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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

mvw
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Andrew
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  • What are the functions you've thought of? Since there are three $y$ such that $4y^3+y = 3y$, there are at least three solutions which are constant functions $f(x) = y$. –  Mar 14 '15 at 19:57
  • Obviously there are the constant functions which can be easily found with the quadratic formula, but the other function I would rather keep to myself. As a challenge from me to the readers. – Andrew Mar 14 '15 at 20:00
  • Sorry i was unclear, there are the 3 constant functions and then there is at least one other non-constant function that satisfies the equation. Sorry for the confusion. – Andrew Mar 14 '15 at 20:09
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    It would be helpful to let us know what are these functions. –  Mar 14 '15 at 20:09
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    Ok so the constant function is 0 (the other 2 values are imaginary) but the point is that you should imagine that this is an Olympiad problem and try to solve it, I am doing this at present. For this reason I believe it would be unethical to tell you what the other function is. (By the way,I thought this up, I was not given this problem.) – Andrew Mar 14 '15 at 20:14
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    You can construct a non-constant solution by setting $f(x):=a$ for $x<0$, $f(0):=b$ and $f(x):=c$ for $x>0$, where $a,b,c$ are (not necessarily distinct) solutions of the equation $4X^3-2X=0$. – sranthrop Mar 14 '15 at 20:14
  • @Andrew. You intentions are dubious. You can expect help proving that the only functions which satisfy the equation if you won't state what those solutions are – Tim Raczkowski Mar 14 '15 at 20:15
  • @Andrew: There are three constant solutions, since the equation $4X^3+X=3X$ (which is equivalent to $2X(2X^2-1)=0$) has three distinct real solutions. – sranthrop Mar 14 '15 at 20:17
  • Ok fine you have brought me to the point where I am forced to show my function, because my honesty and integrity are doubted. I honestly thought that you would enjoy the challenge, but alas my hopes were defeated. So here we go, f(x)=sin(x) satisfies the equation. You leave me with nothing, however I thank you for your honesty. – Andrew Mar 14 '15 at 20:20
  • No. The three solutions are $f(x)=0$ and $f(x)=\pm\frac{1}{\sqrt{2}}$. Are you sure that there aren't any constraints regarding any regularity? – sranthrop Mar 14 '15 at 20:25
  • Wow, my maths is failing today thanks. Whoops sorry you are correct, for some reason I forget to divide by the y term on the one side. there are 3 constant solutions. $f(x)=0, f(x)=\frac{-1}{\sqrt{2}} or f(x)=\frac{-1}{\sqrt{2}}$ – Andrew Mar 14 '15 at 20:26
  • Certainly there are more functions. For instance, $0$ on the rationals and $1/\sqrt 2$ on the irrationals. – Yoni Rozenshein Mar 14 '15 at 20:28
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    There are infinitely many solutions. Note that if $f$ is a solution, then $x\mapsto f(ax)$ is a solution for each $a\in\mathbb R$. – sranthrop Mar 14 '15 at 20:45
  • Andrew: If you're interested, I just added a section about requiring analytic solution functions. – coffeemath Mar 15 '15 at 03:29
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    An easy to construct non-trivial non-analytic but continuous solution is $$\sin\big(x\cos(2\pi\log_3x)\big) \quad\text{ or more general }\quad \sin\big(x \varphi(\log_3x)\big)$$ where $\varphi(x)$ is any continuous periodic function of period $1$. – achille hui Mar 15 '15 at 04:02

4 Answers4

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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.

coffeemath
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  • Thank you for seeing this, the next part is to show that this is the only non-trivial function. (i.e. not constant or made up of constant functions) – Andrew Mar 14 '15 at 20:33
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    Beat me to it. The fact that $\sin$ solves the equation makes me think that proving that there is a finite number of solutions and finding all of them will be kind of hard. – chubakueno Mar 14 '15 at 20:33
  • This is a special case of what I proposed below. The convolution part is so easy for sines, since they are diracs in fourier space. – mathreadler Mar 14 '15 at 20:47
  • @chubakueno, there are infinitely many solutions in that $f(x)=\sin\omega x$ satisfies the identity for all $\omega$. – Barry Cipra Mar 14 '15 at 21:06
  • @Andrew It seems you ask for $f$ being continous as well. – mvw Mar 14 '15 at 21:08
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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...

Censi LI
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  • Isn't this just saying "find all solutions to the resulting equations?" I mean, you could say the same thing for any functional equation but it gives nothing toward actually solving it. – Alex R. Mar 17 '15 at 19:40
  • @AlexR. What you said is right for general case. But for this special case, if we reduce the problem to each obit, we need only solve a cubic equation, I don't see any problem towards solving it... By the way, if continuity is not imposed, I do think any functional equation is not very hard. – Censi LI Mar 17 '15 at 20:04
  • For my personal curiosity then, how would this work for: http://math.stackexchange.com/questions/957705/solving-the-functional-equation-fxy-ffxfy

    which has only one solution, continuous or otherwise. I guess I'm just confused how it's easy as you claim without the usual clever manipulations and test values that go into solving these kinds of things.

     – Alex R. Mar 17 '15 at 20:12
  • @AlexR. I'm sorry I didn't make it clear. By a functional equation I mean an equation of the form $g(f(h(x)))=0$, where $g$ and $h$ is given and $f$ is unknown. So there is differential operator nor iterated function. Otherwise my method is not applicable. – Censi LI Mar 17 '15 at 20:20
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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 $x\in(-\infty,0)$.

Censi LI
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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.

mathreadler
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