Smooth interior data do not repair incompatible Dirichlet corner values

Given: The Axiom of Choice; the square $\mathrm{\Omega }=(0,1{)}^2$; and the boundary datum $g(x,0)=1,g(0,y)=0,g(1,y)=0,g(x,1)=0(0<x,y<1),$ each of the four functions being constant and therefore smooth on its open side.

The refuted assertion concerns the Laplace Dirichlet problem $-\mathrm{\Delta }u=0$ in $\mathrm{\Omega }$ with the prescribed sidewise boundary values; this is the uniformly elliptic divergence-form convention with $a^{ij}={\delta }^{ij}$ and $b=c=0$ (Uniformly elliptic divergence-form operators and their sesquilinear forms, Weak Dirichlet solutions for a divergence-form operator).

Assume the Axiom of Choice. The square $Q=(0,1{)}^2$ is an $H^2$ extension domain by explicit reflection. For $h\in H^2(0,1)$ extend across $0$ by $3h(-x)-2h(-2x)$ on $(-1\mathrm{/}2,0)$ and across $1$ by $3h(2-x)-2h(3-2x)$ on $(1,3\mathrm{/}2)$, retaining $h$ on $[0,1]$. For an $H^2$ interval class, the opened one-dimensional representative corollary applied to $h$ and $h^{\mathrm{\prime }}$ supplies continuous endpoint values. These formulas match the value and first derivative at each join, and affine changes of variables bound the $H^2$ norm on the enlarged interval. To apply the formula to $u\in H^2(Q)$, Fubini and the weak-derivative identities tested against products of one-dimensional smooth tests show that almost every coordinate slice of $u$ is $H^2$, and that the slices of its transverse first derivative are $H^1$. One can choose a common null set by using a countable dense family of interval tests; passage to any test follows by the L2 bounds. On these slices the reflected formulas match the function and its normal first derivative, so integration by parts on the joined intervals has no interface terms. Transverse weak derivatives commute with the reflection by affine change of variables in the tensor test identities; for the mixed derivative only the H1 matching of the transverse first-derivative slices is needed. Thus each coordinate operation bounds all pure and mixed weak derivatives through order two. Applying them successively gives a bounded extension from $H^2(Q)$ to $H^2((-1\mathrm{/}2,3\mathrm{/}2{)}^2)$; multiplying by a smooth cutoff equal to $1$ on $\bar{Q}$ and supported in the larger rectangle, then extending by zero, gives an $H^2({\mathbb{R}}^2)$ extension. Thus $Q$ is a bounded extension domain. Since $2\cdot 2>2$, Higher-order Sobolev embedding gives a continuous representative on $\bar{Q}$ for every $H^2(Q)$ class. (Sobolev extension domains and extension operators, A Euclidean bump for a compact set inside an open set, Fubini's theorem for L^1 functions on a sigma-finite product)

On a bounded $C^2$ domain the global $H^2$ estimate is available for the zero-trace problem; it presupposes a compatible datum and does not by itself produce one. (Global $H^2$ Dirichlet regularity, Regularity estimates do not create boundary compatibility)

The square is bounded Lipschitz but is not $C^1$ at its four corners: the two incident straight edges do not form a single $C^1$ boundary graph. Thus the bounded $C^2$ hypothesis of the global boundary theorem does not apply to this domain. (Bounded C^k domains and boundary charts)

Two continuous functions on $Q$ that agree almost everywhere agree everywhere, since a nonzero difference at one point remains nonzero on an open ball of positive measure. A continuous representative on $\bar{Q}$ that attains the prescribed constant values on the open sides must therefore have equal limits along the two sides at each shared corner.