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$M$ 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.

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  • $\begingroup$ You may need additional assumptions such as $M$ is not empty and perhaps $X$ is separable, $X$ has a strictly convex unit ball... $\endgroup$ Feb 11, 2023 at 13:34

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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.

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  • $\begingroup$ Second challenge: find an example where $A$ does not have a fixed point! $\endgroup$ Feb 11, 2023 at 22:25
  • $\begingroup$ @Gribouillis I added an example of a fixed point free nonexpansive mapping. $\endgroup$
    – Gerd
    Feb 11, 2023 at 22:41
  • $\begingroup$ I don't understand the counterexample, for example if $x = (x_0, x_1, \ldots) = (0, 1, \frac{1}{2}, \frac{1}{3}, \ldots)$, then $\|x\| = 1$ and $A x = x$ with your definition. $\endgroup$ Feb 12, 2023 at 9:07
  • $\begingroup$ @Gribouillis This is not a fixed point. $Ax=x$ means $1-\|x\|=x_1$, $x_1=x_2$, $x_2=x_3$, $\dots$. This means $(x_n)$ is the constant sequence $(1-\|x\| )_n$. $\endgroup$
    – Gerd
    Feb 12, 2023 at 9:12
  • $\begingroup$ So it is $X = c_0(\mathbb{N}^*, \mathbb{R})$. Nice proof and example! $\endgroup$ Feb 12, 2023 at 9:17

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