[proofplan] We first use compact containment to find a positive collar around $K$ that remains inside $U$, so the open-domain mollification is defined near $K$ for small $\varepsilon$. Smoothness follows by rewriting the mollification with the smooth kernel $\eta_\varepsilon$ and differentiating that kernel under the integral, component by component. [Uniform convergence](/page/Uniform%20Convergence) is then a direct consequence of [uniform continuity](/page/Uniform%20Continuity) of $f$ on a compact neighbourhood of $K$ and the normalization of the mollifier. [/proofplan] [step:Choose a compact neighbourhood of $K$ inside $U$] Since $K \subset\subset U$, the set $K$ is compact and lies a positive Euclidean distance from the complement of $U$. Define \begin{align*} \rho:=\operatorname{dist}(K,\mathbb{R}^n \setminus U)=\inf\{|x-y|:x\in K,\ y\in \mathbb{R}^n \setminus U\}. \end{align*} Then $\rho>0$. Define the closed $\rho/2$-neighbourhood of $K$ by \begin{align*} C:=\{y\in \mathbb{R}^n:\operatorname{dist}(y,K)\leq \rho/2\}. \end{align*} The set $C$ is compact and satisfies $C\subset U$. Indeed, compactness follows because $C$ is closed and bounded, while $C\subset U$ follows from the definition of $\rho$: if $y\in C$ and $y\notin U$, then some $x\in K$ satisfies $|y-x|\leq \rho/2$, contradicting $\rho\leq |x-y|$. Let \begin{align*} V:=\{y\in \mathbb{R}^n:\operatorname{dist}(y,K)<\rho/4\}. \end{align*} Then $V$ is an open neighbourhood of $K$. If $0<\varepsilon<\rho/4$, $x\in V$, and $z\in \operatorname{supp}\eta\subset B(0,1)$, choose $a\in K$ with $|x-a|<\rho/4$. Then \begin{align*} |x-\varepsilon z-a|\leq |x-a|+\varepsilon |z|<\rho/4+\rho/4=\rho/2. \end{align*} Hence $x-\varepsilon z\in C\subset U$. Therefore $f_\varepsilon(x)$ is well-defined for every $x\in V$ whenever $0<\varepsilon<\rho/4$. [guided] The only subtle point in mollifying on an [open set](/page/Open%20Set) is that $f$ is not defined outside $U$. We therefore need a uniform amount of room around $K$ before writing down the integral. Compact containment gives exactly this room. Define \begin{align*} \rho:=\operatorname{dist}(K,\mathbb{R}^n \setminus U)=\inf\{|x-y|:x\in K,\ y\in \mathbb{R}^n \setminus U\}. \end{align*} Because $K$ is compact and $U$ is open with $K\subset U$, this distance is positive. Now define \begin{align*} C:=\{y\in \mathbb{R}^n:\operatorname{dist}(y,K)\leq \rho/2\}. \end{align*} This set is closed and bounded, hence compact in $\mathbb{R}^n$. It is also contained in $U$: if $y\in C$ and $y\in \mathbb{R}^n\setminus U$, then for some $x\in K$ we have $|y-x|\leq \rho/2$, which contradicts the definition of $\rho$ as the distance from $K$ to $\mathbb{R}^n\setminus U$. We next choose the open neighbourhood \begin{align*} V:=\{y\in \mathbb{R}^n:\operatorname{dist}(y,K)<\rho/4\}. \end{align*} If $0<\varepsilon<\rho/4$, $x\in V$, and $z\in\operatorname{supp}\eta$, then $|z|<1$. Choosing $a\in K$ with $|x-a|<\rho/4$, the triangle inequality gives \begin{align*} |x-\varepsilon z-a|\leq |x-a|+\varepsilon |z|<\rho/4+\rho/4=\rho/2. \end{align*} Thus $x-\varepsilon z\in C\subset U$. This proves that the formula defining $f_\varepsilon(x)$ only evaluates $f$ at points where $f$ is defined, uniformly for all $x\in V$ and all sufficiently small $\varepsilon$. [/guided] [/step] [step:Differentiate the mollified function by differentiating the kernel] Fix $0<\varepsilon<\rho/4$. For each component index $j\in\{1,\dots,m\}$, let \begin{align*} f_j:U&\to\mathbb{R} \end{align*} denote the $j$th component of $f$. On $V$, the $j$th component of $f_\varepsilon$ is \begin{align*} (f_\varepsilon)_j(x)=\int_{\mathbb{R}^n}\eta(z)f_j(x-\varepsilon z)\,d\mathcal{L}^n(z). \end{align*} Equivalently, using the substitution $y=x-\varepsilon z$, whose Jacobian factor is $\varepsilon^{-n}$ in the inverse direction $z=(x-y)/\varepsilon$, this is \begin{align*} (f_\varepsilon)_j(x)=\int_U \eta_\varepsilon(x-y)f_j(y)\,d\mathcal{L}^n(y), \end{align*} where \begin{align*} \eta_\varepsilon:\mathbb{R}^n&\to\mathbb{R} \end{align*} is defined by \begin{align*} \eta_\varepsilon(w)=\varepsilon^{-n}\eta(w/\varepsilon). \end{align*} For $x\in V$, the integrand is supported in $y\in C$, and $f_j$ is continuous on the compact set $C$, hence bounded there. Let $\alpha=(\alpha_1,\dots,\alpha_n)\in\mathbb{N}_0^n$ be a multi-index. Since $D^\alpha \eta_\varepsilon$ is