Proof of The Convolution of a Continuous Function with a Continuous Kernel is Continuous

Write $\lVert\,\cdot\,\rVert$ for the Euclidean norm, so that $d(z,w)=\lVert z-w\rVert$ by claim 2 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$, let $\bar B(z,r)$ be the closed ball in $(\mathbb{R}^n,d)$, let $\lambda_n$ be Lebesgue measure on the Borel $\sigma$-algebra of $\mathbb{R}^n$, and for $x\in\Omega^{\delta}$ let $h_x$ be the integrand of Convolution of a Continuous Function with a Compactly Supported Continuous Kernel, so that $(f*\rho)(x)=\int_{\mathbb{R}^n}h_x\,d\lambda_n$. By claim 2 of The Convolution Integrand is Continuous, Compactly Supported and Integrable the number $I=\int_{\mathbb{R}^n}|\rho|\,d\lambda_n$ is real, and $I\ge0$ because $|\rho|\ge0$ and by the monotonicity clause of claim 2 of Linearity and Monotonicity of the Lebesgue Integral.

Let $x_0\in\Omega^{\delta}$ and let $\varepsilon>0$ be real. By claim 1 of The $\delta$-Interior of an Open Subset of $\mathbb{R}^n$ is Open there is a real $\eta>0$ with $K=\bar B(x_0,\delta+\eta)\subseteq\Omega$; by claim 2 of A Closed Euclidean Ball is Convex and Compact the set $K$ is compact in the topology of the sets open in $(\mathbb{R}^n,d)$, which is a topology by Metric Open Sets Form a Topology. Put $\varepsilon'=\varepsilon/(1+I)$, a positive real number.

The restriction of $f$ to $K$ is continuous on $K$, since a $\theta$ witnessing continuity of $f$ at a point relative to $\Omega$ also witnesses it relative to the subset $K$ of $\Omega$. Hence by Heine-Cantor Theorem: Continuity on a Compact Subset Implies Uniform Continuity that restriction is uniformly continuous on $K$: there is a real $\theta_0>0$ such that all $z,z'\in K$ with $d(z,z')<\theta_0$ satisfy $|f(z)-f(z')|<\varepsilon'$.

Let $\theta$ be the smaller of $\theta_0$ and $\eta$, and let $x\in\Omega^{\delta}$ satisfy $d(x,x_0)<\theta$. We claim that

$$|h_x(y)-h_{x_0}(y)|\le\varepsilon'|\rho(y)|\qquad\text{for every }y\in\mathbb{R}^n.$$

If $\lVert y\rVert>\delta$ then $\rho(y)=0$, so each of $h_x(y)$ and $h_{x_0}(y)$ is $0$: either the relevant difference lies outside $\Omega$, and the value is $0$ by definition, or it lies in $\Omega$, and the value is a product with the factor $\rho(y)=0$. Hence both sides are $0$. Suppose instead $\lVert y\rVert\le\delta$. Then $\lVert(x_0-y)-x_0\rVert=\lVert y\rVert\le\delta$ by claim 5 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$, so $x_0-y\in\bar B(x_0,\delta)\subseteq K$; and by claim 6 of the same lemma,

$$\lVert(x-y)-x_0\rVert\le\lVert x-x_0\rVert+\lVert y\rVert<\theta+\delta\le\eta+\delta,$$

so $x-y\in K$ as well. Both points therefore lie in $K\subseteq\Omega$, so $h_x(y)=f(x-y)\rho(y)$ and $h_{x_0}(y)=f(x_0-y)\rho(y)$. Moreover $d(x-y,x_0-y)=\lVert x-x_0\rVert<\theta\le\theta_0$, so $|f(x-y)-f(x_0-y)|<\varepsilon'$ and

$$|h_x(y)-h_{x_0}(y)|=|f(x-y)-f(x_0-y)|\,|\rho(y)|\le\varepsilon'|\rho(y)|,$$

as claimed.

By claim 1 of The Convolution Integrand is Continuous, Compactly Supported and Integrable the functions $h_x$ and $h_{x_0}$ are integrable with respect to $\lambda_n$, so by claim 2 of Linearity and Monotonicity of the Lebesgue Integral their difference is integrable with

$$\int_{\mathbb{R}^n}(h_x-h_{x_0})\,d\lambda_n=(f*\rho)(x)-(f*\rho)(x_0),$$

and $|h_x-h_{x_0}|$ is integrable by Integrable Function and the Lebesgue Integral, while $\varepsilon'|\rho|$ is integrable by claim 2 of Linearity and Monotonicity of the Lebesgue Integral. Using the absolute-value bound and then the monotonicity and linearity clauses of that claim,

$$|(f*\rho)(x)-(f*\rho)(x_0)|\le\int_{\mathbb{R}^n}|h_x-h_{x_0}|\,d\lambda_n\le\int_{\mathbb{R}^n}\varepsilon'|\rho|\,d\lambda_n=\varepsilon' I=\frac{\varepsilon I}{1+I}<\varepsilon .$$

On $\mathbb{R}=\mathbb{R}^{1}$ claim 1 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ identifies the Euclidean norm of a real number with its absolute value, so by claim 2 there $d(s,t)=|s-t|$ for real $s,t$. Since therefore $d\bigl((f*\rho)(x),(f*\rho)(x_0)\bigr)=|(f*\rho)(x)-(f*\rho)(x_0)|$, the function $f*\rho$ is continuous at $x_0$ relative to $\Omega^{\delta}$. As $x_0\in\Omega^{\delta}$ was arbitrary, $f*\rho$ is continuous on $\Omega^{\delta}$.