Multiplicative Schwarz projection sweeps contract with explicit overlap constants

For an arbitrary initial error $x$, set $e_{-1}=x$, $d_0={\mathrm{\Pi }}_{0}x$, and $e_0=x-d_0$. For $i=1,\dots ,m$, set $d_i={\mathrm{\Pi }}_i{e}_{i-1}$ and $e_i=e_{i-1}-d_i$. By [F2], each correction is orthogonal to its updated error, so telescoping gives $D:={\sum }_{i=0}^m\mathrm{\parallel }{d}_i{\mathrm{\parallel }}_A^2=\mathrm{\parallel }x{\mathrm{\parallel }}_A^2-\mathrm{\parallel }{e}_m{\mathrm{\parallel }}_A^2\mathrm{.}$ Also $e_{i-1}=x-d_0-{\sum }_{j=1}^{i-1}{d}_j$.

Fix a local index $i\ge 1$ and set $b_i=\mathrm{\parallel }{\mathrm{\Pi }}_{i}x{\mathrm{\parallel }}_A$, $c_i=\mathrm{\parallel }{d}_i{\mathrm{\parallel }}_A$, and $q_i=\mathrm{\parallel }{\mathrm{\Pi }}_i{d}_0{\mathrm{\parallel }}_A$. Since $x=e_{i-1}+d_0+{\sum }_{j<i}{d}_j$ and ${\mathrm{\Pi }}_i$ is an orthogonal projection, $a({\mathrm{\Pi }}_{i}x,e_{i-1})=a({\mathrm{\Pi }}_{i}x,{\mathrm{\Pi }}_i{e}_{i-1})=a({\mathrm{\Pi }}_{i}x,d_i)$ and $a({\mathrm{\Pi }}_{i}x,d_0)=a({\mathrm{\Pi }}_{i}x,{\mathrm{\Pi }}_i{d}_0)$. Thus $b_i^2=a({\mathrm{\Pi }}_{i}x,d_i)+a({\mathrm{\Pi }}_{i}x,{\mathrm{\Pi }}_i{d}_0)+{\sum }_{j<i}a({\mathrm{\Pi }}_{i}x,d_j)$. The first term has absolute value at most $b_i{c}_i$ and the second at most $b_i{q}_i$ by [F4]; the remaining terms are at most $b_i{\sum }_{j<i}{ϵ}_{ij}{c}_j$ by [A2]. If $b_i=0$ the resulting bound is immediate; otherwise divide by $b_i$ to obtain $b_i\le c_i+q_i+{\sum }_{j<i}{ϵ}_{ij}{c}_j\mathrm{.}$

Let $L$ be the strict lower-triangular part of $G$, and let $b,c,q\in {\mathbb{R}}^m$ have entries $b_i,c_i,q_i$. They are nonnegative and $0\le Lc\le Gc$ entrywise. Hence $\mathrm{\parallel }Lc{\mathrm{\parallel }}_2\le \mathrm{\parallel }Gc{\mathrm{\parallel }}_2\le \rho \mathrm{\parallel }c{\mathrm{\parallel }}_2$, and step 2.1 yields $\mathrm{\parallel }b{\mathrm{\parallel }}_2\le (1+\rho )\mathrm{\parallel }c{\mathrm{\parallel }}_2+\mathrm{\parallel }q{\mathrm{\parallel }}_2\mathrm{.}$ Furthermore, by [F2], $\mathrm{\parallel }q{\mathrm{\parallel }}_2^2={\sum }_{i=1}^{m}a({\mathrm{\Pi }}_i{d}_0,d_0)=a({\sum }_{i=1}^m{\mathrm{\Pi }}_i{d}_0,d_0)\le a(T{d}_0,d_0)\le C_1\mathrm{\parallel }{d}_0{\mathrm{\parallel }}_A^2\mathrm{.}$ The first inequality uses that ${\mathrm{\Pi }}_0$ is positive semidefinite; the last is [F3].

By $(r+s{)}^2\le 2r^2+2s^2$, steps 1.1 and 3.1 give ${\sum }_{i=0}^m\mathrm{\parallel }{\mathrm{\Pi }}_{i}x{\mathrm{\parallel }}_A^2=\mathrm{\parallel }{d}_0{\mathrm{\parallel }}_A^2+\mathrm{\parallel }b{\mathrm{\parallel }}_2^2\le (1+2C_1)\mathrm{\parallel }{d}_0{\mathrm{\parallel }}_A^2+2(1+\rho {)}^2{\sum }_{i=1}^m\mathrm{\parallel }{d}_i{\mathrm{\parallel }}_A^2\le KD\mathrm{.}$ The left side is $a(Tx,x)$, which [F3] bounds below by $C_0^{-1}\mathrm{\parallel }x{\mathrm{\parallel }}_A^2$. Thus $D\ge (C_{0}K{)}^{-1}\mathrm{\parallel }x{\mathrm{\parallel }}_A^2$. Since $D\le \mathrm{\parallel }x{\mathrm{\parallel }}_A^2$ by step 1.1, $C_{0}K\ge 1$.

Step 1.1 now gives $\mathrm{\parallel }{E}_{\mathrm{m}\mathrm{u}\mathrm{l}\mathrm{t}}x{\mathrm{\parallel }}_A^2=\mathrm{\parallel }{e}_m{\mathrm{\parallel }}_A^2=\mathrm{\parallel }x{\mathrm{\parallel }}_A^2-D\le (1-\frac{1}{C_{0}K})\mathrm{\parallel }x{\mathrm{\parallel }}_A^2\mathrm{.}$ Taking the supremum over nonzero $x$ proves the operator-norm bound; finiteness of $C_0,K$ makes the contraction factor strictly less than one.