Proof of Mollification Converges Uniformly on Compact Subsets

Order arithmetic is that of Elementary Order Arithmetic in an Ordered Field and Elementary Arithmetic in an Ordered Field, absolute values have the properties of Properties of the Absolute Value in an Ordered Field, and $2=1+1$. By Lebesgue Measure on $\mathbb{R}^n$ the triple formed by $\mathbb{R}^{n}$, its Borel $\sigma$-algebra and $\lambda_{n}$ is a measure space.

Step 1 (proof of claim 1). By A Compact Subset of an Open Set Admits a Uniform Ball Radius, applied to the metric space $(\mathbb{R}^{n},d)$, the open set $\Omega$ and the compact set $K$, there is a real number $r$ with $0<r$ such that $\bar B(x,r)\subseteq\Omega$ for every $x\in K$.

Put $\varepsilon_{1}=r\,\delta^{-1}$; it is positive by claims 7 and 5 of Elementary Order Arithmetic in an Ordered Field. Let $\varepsilon$ satisfy $0<\varepsilon\le\varepsilon_{1}$ and let $x\in K$. Multiplying $\varepsilon\le\varepsilon_{1}$ by the nonnegative element $\delta$ (claim 5 of Elementary Arithmetic in an Ordered Field) gives $\varepsilon\delta\le\varepsilon_{1}\delta=(r\delta^{-1})\delta=r$. Hence every $y$ with $d(x,y)\le\varepsilon\delta$ satisfies $d(x,y)\le r$, so $\bar B(x,\varepsilon\delta)\subseteq\bar B(x,r)\subseteq\Omega$ by Closed Ball in a Metric Space. Thus $x\in\Omega^{\varepsilon\delta}$, and since $x\in K$ was arbitrary, $K\subseteq\Omega^{\varepsilon\delta}$.

Step 2 (choice of $\varepsilon_{0}$). Let $\eta$ satisfy $0<\eta$ and put $\eta'=\eta\cdot 2^{-1}$, so that $0<\eta'$ and $\eta'<\eta$ by claim 8 of Elementary Order Arithmetic in an Ordered Field.

By Uniform Continuity Along a Compact Subset of the Domain, applied to $(\mathbb{R}^{n},d)$, to $\Omega$, to $K$, to $f$ and to $\eta'$, there is a real number $\theta$ with $0<\theta$ such that

$$|f(z)-f(x)|<\eta'\qquad\text{for every }x\in K\text{ and every }z\in\Omega\text{ with }d(x,z)<\theta .$$

Let $\varepsilon_{0}$ be the smaller of $\varepsilon_{1}$ and $\theta\,\delta^{-1}$ (claim 9 of Elementary Order Arithmetic in an Ordered Field); both are positive, so $0<\varepsilon_{0}$.

Now fix $\varepsilon$ with $0<\varepsilon<\varepsilon_{0}$ and fix $x\in K$. From $\varepsilon<\varepsilon_{0}\le\varepsilon_{1}$ we get $\varepsilon\le\varepsilon_{1}$ (claim 2 of Elementary Order Arithmetic in an Ordered Field), so Step 1 gives $K\subseteq\Omega^{\varepsilon\delta}$ and in particular $\bar B(x,\varepsilon\delta)\subseteq\Omega$. From $\varepsilon<\varepsilon_{0}\le\theta\delta^{-1}$ and claim 10 of Elementary Order Arithmetic in an Ordered Field, multiplying by the positive element $\delta$, we get $\varepsilon\delta<(\theta\delta^{-1})\delta=\theta$.

Step 3 (two convolution integrands). By Rescaling a Mollifier Kernel, $\rho_{\varepsilon}$ is a mollifier kernel of radius $\varepsilon\delta$; by conditions 2, 3 and 4 of Mollifier Kernel of Radius $\delta$ on $\mathbb{R}^n$ it satisfies $0\le\rho_{\varepsilon}(y)$ for every $y$, it vanishes at every $y$ with $\varepsilon\delta<\lVert y\rVert$, and it is integrable with $\int_{\mathbb{R}^{n}}\rho_{\varepsilon}\,d\lambda_{n}=1$. By condition 1 and claim 3 of Euclidean Space is Open in Itself, and $C^k$ Maps are Continuous it is continuous on $\mathbb{R}^{n}$ as a map into $(\mathbb{R},d_{\mathbb{R}})$.

Let $h_{x}:\mathbb{R}^{n}\to\mathbb{R}$ be the integrand of Convolution of a Continuous Function with a Compactly Supported Continuous Kernel for $f$ and the kernel $\rho_{\varepsilon}$ of radius $\varepsilon\delta$ at the point $x$, that is, $h_{x}(y)=f(x-y)\rho_{\varepsilon}(y)$ for those $y$ with $x-y\in\Omega$ and $h_{x}(y)=0$ for all other $y$; thus $(f*\rho_{\varepsilon})(x)=\int_{\mathbb{R}^{n}}h_{x}\,d\lambda_{n}$, and $h_{x}$ is integrable by claim 1 of The Convolution Integrand is Continuous, Compactly Supported and Integrable.

