Convergence for $n+1$ iteration in composition of functions.

Question

The problem is if $f_n(x)$ converges to $y$, the next iteration, $f_{n+1}(x)$ converges to $f(y)$?

$x,y$ are fixed values, but no fixed points.

$f^n$ is the n-composition of $f$

i was studing dynamical systems, and in a problem i conjetured this for a proof, but i can't prove it, can you tell me it is true?, or anyone can help with a hint or counterexample, thanks!

Answer 1

As $n$ approaches infinity, $(n+1)$ also approaches infinity.

Conversely, as $(n+1)$ approaches infinity, $n$ also approaches infinity.

It follows that

$$f^n(x) \rightarrow y,\;\text{as}\;n \rightarrow \infty$$

$$\text{if and only if}$$

$$f^{n+1}(x) \rightarrow y,\;\text{as}\;n \rightarrow \infty$$

However, if $f$ is continuous at $y$, you also get $f(y) = y$.

Answer 2

$f^{n+1}(x) = f(f^n(x))$. Presumably $f$ is continuous at $y$. Use the definition of continuity.