[proofplan] We construct a counterexample with the unweighted choice $w\equiv 1$, so the weighted [Sobolev space](/page/Sobolev%20Space) is the ordinary Sobolev space. The domain is the union of two open balls tangent at one boundary point, and the target Sobolev function is constant $0$ on one ball and constant $1$ on the other. This function has weak gradient zero because there is no interior interface, but smooth functions on the closure have a single value at the tangent point. Morrey's boundary continuity for $p>n$ forces any smooth Sobolev approximation to converge at that tangent point to both $0$ and $1$, giving a contradiction. [/proofplan]

[step:Choose two tangent balls and the unweighted Sobolev norm] Fix $n=2$ and $p=3$. Let $e_1=(1,0)\in\mathbb R^2$, and define the open balls \begin{align*} B_-:=B(-e_1,1), \qquad B_+:=B(e_1,1). \end{align*} Set \begin{align*} U:=B_-\cup B_+. \end{align*} Then $U\subset\mathbb R^2$ is open, $B_-\cap B_+=\varnothing$, and \begin{align*} \overline{B_-}\cap\overline{B_+}=\{0\}. \end{align*} Define the measurable weight $w:U\to(0,\infty)$ by $w(x)=1$ for all $x\in U$. Thus $w<\infty$ $\mathcal L^2$-a.e. and $W^{1,3}(U,w)=W^{1,3}(U)$ with the usual norm. [/step]

[step:Define a Sobolev function with different component values] Define \begin{align*} u:U&\to\mathbb R \end{align*} by setting $u(x)=0$ for $x\in B_-$ and $u(x)=1$ for $x\in B_+$. Since $B_-$ and $B_+$ are disjoint open components of $U$, the restrictions $u|_{B_-}$ and $u|_{B_+}$ are constant functions. Hence \begin{align*} \int_U |u|^3\,d\mathcal L^2(x)<\infty \end{align*} because $U$ is bounded, and the weak gradient satisfies $\nabla u=0$ $\mathcal L^2$-a.e. in $U$. Indeed, if $\varphi\in C_c^\infty(U)$ and $i\in\{1,2\}$, then $\varphi|_{B_-}\in C_c^\infty(B_-)$ and $\varphi|_{B_+}\in C_c^\infty(B_+)$. Since constants have zero [weak derivative](/page/Weak%20Derivative) on each ball, \begin{align*} \int_U u\,\partial_{x_i}\varphi\,d\mathcal L^2(x) = \int_{B_+}\partial_{x_i}\varphi\,d\mathcal L^2(x) = 0. \end{align*} Thus $\partial_{x_i}u=0$ in the weak sense on $U$ for each $i$, so $u\in W^{1,3}(U)$. [/step]

[step:Use Morrey boundary continuity on each ball]Assume, for contradiction, that $C^\infty(\overline U)\cap W^{1,3}(U)$ is dense in $W^{1,3}(U)$. Then there exists a sequence $(u_k)_{k=1}^{\infty}$ in $C^\infty(\overline U)\cap W^{1,3}(U)$ such that \begin{align*} \|u_k-u\|_{W^{1,3}(U)}\to 0. \end{align*} Restricting the norm to each component gives \begin{align*} \|u_k\|_{W^{1,3}(B_-)}\to 0 \end{align*} and \begin{align*} \|u_k-1\|_{W^{1,3}(B_+)}\to 0. \end{align*} We apply [Morrey's Inequality](/theorems/62) on the bounded Lipschitz domains $B_-$ and $B_+$. Since $p=3>2=n$, the theorem gives continuous representatives on the closures and, in particular, bounded point-evaluation maps at the boundary point $0$. Therefore there are constants $C_->0$ and $C_+>0$ such that for every $v\in W^{1,3}(B_-)$ with continuous representative $\widetilde v\in C(\overline{B_-})$, \begin{align*} |\widetilde v(0)|\le C_-\|v\|_{W^{1,3}(B_-)} \end{align*} and for every $v\in W^{1,3}(B_+)$ with continuous representative $\widetilde v\in C(\overline{B_+})$, \begin{align*} |\widetilde v(0)|\le C_+\|v\|_{W^{1,3}(B_+)}. \end{align*} Applying the first estimate to $v=u_k|_{B_-}$ gives \begin{align*} |u_k(0)|\le C_-\|u_k\|_{W^{1,3}(B_-)}\to 0. \end{align*} Applying the second estimate to $v=u_k|_{B_+}-1$ gives \begin{align*} |u_k(0)-1|\le C_+\|u_k-1\|_{W^{1,3}(B_+)}\to 0. \end{align*}[/step]

