A zero-boundary solution below the CR trace regularity threshold

On $0<x<1\mathrm{/}4$, writing $c(y)=y(1-y)$, one has ${\mathrm{\partial }}_{x}u=\frac{3}{4}c(y)x^{-1\mathrm{/}4},{\mathrm{\partial }}_{y}u=(1-2y)x^{3\mathrm{/}4}\mathrm{.}$ Everywhere else the cutoff and its derivative are bounded. Since $u$ and ${\mathrm{\partial }}_{y}u$ are square integrable and $\mathrm{\mid }{\mathrm{\partial }}_{x}u{\mathrm{\mid }}^2\le C(1+x^{-1\mathrm{/}2})$, the weak first derivatives lie in $L^2(\mathrm{\Omega })$; the classical interior derivatives give the distributional derivatives because test functions are compactly supported away from the boundary. Thus $u\in H^1(\mathrm{\Omega })$.

To prove the zero-boundary condition, let ${\theta }_{ϵ}(t):=\sigma ((t-ϵ)\mathrm{/}ϵ)$. By [F1], it is smooth, zero for $t\le ϵ$, and one for $t\ge 2ϵ$. Since ${\sigma }^{\mathrm{\prime }}$ is continuous and vanishes outside $[0,1]$, $\mathrm{\parallel }{\theta }_{ϵ}^{\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{\infty }}\le C\mathrm{/}ϵ$. Set ${\eta }_{ϵ}(x,y)={\theta }_{ϵ}(x){\theta }_{ϵ}(y){\theta }_{ϵ}(1-y)$. Each ${\eta }_{ϵ}u$ is smooth and compactly supported in $\mathrm{\Omega }$, since the $x$ cutoff removes the only nonsmooth boundary point and $\chi$ vanishes near $x=1$. Write $u-{\eta }_{ϵ}u=(1-{\theta }_{ϵ}(x))u+{\theta }_{ϵ}(x)(1-{\theta }_{ϵ}(y){\theta }_{ϵ}(1-y))u$. In the first term, all support lies in $x<2ϵ$; the squared $L^2$ and derivative errors are bounded by $C{ϵ}^{2(3\mathrm{/}4)-1}=C{ϵ}^{1\mathrm{/}2}$, including the ${ϵ}^{-1}u$ cutoff derivative. In the second term, support lies in the two strips of width $2ϵ$ near $y=0,1$. There $\mathrm{\mid }u\mathrm{\mid }\le C{x}^{3\mathrm{/}4}\mathrm{dist}(y,\{0,1\})$ and its first derivatives are bounded by $C{x}^{3\mathrm{/}4}+C{x}^{-1\mathrm{/}4}\mathrm{dist}(y,\{0,1\})$; integrating the strips shows the squared errors and cutoff-derivative terms are $O(ϵ)$. Thus ${\eta }_{ϵ}u\to u$ in $H^1(\mathrm{\Omega })$, so the closure definition gives $u\in H_0^1(\mathrm{\Omega })$.

