# Existence of infinite iteration of functions $f_\infty$?

Given a sequence of functions $$\{f_n\}$$ satisifying an iterated relation such as

• $$f_n(x)=g(x+f_{n-1}(x))$$

• $$f_n(x)=g(xf_{n-1}(x))$$

• $$f_n(x)=g(x/f_{n-1}(x))$$

Where $$g:=f_1$$ is continuous on the interval $$[a, b]$$ (or differentiable on $$(a,b)$$ for stronger assumptions)

Question: How to prove the existence of $$f_\infty(x):=\lim\limits_{n\to \infty}f_n(x)$$?

AND Are there any methods to prove such $$f_\infty$$ does not exist?

The question comes from the problems

Let $$g(x)=\sin x$$.

I "proved" the $$1^{\rm{st}}$$ and the $$3^{\rm{rd}}$$ integral by assuming the exsistence of $$f_\infty$$.

@Sangchul Lee think $$f_\infty$$ in the $$2^{\rm{nd}}$$ integral does not exist due to the chaotic behavior.

If $$f_\infty$$ in the $$2^{\rm{nd}}$$ integral exists, then $$L=\int_0^\alpha \sin y\,\mathrm d\left(\frac y{\sin y}\right) =1.86006...$$ where $$\alpha=2.31373...$$ is the positive root of $$\dfrac t{\sin t}= \pi$$.

Some thoughts so far:

If we could prove $$f(t)=g(x_0+t)$$ is a contraction mapping on $$[a,b]$$ for every $$x_0\in[a,b]$$, that is, if $$t_0$$ (depending on $$x_0$$) is the only fixed point on $$[a,b]$$, then the result is intuitively true from Banach Fixed Point Theorem (similar to the case $$f(t)=g(x_0t)$$ and $$f(t)=g(x_0/t)$$).

However, we could not apply the theorem for any $$f$$, one example is $$f(t)=\sin(x_0+t)$$ in the $$1^{\rm{st}}$$ integral.

• Where it exists, I have found a general (addition) formula using your method: from $a$ to $b$, $$\small L=af(a)-bf(b)-\frac{f(a)^2-f(b)^2}2+(a-f(a))f(a-f(a))-(b-f(b))f(b-f(b))+F(b-f(b))-F(a-f(a))$$ where $F$ is the antiderivative of $f$. – TheSimpliFire Oct 7 '18 at 18:36
• Where it exists, I have found a general (multiplication) formula using your method: from $a$ to $b$, $$L=\frac b{f(b)}-\frac a{f(a)}-\int_{a/f(a)}^{b/f(b)}\frac{tf'(t)}{f(t)}\,dt$$ and the integral cannot be simplified further for a general function $f$. – TheSimpliFire Oct 7 '18 at 18:42
• Where it exists, I have found a general (division) formula using your method: from $a$ to $b$, $$L=\int_{af(a)}^{bf(b)}f(t)(f(t)+tf'(t))\,dt$$ and the integral cannot be simplified further for a general function $f$. – TheSimpliFire Oct 7 '18 at 18:46

## 1. The first integral

Consider the sequence $$(f_n)$$ of functions on $$[0, \pi]$$ defined recursively by

$$f_0(x) = 0, \qquad f_{n+1}(x) = \sin(x + f_n(x)).$$

We first establish the following simple lemma to guarantee that the sequence remains bounded on a certain region. Throughout this section, we always assume that $$x$$ takes values in $$[0, \pi]$$.

Overview of proof.

1. We establish bounds for $$(f_n(x))$$ which ensures that the sequence does not behave wild.

2. We show that contraction mapping theorem is applicable. Details of the argument will depend on the value of $$x$$.

Step 1. $$f_n(x) \in [0, \pi-x]$$ for all $$n \geq 1$$.

Proof. Recall that $$\sin\theta < \theta$$ for all $$\theta > 0$$. So if $$t \in [0, \pi-x]$$, then $$x+t \in [x, \pi]$$ and hence

$$0 \leq \sin(x+t) = \sin(\pi-x-t) \leq \pi-x-t \leq \pi-x.$$

Since $$f_1(x) = \sin(x) \in [0, \pi-x]$$, inductively applying the above inequality shows the desired claim.

Step 2. $$(f_n(x))$$ converges for each $$x \in [0, \pi]$$.

