How to prove this problem related to fixed point theorem? $M$ is a bounded,convex and closed subset of Banach space $X$, $A:M\rightarrow M$ satisfies:
$$||Ax-Ay||\le ||x-y||$$
for all $x,y\in M$
Show that $\forall \varepsilon>0$,there exist $x\in M$ such that:
$$||Ax-x||\le  \varepsilon$$
The form of this problem is similar to fixed point theorem in Banach space.
 A: Let $M\not=\emptyset$ be a closed, bounded and convex subset of $X$.
Let $\delta \in (0,1)$ and let $b \ge 0$ be such that $\|x\| \le b$ $(x \in M)$. Fix any $x_0 \in M$ (note that $M\not=\emptyset$) and let $B:M \to M$ be defined as $Bx= (1-\delta)Ax + \delta x_0$ (note that convexity of $M$ implies $Bx \in M$ for $x \in M$). Now, for $x,y \in M$ we have
$$\|Bx-By\| =  (1-\delta)\|Ax-Ay\| \le (1-\delta)\|x-y\|.$$
By Banach's fixed point theorem $B$ has a fixed point $z \in M$ (note that $M$ is closed, hence complete).
Now
$$
\|Az-z\| = \|Az-Bz\|=  \|Az - ((1-\delta)Az + \delta x_0)\| = \delta\| Az-x_0\|\le 2\delta b
$$
If $\varepsilon > 0$ is given choose $\delta \in (0,1)$ such that $2\delta b \le  \varepsilon$. Then $\|Az-z\| \le \varepsilon$.
Edit: $A$ may have no fixed point: Let $X=c_0(\mathbb{N},\mathbb{R})$ endowed with the maximum norm, $M$ the closed unit ball and
$$
A:M \to M, \quad Ax=A(x_n)=(1-\|x\|,x_1,x_2, \dots).
$$
Then $A$ is nonexpansive, and $Az=z$ for some $z \in M$, that is
$$
(1-\|z\|,z_1,z_2, \dots)=(z_1,z_2,z_2,\dots),
$$
leads to $1-\|z\|=z_n$ $(n \in \mathbb{N})$.
Since $z_n \to 0$ $(n \to \infty)$ this forces $z_n = 0$ $(n \in \mathbb{N})$. Then $1=1-\|z\|=0$, a contradiction.