Choose $0<\delta <1\mathrm{/}12$ so small that $c(y^{\mathrm{\prime }})\mathrm{/}c(y)\le 1.05$ whenever $0<x<\delta$, $1\mathrm{/}3\le y\le 2\mathrm{/}5$, and $y\le y^{\mathrm{\prime }}\le y+x$. For the pairs below, $x^{\mathrm{\prime }}\le 3x<1\mathrm{/}4$, so both points lie where $\chi =1$; also $c(y)$ is bounded below by a positive constant on this $y$ interval. For $2x\le x^{\mathrm{\prime }}\le 3x$ and $y\le y^{\mathrm{\prime }}\le y+x$, the values of ${\mathrm{\partial }}_{x}u(x,y)$ and ${\mathrm{\partial }}_{x}u(x^{\mathrm{\prime }},y^{\mathrm{\prime }})$ then differ by at least $\frac{3}{4}c(y)x^{-1\mathrm{/}4}(1-1.052^{-1\mathrm{/}4})\mathrm{.}$ Thus $\mathrm{\mid }{\mathrm{\partial }}_{x}u(x,y)-{\mathrm{\partial }}_{x}u(x^{\mathrm{\prime }},y^{\mathrm{\prime }})\mathrm{\mid }\ge C{x}^{-1\mathrm{/}4}$. Also $\mathrm{\mid }(x,y)-(x^{\mathrm{\prime }},y^{\mathrm{\prime }})\mathrm{\mid }\le \sqrt{5}x$. By [F2], restricting the $H^{1\mathrm{/}2}$ Gagliardo integral to these pairs gives a lower bound $C{\int }_0^{\delta }{x}^{-1\mathrm{/}2}{x}^{-3}{x}^{2}dx=C{\int }_0^{\delta }{x}^{-3\mathrm{/}2}dx=\mathrm{\infty },$ where the two factors of $x$ are the lengths of the $x^{\mathrm{\prime }}$ and $y^{\mathrm{\prime }}$ intervals. Thus ${\mathrm{\partial }}_{x}u\mathrm{\notin }{H}^{1\mathrm{/}2}(\mathrm{\Omega })$.

Let $D=\mathrm{diam}(\mathrm{\Omega })=\sqrt{2}$. For $1\mathrm{/}2<r<1$ and $x\mathrm{\ne }y$, $\mathrm{\mid }x-y{\mathrm{\mid }}^{-3}=\mathrm{\mid }x-y{\mathrm{\mid }}^{2r-1}\mathrm{\mid }x-y{\mathrm{\mid }}^{-2-2r}\le D^{2r-1}\mathrm{\mid }x-y{\mathrm{\mid }}^{-2-2r}$. Integrating gives $[q{]}_{H^{1\mathrm{/}2}(\mathrm{\Omega })}^2\le D^{2r-1}[q{]}_{H^r(\mathrm{\Omega })}^2$ for every $q\in H^r(\mathrm{\Omega })$, so $H^r(\mathrm{\Omega })\subset H^{1\mathrm{/}2}(\mathrm{\Omega })$. If $u\in H^{1+r}(\mathrm{\Omega })$ for such an $r$, [F4] would imply ${\mathrm{\partial }}_{x}u\in H^r(\mathrm{\Omega })$ and hence $H^{1\mathrm{/}2}(\mathrm{\Omega })$, contradicting step 2.1. Thus no exponent $r\in (1\mathrm{/}2,1)$ satisfies the refuted claim.

It remains to exclude $r=1$. For $0<x<1\mathrm{/}4$, the cutoff is one and $-\mathrm{\Delta }u=\frac{3}{16}c(y)x^{-5\mathrm{/}4}+2x^{3\mathrm{/}4}$. On $1\mathrm{/}3\le y\le 2\mathrm{/}5$, $c(y)$ is bounded below by a positive number, so this classical Laplacian is bounded below by $c_0{x}^{-5\mathrm{/}4}$ for some $c_0>0$; its square has divergent integral at $x=0$. Since the distributional Laplacian agrees with this classical function in the open square, it has no $L^2$ representative. By [F5], $u\mathrm{\notin }{H}^2(\mathrm{\Omega })=W^{2,2}(\mathrm{\Omega })$. The example still defines a bounded weak functional on $H_0^1(\mathrm{\Omega })$ by $v\mapsto {\int }_{\mathrm{\Omega }}\mathrm{\nabla }u\cdot \mathrm{\nabla }v$, using Holder [F5]; it does not supply the $L^2$ source assumed by the method. Since the example has no $H^{1+r}$ regularity with $r>1\mathrm{/}2$, it does not satisfy the trace hypothesis in [F3], and no normal-gradient trace conclusion is claimed.

Step 3.1 excludes every $r\in (1\mathrm{/}2,1)$, and step 3.2 excludes $r=1$. These are all exponents in the asserted range, so the counterexample refutes the Statement.