• Case $$x < \pi-1$$. In this case, by the mean value theorem, there exists $$\xi \in [0, 1]$$ such that

\begin{align*} \left|f_{n+1}(x) - f_n(x)\right| &= \left|\cos(x+\xi)\right| \left|f_n(x) - f_{n-1}(x)\right| \\ &\leq r \left|f_n(x) - f_{n-1}(x)\right|, \end{align*}

where $$r = \max\{ \lvert \cos(x+t)\rvert : t \in [0, 1] \}$$. By the assumption, we check that $$r < 1$$, and the claim follows from the contraction mapping theorem.

• Case $$\pi-1 \leq x < \pi$$. In this case, $$t \mapsto \sin(x+t)$$ is a strictly decreasing function on $$[0, \pi-x]$$. This has two consequences.

(1) Since $$f_0(x) \leq f_2(x)$$, this implies that $$f_{2n}(x) \leq f_{2n+2}(x)$$ and $$f_{2n+1}(x) \leq f_{2n-1}(x)$$. So both the even-th terms and the odd-th terms converge.

(2) Since $$f_0(x) \leq f_1(x)$$, it follows that $$f_{2n}(x) \leq f_{2n+1}(x) \leq f_1(x)$$.

Combining altogether, $$(f_n(x))$$ is bounded between $$0$$ and $$\sin(x) = \sin(\pi-x) < \pi-x$$. So, as in the previous case, there exists $$\xi \in [0, \sin(x)]$$ such that

$$\left|f_{n+1}(x) - f_n(x)\right| = \left|\cos(x+\xi)\right| \left|f_n(x) - f_{n-1}(x)\right| \leq r \left|f_n(x) - f_{n-1}(x)\right|,$$

where $$r = \max\{ \lvert \cos(x+t)\rvert : t \in [0, \sin(x)] \}$$. Since $$r < 1$$, we can still apply the contraction mapping theorem.

• Case $$x = \pi$$. This case is trivial.

Therefore $$(f_n(x))$$ converges for all $$x \in [0, \pi]$$.

## 2. The third integral

Let $$(f_n)$$ be the sequence of functions on $$(0, \pi/2]$$ defined by

$$f_0(x) = 1, \qquad f_{n+1}(x) = \sin(x/f_n(x)).$$

We assume that $$x \in (0, \pi/2]$$ henceforth.

Overview of proof.

1. We establish bounds of $$(f_n(x))$$ that ensures that the iteration behaves well.

2. We prove that $$(f_{2n+1}(x))$$ is increasing in $$n$$ and $$(f_{2n}(x))$$ is decreasing in $$n$$, and so, both $$\alpha(x) := \lim_{n\to\infty} f_{2n+1}(x)$$ and $$\beta(x) := \lim_{n\to\infty} f_{2n}(x)$$, although it is not yet known whether they coincide.

3. Both $$\alpha$$ and $$\beta$$ are solutions of a certain functional equation. We show that, under an appropriate condition, this equation has a unique solution. This tells that $$\alpha = \beta$$, hence the sequence $$(f_n(x))$$ converges.

Step 1. $$f_n(x) \in [\sin x, 1]$$ for all $$n \geq 1$$.

Proof. If $$t \in [\sin x, 1]$$, then

$$\sin x \leq \sin \left(\frac{x}{t}\right) \leq \sin \left(\frac{x}{\sin x}\right) \leq \sin \left(\frac{\pi}{2}\right) = 1.$$

Therefore the claim follows by mathematical induction.

Step 2. $$(f_n(x))$$ converges.

For each $$x$$, consider $$g_x(t) = \sin(x/t)$$. Then $$h_x$$ is a strictly decreasing function on $$[\sin(x), 1]$$. Together with $$f_1(x) = \sin x \leq f_2(x) \leq 1 = f_0(x)$$, this implies that

$$f_1(x) \leq f_3(x) \leq \cdots \leq f_{2n+1}(x) \leq f_{2n}(x) \leq \cdots \leq f_2(x) \leq f_0(x).$$

So it follows that both $$(f_{2n+1}(x))$$ and $$(f_{2n}(x))$$ converge. Let $$\alpha(x) := \lim_{n\to\infty} f_{2n+1}(x)$$ and $$\beta(x) := \lim_{n\to\infty} f_{2n}(x)$$. So it remains to prove that $$\alpha(x) = \beta(x)$$.

