Additive Schwarz condition bound from stable decomposition and synthesis

The stable-decomposition hypothesis implies that the local spaces span $V$, so [F3] makes $B$ SPD. Each ${\mathrm{\Pi }}_i$ is $A$-self-adjoint by [F2], hence $T=BA={\sum }_i{\mathrm{\Pi }}_i$ is $A$-self-adjoint. Moreover $a(Tx,x)={\sum }_i\mathrm{\parallel }{\mathrm{\Pi }}_{i}x{\mathrm{\parallel }}_A^2\ge 0$; if this is zero, then every ${\mathrm{\Pi }}_{i}x=0$. For any $v_i\in V_i$, [F2] gives ${\mathrm{\Pi }}_i{v}_i=v_i$, so $a(x,v_i)=a({\mathrm{\Pi }}_{i}x,v_i)=0$. The spaces span $V$, hence $x=0$ and $T$ is positive.

Fix $x\mathrm{\ne }0$ and choose the decomposition $x={\sum }_i{R}_i{z}_i$ from the stable-decomposition hypothesis. Orthogonal projection gives $\mathrm{\parallel }x{\mathrm{\parallel }}_A^2={\sum }_{i}a(x,R_i{z}_i)={\sum }_{i}a({\mathrm{\Pi }}_{i}x,R_i{z}_i)\le {\left({\sum }_i\mathrm{\parallel }{\mathrm{\Pi }}_{i}x{\mathrm{\parallel }}_A^2\right)}^{1\mathrm{/}2}{\left({\sum }_i{z}_i^{\mathsf{T}}{A}_i{z}_i\right)}^{1\mathrm{/}2}\mathrm{.}$ The last inequality is [F5] in the product inner product on the finite tuple space. The assumed bound on the second factor yields $a(Tx,x)={\sum }_i\mathrm{\parallel }{\mathrm{\Pi }}_{i}x{\mathrm{\parallel }}_A^2\ge C_0^{-1}\mathrm{\parallel }x{\mathrm{\parallel }}_A^2$, proving the lower Rayleigh bound.

For each $x$, write ${\mathrm{\Pi }}_{i}x=R_i{y}_i$ with $y_i=A_i^{-1}{R}_i^{\mathsf{T}}Ax$. The synthesis hypothesis and [F2] give $\mathrm{\parallel }Tx{\mathrm{\parallel }}_A^2={\parallel {\sum }_i{R}_i{y}_i\parallel }_A^2\le C_1{\sum }_i\mathrm{\parallel }{\mathrm{\Pi }}_{i}x{\mathrm{\parallel }}_A^2=C_{1}a(Tx,x)\mathrm{.}$ Set $q=a(Tx,x)\mathrm{/}\mathrm{\parallel }x{\mathrm{\parallel }}_A^2$. [F5] gives $q\mathrm{\parallel }x{\mathrm{\parallel }}_A\le \mathrm{\parallel }Tx{\mathrm{\parallel }}_A$, while the displayed inequality gives $\mathrm{\parallel }Tx{\mathrm{\parallel }}_A^2\le C_{1}q\mathrm{\parallel }x{\mathrm{\parallel }}_A^2$. If $q>0$, these imply $q\le C_1$; if $q=0$, the same upper bound is immediate.

By [F4] and the assumed lower order bound, every ${\tilde{A}}_i$ is SPD. For every SPD matrix $M$ and vector $y$, $y^{\mathsf{T}}{M}^{-1}y={\max }_z(2y^{\mathsf{T}}z-z^{\mathsf{T}}Mz),$ because completing the square subtracts $(z-M^{-1}y{)}^{\mathsf{T}}M(z-M^{-1}y)$. Since ${\tilde{A}}_i⪯\beta A_i$, this maximum for ${\tilde{A}}_i$ is at least the maximum for $\beta A_i$, namely ${\beta }^{-1}{y}^{\mathsf{T}}{A}_i^{-1}y$; since $\alpha A_i⪯{\tilde{A}}_i$, it is at most the maximum for $\alpha A_i$, namely ${\alpha }^{-1}{y}^{\mathsf{T}}{A}_i^{-1}y$. Thus ${\beta }^{-1}{A}_i^{-1}⪯{\tilde{A}}_i^{-1}⪯{\alpha }^{-1}{A}_i^{-1}$. Summing after congruence by $R_i$ yields ${\beta }^{-1}B⪯\tilde{B}⪯{\alpha }^{-1}B$.

Step 1.4 establishes that each inverse in $\tilde{B}$ is SPD. By the stable-decomposition hypothesis, the local spaces span $V$; hence if $x\mathrm{\ne }0$ then $R_i^{\mathsf{T}}x\mathrm{\ne }0$ for some $i$, since otherwise $x$ would be Euclidean-orthogonal to their span $V$. Therefore $x^{\mathsf{T}}\tilde{B}x={\sum }_i(R_i^{\mathsf{T}}x{)}^{\mathsf{T}}{\tilde{A}}_i^{-1}(R_i^{\mathsf{T}}x)>0$, so $\tilde{B}$ is symmetric positive definite. It follows that $\tilde{B}A$ is $A$-self-adjoint and positive, because $A\tilde{B}A$ is symmetric positive definite; the inexact Rayleigh quotient and condition number are therefore defined. The matrix comparison in step 1.4 bounds $a(\tilde{B}Ax,x)=x^{\mathsf{T}}A\tilde{B}Ax$ below by ${\beta }^{-1}{x}^{\mathsf{T}}ABAx$ and above by ${\alpha }^{-1}{x}^{\mathsf{T}}ABAx$. The exact bounds in steps 1.2 and 1.3 therefore give the stated inexact minimum, maximum, and condition-ratio bounds.