Recently I started studying functional equations. Now I'm trying to find all solutions to the following functional equation:
Unfortunately, I was able to show only few things about this equation, moreover, I wasn't even able to find $f(0)$ using substitution method. What I got (only by guessing) that there are at least two solutions: $f(x)=0$ and $f(x)=x$. But I don't know if there are any others. Any help would be very appreciated.

  • 1
    $\begingroup$ If you assume that $ f $ is differentiable, then differentiating the equation you get $ 6 f ( 2 x ) ^ 2 f ' ( 2 x ) = 4 f ' ( 4 x ) \big( f ( x ) ^ 2 + x f ( x ) \big) + f ( 4 x ) \big( 2 f ( x ) f ' ( x ) + f ( x ) + x f ' ( x ) \big) $, which by letting $ x = 0 $ gives $ f ( 0 ) = 0 $. $\endgroup$ May 26, 2017 at 16:18

2 Answers 2


Assuming that $\;f(x)\;$ has a power series expansion $\;f(x)=\sum_{n=0}^\infty c_n x^n,\;$ then the functional equation rewritten as $\;f(4x)(f(x)^2+xf(x))-f(2x)^3 = 0\;$ implies that $\; c_0^2 x +O(x)^2 = 0\;$ and thus $\;c_0=0.\;$ Assuming $\;c_0=0,\;$ the functioional equation implies that $\;0=4c_1^2(c_1-1)x^3 +O(x)^4\;$ and thus $\;c_1=0\;$ or $\;c_1=1.\;$ In both cases, $\;c_n=0\;$ for all $\;n\!>\!1,\;$ thus $\;f(x)=0\;$ or $\;f(x)=x\;$ for all $\;x$.


Probably late, and could have been a comment since this "answer" doesn't solve the equation, but:

Let $\mathcal{A} := \{A \subset \mathbb{R} \mid x \in A \,\, \Leftrightarrow \,\, 2x \in A\}$, and let $f$ be any solution of the the equation.
Define $g = g_A := f \cdot 𝟙_A$ for any $A \in \mathcal{A}$. Let's show that all $g$ of this form are again solutions of the equation, and thus that there are infinitely many (non-continuous) solutions thanks to $x \mapsto x$ being a solution.

Let $x \in \mathbb{R}$.

  • If $x \in A$, then $2x \in A$ and $4x \in A$, therefore: $$g(2x)^3 = f(2x)^3 = f(4x)\left(f(x)^2 + xf(x)\right) = g(4x)\left(g(x)^2 + xg(x)\right)$$
  • If $x \not\in A$, then $2x \not\in A$ and $4x \not\in A$, therefore: $$g(2x)^3 = 0 = g(4x)\left(g(x)^2 + xg(x)\right)$$

These are the only two possible outcomes, and so $g$ is indeed solution of the equation.

Note I went with one element of $\mathcal{A}$ and one solution $f$, but by the same type of argument you can show that any finite sum $\sum_{k=1}^{n} f_k \cdot 𝟙_{A_k}$ with $f_1,\dots,f_n$ solutions and $A_1,\dots,A_n$ pairwise disjoint elements of $\mathcal{A}$ is a solution of the equation, it's just that currently we only have one interesting solution ($x \mapsto x$).

Basically, the idea is that this equation for a fixed $x$ doesn't give any information for any $y$ that's not $2^n x$, $n \in \mathbb{Z}$ without any further hypothesis like continuity or differentiability, so as long as you have solutions along these different "spaces" you can always combine them to get more solutions.
For example, if you take a solution $f$ and any real $a$, then $x \mapsto \begin{cases} f(x) &\,\,\text{if}\,\, x \neq 0 \\ a &\,\,\text{if}\,\, x = 0\end{cases}$ is still a solution, and intuitively it's because without continuity the value at $0$ is not impacted by nor impacts the rest of the function.

I guess some hope could be to somehow show that $x \mapsto x$ is the only continuous non-zero solution? Since then the corresponding $g_A$ from above should be exactly the solutions, though that would remain to be proven too...


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