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I recently read an article in which the author describes how to find some functions that obey to certain recursion relationships. If we want to find a function that satisfies, for example, $f(x^a) = x \cdot f(x)$, then the author explains we can proceed as follows:

Consider $ x > a > 1$ . Then $f(x) = f((x^{(1/a)})^a) = x^{1/a}f(x^{(1/a)})$

$= x^{1/a}f((x^{(1/a^2)})^a) = x^{1/a + 1/a^2}f(x^{(1/a^2)}) = x^{1/a + 1/a^2 + ... + 1/a^n}f(x^{1/a^n})$ .

We know that the limit of $1/a^n$ is zero, when $n$ tends to infinity. The equality $ r + r^2 + r^3 + ... = r/(1-r)$ is also useful. When we finally set $f(1) = 1$, we may write:

$f(x) = x^{\frac{1/a}{1-1/a}} f(x^0) = x^{\frac1{a-1}}$.

Now, the author and I wonder if a function can be found that satisifies the recurrence relationship $f(\log(x)) = x \cdot f(x)$ . For me, the main motivator for asking this question is plain curiosity. As always, pointers to relevant literature are very much appreciated.



NB: log(x) is the natural logarithm of x.

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Have you tried using $\log(x)=\lim_{h \to 0} \frac{x^h-1}{h}$ ? – Raskolnikov Nov 19 '10 at 15:49
@ Raskolnikov: being unaware of that equality, I didn't. Thanks for the suggestion. – Max Muller Nov 19 '10 at 15:54
Obviously the same technique won't work. $x = 1$ is a fixed point of the recurrence: $1^a = 1$. So you can run a iteration that attracts $x$ to 1. $log x$ however, has not the same property (i.e. we have $x^a - x = 0$ has a solution $x = 1$, $\log x - x = 0$ does not have solution on the reals). So iterations will not stabilize. – Willie Wong Nov 19 '10 at 15:55
$f(x)=0$ for all $x$ – anonymous Nov 19 '10 at 15:57
up vote 7 down vote accepted

Claim given any function $g:(0,1]\to \mathbb{R}$, there exists a unique extension $f:\mathbb{R}\to\mathbb{R}$ satisfying your relation.

Proof. For $x > 1$, there exists a unique $n\in\mathbb{N}$ such that taken the $n$-fold logarithm of $x$ gives you a number in $(0,1]$. And for $x \leq 0$, $e^x \in (0,1]$. So by iteration the function is well-defined.

Edit Okay, let me do this explicitly. Fix your favourite function $g(x)$ on $(0,1]$. Mine happens to be the Cantor function. It doesn't matter at all for the construction what this function is.

Let your $f(x)$ be defined piecewise. For $x\in (0,1]$ define $f(x) = g(x)$. For $x\leq 0$, define $f(x) = e^{x} g(e^{x})$. Notice that $e^x$ for $x\leq 0$ is a number in $(0,1]$.

For $1 < x \leq e$, define $f(x) = \frac1x g(\log x)$. For $e < x \leq e^e$, let $f(x) = \frac1x f(\log x) = \frac1{x\log x} g(\log \log x)$. For $e^e < x \leq e^{e^e}$, let $f(x) = \frac1x f(\log x) = \frac1{x \cdot \log x \cdot \log\log x} g(\log \log \log x)$. And so on.

If your favourite function is $g(x) = 0$, then when you run this procedure you get Chandru's example where $f(x) = 0$ everywhere. If your favourite function is $g(x) = 1$, you have that $f(x) = e^{x}$ for $x \leq 0$, $f(x) = 1$ for $0 < x \leq 1$, $f(x) = \frac1x$ for $1 < x \leq e$, $f(x) = \frac{1}{x\log x}$ for $e < x \leq e^e$ and so on and so forth.

For whatever function you choose to start with as $g(x)$ defined on $(0,1]$, you get one corresponding function $f(x)$ that solves your recurrence relation. Since there are uncountably many functions on $(0,1]$, there are also uncountably many possible solutions to your recurrence relation.

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@ Willie Wong: Thanks. ok, but this proofs a well-defined function exists. The question that remains is: what does the function look like? – Max Muller Nov 19 '10 at 16:13
The function can look like pretty much anything you want. You have complete freedom to prescribe $g$. – Willie Wong Nov 19 '10 at 16:24
Now, if you impose a little bit more requirements: if you want $f$ to be continuous, it is simple to check that firstly, you need $\lim_{x\to 0}g(x) = g(1)$. This is the only requirement, and will imply that $f\to 0$ as $x\to \pm\infty$. – Willie Wong Nov 19 '10 at 16:28
@ Willie Wong: I feel a bit ashamed, but I have to say I don't understand your method. Could you please give an explicit example of a function f(x) that satisfies the relationship and how you arrived at it? – Max Muller Nov 19 '10 at 16:42
For every polynomial $h(x)$ of degree at least $2$ such that $h(0)=h(1)$, the polynomial $g(x)=h(x)+h'(0)-h'(1)$ has the property that the function $f$ defined in the answer is continuous and differentiable everywhere. For example, take $g(x) = x^2 - x - 2$. – Greg Martin Feb 3 '14 at 8:54

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