[guided]The important point is that convergence in $W^{1,3}$ on either ball controls the boundary value at the tangent point. This is not an interior-ball argument around $0$, because no open ball centered at $0$ lies inside $U$. Instead we work separately on the two smooth pieces $B_-$ and $B_+$. We first restrict the assumed convergence to the left ball. Since the Sobolev norm on $U=B_-\cup B_+$ is the sum of the component contributions, the convergence \begin{align*} \|u_k-u\|_{W^{1,3}(U)}\to 0 \end{align*} implies \begin{align*} \|u_k-u\|_{W^{1,3}(B_-)}\to 0. \end{align*} But $u=0$ on $B_-$, so this is exactly \begin{align*} \|u_k\|_{W^{1,3}(B_-)}\to 0. \end{align*} Similarly, since $u=1$ on $B_+$, we get \begin{align*} \|u_k-1\|_{W^{1,3}(B_+)}\to 0. \end{align*} Now we use [Morrey's Inequality](/theorems/62). Its hypotheses are satisfied on each component: $B_-$ and $B_+$ are bounded Lipschitz domains in $\mathbb R^2$, and the exponent is $p=3>2=n$. Morrey's theorem gives a continuous representative on the closure of each ball, and point evaluation at any point of the closure is a bounded linear functional on $W^{1,3}$ for that ball. In particular, there are constants $C_->0$ and $C_+>0$ such that \begin{align*} |\widetilde v(0)|\le C_-\|v\|_{W^{1,3}(B_-)} \end{align*} for $v\in W^{1,3}(B_-)$ and \begin{align*} |\widetilde v(0)|\le C_+\|v\|_{W^{1,3}(B_+)} \end{align*} for $v\in W^{1,3}(B_+)$. Because each $u_k$ lies in $C^\infty(\overline U)$, it has one continuous value at the shared closure point $0\in\overline{B_-}\cap\overline{B_+}$. Thus the continuous representative of $u_k|_{B_-}$ takes the value $u_k(0)$ at $0$, and the continuous representative of $u_k|_{B_+}$ takes the same value $u_k(0)$ at $0$. Applying the left estimate to $u_k|_{B_-}$ yields \begin{align*} |u_k(0)|\le C_-\|u_k\|_{W^{1,3}(B_-)}\to 0. \end{align*} Applying the right estimate to $u_k|_{B_+}-1$ yields \begin{align*} |u_k(0)-1|\le C_+\|u_k-1\|_{W^{1,3}(B_+)}\to 0. \end{align*} So the same real number sequence $(u_k(0))$ must converge both to $0$ and to $1$.[/guided]

[step:Derive the contradiction from the single shared boundary value] From the left component estimate, \begin{align*} u_k(0)\to 0. \end{align*} From the right component estimate, \begin{align*} u_k(0)\to 1. \end{align*} A real sequence has at most one limit, so this is impossible. Therefore no sequence in $C^\infty(\overline U)\cap W^{1,3}(U)$ can converge to $u$ in $W^{1,3}(U)$. Since $w\equiv 1$, this says precisely that \begin{align*} C^\infty(\overline U)\cap W^{1,3}(U,w) \end{align*} is not dense in $W^{1,3}(U,w)$. This proves the existential statement. [/step]