Taking limit to the recursive formula, it is clear that

$$\beta(x) = g_x(\alpha(x)), \qquad \alpha(x) = g_x(\beta(x)).$$

So both $$\alpha$$ and $$\beta$$ solve the functional equation $$f(x) = g_x(g_x(f(x)))$$.

Now let $$f : (0, \pi/2] \to (0, 1]$$ be any solution of this functional equation satisfying the bound $$\sin x \leq f(x) \leq 1$$. By writing $$y = f(x)$$, we find that $$x/\sin(x/y) \in [x, x/\sin x] \subseteq [0, \pi/2]$$ and hence

\begin{align*} y = \sin(x/\sin(x/y)) &\quad\Longleftrightarrow \quad \arcsin(y) = \frac{x}{\sin(x/y)} = \frac{y}{\operatorname{sinc}(x/y)} \\ &\quad\Longleftrightarrow \quad \operatorname{sinc}(x/y) = \frac{y}{\arcsin(y)} \end{align*}

Note that $$\frac{x}{y} = \frac{x}{f(x)} \leq \frac{x}{\sin x} \leq \frac{\pi}{2}$$ and $$\operatorname{sinc}$$ is injective on $$(0, \pi/2]$$. If we denote the inverse of $$\operatorname{sinc}$$ restricted onto $$(0, \pi/2]$$ by $$\operatorname{sinc}^{-1}$$, then

\begin{align*} y = \sin(x/\sin(x/y)) &\quad\Longleftrightarrow \quad x = y \operatorname{sinc}^{-1}\left(\frac{y}{\arcsin(y)}\right). \end{align*}

This implies that $$f$$ is injective and its inverse is explicitly given by the formula above. So the functional equation with the prescribed bound uniquely determines $$f$$. Therefore $$\alpha = \beta$$ and the claim follows.

• Fantastic! This shows the contraction mapping theorem and the monotone convergence theorem could be applied in this question. For a more general case, i.e. other functions instead of $\sin x$, are there any other ways to prove the existence of $f_\infty$? – Tianlalu Oct 8 '18 at 8:26
• Alternatively, are there any methods to prove such $f_\infty$ does not exist? We may use $\sin(x\sin(x\cdots))$ as an example. – Tianlalu Oct 9 '18 at 1:57
• @Tianlalu, There is a whole theory, called dynamical system theory, to analyze such behavior. However, I have little exposure to this area... – Sangchul Lee Oct 9 '18 at 2:00

From the definition, you have

$$f_1(x)=g(x)$$

then

$$f_2(x)=g(x+g(x)), \\f_3(x)=g(x+g(x+g(x))), \\\cdots$$

which is an "ordinary" sequence for a given $$x$$.

You can write it as

$$a_n=g(x+a_{n-1}),\\a_0=0$$ and use the fixed-point theorem.

For instance, with $$g(x):=\dfrac x2$$,

$$a_1=\frac x2, \\a_2=\frac{3x}4, \\a_3=\frac{7x}8, \\$$ which converges pointwise to $$a_\infty=x$$.

For $$g(x):=\sin x$$,

$$f_n(x)=\sin(x+f_{n-1}(x))$$

can be written

$$a_n=\sin(x+a_{n-1}),\\a_0=0.$$

If it converges, it will converge to $$a=\sin(x+a)$$, that has solutions for all $$x$$, and the convergence conditions are given by the fixed-point theorem.

As $$|(\sin a)'|<1$$ for all $$a\ne k\pi$$, the fixed-point is attractive almost everywhere. And as then next iterate of $$a=k\pi$$ is $$\sin x$$, we don't remain stuck.

• How about $f_1(x)=\sin(x)$? – Tianlalu Oct 7 '18 at 18:23
• OK, but how is this even a proof? You've given only the definition and an example. This doesn't prove the existence of a limit for all functions. – TheSimpliFire Oct 7 '18 at 18:24
• @TheSimpliFire: that was not intended in my answer. (And of course the claim is not true for all $g$.) – Yves Daoust Oct 7 '18 at 18:25
• We couldn't apply the fixed-point theorem directly because $|\cos(x+\xi)|=1$ when $x+\xi=\pi$, which means $f$ may not be a contraction mapping. However, @Sangchul Lee has proved this to be true by considering different cases. – Tianlalu Oct 8 '18 at 8:20
• @Tianlalu: I already commented about that. – Yves Daoust Oct 8 '18 at 8:23