A smaller constant test space gives a nontrivial practical DPG kernel

Choose $W_h=\mathrm{span}\{(e_1,0),(e_2,0),(x_1{e}_1,0),(x_2{e}_2,0)\}$ in the vector/scalar volume trial variables; these four fields are linearly independent as $L^2$ pairs. The constant test space has dimension $2+1=3$.

In a fixed basis of the constant test space, the Gram matrix is positive definite, and [F2] returns $G^{-1}g(w)$ in that basis. The right-hand side $g(w)$ is linear in $w$, so the restricted practical trial-to-test map is linear from the four-dimensional trial subspace into the three-dimensional test space. The images of four independent basis vectors are therefore linearly dependent; the corresponding nonzero coefficient vector is in the kernel. The normal matrix is $B_r^{\ast }{B}_r$, so this vector has zero quadratic energy; if a solution exists, adding any multiple produces another trial solution.