The $H^2$ estimate needs the $L^2$ kernel term without injectivity

Given: Countable Choice; $\mathrm{\Omega }=B_1(0)\subset {\mathbb{R}}^2$, $a(r)=\frac{1}{4}-\frac{1}{8}{r}^2$, $A(x)=a(\mathrm{\mid }x\mathrm{\mid })I$, $Lw=-\mathrm{div}(A\mathrm{\nabla }w)-w$, zero datum $f=0$, and $u(x)=1-\mathrm{\mid }x{\mathrm{\mid }}^2$.

A weak Dirichlet solution is a function $w\in H_0^1(B_1)$ satisfying the form identity ${\int }_{B_1}(A\mathrm{\nabla }w\cdot \overline{\mathrm{\nabla }v}-w\bar{v})dx={\int }_{B_1}f\bar{v}dx$ for all $v\in H_0^1(B_1)$ (Weak Dirichlet solutions for a divergence-form operator, Uniformly elliptic divergence-form operators and their sesquilinear forms).

For $0\le r\le 1$, $\frac{1}{8}\le a(r)\le \frac{1}{4}$. Thus $A$ is smooth, bounded, and uniformly elliptic with ellipticity constant $\theta =1\mathrm{/}8$; the lower-order coefficients are bounded, with $b=0$ and $c=-1$ (Uniformly elliptic divergence-form operators and their sesquilinear forms).

The function $u=1-\mathrm{\mid }x{\mathrm{\mid }}^2$ is smooth on $\overline{B_1}$, zero on $\mathrm{\partial }{B}_1$, and nonzero, so $u\in H^2(B_1)$. To prove $u\in H_0^1(B_1)$ directly, choose smooth radial cutoffs ${\eta }_{\epsilon }$ equal to one for $r\le 1-2\epsilon$ and zero for $r\ge 1-\epsilon$, with $\mathrm{\mid }D{\eta }_{\epsilon }\mathrm{\mid }\le C\mathrm{/}\epsilon$, using a rescaled fixed smooth step. Then ${\eta }_{\epsilon }u\in C_c^{\mathrm{\infty }}(B_1)$; on the boundary strip $\mathrm{\mid }u\mathrm{\mid }\le C\epsilon$, $\mathrm{\mid }Du\mathrm{\mid }\le 2$, and its area is at most $C\epsilon$. Thus $\mathrm{\parallel }{\eta }_{\epsilon }u-u{\mathrm{\parallel }}_{H^1}^2\le C\epsilon \to 0$. Consequently $u\in H^2(B_1)\cap H_0^1(B_1)$ and $\mathrm{\parallel }u{\mathrm{\parallel }}_{H^2(B_1)}>0$. With $r=\mathrm{\mid }x\mathrm{\mid }$, $\mathrm{\nabla }u=-2x$ and $a^{\mathrm{\prime }}(r)=-r\mathrm{/}4$, whence $-\mathrm{div}(a(r)\mathrm{\nabla }u)=4a(r)+2r{a}^{\mathrm{\prime }}(r)=1-r^2=u\mathrm{.}$ Therefore $Lu=0$ pointwise.

On bounded $C^2$ domains in dimensions $n\ge 2$, the global theorem Global $H^2$ Dirichlet regularity includes an $L^2$ term on the right, while the estimate without that term is supplied under a trivial-kernel hypothesis by The global $H^2$ estimate without the $L^2$ term under uniqueness.