Let $U\subset \Bbb R^d$ be bounded with Lipschitz boundary $K\subset \bar{U}$ be compact. The capacity of $K$ in $U$ is defined by \begin{align*} \text{Cap}_{p}(K, U) := \inf \left\{ \int_U |\nabla \varphi|^p \, dx : \varphi \in C_c^\infty(\bar{U}), \, \varphi \geq 1 \text{ on } K \right\}. \end{align*}
Given a closed subset $F\subset \Bbb R^d$, we consider the set of smooth functions vanishing in a neighborhood of $F$, that is, \begin{align} C_{c,F}^\infty(U)\Big\{ u\in C_c^\infty(\bar{U}) \,\,|\,\, \text{there is $\delta>0$ such that $u=0$ on $F_\delta$}\Big\} \end{align} where $F_\delta=\{x\in \Bbb R^d\,\,|\,\, dist(x,F)<\delta\}$ is the $\delta$-tubular neighborhood of $F$. Next, we introduce the space $W^{1,p}_F(U)$, $1<p<\infty$ the space of functions that vanish on $F$, as the closure of $C_{c,F}^\infty(U)$ in the usual Sobolev space $W^{1,p}(U)$ that is \begin{align} W^{1,p}_F(U)= \overline{C_{c,F}^\infty(U)}^{W^{1,p}(U)}, \end{align} Question: if $\text{Cap}(K,U)>0$ show that $1\notin W^{1,p}_K(U)$. Any reference, answer or comment is greatly appreciated.
In the simple situation where $F\subset \partial U$ has positive Hausdorff measure(the restriction of the Lebesgue measure on $\partial U$), i.e., $\mathcal{H}(F)>0$ then by trace theorem we know that $1\notin W^{1,p}_F(U).$ In particular if $F= \partial U$ then we have $W^{1,p}_F(U)= W^{1,p}_0(U)$.
Note that if $\mathcal{H}(K)>0$ implies $\text{Cap}(K,U)>0$ but the converse is not true. For instance when $K$ is a Cantor dust we have $\text{Cap}(K,U)>0$ and $\mathcal{H}(K)=0$.
