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

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.

• Can try to get $f \circ g$ injective ?
– EDX
May 5 '20 at 16:54
• but $f(g(x))$ would be a constant. Wait, that kinda proves $g$ is not injective. :( May 5 '20 at 17:02
• 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. May 5 '20 at 17:19
• I would suspect that a better direction is to show that ${f(x) \over x}$ is a constant? May 5 '20 at 17:20
• 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)
– EDX
May 5 '20 at 17:20

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.

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

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: $$0 which can be satisfied iff $$\lambda(x)=\lambda(y)$$. Thus, $$f(x) =\lambda x$$ for some $$\lambda>0$$.