Ambient-smooth density fails on a slit disc

The set $\mathrm{\Omega }$ is open, because it is the intersection of the open disc $B(0,1)$ with the open set $\{y\mathrm{\ne }0\}\cup \{x<0\}$, and it is bounded; the branch $u$ is of class $C^{\mathrm{\infty }}$ on $\mathrm{\Omega }$ with $0<u<2\pi$ and $\mathrm{\nabla }u(x,y)=(\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2}),\mathrm{\mid }\mathrm{\nabla }u(x,y)\mathrm{\mid }=\frac{1}{r},r=\sqrt{x^2+y^2}\mathrm{.}$

Endpoint estimate. Let $g\in C^1([0,\epsilon ])$ and let $1\le p<\mathrm{\infty }$. For every $s\in (0,\epsilon )$ the fundamental theorem gives $g(0)=g(s)-{\int }_0^s{g}^{\mathrm{\prime }}(t)dt$, so $\epsilon \mathrm{\mid }g(0)\mathrm{\mid }\le {\int }_0^{\epsilon }\mathrm{\mid }g\mathrm{\mid }+\epsilon {\int }_0^{\epsilon }\mathrm{\mid }{g}^{\mathrm{\prime }}\mathrm{\mid }$, and [F5] applied to the two summands yields $\mathrm{\mid }g(0){\mathrm{\mid }}^p\le C(\epsilon ,p)({\int }_0^{\epsilon }\mathrm{\mid }g{\mathrm{\mid }}^p+{\int }_0^{\epsilon }\mathrm{\mid }{g}^{\mathrm{\prime }}{\mathrm{\mid }}^p)$ with $C(\epsilon ,p)=2^{p-1}\max \{{\epsilon }^{-1},{\epsilon }^{p-1}\}$; the same estimate holds on $(-\epsilon ,0)$ for the endpoint $0$.

Integrability. Since $\mathrm{\mid }u\mathrm{\mid }\le 2\pi$ and $\mathrm{\Omega }\subseteq B(0,1)$ has finite area, ${\int }_{\mathrm{\Omega }}\mathrm{\mid }u{\mathrm{\mid }}^p<\mathrm{\infty }$; by [F3], [F7] and $\mathrm{\mid }\mathrm{\nabla }u\mathrm{\mid }=1\mathrm{/}r$ of step 1.1, ${\int }_{\mathrm{\Omega }}\mathrm{\mid }\mathrm{\nabla }u{\mathrm{\mid }}^p={\int }_{\mathrm{\Omega }}{r}^{-p}\le {\int }_0^{1}2\pi r^{1-p}dr=\frac{2\pi }{2-p}<\mathrm{\infty },$ because $p<2$ makes the one-dimensional integral converge at $r=0$.

Membership. The function $u$ is $C^{\mathrm{\infty }}$ on the open set $\mathrm{\Omega }$, so by [F1] its classical partial derivatives of step 1.1 are its weak derivatives; by step 2.1 they lie in $L^p(\mathrm{\Omega })$ together with $u$, and the membership criterion of [F2] gives $u\in W^{1,p}(\mathrm{\Omega })$ for every $1\le p<2$.

