A reentrant corner prevents automatic H² dual regularity

Put $\alpha =3\pi \mathrm{/}2$ and $\lambda =2\mathrm{/}3$. The domain is bounded Lipschitz: at the origin, the rotation $s=(x_1+x_2)\mathrm{/}\sqrt{2}$, $t=(x_2-x_1)\mathrm{/}\sqrt{2}$ writes it locally as $t>-\mathrm{\mid }s\mathrm{\mid }$; every other boundary point lies on a line segment or at a right-angle corner, where a rotation gives a Lipschitz graph. Define $\chi (r)=1-\sigma (4r-1)$, so $\chi =1$ for $r\le 1\mathrm{/}4$ and $\chi =0$ for $r\ge 1\mathrm{/}2$, and set $\mathrm{\Phi }(r,\theta )=r^{\lambda }\sin (\lambda \theta )$ and $u=\chi \mathrm{\Phi }$. Polar differentiation gives ${\mathrm{\partial }}_{rr}\mathrm{\Phi }=\lambda (\lambda -1)r^{\lambda -2}\sin (\lambda \theta )$ and $\mathrm{\Delta }\mathrm{\Phi }=0$. Also $\mathrm{\mid }u\mathrm{\mid }\le C{r}^{2\mathrm{/}3}$ and $\mathrm{\mid }\mathrm{\nabla }u\mathrm{\mid }\le C{r}^{-1\mathrm{/}3}$ near the corner. By [F5], ${\int }_{r<1\mathrm{/}4}(\mathrm{\mid }u{\mathrm{\mid }}^2+\mathrm{\mid }\mathrm{\nabla }u{\mathrm{\mid }}^2)dx\le C{\int }_0^{1\mathrm{/}4}(r^{4\mathrm{/}3}+r^{-2\mathrm{/}3})rdr<\mathrm{\infty }$; on the remaining support $u$ is smooth with bounded derivatives by [F3, F4]. Thus $u$ and its classical first derivatives are in $L^2(\mathrm{\Omega })$.

Let ${ϵ}_n=1\mathrm{/}(100(n+1))$, ${\eta }_n(r)=\sigma ((r-{ϵ}_n)\mathrm{/}{ϵ}_n)$, and ${\kappa }_n(\theta )=\sigma ((\theta -{ϵ}_n)\mathrm{/}{ϵ}_n)\sigma ((\alpha -\theta -{ϵ}_n)\mathrm{/}{ϵ}_n)$. Then $u_n={\eta }_n{\kappa }_{n}u$ belongs to $C_c^{\mathrm{\infty }}(\mathrm{\Omega })$: it vanishes for $r\le {ϵ}_n$, for $r\ge 1\mathrm{/}2$, and within angular distance ${ϵ}_n$ of either boundary ray. The cutoff derivatives satisfy $\mathrm{\mid }\mathrm{\nabla }{\eta }_n\mathrm{\mid }\le C\mathrm{/}{ϵ}_n$ and $\mathrm{\mid }\mathrm{\nabla }{\kappa }_n\mathrm{\mid }\le C\mathrm{/}({ϵ}_{n}r)$ on their transition regions by [F3, F4].

The radial cutoff region contributes $O({ϵ}_n^{10\mathrm{/}3})$ to the squared function error and $O({ϵ}_n^{4\mathrm{/}3})$ to the squared gradient error; also ${\int }_{{ϵ}_n<r<2{ϵ}_n}\mathrm{\mid }u\mathrm{\nabla }{\eta }_n{\mathrm{\mid }}^2=O({ϵ}_n^{4\mathrm{/}3})$. In the angular strips of width $2{ϵ}_n$, $\mathrm{\mid }u\mathrm{\mid }\le C{r}^{2\mathrm{/}3}\min (\theta ,\alpha -\theta )$ and $\mathrm{\mid }\mathrm{\nabla }u\mathrm{\mid }\le C{r}^{-1\mathrm{/}3}$; hence $\mathrm{\mid }u\mathrm{\nabla }{\kappa }_n\mathrm{\mid }\le C{r}^{-1\mathrm{/}3}$. Polar integration [F5] bounds the squared function and gradient errors on those strips by $O({ϵ}_n)$. These estimates give $u_n\to u$ in $H^1(\mathrm{\Omega })$, so [F1] gives $u\in H_0^1(\mathrm{\Omega })$.

