Does an infinite composition of functions converge? Can we determine it?

Question

I’m in 6th grade and this is one question I have about infinite function compositions:

I recently thought about function composition; things in the form $a_1(a_2(…(a_n(x))…)$ which are things like $e^{2x}$ or $\sin(\sin(x^2))$.

After that, I thought about infinite function compositions of the same function; $f(f(f(f(f(f(…$. However, this doesn’t really make sense as we will never reach $f(x)$, so I consider an infinite composition as something in the form $\cdots f(f(x))))\cdots$, where there are infinite compositions to the left.

Consider $\sin(x)$ where we infinitely many times repeatedly apply the sine function to whatever the previous thing was; this is infinite function composition, which we could write as $\cdots(\sin(\sin(x)))\cdots$ where $)))\cdots$ denotes the infinite number of parenthesis.

Are there any mathematical tools to determine if an infinite composition (like I described) will converge for some $x$? Can it even converge?

Answer 1

Yes. For a fixed value of $x$, say $x_0$, you obtain a sequence, $$\begin{align*} a_0 &= x_0\\ a_1 &= f(a_0)\\ a_2 &= f(a_1)\\ &\vdots\\ a_{n+1}&= f(a_n)\\ &\vdots \end{align*}$$ There is a well-developed tool to determine if this sequence converges to a particular value: limits. So we can ask if $\lim_{n\to\infty} a_n$ exists, and if so what it is. We could then define a function $f^{(\infty)}$ to have domain all values of $x$ for which the limit exists, and define it to take as value that limit when it exists. This is called "pointwise convergence."

There are other ways of trying to measure the "distance" between two functions. For example, you can consider the area between them by integrating $|f-g|$. (This is, roughly, called the $L_1$-distance; not exactly, but essentially). You could then consider the sequence $f^{(n)}$ and determine if there is a function $F$ such that the distance of $f^{(n)}$ to $F$ approaches $0$ as $n$ goes to infinity, and obtain a "limit" that way. Or you can define a distance using the "sup norm": $$\lVert f-g\rVert = \sup\{|f(a)-g(a)|\mid a\in\mathbb{R}\}$$ and proceed as above. There are other ways for sequences of functions to converge ("in measure", "almost everywhere", "uniformly", "almost uniformly", $L_p$-distance, and more), and each of them will give you a different way of deciding whether the sequence of functions converges or not.

In addition, one need not have a "target" limit in mind to investigate the question: one can determine if the sequence is a Cauchy sequence, and if so there are results that generally tell us that if the sequence is a Cauchy sequence, then it converges.

Answer 2

There should be many examples of an infinite composition also known as infinite 'iteration' that 'converges'.

The only one I can think of right now is the Hahn Banach fixed point theorem:

Theorem Let $(X,d)$ be a non empty complete metric space, and let $\phi : X \to X$ be a contraction (there is $\alpha \in (0,1) \subset \mathbb{R}$ s.t. $d(\phi(x),\phi(y)) \leq \alpha d(x,y)$ for any $x,y \in X$). Then the following is true:

1. $\phi$ has a unique fixed point $x^*$, (there is $x^* \in X$ s.t. $\phi(x^*) = x^*$).

2. For any $x_0 \in X$ the sequence ${(\phi^k(x_0))}$ converges to $x^* \in X$ [this is your infinite composition!], and $\displaystyle d(\phi^k(x_0),x^*) \leq \frac{\alpha^k}{1-\alpha} d(\phi(x_0),x_0),$ where $\alpha$ is the constant of the contracton $\phi$, and $\phi^k$ is the $k$-iteration for $\phi$ for $k \in \mathbb{N}$, with $\phi^0 := id_{X}$.

So basically, this theorem says that under those conditions

$$\ldots\phi(\ldots\phi(\phi(\phi(x_0))))\ldots)\ldots = \lim_{k \to \infty} \phi^k(x_0) = x^*,$$ for any $x_0 \in X$ and for a unique $x^* \in X$.