# Iterative application of continuous function $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $\lim_{n \to \infty}f^{n}(x)$ exists $\forall x$.

The question Let $$f: \mathbb{R} \rightarrow \mathbb{R}$$ be a continuous function such that $$\lim_{n \to \infty}f^{n}(x)$$ exists $$\forall x$$. Define $$S = \{\lim_{n \to \infty}f^{n}(x): x \in \mathbb{R}\}$$ and $$T = \{x \in \mathbb{R}: f(x)=x\}$$. Then conclude what can be said about the inclusion relation between $$S$$ and $$T$$, either $$S$$ is properly contained in $$T$$ or vice versa, or $$S=T$$.

Thoughts

I somehow feel that the answer should be $$S$$ is contained in $$T$$, as the sequence will converge to a stable point, and not every stable point needs to be an attracting point. Though I am not sure how to prove this. When the function is contractive, there is Banach fixed point theorem to help, but when it's just continuous I don't know how to conclude anything concretely.

• You can't really say anything at all unless $f$ is continuous. Commented Apr 29 at 15:28
• For example, your intuition/thoughts are true only if $f$ is continuous. Commented Apr 29 at 15:32
• In the title I mentioned that f is continuous, I added it to the body as well now. Any thoughts now? @ThomasAndrews
– Debu
Commented Apr 29 at 15:33
• It seems to me that $S=T$: If $x \in T$ then trivially $\lim_{n \to \infty}f^n(x)=x$, hence $x \in S$. If $y \in S$, then for some $x$ we have $y= \lim_{n \to \infty}f^n(x)$. Since $f$ is continuous $y$ is a fixed point of $f$. Hence $y \in T$.
– Gerd
Commented Apr 29 at 15:36
• The only way to have $S\subsetneq T$ is if you define $S$ differently, like: $$S=\{\lim f^n(x)\mid x\notin T\}.$$ Commented Apr 29 at 15:42

I claim that $$S = T$$. Let us prove the inclusion $$\supseteq$$ first. If $$x \in T$$, then $$f(x) = x$$ and inductively $$f^n(x) = x$$. It follows that
$$\lim_{n \to \infty} f^n(x) = x,$$ therefore $$x$$ appears as a limit of iterating $$f$$ and consequently, $$x \in S$$. For the converse direction, let $$L \in S$$, so there exists $$x \in \mathbb{R}$$ such that $$\lim_{n\to\infty}f^n(x) = L$$. We want to show that $$f(L) = L$$, so that we may conclude that $$L \in T$$. We have
$$f(L) = f(\lim_{n \to \infty} f^{n}(x)) = \lim_{n\to \infty} f(f^n(x)) = \lim_{n \to \infty} f^{n+1}(x) = L,$$ which finishes the proof. Note that for the second equality, continuity of $$f$$ was needed.