Define $f=-\mathrm{\Delta }u$ on $\mathrm{\Omega }$. Since $\chi =1$ and $\mathrm{\Phi }$ is harmonic for $r<1\mathrm{/}4$, while the derivatives of $\chi$ are supported in the compact annulus $1\mathrm{/}4\le r\le 1\mathrm{/}2$, $f$ is bounded and supported in that annulus. Thus $f\in L^2(\mathrm{\Omega })$.

On $0<r<1\mathrm{/}4$, the classical radial second derivative is a unit-direction contraction of the Cartesian Hessian. If $u\in H^2(\mathrm{\Omega })$, [F10] identifies its weak Hessian with the classical Hessian on $\mathrm{\Omega }$, and $\mathrm{\mid }{e}_r^{\mathsf{T}}{D}^{2}u{e}_r\mathrm{\mid }\le \mathrm{\mid }{D}^{2}u{\mathrm{\mid }}_{\mathrm{F}}$ makes the radial derivative square-integrable. But [F5] gives ${\int }_{0<r<1\mathrm{/}4}\mathrm{\mid }{\mathrm{\partial }}_{rr}u{\mathrm{\mid }}^{2}dx=\mathrm{\mid }\lambda (\lambda -1){\mathrm{\mid }}^2({\int }_0^{3\pi \mathrm{/}2}{\sin }^2(\lambda \theta )d\theta )\left({\int }_0^{1\mathrm{/}4}{r}^{2\lambda -3}dr\right)=\mathrm{\infty },$ since the angular integral is $3\pi \mathrm{/}4$ and $2\lambda -3=-5\mathrm{/}3<-1$. This contradicts square-integrability of the Hessian, so $u\mathrm{\notin }{H}^2(\mathrm{\Omega })$.

Set $F(v)={\int }_{\mathrm{\Omega }}f\bar{v}$. By [F6] and $\mathrm{\parallel }v{\mathrm{\parallel }}_{L^2}\le \mathrm{\parallel }v{\mathrm{\parallel }}_{H^1}$, $F$ is bounded and conjugate-linear, so $F\in H^{-1}(\mathrm{\Omega })$ by [F9]. For every test $v\in C_c^{\mathrm{\infty }}(\mathrm{\Omega })$, support is away from $\mathrm{\partial }\mathrm{\Omega }$ and classical integration by parts gives ${\int }_{\mathrm{\Omega }}\mathrm{\nabla }u\cdot \overline{\mathrm{\nabla }v}=F(v)$. [F10] gives the weak-derivative interpretation, and both sides are continuous in the $H^1$ norm by [F6]; density [F1] extends the identity to all $v\in H_0^1(\mathrm{\Omega })$. Thus $u$ is a zero-Dirichlet weak solution by [F8].

Let $\tilde{u}\in H_0^1(\mathrm{\Omega })$ be any other zero-Dirichlet weak solution for this $f$. By [F8], subtracting the two variational identities and testing with $w=u-\tilde{u}$ gives $\mathrm{\parallel }\mathrm{\nabla }w{\mathrm{\parallel }}_{L^2}^2=0$ by [F6]. Since $\mathrm{\Omega }\subset \{-1<x_1<1\}$, [F7] gives $\mathrm{\parallel }w{\mathrm{\parallel }}_{L^2}\le 2\mathrm{\parallel }\mathrm{\nabla }w{\mathrm{\parallel }}_{L^2}=0$. Thus $u=\tilde{u}$ as a Sobolev class, so an $H^2$ weak solution would force $u\in H^2(\mathrm{\Omega })$.

Step 4.1 shows that no other weak solution can lie in $H^2$, while step 2.3 excludes $u$ itself. Thus this $L^2$ forcing on a bounded Lipschitz domain has no $H^2$ zero-Dirichlet weak solution, refuting the universal statement. The Aubin–Nitsche estimate remains conditional on its actual adjoint regularity and approximation hypotheses [F11].