\le\rho(\alpha y)$, and $0\le\mu$ by Step 0. Claim 5 of Elementary Arithmetic in an Ordered Field gives $\mu\cdot 0\le\mu\,\rho(\alpha y)$, and $\mu\cdot 0=0$ by claim 1 of Zero Products and Elementary Identities in a Field; hence $0\le\rho_{\varepsilon}(y)$.

Step 3 (support). Let $y\in\mathbb{R}^{n}$ satisfy $\varepsilon\delta<\lVert y\rVert$. 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 and Step 0,

$$\lVert\alpha y\rVert=|\alpha|\,\lVert y\rVert=\alpha\,\lVert y\rVert .$$

Multiplying the inequality $\varepsilon\delta<\lVert y\rVert$ by the positive element $\alpha$ (claim 10 of Elementary Order Arithmetic in an Ordered Field) gives $\alpha(\varepsilon\delta)<\alpha\lVert y\rVert$, and $\alpha(\varepsilon\delta)=(\alpha\varepsilon)\delta=\delta$. Hence $\delta<\lVert\alpha y\rVert$, so $\rho(\alpha y)=0$ by condition 3 for $\rho$, and $\rho_{\varepsilon}(y)=\mu\cdot 0=0$ by claim 1 of Zero Products and Elementary Identities in a Field.

Step 4 (unit mass). We first record that $\rho$ is measurable with respect to $\mathcal{B}(\mathbb{R}^{n})$ and the Borel $\sigma$-algebra of the real line. Indeed, $\rho$ is smooth on $\mathbb{R}^{n}$, so claim 3 of Euclidean Space is Open in Itself, and $C^k$ Maps are Continuous makes it continuous at every point of $\mathbb{R}^{n}$ as a map from $(\mathbb{R}^{n},d_{E})$ into $(\mathbb{R},d_{\mathbb{R}})$; by condition 3 for $\rho$ and claim 2 of Compact Support on $\mathbb{R}^n$ Means Vanishing Outside a Bounded Set, applied with the positive real number $\delta$, the map $\rho$ is compactly supported, the topology on $\mathbb{R}^{n}$ being that of the sets open in $(\mathbb{R}^{n},d_{E})$, which is a topology by The Open Subsets of a Metric Space Form a Topology; and claim 2 of A Continuous Compactly Supported Function on $\mathbb{R}^n$ is Bounded and Integrable then gives the asserted measurability. By condition 4 for $\rho$, the map $\rho$ is integrable with respect to $\lambda_{n}$ and $\int_{\mathbb{R}^{n}}\rho\,d\lambda_{n}=1$.

Apply claim 3 of Scaling of Lebesgue Measure and the Lebesgue Integral on $\mathbb{R}^n$ with the nonzero real number $\alpha$ and with $\rho$ in the role of $f$: the map $y\mapsto\rho(\alpha y)$ is integrable with respect to $\lambda_{n}$ and

$$\int_{\mathbb{R}^{n}}\rho(\alpha y)\,d\lambda_{n}(y)=|\alpha|^{-n}\int_{\mathbb{R}^{n}}\rho\,d\lambda_{n}=|\alpha|^{-n},$$

where $|\alpha|^{-n}$ denotes the multiplicative inverse of $|\alpha|^{n}$. By Step 0, $|\alpha|^{n}=\mu$ and the multiplicative inverse of $\mu$ is $\varepsilon^{n}$, so

$$\int_{\mathbb{R}^{n}}\rho(\alpha y)\,d\lambda_{n}(y)=\varepsilon^{n}.$$

By Lebesgue Measure on $\mathbb{R}^n$ the triple $\bigl(\mathbb{R}^{n},\mathcal{B}(\mathbb{R}^{n}),\lambda_{n}\bigr)$ is a measure space. Write $\sigma$ for the map $y\mapsto\rho(\alpha y)$, which has just been shown integrable. Claim 2 of Linearity and Monotonicity of the Lebesgue Integral, applied with $f=g=\sigma$, with $a=\mu$ and with $b=0$, shows that $\mu\sigma+0\sigma$ is integrable with respect to $\lambda_{n}$ and that

$$\int_{\mathbb{R}^{n}}\bigl(\mu\sigma+0\sigma\bigr)\,d\lambda_{n}=\mu\,\varepsilon^{n}+0\cdot\varepsilon^{n}.$$

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

All four conditions of Mollifier Kernel of Radius $\delta$ on $\mathbb{R}^n$ hold for $\rho_{\varepsilon}$ with radius $\varepsilon\delta$, so $\rho_{\varepsilon}$ is a mollifier kernel of radius $\varepsilon\delta$ on $\mathbb{R}^{n}$. $\blacksquare$