Matching patch geometry does not repair a field trace jump

The two constants are square integrable on their bounded rectangles, hence $v\in L^2(D)$. Choose $\eta \in C_c^{\mathrm{\infty }}((0,1))$ with ${\int }_0^1\eta (y)dy\mathrm{\ne }0$, and choose $\psi \in C_c^{\mathrm{\infty }}((-1,1))$ with $\psi (0)=1$. For $0<\epsilon <1$, set ${\varphi }_{\epsilon }(x,y)=\psi ((x-1)\mathrm{/}\epsilon )\eta (y)\in C_c^{\mathrm{\infty }}(D)$.

Since $v$ vanishes for $x<1$ and equals one for $x>1$, a change of variable $t=(x-1)\mathrm{/}\epsilon$ gives ${\int }_{D}v{\mathrm{\partial }}_x{\varphi }_{\epsilon }dxdy=\left({\int }_0^1\eta \right){\int }_1^2{\epsilon }^{-1}{\psi }^{\mathrm{\prime }}((x-1)\mathrm{/}\epsilon )dx=-\psi (0){\int }_0^1\eta ,$ which is a fixed nonzero number. If an $L^2$ weak derivative $g={\mathrm{\partial }}_{x}v$ existed, [F2] and Cauchy--Schwarz would make the absolute value at most $\mathrm{\parallel }g{\mathrm{\parallel }}_2\mathrm{\parallel }{\varphi }_{\epsilon }{\mathrm{\parallel }}_2$. But $\mathrm{\parallel }{\varphi }_{\epsilon }{\mathrm{\parallel }}_2=\sqrt{\epsilon }\mathrm{\parallel }\psi {\mathrm{\parallel }}_2\mathrm{\parallel }\eta {\mathrm{\parallel }}_2\to 0$, contradiction. Therefore $v\mathrm{\notin }{H}^1(D)$. Its two constant traces across $x=1$ are 0 and 1, violating [F1] even though the geometry is matching; the sufficient condition of [F3] does not apply.