A functional equation: $4f(x)^3 +f(3x)=3f(x)$

Question

Find all functions $f:\mathbb{R}\to\mathbb{R}$ such that

$$4f(x)^{3}+f(3x)=3f(x)$$

I know of 2 functions that satisfy the equation but I do not know how to prove that they are the only ones.

Thanks

Andrew

Answer 1

One solution is $f(x)=\sin(x),$ from the identity $$\sin(3x)=3\sin(x)-4\sin^3(x)\tag{1}$$ which can be checked using the addition formula for sine.

Analytic solutions are (I think) only constants and those of the form $\pm \sin(kx).$ Sketch: Assume $f$ is a nonconstant solution. That $f(0)=0$ follows from the relation, and (this needs work but I think it could be shown) $f$ must be zero for some positive $x.$ Since generally if $f(x)$ is a solution so is $\pm f(kx),$ we may assume $f$ has its first positive zero at $\pi,$ and $f>0$ on $(0,\pi).$ Then the idea is to use the formula $(1)$ to produce the sequence $\pi,\pi/3, \pi/9,...,\pi/3^r,...$ on which $f$ has the same value as $\sin(x).$ We then would have two analytic functions agreeing on a convergent sequence in their domains and could conclude $f(x)=\sin(x)$ for all $x$. The use of the polynomial $g(x)=3x-4x^3$ for this, in getting values of $f(x/3)$ uniquely from values of $f(x),$ is not difficult since $g$ is strictly increasing on $[0,1/2]$ with range $[0,1]$ (one doesn't need $g$ beyond $[0,1/2]$ Of course the first value $f(\pi/3)=\sqrt{3}/2$ comes directly from $f(\pi)=0.$ After that $\pi/9 \approx .349$ is already in the interval $[0,1/2]$ on which $g$ is increasing.

Answer 2

If you don't require that $f$ be continous, this question is rather easy. Let $O$ denotes the multipicative subgroup of $\mathbb R$ generated by $3$, that is, $O:=\{3^n|n\in\mathbb Z\}$. Consider the action of $O$ on $\mathbb R$, and then solve it orbit by orbit...

Answer 3

I'm thinking of the Fast Fourier transform. Frequency shift (3x) is easily expressible in FFT domain (1->1 mapping). Multiplication is convolution. Then we get a set of equations. Those convolutions might be quite nasty though.

Answer 4

I can show that every continuous function $f:[1,3]\mapsto(-1,1)$ which satisfies $4f(1)^3 +f(3)=3f(1)$ and $f(1)\in[-\frac 12,\frac 12]$ can be extended to a solution:

First, define the value of $f$ over $(3^n,3^{n+1}](n=1,2,3,\cdots)$ recursively by the formula $f\left(x\right)=3f\left(\frac x3\right)-4f\left(\frac x3\right)^3$;

Second, note that $g:[-\frac 12,\frac 12]\to[-1,1]\quad x\mapsto 3x-4x^3$ is a homeomorphism, denote its inverse by $h$. Define the value of $f$ over $[3^{-(n+1)},3^{-n})(n=0,1,2,3,\cdots)$ recursively by the formula $f\left(x\right)=g\left(f\left(3x\right)\right)$;

Third, note that by now we've extend $f$ to a function over $(0,+\infty)$ that satisfies $4f(x)^3 +f(3x)=3f(x)$, that is to say, we have $\frac{f(\frac x3)}{f(x)}=\frac 1{3-4f(\frac x3)^2}$. Then by the construction of $h$, we have $\forall x\in(0,1),\ 0<\frac{f(\frac x3)}{f(x)}<1-\epsilon$ for a certain $\epsilon>0$, which implies $f(x)\to0\ (x\to0)$. So we can define $f(0)=0$ and $f(x)=-f(-x)$ for $x\in(-\infty,0)$.