Upper strip. Suppose ${\phi }_j\in C^{\mathrm{\infty }}({\mathbb{R}}^2)$ satisfy $\mathrm{\parallel }{\phi }_j{\mathrm{\mid }}_{\mathrm{\Omega }}-u{\mathrm{\parallel }}_{W^{1,p}(\mathrm{\Omega })}\to 0$. Fix $0<\epsilon <1\mathrm{/}4$ and put $S_{+}=(1\mathrm{/}4,3\mathrm{/}4)\times (0,\epsilon )\subset \mathrm{\Omega }$. The function $u=\arctan (y\mathrm{/}x)$ extends $C^1$ to the closed strip, with $u(x,0)=0$. For each $j$, apply the endpoint estimate of step 1.2 to $g_j(x,y)={\phi }_j(x,y)-u(x,y)$ on each vertical section of $S_{+}$ and integrate in $x$ using [F4]. Since $\epsilon$ is fixed, its constant is fixed, and ${\int }_{1\mathrm{/}4}^{3\mathrm{/}4}\mathrm{\mid }{\phi }_j(x,0){\mathrm{\mid }}^{p}dx\le C(\epsilon ,p){\int }_{S_{+}}(\mathrm{\mid }{\phi }_j-u{\mathrm{\mid }}^p+\mathrm{\mid }{\mathrm{\partial }}_y{\phi }_j-{\mathrm{\partial }}_{y}u{\mathrm{\mid }}^p)\le C(\epsilon ,p)\mathrm{\parallel }{\phi }_j-u{\mathrm{\parallel }}_{W^{1,p}(\mathrm{\Omega })}^p⟶0.$

Lower strip. On $S_{-}=(1\mathrm{/}4,3\mathrm{/}4)\times (-\epsilon ,0)$ the function $u=2\pi -\arctan (\mathrm{\mid }y\mathrm{\mid }\mathrm{/}x)$ extends $C^1$ to the closed strip with $u(x,0)=2\pi$. Applying step 1.2 to $g_j={\phi }_j-u$ on each vertical section and integrating in $x$ gives, for the same fixed $\epsilon$, ${\int }_{1\mathrm{/}4}^{3\mathrm{/}4}\mathrm{\mid }{\phi }_j(x,0)-2\pi {\mathrm{\mid }}^{p}dx\le C(\epsilon ,p){\int }_{S_{-}}(\mathrm{\mid }{\phi }_j-u{\mathrm{\mid }}^p+\mathrm{\mid }{\mathrm{\partial }}_y{\phi }_j-{\mathrm{\partial }}_{y}u{\mathrm{\mid }}^p)\le C(\epsilon ,p)\mathrm{\parallel }{\phi }_j-u{\mathrm{\parallel }}_{W^{1,p}(\mathrm{\Omega })}^p⟶0.$

Contradiction. For every $j$ the elementary inequality $\mathrm{\mid }2\pi {\mathrm{\mid }}^p\le 2^{p-1}(\mathrm{\mid }{\phi }_j(x,0){\mathrm{\mid }}^p+\mathrm{\mid }{\phi }_j(x,0)-2\pi {\mathrm{\mid }}^p)$ holds pointwise on $(1\mathrm{/}4,3\mathrm{/}4)$, so $\frac{1}{2}(2\pi {)}^p={\int }_{1\mathrm{/}4}^{3\mathrm{/}4}\mathrm{\mid }2\pi {\mathrm{\mid }}^{p}dx\le 2^{p-1}({\int }_{1\mathrm{/}4}^{3\mathrm{/}4}\mathrm{\mid }{\phi }_j(x,0){\mathrm{\mid }}^{p}dx+{\int }_{1\mathrm{/}4}^{3\mathrm{/}4}\mathrm{\mid }{\phi }_j(x,0)-2\pi {\mathrm{\mid }}^{p}dx)\underset{j\to \mathrm{\infty }}{\overset{}{\to }}0$ by steps 4.1 and 4.2, which is impossible since $(2\pi {)}^p\mathrm{/}2>0$.

Therefore no sequence of globally smooth functions converges to $u$ in $W^{1,p}(\mathrm{\Omega })$ for any $1\le p<2$, so ambient smooth restrictions are not dense on this bounded open set. By contrast [F6] gives that density for every bounded $C^k$ domain, so the conclusion cannot be extended to arbitrary open sets. Indeed $\mathrm{\Omega }$ is not a bounded $C^1$ domain in the graph sense: near a slit point $(x_0,0)$ with $0<x_0<1$, its complement is only a line segment and has empty interior, so $\mathrm{\Omega }$ is dense on both sides of that segment. A one-sided graph domain has a nonempty open complementary side in every sufficiently small chart neighbourhood, which rules out such a chart here.