Douglas–Rachford shadows converge to a zero of the operator sum

Choose $b\in B(p)$ and $a\in A(p)$ with $a+b=0$. The resolvent identities give $J_{\rho B}(p+\rho b)=p$ and $J_{\rho A}(p-\rho b)=p$. At $w_{\ast }=p+\rho b$, the second input is $2p-w_{\ast }=p-\rho b$, so the update has $p_B=q=p$ and $T(w_{\ast })=w_{\ast }$. Thus $T$ has a fixed point.

Apply [F3] using firm nonexpansiveness [F2]. It follows that the states $w_k=T^k(w_0)$ converge to some $w_{\ast }^{\mathrm{\prime }}$ with $T(w_{\ast }^{\mathrm{\prime }})=w_{\ast }^{\mathrm{\prime }}$.

Put $p^{\mathrm{\prime }}=J_{\rho B}(w_{\ast }^{\mathrm{\prime }})$ and $q^{\mathrm{\prime }}=J_{\rho A}(2p^{\mathrm{\prime }}-w_{\ast }^{\mathrm{\prime }})$. Fixedness says $q^{\mathrm{\prime }}=p^{\mathrm{\prime }}$. The resolvent inclusions yield $w_{\ast }^{\mathrm{\prime }}-p^{\mathrm{\prime }}\in \rho B(p^{\mathrm{\prime }})$ and $p^{\mathrm{\prime }}-w_{\ast }^{\mathrm{\prime }}\in \rho A(p^{\mathrm{\prime }})$; adding them gives $0\in (A+B)(p^{\mathrm{\prime }})$.

The shadow $p_k=J_{\rho B}(w_k)$ converges to $p^{\mathrm{\prime }}$ because the resolvent is 1-Lipschitz by [F4] and $w_k\to w_{\ast }^{\mathrm{\prime }}$. Thus the shadow converges to a zero of the sum.