Given the mapping
$$f(x, y) = (\cos(2x - y), \frac{1}{2} \cdot \ln(1 + x^2 + y^2))$$
and the region $R = [-1, 1] \times [-1, 1]$, I want to know if I can use contraction mapping iteration in order to find a fixed point.
I'm pretty sure the answer is NO because I don't think $\ln(1 + x^2 + y^2)/2$ is a contraction; however, I'm having a difficult time showing this since the function is multivariate.
I've seen a similar post here, but I'm still not able to make much progress.
Define $g(x, y) = \ln(1 + x^2 + y^2)/2$. To show $g$ is not a contraction, I need to show that there is no constant $C \in (0, 1)$ for which
$$|g(p_1) - g(p_2)| \leq C|p_1 - p_2|$$ for all points $p_1, p_2 \in R$. I am not so sure about how to prove this claim though, and I would really aprpeciate any help. I've tried many things (e.g. taking a limit), but I have not had any luck.
$g$ is a contraction; the mean value theorem in $2$ dimensions gives $$ |g(q) - g(p)| \leq |\nabla g((1 - t)p + tq)| |q - p| $$ where $p, q$ play the role of $p_1, p_2 \in \mathbb [-1, 1]^2$. Here $\nabla g: \mathbb [-1, 1]^2 \rightarrow \mathbb R^2$ is the gradient: $$ \nabla g(x, y) = \left(\frac{\partial g}{\partial x}(x, y), \frac{\partial g}{\partial y}(x, y)\right) = \frac{1}{2}\left(\frac{2x}{1 + x^2 + y^2}, \frac{2y}{1 + x^2 + y^2}\right).$$ The norm of the gradient is $$ |\nabla g(x, y)| = \frac{1}{2}\sqrt{\frac{4x^2}{(1 + x^2 + y^2)^2} + \frac{4y^2}{(1 + x^2 + y^2)^2}} = \frac{\sqrt{x^2 + y^2}}{1 + x^2 + y^2}. $$ If you take $z = x^2 + y^2$, you should be able to find a bound for $|\nabla g(x, y)| = |\nabla g(z)|$.