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

Question

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

• $\displaystyle\int_0^\pi\sin(x+\sin(x+\sin(x+\cdots)))\,\mathrm dx=2$ and

• $\displaystyle\int_0^\pi\sin(x\sin(x\sin(x\cdots)))\,\mathrm dx$ and

• $\displaystyle\int_{-\frac\pi2}^{\frac\pi2}\sin\frac x{\sin\frac x{\sin\frac x{\sin\cdots}}}\,\mathrm dx=\frac34 \pi$.

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.

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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 $\dfrac t{\sin t}= \pi$.

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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.

Answer 1

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 $a_\infty=x$.

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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.

Answer 2

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.