Boundary $H^2$ regularity needs domain regularity

The singular factor is harmonic. For $v=r^{\alpha }\sin (\alpha \theta )$ one has $v_{rr}=\alpha (\alpha -1)r^{\alpha -2}\sin (\alpha \theta )$, $v_r=\alpha r^{\alpha -1}\sin (\alpha \theta )$ and $v_{\theta \theta }=-{\alpha }^2{r}^{\alpha }\sin (\alpha \theta )$; substituting into the polar formula of [F3] gives $\mathrm{\Delta }v=(\alpha (\alpha -1)+\alpha -{\alpha }^2)r^{\alpha -2}\sin (\alpha \theta )=0$. It vanishes on both radial edges.

The cutoff solution lies in $H_0^1$. The cutoff $u=\chi v$ has the same $H^1$ singularity near $0$, is zero near $r=1$, and vanishes on the two radial edges. For $ϵ>0$ choose a smooth radial cutoff ${\rho }_{ϵ}$ that is zero for $r\le ϵ$, one for $r\ge 2ϵ$, and satisfies $\mathrm{\mid }D{\rho }_{ϵ}\mathrm{\mid }\le C\mathrm{/}ϵ$. For $\delta >0$ choose a smooth angular cutoff ${\eta }_{\delta }$ that vanishes within angular distance $\delta$ of the two radial edges, equals one beyond distance $2\delta$, and satisfies $\mathrm{\mid }D{\eta }_{\delta }\mathrm{\mid }\le C\mathrm{/}(r\delta )$ in its transition strips. Then $u_{ϵ,\delta }={\rho }_{ϵ}{\eta }_{\delta }u$ lies in $C_c^{\mathrm{\infty }}(S_{\omega })$. Near the vertex $\mathrm{\mid }u\mathrm{\mid }\le C{r}^{\alpha }$ and $\mathrm{\mid }Du\mathrm{\mid }\le C{r}^{\alpha -1}$, so polar integration bounds the squared $H^1$ error from ${\rho }_{ϵ}$ by $C{ϵ}^{2\alpha }$. For fixed $ϵ$, the error from ${\eta }_{\delta }$ tends to zero as $\delta \downarrow 0$: near each edge $\mathrm{\mid }u\mathrm{\mid }\le C{r}^{\alpha }d(\theta )$ and $\mathrm{\mid }Du\mathrm{\mid }\le C{r}^{\alpha -1}$, and the derivative-cutoff term has squared integral at most $C_{ϵ}\delta$. Choose $\delta =\delta (ϵ)$ so this second error tends to zero as $ϵ\downarrow 0$. Thus $u_{ϵ,\delta (ϵ)}\to u$ in $H^1$, proving $u\in H_0^1(S_{\omega })$.

The singular factor lies in $H^1$ but not $H^2$. Its polar derivatives give $\mathrm{\mid }\mathrm{\nabla }v{\mathrm{\mid }}^2={\alpha }^2{r}^{2\alpha -2}$ and $\mathrm{\mid }v\mathrm{\mid }\le r^{\alpha }$, so by [F4] ${\int }_{S_{\omega }}(\mathrm{\mid }v{\mathrm{\mid }}^2+\mathrm{\mid }\mathrm{\nabla }v{\mathrm{\mid }}^2)dx\le \omega (\frac{1}{2\alpha +2}+\frac{{\alpha }^2}{2\alpha })<\mathrm{\infty }\mathrm{.}$ Thus $v\in H^1(S_{\omega })$. Its radial second derivative has squared integral ${\int }_{S_{\omega }}\mathrm{\mid }{v}_{rr}{\mathrm{\mid }}^{2}dx={\alpha }^2(1-\alpha {)}^2({\int }_0^{\omega }{\sin }^2(\alpha \theta )d\theta )\left({\int }_0^1{r}^{2\alpha -3}dr\right)=+\mathrm{\infty },$ since $0<\alpha <1$. If all Cartesian second derivatives were in $L^2$, then $v_{rr}=D^{2}v[e_r,e_r]$ would be in $L^2$ as well (the radial direction $e_r$ is a unit vector), a contradiction. Hence $v\mathrm{\notin }{H}^2(S_{\omega })$.

The forcing is square-integrable. The function $u=\chi v$ is smooth in the sector, and $f:=-\mathrm{\Delta }u$ vanishes wherever $\chi$ is constant because $\mathrm{\Delta }v=0$. The derivatives of $\chi$ are supported in the annulus $1\mathrm{/}2\le r\le 3\mathrm{/}4$, where $v$ and its derivatives are bounded. Therefore $f\in L^2(S_{\omega })$.

The uncut factor is locally weakly harmonic. For any $\phi \in C_c^{\mathrm{\infty }}(S_{\omega })$, its support lies in a compact subset of the open sector where $v$ is smooth. Integration by parts there and $\mathrm{\Delta }v=0$ give ${\int }_{S_{\omega }}\mathrm{\nabla }v\cdot \overline{\mathrm{\nabla }\phi }dx=0$. Thus $v$ is the local weak solution recorded in the statement.

The weak equation and boundary condition. For every $\phi \in C_c^{\mathrm{\infty }}(S_{\omega })$, integration by parts on a neighborhood of its compact support gives ${\int }_{S_{\omega }}\mathrm{\nabla }u\cdot \overline{\mathrm{\nabla }\phi }dx={\int }_{S_{\omega }}f\bar{\phi }dx$. By step 1.2, $u\in H_0^1(S_{\omega })$; both sides are continuous in the $H_0^1$ norm because $f\in L^2$ and the principal form is bounded. Density extends the identity to every $H_0^1$ test. Thus $u$ is a zero-boundary weak Dirichlet solution of $-\mathrm{\Delta }u=f$.

Failure of $H^2$. On $B_{1\mathrm{/}2}(0)\cap S_{\omega }$, $u=v$, so the divergent radial second-derivative integral of step 2.1 also occurs for $u$. As there $u_{rr}=D^{2}u[e_r,e_r]$, this precludes $u\in H^2(S_{\omega })$.

Lipschitz is not enough. By [F5], $S_{\omega }$ is bounded Lipschitz but not $C^1$ at its vertex. Steps 1.2 and 2.2--3.2 give a zero-boundary weak solution with $f\in L^2$, while step 3.3 shows that it is not in $H^2$. The $C^2$ hypothesis of [F8] therefore cannot be replaced by Lipschitz regularity, even for the Laplacian, smooth forcing and zero boundary data.