A broken polynomial with a nonzero jump is not globally H1

The traces at $x=1\mathrm{/}2$ are $v^{-}=1$ and $v^{+}=0$, so [F2] gives $[v]=1$. For every smooth compactly supported $\phi$, direct integration on the two intervals gives ${\int }_0^{1}v{\phi }^{\mathrm{\prime }}={\int }_0^{1\mathrm{/}2}{\phi }^{\mathrm{\prime }}=\phi (1\mathrm{/}2)$.

If an $L^2$ weak derivative $g$ existed, [F1] would give $\phi (1\mathrm{/}2)=-{\int }_0^{1}g\phi$. Use the explicit bump $\eta (t)=\exp (1-(1-4t^2{)}^{-1})$ for $\mathrm{\mid }t\mathrm{\mid }<1\mathrm{/}2$ and $\eta (t)=0$ otherwise; it is smooth because the exponential is flat at the endpoints, is supported in $[-1\mathrm{/}2,1\mathrm{/}2]\subset (-1,1)$, and satisfies $\eta (0)=1$. Let ${\phi }_{\epsilon }(x)=\eta ((x-1\mathrm{/}2)\mathrm{/}\epsilon )$ for $0<\epsilon <1\mathrm{/}2$. Then ${\phi }_{\epsilon }\in C_c^{\mathrm{\infty }}(0,1)$, ${\phi }_{\epsilon }(1\mathrm{/}2)=1$, and $\mathrm{\parallel }{\phi }_{\epsilon }{\mathrm{\parallel }}_{L^2}=\sqrt{\epsilon }\mathrm{\parallel }\eta {\mathrm{\parallel }}_{L^2(-1,1)}\to 0$. The integral Cauchy–Schwarz inequality [F4] forces $\mathrm{\mid }\int g{\phi }_{\epsilon }\mathrm{\mid }\le \mathrm{\parallel }g{\mathrm{\parallel }}_2\mathrm{\parallel }{\phi }_{\epsilon }{\mathrm{\parallel }}_2\to 0$, contradicting step 1.1. Thus no such $g$ exists and $v\mathrm{\notin }{W}^{1,2}(0,1)$.