Let $f$ be a Lipschitz continuous function, $(a_{ij})_{1\le i,j\le s}$ and $(b_i)_{1\le i \le s}$ some real numbers, $x_0 \in \mathbb{R}$ and $h$ an arbitrary small positive number. Under some regularity hypothesis on $f$, for each $i$, there exists a unique $k_i \in \mathbb{R}$ such that : $k_i=f\big(x_0+h\sum\limits_{j=1}^s a_{ij}k_j\big)$.
To compute numerically these $k_i$, we use the fixed point algorithm : $k_i^{m+1}=f\big ( x_0+h\sum\limits_{j=1}^s a_{ij}k_j^m\big)$ with the initial conditions $\forall i, k_i^0=f(x_0)$. I would like to prove that, for a large number $m$, this scheme does converge to our initial $(k_i)_i$. Here's what I could demonstrate :
$\forall 1 \le i \le s, \quad |k_i^m-k_i|\le (\alpha h)^m \sum\limits_{j_1,j_2,…,j_m=1}^s |a_{ij_1}a_{j_1j_2}…a_{j_{m-1}j_m}|\hspace{2mm}|f(x_0)-k_{j_m}|$
Knowing that $k_i=_{h\to0}f(x_0)+O(h)$ for all $i$, we can write (in a neighbouring of 0) :$\forall 1 \le i \le s, \quad |k_i^m-k_i|\le Ch(\alpha h)^m \sum\limits_{j_1,j_2,…,j_m=1}^s |a_{ij_1}a_{j_1j_2}…a_{j_{m-1}j_m}|$
Using this form, I don't think I can conclude. I should find a way to bound the sum independently from $m$. How to ?
The only way I see is to use the infinity norm over $(a_{ij})$. But then, for all $i$, $|k_i^m-k_i|\le Ch(\alpha \|a\|_{\infty} s h)^m$
which means that a sufficient condition to have convergence of the scheme is to suppose $h<1/\alpha \|a\|_{\infty} s$.
Are there less restrictive conditions ?