Let $\phi:\mathbb{R}\rightarrow\mathbb{R}$ be a function (not necessarily continuous).

Let $\phi_0(x)=\phi(x)$ and $\forall k\in\mathbb{N},\phi_{k+1}(x)=\phi(x\cdot\phi_k(x))$.

  • 1. Let $k\in\mathbb{N}^*$. What can be said of the set of functions $\phi$ verifying $\forall x\in\mathbb{R},\phi(x)=\phi_k(x)$ ?

  • 2. What can be said of the set of functions $\phi$ verifying $\forall x\in\mathbb{R},\exists k\in\mathbb{N}^*,\phi(x)=\phi_k(x)$ ?

  • 3. Let $x_1,\dots,x_n\in\mathbb{R}$. . Can we build non-trivial (as in, not mapping the $x_i$ to $0$ or terms that, added, make $0$) functions $\phi$ such that $\displaystyle\sum_{k=0}^\infty\sum_{i=1}^n\phi_k(x_i)$ is convergent ?

  • 4. For what functions $\phi$ is the sum $\displaystyle\sum_{k=0}^\infty\phi_k(x)$ convergent for all $x\in\mathbb{R}$? Absolutely convergent ?

Context :

In a dream I had this morning, I was passing an oral exam and the examiner handed me the exercise sheet, on which I could read a line : $\phi(x_1)+\phi(x_2)+...+\phi(x_1\phi(x_1))+\phi(x_2\phi(x_2)+\dots$. Unfortunately, I woke up right after and hence didn't have time to read the rest of the paper. However, as I found it interesting , I wrote that line down to remember it. Thinking back about it during the day, the questions above arose, and I'd be very interested in having them solved.

  • 3
    $\begingroup$ Can you modify the title to be more expressive of the actual question? $\endgroup$ – user98602 Oct 30 '14 at 17:23
  • 1
    $\begingroup$ @MikeMiller Please suggest a better title, I put this one because I did not really know what to put. $\endgroup$ – Hippalectryon Oct 30 '14 at 17:23
  • $\begingroup$ The third question has a problem: $\sigma_k$ has a sum with variable $k$ in it - it's not clear what $\sigma_k$ is actually meant to be. $\endgroup$ – Milo Brandt Nov 4 '14 at 22:21
  • $\begingroup$ @Meelo Thanks for noticing, I edited it $\endgroup$ – Hippalectryon Nov 4 '14 at 22:32

I can say a few things about part 4). For the sum to converge for some $x_0$ it is necessary that

$$\lim_{k\to\infty} \phi_k(x_0) = 0$$

If $\phi(x)$ is a continuous function everywhere, then $\phi_k(x)$ is as well and we can take the following limit

$$\lim_{k\to\infty} \phi_{k+1}(x_0) = 0 = \lim_{k\to\infty} \phi(x_0\phi_k(x_0)) = \phi(x\cdot 0) = \phi(0)$$

Therefore $\phi$ needs to be $0$ at the $x = 0$. Now suppose that $\phi(x)$ is a power series converging for all real numbers. We can say

$$\phi_0(x) = xg_0(x)$$

where $g_0(x)$ is given by a power series starting with the $x^0$ term. Iterating the recurrence, keeping track of the constant coefficient and using induction we have that

$$\phi_k(x) = x^{k+1}g_k(x)$$

where $g_k(x)$ starts with an $x^0$ term. Now, let's try some functions $g_0(x)$. The easiest one to try is $g_0(x) = 1$. Then

$$\phi_k(x) = x^{k+1}$$


$$\sum_{k=0}^{\infty} \phi_k(x) = \sum_{k=0}^\infty x^{k+1} = \frac{x}{1-x}$$

While $\phi(x)$ is convergent everywhere, the sum of the iterates leaves us with a function with finite radius of convergence. In general you can get the iterated function $g_k$ by

$$g_{k+1}(x) = g_k(x)g_k(x^{k+2}g_k(x))$$

Here's another easy example. Let $g_0(x) = x^{p_0}$ where $p_0$ is a positive integer. Then $g_k(x) = x^{p_k}$ for some $p_k$ given by

$$g_{k+1}(x) = x^{p_{k+1}} = x^{p_k}(x^{k+2}x^{p_k})^{p_k} = x^{p_k^2+(k+3)p_k}$$


$$\sum_{k=0}^\infty \phi_k(x) = \sum_{k=0}^\infty x^{p_k + k+1}$$

$$\frac{p_{k+1}+k+2}{p_k + k+1} = \frac{p_k^2+(k+3)p_k + k+2}{p_k + k + 1}$$

which diverges as $k \to \infty$ since $p_k$ grows without bound, so the sum of the iterates is a Lacunary function with radius of convergence $1$. It's probably worth guessing that the radius of convergence for the sum of iterates is finite when $\phi(x)$ is entire.


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