5

Find all strictly increasing functions that satisfy $$f(f(x))=3x\text.$$

Obviously $\sqrt{3} x$ is an answer, but I am having a hard time proving that is the only solution. Any ideas or solutions? Thanks for the help!

Mohsen Shahriari
  • 6,718
Baker5680
  • 451
  • 1
    You have a hard time proving that $f(x)=\sqrt{3}x$ is the only solution; but are you sure it is the only solution? – Servaes Sep 03 '19 at 20:49
  • 1
    $f^{-1}(3x) = 3f^{-1}(x)$ – amsmath Sep 03 '19 at 20:53
  • 2
    Connected : https://math.stackexchange.com/q/2618593 – Jean Marie Sep 03 '19 at 20:53
  • 1
    The statement implies at least that $f$ is bijective and strictly increasing so it is continuous, hence a homeomorphism from $\mathbb{R}$ to itself.

    Besides, note that $f(3x)=3f(x)$, from which we deduce that $f(0)=0$.

     – Aphelli Sep 03 '19 at 21:05
  • 1
    @Mindlack Why bijective? I don't see that in the statement. – amsmath Sep 03 '19 at 21:14
  • 1
    @amsmath: $x=f(y)$, where $y=f(x/3)$, so $f$ is surjective. If $f(x)=f(y)=z$, $x=f(z)/3=y$, so $f$ injective. – Aphelli Sep 03 '19 at 21:16
  • 1
    @Mindlack Injective is clear from strictly increasing anyways, but the surjectivity was not clear to me. But it's obvious. Thanks. – amsmath Sep 03 '19 at 21:18
  • 3
    @Servaes: it is strictly increasing + surjective that implies continuity. Basically, if $y_{\pm}=\lim_{x^{\pm}}f$ and $y_- < y^+$ then any point strictly between $y_-$ and $y_+$ can only be reached as $f(x)$. This goes against $f$ being surjective. – Aphelli Sep 03 '19 at 21:21
  • 1
    Condsider $g(x)=\frac{f(x)}x$. Then $g(3x) = g(x)$.... – amsmath Sep 03 '19 at 21:26
  • 1
    @amsmath How did you get that? – Kavi Rama Murthy Sep 03 '19 at 23:37
  • 1
    @KaviRamaMurthy $g(3x) = \frac{f(3x)}{3x} = \frac{3f(x)}{3x} = g(x)$. – amsmath Sep 04 '19 at 00:42
  • 1
    It seems to me that $\sqrt 3x$ is not the only increasing function satisfying the functional equation. It could be that $f(x)$ buzzes around the line $\sqrt 3x$. At least, if, e.g., $f(x) > \sqrt 3x$, then $f(f(x)) < \sqrt 3f(x)$, that is, if $y = f(x)$ is above the line, then the next value $f(y)$ is below the line. – amsmath Sep 04 '19 at 00:48
  • 1
    In any case, it is clear that for $x>0$ we have $x<f(x)<3x$ for if $f(x)\le x$, then $3x = f(f(x))\le f(x)\le x$ and if $f(x)\ge 3x$, then $3x = f(f(x))\ge f(3x) = 3f(x)\ge 9x$, which are both contradictions. – amsmath Sep 04 '19 at 01:04

1 Answers1

3

$x\mapsto\sqrt 3x$ is not the only solution. In fact, there are infinitely many. Obviously $f|_{\mathbb R^-}$ and $f|_{\mathbb R^+}$ are independent of each other. So, I'll only consider $f$ on $[0,\infty)$. To construct an alternative solution, let $f : [1,\sqrt 3]\to[\sqrt 3,3]$ be any strictly increasing continuous function with $f(1) = \sqrt 3$ and $f(\sqrt 3) = 3$. Now, for $x\in [\sqrt 3,3]$ set $f(x) := 3f^{-1}(x)$. Then this part of $f$ maps $[\sqrt 3,3]$ to $[3,3\sqrt 3]$. This defines an increasing continuous function from $[1,3]$ to $[\sqrt 3,3\sqrt 3]$.

Now for $x\in [3^k,3^{k+1}]$, $k\in\mathbb Z$, set $f(x) := 3^kf(3^{-k}x)$ and finally $f(0) := 0$. Thus, we have defined a continuous and increasing function on all of $[0,\infty)$. Let us show that it satisfies the functional equation.

To begin with, for $x\in [1,\sqrt 3]$ we have $f(f(x)) = 3f^{-1}(f(x)) = 3x$. Let $x\in [3^k,3^{k}\sqrt 3)$. Then, since $f(3^{-k}x)\in [\sqrt 3,3)$, we have $f(x) = 3^kf(3^{-k}x)\in [3^k\sqrt 3,3^{k+1})$ and so $$ f(f(x)) = 3^kf(3^{-k}f(x)) = 3^kf(f(3^{-k}x)) = 3x. $$ Finally, for $x\in [3^k\sqrt 3,3^{k+1})$ (including $k=0$), we have $3^{-k}x\in [\sqrt 3,3)$ and thus $f(3^{-k}x)\in [3,3\sqrt 3)$, i.e., $f(x) = 3^kf(3^{-k}x)\in [3^{k+1},3^{k+1}\sqrt 3)$. Thus, $$ f(f(x)) = 3^{k+1}f(3^{-k-1}f(x)) = 3^{k+1}f(3^{-k-1}3^kf(3^{-k}x)) = 3^{k+1}f(3^{-1}3f^{-1}(3^{-k}x)) = 3x. $$

Remark. Instead of $[1,\sqrt 3]$ you can start with any interval $[a,\sqrt 3a]$ and map it increasingly onto $[\sqrt 3a,3a]$. In other words, you can scale everything by $a>0$. Then you have found all possible solutions. And, of course, to make it more general, you can replace $3$ by any positive number.

amsmath
  • 10,633