smooth and compactly supported, differentiating under the integral gives \begin{align*} D^\alpha (f_\varepsilon)_j(x)=\int_U D^\alpha \eta_\varepsilon(x-y)f_j(y)\,d\mathcal{L}^n(y) \end{align*} for every $x\in V$. The integrand is dominated locally in $x$ by a bounded multiple of $\mathbb{1}_C(y)$, because $f_j$ is bounded on $C$ and $D^\alpha\eta_\varepsilon$ is bounded with compact support. Therefore the displayed derivative depends continuously on $x$. Since this holds for every component $j$ and every multi-index $\alpha$, the map \begin{align*} f_\varepsilon:V&\to\mathbb{R}^m \end{align*} is smooth. [/step] [step:Use uniform continuity on the compact neighbourhood to control the error] Since $f:C\to\mathbb{R}^m$ is continuous and $C$ is compact, $f$ is uniformly continuous on $C$. Thus, for every $\gamma>0$, there exists $\delta\in(0,\rho/2)$ such that whenever $p,q\in C$ and $|p-q|<\delta$, one has \begin{align*} |f(p)-f(q)|<\gamma. \end{align*} Fix $\gamma>0$ and choose such a $\delta$. If $0<\varepsilon<\delta$ and $x\in K$, then for every $z\in\operatorname{supp}\eta$ we have $x\in C$, $x-\varepsilon z\in C$, and \begin{align*} |x-\varepsilon z-x|=\varepsilon |z|<\delta. \end{align*} Using the normalization and non-negativity of $\eta$, we estimate \begin{align*} |f_\varepsilon(x)-f(x)|=\left|\int_{\mathbb{R}^n}\eta(z)\bigl(f(x-\varepsilon z)-f(x)\bigr)\,d\mathcal{L}^n(z)\right|. \end{align*} By the triangle inequality for the Euclidean norm in $\mathbb{R}^m$ and the non-negativity of $\eta$, \begin{align*} |f_\varepsilon(x)-f(x)|\leq \int_{\mathbb{R}^n}\eta(z)|f(x-\varepsilon z)-f(x)|\,d\mathcal{L}^n(z). \end{align*} The integrand is bounded above by $\eta(z)\gamma$ on $\operatorname{supp}\eta$ and vanishes outside $\operatorname{supp}\eta$, so \begin{align*} |f_\varepsilon(x)-f(x)|\leq \gamma\int_{\mathbb{R}^n}\eta(z)\,d\mathcal{L}^n(z)=\gamma. \end{align*} Taking the supremum over $x\in K$ gives \begin{align*} \sup_{x\in K}|f_\varepsilon(x)-f(x)|\leq \gamma \end{align*} for every $0<\varepsilon<\delta$. [guided] The convergence is uniform because the same continuity modulus works for every point of $K$. The compact set on which we use continuity is not just $K$, but the slightly larger compact set $C$, because the mollifier samples nearby points $x-\varepsilon z$ rather than only the point $x$ itself. Since $f:C\to\mathbb{R}^m$ is continuous and $C$ is compact, $f$ is uniformly continuous on $C$. Therefore, given $\gamma>0$, there exists $\delta\in(0,\rho/2)$ such that for all $p,q\in C$, \begin{align*} |p-q|<\delta \implies |f(p)-f(q)|<\gamma. \end{align*} Now take $0<\varepsilon<\delta$ and $x\in K$. For every $z\in\operatorname{supp}\eta$, we have $|z|<1$, hence \begin{align*} |x-\varepsilon z-x|=\varepsilon |z|<\delta. \end{align*} Also $x\in K\subset C$ and $x-\varepsilon z\in C$ by the neighbourhood construction. Uniform continuity therefore gives \begin{align*} |f(x-\varepsilon z)-f(x)|<\gamma \end{align*} for every $z\in\operatorname{supp}\eta$. We now average this pointwise estimate against the mollifier. Since $\eta\geq 0$ and its integral is $1$, the triangle inequality gives \begin{align*} |f_\varepsilon(x)-f(x)|\leq \int_{\mathbb{R}^n}\eta(z)|f(x-\varepsilon z)-f(x)|\,d\mathcal{L}^n(z). \end{align*} The integrand is at most $\eta(z)\gamma$ on the support of $\eta$, so \begin{align*} |f_\varepsilon(x)-f(x)|\leq \gamma\int_{\mathbb{R}^n}\eta(z)\,d\mathcal{L}^n(z)=\gamma. \end{align*} The same bound holds for every $x\in K$, so \begin{align*} \sup_{x\in K}|f_\varepsilon(x)-f(x)|\leq \gamma \end{align*} whenever $0<\varepsilon<\delta$. [/guided] [/step] [step:Conclude smoothness near $K$ and uniform convergence on $K$] The first step produced an open neighbourhood $V$ of $K$ such that $f_\varepsilon$ is well-defined on $V$ for every $0<\varepsilon<\rho/4$, and the second step proved that each such $f_\varepsilon$ is smooth on $V$. The third step shows that for every $\gamma>0$ there exists $\delta>0$ such that \begin{align*} 0<\varepsilon<\delta \implies \sup_{x\in K}|f_\varepsilon(x)-f(x)|\leq \gamma. \end{align*} This is precisely \begin{align*} \sup_{x \in K}|f_\varepsilon(x)-f(x)|\to 0 \end{align*} as $\varepsilon\to 0$. Thus the componentwise mollifications are smooth on a neighbourhood of $K$ for all sufficiently small $\varepsilon>0$ and converge uniformly to $f$ on $K$. [/step]