# Show that $\exists u_0 \in C : g(u_0) = u_0$, if $g$ is nonexpansive over a Banach subspace.

Exercise :

Let $$X$$ be a Banach space, $$C \subseteq X$$ compact and convex and $$g : C \to C$$ a nonexpansive operator. Show that $$\exists u_0 \in C : g(u_0) = u_0$$.

Thoughts :

In a previous exercise, I have proven that if $$X$$ is Banach, $$g: C \to C$$ is non expansive and $$C \subseteq X$$ is a closed, convex and bounded set, then it is : $$\inf \{\|u-g(u)\| : u \in C\} = 0$$.

That essentialy tells us that we can find a $$u_0$$ such that $$g(u_0) = u_0$$ as they come abstractly close, but as correctly pointed in the comments, closed and bounded doesn't always imply compactness (infinite dimensional cases).

Now, coming to the specific excercise, how would one work with compactness ? There really seems a connection between the two exercises but I can't yield an intuition.

• "I guess $C$ being closed (convex) and bounded equals compact and convext as well." Closed and bounded does not imply compact in infinite dimensional Banach spaces. – MisterRiemann Mar 17 at 15:20
• @MisterRiemann Completely missed out on that. True that though, how would that affect a potential approach ? Without making use of what I proved, I can't see any other way around it. – Rebellos Mar 17 at 15:23
• On the other hand, compact does imply closed and bounded, so that you may use what you proved, since you're assuming that $C$ is compact in the first place. I apologize if I confused you. – MisterRiemann Mar 17 at 15:33

If $$C$$ is compact then it is closed and bounded. From your previous exercise we know that there must exist a sequence $$(u_n)\subset C$$ such that $$\|u_n-g(u_n)\|\to 0$$. Now as $$C$$ is compact there exists some subsequence $$(u_{n_k})$$ converging to $$u\in C$$. By the uniqueness of limits, and continuity of $$g$$, we thus have $$0=\lim_{k\to\infty}\|g(u_{n_k})-u_{n_k}\|=\|g(u)-u\|.$$ Thus $$g(u)=u$$ and we are done.