Conditions for convergence of fixed-point iterations (not necessarily to a unique fixed-point)

Question

Let $X \in R^n$ be a compact convex set, and $f:X \to X$ be a continuous function. Then, can we say that from all $x_0 \in X$, the fixed-point iterations $x_{k+1}=f(x_k)$ to converge to some fixed-point $\bar{x}(x_0) \in X$?

If not, what are the conditions that $f$ must satisfy such that the iterations $x_{k+1}=f(x_k)$ always converge to some fixed-point $\bar{x}(x_0) \in X$ starting from any $x_0 \in X$?

Please note that the fixed-point $\bar{x}(x_0)$ need not be unique.

If I understand correctly, the Brouwer fixed-point theorem states that there exists atleast one $\tilde{x} \in X$ satisfying $\tilde{x} =f(\tilde{x})$, but does it say something about the convergence of fixed-point iterations?

Answer 1

That's not true. For example, let $X$ be the closed unit ball and $f$ be a non-trivial rotation. Then unless $x_0$ is the origin (which is the unique fixed point of $f$), the sequence $x_{k+1} = f(x_k)$ is not convergent.

Of course if $f$ is a contraction, then any such sequence converges to the unique fixed point.

Answer 2

Similar to Arctic Char's answer.

No. What about $\ X=\ $ unit circle in $\ \mathbb{R}^2\ $ and $\ f\ $ is reflection in the $\ y-$axis. $\ X\ $ is compact and convex, and $\ f\ $ is continuous. However, $\ f\ $ repeatedly acting on $\ x_0=(0,-1)\ $ looks like: $\ (0,-1)\to(0,1)\to (0,-1)\to(0,1)\to\ldots\ $ which does not converge.