The alternating Q1/P0 checkerboard pressure defeats the discrete inf-sup condition

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.