The triangle area is $1\mathrm{/}2$, and ${\int }_{K}x={\int }_0^{1}x(1-x)dx=1\mathrm{/}6$, so ${\int }_{K}q=1\mathrm{/}6-(1\mathrm{/}3)(1\mathrm{/}2)=0$. Also $\mathrm{div}v=2x$.

Direct integration gives ${\int }_K{x}^2={\int }_0^1{x}^2(1-x)dx=1\mathrm{/}12$; hence $b_K(v,q)=-2({\int }_K{x}^2-\frac{1}{3}{\int }_{K}x)=-2(\frac{1}{12}-\frac{1}{18})=-\frac{1}{18}$. This nonzero local pairing detects that polynomial pressure mode against this polynomial velocity; no boundary condition or global mesh stability is asserted.

Since $\mathrm{\nabla }q$ is constant, $\mathrm{div}(b_K\mathrm{\nabla }q)=\mathrm{\nabla }{b}_K\cdot \mathrm{\nabla }q$. Apply [F3] to $b_K\mathrm{\nabla }q$; the boundary flux vanishes because $b_K=0$ on $\mathrm{\partial }K$. Thus $-{\int }_{K}q\mathrm{div}(b_K\mathrm{\nabla }q)=-{\int }_{\mathrm{\partial }K}q{b}_K\mathrm{\nabla }q\cdot n+{\int }_K{b}_K\mathrm{\mid }\mathrm{\nabla }q{\mathrm{\mid }}^2={\int }_K{b}_K\mathrm{\mid }\mathrm{\nabla }q{\mathrm{\mid }}^2$.

Because $q$ is nonconstant, $\mathrm{\mid }\mathrm{\nabla }q{\mathrm{\mid }}^2>0$; because $b_K>0$ in the interior, the integrand is positive on a set of positive volume. Its integral is therefore strictly positive. The calculation concerns one cell contribution only.

Define $q=+1$ on the lower-left and upper-right cells and $q=-1$ on the other two. Equal cell areas show ${\int }_{D}q=0$, and $q\mathrm{\ne }0$. Let $\eta$ be the one-dimensional hat with values $0,1,0$ at $0,1\mathrm{/}2,1$. The only interior Q1 node is the center, so each scalar velocity is a multiple of $\varphi (x,y)=\eta (x)\eta (y)$.

Let $I_L=[0,1\mathrm{/}2]$, $I_R=[1\mathrm{/}2,1]$, and similarly $J_B,J_T$. Direct integration gives ${\int }_{I_L}{\eta }^{\mathrm{\prime }}=1$, ${\int }_{I_R}{\eta }^{\mathrm{\prime }}=-1$, and ${\int }_{J_B}\eta ={\int }_{J_T}\eta =1\mathrm{/}4$. The alternating signs factor, so ${\int }_{D}q{\mathrm{\partial }}_x\varphi =(1-(-1))(1\mathrm{/}4-1\mathrm{/}4)=0$. Interchanging $x,y$ gives ${\int }_{D}q{\mathrm{\partial }}_y\varphi =0$.

Every admissible velocity is $(a\varphi ,b\varphi )$ for constants $a,b$, so step 2.1 gives $b(v,q)=0$ for every velocity while $q$ is a nonzero mean-zero pressure. Its supremum in the discrete inf-sup quotient is zero; nonnegativity of that quotient forces the discrete inf-sup constant to be $0$. Thus this family is not uniformly inf-sup stable.

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.