# 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.

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

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.

• Second challenge: find an example where $A$ does not have a fixed point! Feb 11, 2023 at 22:25
• @Gribouillis I added an example of a fixed point free nonexpansive mapping.
– Gerd
Feb 11, 2023 at 22:41
• 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. Feb 12, 2023 at 9:07
• @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 $(1-\|x\| )_n$.
– Gerd
Feb 12, 2023 at 9:12
• So it is $X = c_0(\mathbb{N}^*, \mathbb{R})$. Nice proof and example! Feb 12, 2023 at 9:17