7
$\begingroup$

Find all strictly monotone functions $f:(0,+\infty) \to (0,+\infty)$ such that $$f\left(\frac{x^2}{f(x)}\right)=x.$$

My try: it is clear that $f$ is surjective. And because it is monotone it must also be injective. Therefore we can take $f^{-1}$ from both sides: $x^2 = f(x) \cdot f^{-1}(x)$. We can take $x = f(y)$ (because of surjectivity) and get that: $\frac{f(y)}{y} = \frac{f(f(y))}{f(y)}$. So, if we define $g(x) = \frac{f(x)}{x}$ we have that $g(y) = g\big(f(y)\big)$ and I was hoping to prove that $g$ is injective so we would have $f(x) = x$ only. But I couldn't figure that last step. There may be a better way to deal with this problem.

EDIT: There is another solution on AOPS, problem 312.

$\endgroup$
8
  • $\begingroup$ Can try to get $f \circ g$ injective ? $\endgroup$
    – EDX
    May 5 '20 at 16:54
  • 2
    $\begingroup$ but $f(g(x))$ would be a constant. Wait, that kinda proves $g$ is not injective. :( $\endgroup$ May 5 '20 at 17:02
  • 1
    $\begingroup$ Note that for any $\lambda>0$ the function $f(x) = \lambda x$ satisfies the condition, so you will not be able to conclude that $f$ is the identity. $\endgroup$
    – copper.hat
    May 5 '20 at 17:19
  • 1
    $\begingroup$ I would suspect that a better direction is to show that ${f(x) \over x}$ is a constant? $\endgroup$
    – copper.hat
    May 5 '20 at 17:20
  • 1
    $\begingroup$ Hint: If $g$ is not constant. 1. $f$ is strictly decreasing so $g$ is injective absurde 2. $f$ strictly increasing (try to compare to $x$, seeking idea on that) $\endgroup$
    – EDX
    May 5 '20 at 17:20
2
$\begingroup$

Consider $h : x \mapsto \ln\big(f(e^x)\big) $. You need to prove that $h(x)+h^{-1}(x)=2x$ (I leave it to you because it is simple). we have that $h$ is increasing (also easy to prove by contradiction or another way).

Now consider $ n \in \mathbb N$ and define $r_n:= h^n(x)$ and $s_n:=h^{-n}(x)$.

We have: $$r_{n+1}+r_{n-1}=h(r_n)+h^{-1}(r_n)=2r_n \text,$$ and similarly $$s_{n-1} +s_{n+1}=2s_n\text.$$

Therefore: $$ r_n= \lambda(x) + \mu(x)n $$ (and $ s_n= \alpha(x) + \beta(x)n $).

Now, let's prove that $h$ is continuous: let $x , y \in \mathbb R $ such that $x>y$. $h(x) - h(y) < h(x) - h(y) + h^{-1}(x) - h^{-1}(y)$ , because $h^{-1}$ is also increasing. Therefore $h(x) - h(y) < 2(x-y)$ or $|h(x) - h(y) |< 2|x-y|$. Thus $h$ is continuous.

I wasn't able to proceed from here, but given the continuity you can use the linked post's answer by Martin R.

$\endgroup$
11
  • $\begingroup$ I feel like it misses solutions like $f(x) = 2x$, but it got some good stuff. I think if $h(x) < x$ we have $r_{n+1} < r_n$ $\endgroup$ May 5 '20 at 18:20
  • $\begingroup$ It must have, because $f(x) = \lambda x$ for $\lambda >0$ are also solutions. $\endgroup$
    – copper.hat
    May 5 '20 at 18:23
  • $\begingroup$ yeah if $h(x) < x$ $(r_n)_n$ is decreasing. $\endgroup$ May 5 '20 at 18:25
  • 1
    $\begingroup$ @hellofriends: Generally a good idea to wait a day or two before accepting. $\endgroup$
    – copper.hat
    May 5 '20 at 18:27
  • 1
    $\begingroup$ @copper.hat what do you think: math.stackexchange.com/questions/1336481/… $\endgroup$ May 5 '20 at 18:35
1
$\begingroup$

Starting from $\frac{f(f(y))}{f(y)} = \frac{f(y)}{y}$:

For each fixed $y>0$, let $y_n=f^n(y)$, $n\in {\Bbb Z}$ be its orbit. The above condition on $f$ yields: $$ \forall n \in {\Bbb Z}: \frac{y_{n+1}}{y_n}=\frac{y_n}{y_{n-1}} =: \lambda(y)$$ for some $\lambda(y)>0$ independent of $n$. Thus, $f^n(y) = y \lambda(y)^n$ for all $n\in {\Bbb Z}$. Monotonicity of $f$, whence of $f^n$, implies: $$ 0<x<y \Rightarrow \forall n \in {\Bbb Z} : 0 < x \lambda(x)^n < y \lambda(y)^n$$ which can be satisfied iff $\lambda(x)=\lambda(y)$. Thus, $f(x) =\lambda x$ for some $\lambda>0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.