Example of nonexpansive mapping.

Question

I am trying to construct some examples of the nonexpansive mapping $T$ from $R^2$ to $R^2$ such that $T$ should have fixed points more than one. But I could not construct. Can somebody help me? Please.

A mapping $T:X \to Y$ is called nonexpansive if $||Tx-Ty|| \leq ||x-y||$ for all $x,y$ in $X$.

Answer 1

For instance, $f(x)=(\arctan x_1,\arctan x_2)$.

Answer 2

Any linear map

$T:\Bbb R^2 \to \Bbb R^2 \tag 1$

will be nonexpansive by this definition provided

$\Vert T \Vert \le 1, \tag 2$

since then

$\Vert Tx - Ty \Vert = \Vert T(x - y) \Vert \le \Vert T \Vert \Vert x - y \Vert = \Vert x - y \Vert. \tag 2$

Examples include the identity map $I$ as well as

$J = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}, \tag 2$

$O(\theta) = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}; \tag 3$

there are many more.

Answer 3

Let $P(x)$ be a nonexpansive mapping from $R$ to $R$ with the fixed points set $A$. Then $T(x,y)=(P(x),P(y))$ is a nonexpansive mapping from $R^2$ to $R^2$ with the fixed points set $A \times A$.

Answer 4

If you only require $||Tx-Ty|| \le ||x-y||\quad \forall x,y \in X$, you can take T to be the identity map. Then every point in $X$ is a fixed pioint and $||Tx-Ty|| = ||x-y||\quad \forall x,y \in X$.