A continuous P1 pressure gives a nonconservative primal flux

Let $K_1=\{0<y<x<1\}$ and $K_2=\{0<x<y<1\}$. The affine formulas $p_1(x,y)=y$ and $p_2(x,y)=x$ agree on $x=y$ and attain the four stated vertex values, so they define a continuous piecewise-P1 function. Their negative gradients are $q_1=(0,-1)$ and $q_2=(-1,0)$, each with classical divergence zero. With fixed diagonal normal $n=(1,-1)\mathrm{/}\sqrt{2}$, $q_1\cdot n=1\mathrm{/}\sqrt{2}$ and $q_2\cdot n=-1\mathrm{/}\sqrt{2}$, so the jump magnitude is $\sqrt{2}$.

For $\varphi \in C_c^{\mathrm{\infty }}(D)$, integration by parts on the two triangles shows that the distributional divergence of the piecewise field is the nonzero line measure on the diagonal with density $q_1\cdot n-q_2\cdot n=\sqrt{2}$ in this orientation (the distributional pairing is the negative of the summed outward boundary flux). To see it cannot be an $L^2$ function, choose a nonnegative smooth $\psi$ supported in the open diagonal with positive line integral and a smooth transverse cutoff $\eta$ with $\eta (0)=1$; set ${\varphi }_{\epsilon }(y,t)=\psi (y)\eta (t\mathrm{/}\epsilon )$ in tubular coordinates, extended by zero. The surface pairing is the fixed nonzero number $\sqrt{2}\int \psi$, whereas $\mathrm{\parallel }{\varphi }_{\epsilon }{\mathrm{\parallel }}_{L^2(D)}=O({\epsilon }^{1\mathrm{/}2})$. If the distribution were represented by $g\in L^2(D)$, [F2] would give $\mathrm{\mid }\int g{\varphi }_{\epsilon }\mathrm{\mid }\le \mathrm{\parallel }g{\mathrm{\parallel }}_2\mathrm{\parallel }{\varphi }_{\epsilon }{\mathrm{\parallel }}_2\to 0$, a contradiction. By [F1] the field is not in H(div).