Is there a solution to the functional equation $f(x)=2\log(x)^2f\left(x^\frac{3}{8}\right)^2f\left(x^\frac{1}{4}\right)^2$?

Question

I was going through my old notebooks and I found a sheet of paper with this problem on it. I thought it would be a shame to let such an unreasonably difficult question go to waste, so I decided I would share it. The problem simply states:

Solve for $f$: $$f(x)=2\log(x)^2f\left(x^\frac{3}{8}\right)^2f\left(x^\frac{1}{4}\right)^2$$

No other information or context is given, but I'm assuming that $f$ is a complex valued function of a single real or complex variable (since evaluating the function for negative $x$ would require $f(x)$ to be complex), and that $\log$ is the natural logarithm (since no one would use it for log base-10 if they were talking about complex functions).

For curiosity's sake, I present it as a challenge to either solve for $f$ or prove that a solution does not exist there is one and only one solution (at least one solution exists, courtesy of Chrystomath). My own attempts at solving have been... unsuccessful.

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Edit:

In their answer, Chrystomath provides a solution:

$$f(x)=a\log(x)^{-\frac{2}{3}}\quad:\quad a\in\left\{z\in\mathbb{C}\Bigg| z^9=\frac{6^4}{8^9}\right\}$$

Which is quite possibly the most multivalued multivalued function I've ever seen.

I don't know whether or not the solution is unique, and I would still be interested in any other solutions.

Answer 1

One possible solution. Try $f(x)=a(\ln x)^k$, then $$a(\ln x)^k=2(\ln x)^2a^2(\frac{3}{8})^{2k}(\ln x)^{2k}a^2(\frac{1}{4})^{2k}(\ln x)^{2k}=2a^4(\frac{3}{32})^{2k}(\ln x)^{2(1+2k)}$$ from which $a$ and $k=-2/3$ can be found.

Answer 2

Here, we will only seek for real-valued solutions on an appropriate domain. We transform the equation by substituting

$$ f(x) = a \left| \log x \right|^k \exp( g(\log x) ),$$

where $k$ and $a$ are to be chosen later. Writing $u = \log x$ and plugging this to the functional equation,

$$ a \left| u \right|^k \exp( g(u) ) = 2 a^4 \bigl(\tfrac{3}{32}\bigr)^{2k} \left| u \right|^{4k+2} \exp\bigl( 2g\bigl(\tfrac{3}{8}u\bigr) + 2g\bigl(\tfrac{1}{4}u\bigr) \bigr). $$

As in @Chrystomath's solution, we can choose $k$ and $a$ as

$$ k = -\frac{2}{3} \qquad \text{and} \qquad a = 3^{4/9}2^{-23/9}. $$

Then the above equation reduces to

$$ g(u) = 2g(\tfrac{3}{8}u) + 2g(\tfrac{1}{4}u), \tag{*} $$

Now we analyze this equation.

1. @Chrystomath's solution corresponds to the trivial solution $g(u) \equiv 0$.

2. Setting $g(u) = c \left| u\right|^{\alpha}$ and plugging to $\text{(*)}$, the equation boils down to $$2\left( (3/8)^{\alpha} + (1/4)^{\alpha} \right) = 1$$ This equation has a unique real root $\alpha \approx 1.195882190\cdots$. So, with this $\alpha$ and for any $c$, $$ f(x) = a \left| \log x \right|^{-2/3} e^{c \left| \log x \right|^{\alpha}}$$ also solves the functional equation.

3. $\text{(*)}$ can relate the values of $g(u)$ and $g(u')$ only when $u$ and $u'$ satisfy $u' = \left(\frac{3}{8}\right)^m\left(\frac{1}{4}\right)^n u$ for some $m, n \in \mathbb{Z}$. So by using the Axiom of Choice, we can find infinitely many solutions of $\text{(*)}$.