Exact affine data under weak imposition

The function $u=x$ is affine, hence belongs to $V_h$ for every $k\ge 1$, and it has the prescribed trace $g$. Apply [F2] to $G=w_h\mathrm{\nabla }u$: since $\mathrm{\Delta }u=0$, its divergence is $\mathrm{\nabla }{w}_h\cdot \mathrm{\nabla }u$, and its boundary flux is $w_h{\mathrm{\partial }}_{n}u$. Thus the volume term in $a_h^0(u,w_h)$ cancels its first boundary-flux term. The remaining term is ${\gamma }_0{\sum }_F{h}_F^{-1}{\int }_{F}g{w}_h$, exactly ${\mathrm{\ell }}_h^0(w_h)$ because $f=0$.

For $\theta =1$, the extra form term is $-{\sum }_F{\int }_{F}u{\mathrm{\partial }}_n{w}_h$, and the extra load term is $-{\sum }_F{\int }_{F}g{\mathrm{\partial }}_n{w}_h$; these agree since $u{\mathrm{\mid }}_F=g$. Thus $u$ satisfies both weak equations. Coercivity [F3] implies uniqueness: the difference $d_h$ of two solutions obeys $a_h^{\theta }(d_h,w_h)=0$ for every $w_h$, so testing $w_h=d_h$ gives ${\lambda }_{\theta }\mathrm{\parallel }{d}_h{\mathrm{\parallel }}_{E,h}^2\le 0$ and hence $d_h=0$. Therefore $u_h^0=u_h^1=u$. For strong imposition, $u\in V_h$ has the required boundary trace and satisfies the Poisson weak equation on each test in $Y_h$: the simplex Green identity [F2] and the zero boundary restriction [F4] make its boundary flux term vanish. Thus it is the strong Galerkin solution as well.