Exact Schwarz local corrections are energy-orthogonal projections

Fix $i\in \mathcal{I}$ with $i\mathrm{\ne }0$. Multiplication using [F1] gives ${\mathrm{\Pi }}_i^2=R_i{A}_i^{-1}(R_i^{\mathsf{T}}A{R}_i)A_i^{-1}{R}_i^{\mathsf{T}}A={\mathrm{\Pi }}_i\mathrm{.}$ By [F2], $A_i^{-1}$ is symmetric; [F4] therefore gives ${\mathrm{\Pi }}_i^{\mathsf{T}}A=A{\mathrm{\Pi }}_i$. Its range is contained in $\mathrm{Ran}(R_i)$, and ${\mathrm{\Pi }}_i{R}_i=R_i$, so its range is exactly $V_i$.

If $0\in \mathcal{I}$, write ${\mathrm{\Pi }}_0=I-C$ with $C$ from [F3]. Since $C^2=C$ and $C$ is $A$-self-adjoint, $I-C$ is also idempotent and $A$-self-adjoint. Moreover, $\mathrm{Ran}(I-C)=\ker C=\mathrm{Ran}(P)$ because $C$ projects onto $\mathrm{Ran}(P{)}^{{\perp }_A}$. Thus ${\mathrm{\Pi }}_0$ is the $A$-orthogonal projection onto $V_0$. If $0\mathrm{\notin }\mathcal{I}$, there is no coarse projection to check.

For every present index $i$, idempotence and $A$-self-adjointness give $a({\mathrm{\Pi }}_{i}x,(I-{\mathrm{\Pi }}_i)x)=a(x,{\mathrm{\Pi }}_i(I-{\mathrm{\Pi }}_i)x)=0$. Expanding the squared norm of $x={\mathrm{\Pi }}_{i}x+(I-{\mathrm{\Pi }}_i)x$ proves the Pythagorean identity. Also $a({\mathrm{\Pi }}_{i}x,x)=a({\mathrm{\Pi }}_{i}x,{\mathrm{\Pi }}_{i}x)$ by the same orthogonality.