I need to prove the convergence of a coupled iterative scheme. The scheme results from the discretization of a problem from function space to a finite-dimensional vector space, and is as follows:
$AX = b \times\max(Y - X - a, 0)$ and $AY = b \times\max(X + a - Y, 0)$ where $A \in \mathbb R^{n\times n}$; $X, Y, a\in \mathbb R^{n}$ and $b$ is a real scalar.
Take the first equation $AX = b \times\max(Y - X - a, 0)$ and pin $Y$, I write $X = f(X)$. I am able to show $f$ is a contraction, and apply the Banach fixed point theorem. This I can do since I can show for $(X, d)$ to be a complete metric space, where $d$ is the $p$-norm on $R^n$; and I am able to compute the bound of the induced matrix norm for the inverse of $-A$, which is 'correctly' bounded (i.e. $X = -A^{-1}(b \times \min(X, Y-a) - b \times (Y-a))$ contracts). Similar for $AY =... $
Thus for respectively $X$, and $Y$ I am able to compute its fixed point (by iteration) given the value of the other. Therefore if I initialise $X$ and $Y$ and iterate the fixed point computations, my overall scheme must go 'somewhere'. Intuitively (and by computer) it makes sense for both $X$ and $Y$ to converge, however I can not seem to prove it.
The scheme would result in a sequence of fixed points, and contractions for both $X$, and $Y$. I opted for Theorem $1.2$ from 'Lectures On Some Fixed Point Theorems Of Functional Analysis' by F.F. Bonsall which is as follows:
Let $E$ be a complete metric space, and let $T$ and $T_n(n =1, 2, \cdots)$ be contraction mappings of $E$ into itself with the same Lipschitz constant $K < 1$, and with fixed points $u$ and $u_n$ respectively. Suppose that $\lim_{n\to ∞} T_nx = Tx$ for every $x \in E$. Then $\lim_{n\to ∞} u_n=u $.
Besides the technical details (quite a strong/restrictive Lipschitz constant condition, which might not hold in my case), I would not be able to use the result since my contractions for either $X$ or $Y$ depend on the other. Thus I would need to assume what I try to prove for "suppose that $\lim_{n\to ∞} T_nx = Tx$" to make sense.
I also thought of doing just one iteration for each fixed point search, and then switch equations. This seems to converge similarly (by computer) though I would think is harder to prove (I do not know what I have at hand after a single fixed point iteration, whereas if I continue I would have the fixed point which opens doors..?).
I feel like it is not necessarily a very hard problem, however I seem to be stuck in circle-reasoning. Is there any work on these type of problems I am unaware of? Or maybe a completely different route towards a solution?