Are piece-wise contraction mappings over $[0, 1]$ also contraction mappings? and is there an associated fixed-point theorem for such functions?
More formally, let $[0, 1]$ be the unit interval of real numbers and $[0, 1] \times [0, 1] = \bigcup_{i = 1}^N \mathbb{D}_i$ where $\mathbb{D}_i$ is the product of two sub-intervals of $[0, 1]$ and $\mathbb{D}_i \cap \mathbb{D}_j = \emptyset$ for all $i \neq j \in [1, N]$.
Furthermore, let $f_i : [0, 1]^2 \to [0, 1], i \in [1, N]$ be contraction mappings (i.e. $d\left(f(x_1, x_2), f(y_1, y_2)\right) < d(x_1, y_1) + d(x_2, y_2)$ for some complete metric $d$ on $[0, 1]$).
Is it true that a continuous piece-wise function $g(x, y)$ (as below) is a contraction mapping? $g(x, y) = \left\{\begin{array}{ll} f_1(x, y) & x,y \in \mathbb{D}_1 \\ ... & \\ f_N(x, y) & x,y \in \mathbb{D}_N \\ \end{array}\right.$
Is it true that $g(x,y)$ have a fixed-point for each $x, y$?