I found an old problem from notes, which I was not able to solve.
Assume that we have a given (arbitrary) norm $\| \cdot\|$ on $K$ and function $g:K \times K \rightarrow K \times K$ for some compact and convex subset $K\subset \mathbb{R}^n$, such that
$$
g(x, y) = \pmatrix{g_1(x,y) \\ g_2(x, y)}
$$
where $g_1, g_2$ are Lipschitz in their entries with constants $\gamma_1, \gamma_2, \gamma_3, \gamma_4 < 1$.
In other words, assume
$$
\| g_1 (x, y) – g_1(x', y') \| \leq \gamma_1 \| x – x'\| + \gamma_2 \| y – y'\| \\
\| g_2 (x, y) – g_2(x', y') \| \leq \gamma_3 \| x – x'\| + \gamma_4 \| y – y'\|
$$
for all $x,y,x', y' \in K$.
Prove that the iterates
$$
(x_{n+1}, y_{n+1}) := g(x_{n}, y_{n})
$$
converge to a fixed point of $g$.
It seems that the existence of a fixed point follows from Brouwer's fixed point theorem, but I am stuck with convergence.
I have tried finding an appropriate metric on the space $K\times K$ that makes the function a contraction, i.e, a function $f: \mathbb{R}^{\geq 0} \times \mathbb{R}^{\geq 0} \rightarrow \mathbb{R}^{\geq 0}$ that makes
$$
d((x, y), (x', y')) := f(\| x – x'\|, \|y – y'\|)
$$
a complete metric on $K \times K$ and $g$ a contraction, but was not successful so far.
Is the problem possible to solve?
Best Answer
Consider $g:[-1,1]^2\to [-1,1]^2$, $g(x,y)=(-(x+y)/2,-(x+y)/2)$. Then $$ |g_k(x,y)-g_k(x',y')| \le \frac{1}{2}|x-x'| + \frac{1}{2}|y-y'| \quad (k=1,2). $$ Starting the iteration at $(1,1)$ leads to the sequence $$ (1,1), ~ (-1,-1), ~ (1,1), ~ (-1,-1) \dots $$ which is not convergent.