Let $F:\Omega\to\mathbb{R}$ be given by $F(z)=f(z)+(-f(x))$. By claims 1, 2 and 5 of Continuity of Sums and Products of Real-Valued Functions on a Metric Space, $F$ is continuous on $\Omega$ as a map into $(\mathbb{R},d_{\mathbb{R}})$. Let $g_{x}:\mathbb{R}^{n}\to\mathbb{R}$ be the integrand of Convolution of a Continuous Function with a Compactly Supported Continuous Kernel for $F$ and the same kernel at the same point, that is, $g_{x}(y)=F(x-y)\rho_{\varepsilon}(y)$ for those $y$ with $x-y\in\Omega$ and $g_{x}(y)=0$ for all other $y$.

Step 4 ($g_{x}=h_{x}+(-f(x))\rho_{\varepsilon}$ pointwise). Let $y\in\mathbb{R}^{n}$.

If $x-y\in\Omega$, then by distributivity in the field $\mathbb{R}$,

$$g_{x}(y)=\bigl(f(x-y)+(-f(x))\bigr)\rho_{\varepsilon}(y)=f(x-y)\rho_{\varepsilon}(y)+(-f(x))\rho_{\varepsilon}(y)=h_{x}(y)+(-f(x))\rho_{\varepsilon}(y).$$

If $x-y\notin\Omega$, then $\varepsilon\delta<\lVert y\rVert$. Indeed, $x-(x-y)=y$ in the real vector space $\mathbb{R}^{n}$, so claim 2 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square gives $d(x,x-y)=\lVert x-(x-y)\rVert=\lVert y\rVert$; were $\lVert y\rVert\le\varepsilon\delta$, then $x-y$ would lie in $\bar B(x,\varepsilon\delta)\subseteq\Omega$, contrary to assumption, and the order on $\mathbb{R}$ being total this leaves $\varepsilon\delta<\lVert y\rVert$. Hence $\rho_{\varepsilon}(y)=0$, so both sides of the asserted identity are $0$, using claim 1 of Zero Products and Elementary Identities in a Field.

Step 5 (the difference as an integral). By claim 2 of Linearity and Monotonicity of the Lebesgue Integral, applied to the integrable functions $h_{x}$ and $\rho_{\varepsilon}$ with $a=1$ and $b=-f(x)$, the function $1\cdot h_{x}+(-f(x))\rho_{\varepsilon}$ is integrable with respect to $\lambda_{n}$ and

$$\int_{\mathbb{R}^{n}}\bigl(1\cdot h_{x}+(-f(x))\rho_{\varepsilon}\bigr)\,d\lambda_{n}=1\cdot\int_{\mathbb{R}^{n}}h_{x}\,d\lambda_{n}+(-f(x))\int_{\mathbb{R}^{n}}\rho_{\varepsilon}\,d\lambda_{n}.$$

By Step 4 the function on the left is $g_{x}$, and the right-hand side is $(f*\rho_{\varepsilon})(x)+(-f(x))\cdot 1=(f*\rho_{\varepsilon})(x)-f(x)$. Hence

$$\int_{\mathbb{R}^{n}}g_{x}\,d\lambda_{n}=(f*\rho_{\varepsilon})(x)-f(x).$$

Step 6 (the bound). Let $z\in\bar B(x,\varepsilon\delta)$. Then $z\in\Omega$ by Step 2, and $d(x,z)\le\varepsilon\delta<\theta$ by Closed Ball in a Metric Space and Step 2, so $d(x,z)<\theta$ by claim 2 of Elementary Order Arithmetic in an Ordered Field. Since $x\in K$, Step 2 gives $|F(z)|=|f(z)-f(x)|<\eta'$, hence $|F(z)|\le\eta'$.

Thus $\eta'$ is a nonnegative real number bounding $|F|$ on $\bar B(x,\varepsilon\delta)$, and $\bar B(x,\varepsilon\delta)\subseteq\Omega$. Claim 2 of The Convolution Integrand is Continuous, Compactly Supported and Integrable, applied to $\Omega$, to $F$, to the kernel $\rho_{\varepsilon}$ with radius $\varepsilon\delta$, to the point $x$ and to $M=\eta'$, therefore gives

$$\Bigl|\int_{\mathbb{R}^{n}}g_{x}\,d\lambda_{n}\Bigr|\le\eta'\int_{\mathbb{R}^{n}}|\rho_{\varepsilon}|\,d\lambda_{n}.$$

For every $y$ we have $0\le\rho_{\varepsilon}(y)$, and by claim 1 of Properties of the Absolute Value in an Ordered Field the value $|\rho_{\varepsilon}(y)|$ is $\rho_{\varepsilon}(y)$ or $-\rho_{\varepsilon}(y)$; in the second case $0\le-\rho_{\varepsilon}(y)$ gives $\rho_{\varepsilon}(y)\le 0$ by claim 4 of Elementary Order Arithmetic in an Ordered Field and hence $\rho_{\varepsilon}(y)=0=|\rho_{\varepsilon}(y)|$. So $|\rho_{\varepsilon}|$ is the function $\rho_{\varepsilon}$, and its integral is $1$.

Combining with Step 5,

$$\bigl|(f*\rho_{\varepsilon})(x)-f(x)\bigr|\le\eta'\cdot 1=\eta'<\eta ,$$

so $|(f*\rho_{\varepsilon})(x)-f(x)|<\eta$ by claim 2 of Elementary Order Arithmetic in an Ordered Field. Since $x\in K$ and $\varepsilon$ with $0<\varepsilon<\varepsilon_{0}$ were arbitrary, claim 2 follows. $\blacksquare$