Is every convergent limit of an iteration a fixed point as well?

Question

Let $f(x)$ be a function and suppose $\lim_{n \to \infty}f^n(a)=L$ for some $a$ in the domain of $f$.

What are the sufficient conditions for $L$ being a fixed point of $f$?

Is the continuity of $f$ enough?

Answer 1

If the iterates $f^n(a)$ converge to a point $L$, and $f$ is continuous in $L$, then $L$ is a fixed point of $f$, provided the domain is Hausdorff. The continuity in $L$ means that we can interchange the limit at $L$ with the application of $f$,

$$f(L) = f\left(\lim_{n\to\infty} f^n(a)\right) = \lim_{n\to\infty} f\left(f^n(a)\right) = \lim_{n\to\infty} f^{n+1}(a) = \lim_{m\to\infty} f^m(a) = L.$$

Directly from the definition of continuity at a point, for every neighbourhood $U$ of $f(L)$, there is a neighbourhood $V$ of $L$ with $f(V) \subset U$. By the definition of convergence, there is an $n_V\in \mathbb{N}$ such that $f^n(a) \in V$ for all $n\geqslant n_V$. Then we have $f^n(a) \in V \cap U$ for all $n \geqslant n_V + 1$. If we had $f(L)\neq L$, by the Hausdorff condition, there are two disjoint open neighbourhoods $U_1$ of $f(L)$ and $V_1$ of $L$. Choosing $V = V_1 \cap f^{-1}(U_1)$ then leads to the contradiction

$$f^n(a) \in V \cap U_1 \subset V_1 \cap U_1 = \varnothing$$

for $n > n_V$, so the assumption $f(L)\neq L$ must have been wrong.

The above argument shows that if $f$ is continuous in $L$, then we also have $\lim\limits_{n\to\infty} f^n(a) = f(L)$, so $L$ must be a fixed point whenever the space is such that every convergent sequence has only one limit point. Such spaces are called US-spaces. That is a strictly weaker condition than being Hausdorff, but for the moment, let us stick to Hausdorff spaces.