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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.

Thanks,

Max

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

J. M. ain't a mathematician
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Max Muller
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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
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    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
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    $f(x)=0$ for all $x$ –  Nov 19 '10 at 15:57

1 Answers1

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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.

Willie Wong
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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
  • (post edit) notice that for the case starting with $g(x) = 1$, the function I wrote down above is continuous. – Willie Wong Nov 19 '10 at 17:25
  • I wonder whether there's some additional property that can be used to ensure an 'interesting' unique solution in the same way that the gamma function is the only log-convex function that satisfies its functional equation $g(x+1) = xg(x)$. – Steven Stadnicki Nov 19 '10 at 19:52
  • @Steven: there probably is. But what that additional property is will most likely depend on what you want in the end to be the interesting unique solution. The gamma function is nice because it also has other nice properties. – Willie Wong Nov 19 '10 at 19:59
  • This function can probably be expressed with tetration type functions. We then have a differentiable solution. I will add the tetration tag. – mick Sep 21 '12 at 19:57
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    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 03 '14 at 08:54