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$f$ also has to satisfy several conditions:

  • $f: [0; +\infty) \rightarrow [0; +\infty)$
  • $f(x) \ge x$ and $f(x) \ge 1$
  • $f$ is strictly increasing

These conditions may not be used (this is not a problem from a MO), and I only care about all the values of $f$ that satisfy the functional equation.

I had found a family of $f$: $f(x) = (x^a+1)^{a^{-1}}$, but that solution is bad (for my purposes) and would like to find more $f$-es.

You may introduce more restrictions for $f$ (like differentiable, continuous, etc.), since I want analytic $f$-es.

gbnam8
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    Can you explain a little bit of why $f(x)=(x^a+1)^{a^{-1}}$ isn't good for your purposes? If there are a lot of functions that satisfy this, this might help you get answers that are more useful to you. – Carl Schildkraut Sep 04 '22 at 03:36
  • I just want more degrees of freedom to tweak $f$, that's it. – gbnam8 Sep 04 '22 at 04:12
  • There are infinitely many functions that satisfy this and we don't know what are the motivations behind this. Another example is $e^x$ and we can construct many more. Can you tell us more? – PepeHands Sep 04 '22 at 16:58
  • Basically I wanted a function that combines an (repeatable) set of numbers into one singular one. $f$ in the equation above is a specialization of that function when the set is {x, 1}, and that is enough to determine the whole thing. To make the combine function behave reasonable, it should be transitive (idk if that's the right word), commutative, and linear (scaling the array then apply the function is basically the same as apply the function and then scale it). I want to tweak the behavior of this combine function. – gbnam8 Sep 06 '22 at 12:08
  • The problem with the exponential solution I had above is that it increases too fast when $x$ is small. I want to change that. And I initially thought that this functional equation would be "easy" to solve (for experienced MO people), or there is already a class of function like that. – gbnam8 Sep 06 '22 at 12:11
  • Assuming just that the $f$ are increasing and nonnegative.

    For $y > 0$, \begin{equation} f(yf(f(y) /y)) = f(y) f(f(y) /f(y)) = f(y) f(1). \end{equation} On the other hand \begin{equation} f(y f(1)) = f(yf(y/y)) = yf(f(y) /y). \end{equation} Thus \begin{equation} f(f(y f(1))) = f(y) f(1). \end{equation} So if $f(1)=1$, then $f(y) = y$.

    One program is to look at the inverse function $g(x) = (x^a-1)^{1/a}$. Then consider an $f$ (obeying the functional equation) such that $f(1) = 2^{-a}$. Show $g \circ f$ also satisfies the equation and conclude $f$ has your form.

     – David Perrella Sep 07 '22 at 13:55

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