A weakly divergence-free mini velocity need not be pointwise divergence-free

This mesh has four vertices, no interior vertices, and two cells. [F1] gives $\dim V_h=2\cdot 0+2\cdot 2=4$ and $\dim Q_h=4-1=3$. The linear constraint map from $V_h$ to $Q_h^{\mathrm{\prime }}$ therefore has nonzero kernel by rank-nullity, so choose $0\mathrm{\ne }{v}_h\in Z_h$.

Since all square vertices lie on the boundary, the zero-boundary P1 velocity part is zero. By [F2], $v_h$ consists only of vector bubbles, one supported on each triangle. At least one cell has a nonzero bubble coefficient $a\in {\mathbb{R}}^2$.

On that cell, $\mathrm{div}(b_{K}a)=a\cdot \mathrm{\nabla }{b}_K$. If this polynomial vanished identically, $b_K$ would be constant along every line segment parallel to $a$ inside $K$. The line through the barycenter parallel to $a$ meets $\mathrm{\partial }K$ at two endpoints where $b_K=0$, so constancy would force $b_K$ to vanish at the barycenter, contradicting $b_K>0$ there. Thus this polynomial is nonzero at some point of the cell interior. By continuity it remains nonzero on a small open ball, which has positive measure; hence $\mathrm{div}{v}_h\mathrm{\ne }0$ on a positive-measure subset of that cell.

By [F3], the selected $v_h$ lies in $Z_h$ but not $Z_h^0$. For any exactly divergence-free velocity, $b(v_h,q_h)=0$ follows immediately from [F3]'s pairing, so $Z_h^0\subseteq Z_h$; the inclusion is strict on this mesh.