3
$\begingroup$

Exercise :

Let $X$ be a Banach space and $C \subseteq X$ be closed, convex and bounded. Moreover, let $g:C \to C$ be a non-expansive operator, meaning that : $$\|g(u) - g(v) \| \leq \|u-v\| \; \forall u,v \in C$$ Show that : $$\inf\{\|u-g(u)\|:u \in C\} = 0$$

Attempt :

Since $X$ is Banach and $C$ is closed, this means that $C$ is also Banach.

Since it is $g : C \to C$, this means that $g(C) \subseteq C$. Since $C$ is convex, any such $g_n \in g(C)$ can be written as $$g_n(x) = (1-r_n)y + r_ng(x)$$ with $x,y \in C$ and $\{r_n\}_{n \geq 1} \in [0,1]$ with $r_n \xrightarrow{n \to \infty} 1$.

It also is :

$$\|g_n(x) - g_n(z) \| = r_n \|g(x) - g(z)\| \leq r_n\|x-z\|\; \forall x,z \in C$$

Let $\{x_n\}_{n \geq 1} \in C$ such that $g_n(x_n) = x_n$. Then : $$x_n = (1-r_n)y + r_ng(x_n)$$ But this leads to : $$\|x_n -g(x_n) \| = (1-r_n)\|y-g(x_n)\|$$ But since $C$ is also bounded, $\|y-g(x_n)\|$ will be a finite number and thus : $$\|x_n - g(x_n) \| = (1-r_n)\|y-g(x_n)\| \xrightarrow{n \to \infty} 0$$ $$\implies$$ $$\inf\{\|u-g(u)\|:u \in C\} = 0$$

I would really appreciate any feedback on this attempt, since it was completely based on intuitions (I have never handled something similar before). Any hints, tips or alternative elaborations will be most welcome.

$\endgroup$
  • $\begingroup$ I don't see a question? $\endgroup$ – user370967 Mar 17 at 11:28
  • $\begingroup$ @Math_QED There is a bold text section at the end ? $\endgroup$ – Rebellos Mar 17 at 11:33
  • $\begingroup$ Check the exercice. What are you trying to prove? $\endgroup$ – user370967 Mar 17 at 11:33
  • $\begingroup$ @Math_QED Oh, apologies. The point is to prove $\inf\{\|u-g(u)\|:u \in C\} = 0$ but I probably deleted it by mistake from the exercise's body while typing. $\endgroup$ – Rebellos Mar 17 at 11:35
  • 1
    $\begingroup$ What exactly is $g_n$? $\endgroup$ – user370967 Mar 17 at 11:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.