The conforming RT0/P1 least-squares Poisson minimizer is unique and best approximating

The weak solution supplier in [A1] applies because a bounded polygon is open and bounded in one direction, and $v\mapsto (f,v)$ is bounded on $H_0^1$ by Cauchy–Schwarz and Poincare. It gives $u\in H_0^1$; the distributional equation gives $\sigma =-\mathrm{\nabla }u\in H(div)$ and $div\sigma =f$.

For any $({\sigma }_h,u_h)$ in the trial family, the residual of its difference from $(\sigma ,u)$ equals $({\sigma }_h+\mathrm{\nabla }{u}_h,div{\sigma }_h-f)$ by [F3]. Thus $J_h=\mathrm{\parallel }R(({\sigma }_h,u_h)-(\sigma ,u)){\mathrm{\parallel }}^2$. By [F2], $(z,w{)}_R=(Rz,Rw{)}_{L^2\times L^2}$ is an inner product on the ambient error space.

By Gram–Schmidt choose an orthonormal basis $e_1,\dots ,e_N$ of $R(X_h)$, and define $Pq={\sum }_{j=1}^N(q,e_j)e_j$. Then $q-Pq$ is orthogonal to every $e_j$, so $Pq$ uniquely minimizes distance to $R(X_h)$ by Pythagoras. The finite-dimensional image is closed; injectivity in [F2] pulls $P(R(\sigma ,u))$ back to a unique minimizer and the same identity gives best approximation. The scalar component stays in the stated H1_0 trial factor.