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Find all monotonically increasing functions $f$ on $\left[1,+\infty\right)$ such that

$$x\left( f \left( x^{2} \right) + 1 \right) = f \left( x \right) \left( x^{2}+1 \right) $$

Does there only exist the unique solution $f(x)=x$?

At first time, I think that $f$ is monotonically increasing is necessary and meaningful here.

Thanks for the comments, there are some strange solutions beyond my thought.

And now it seems the property of monotonically increasing is not important, maybe it is because there exists a closed and beautiful form of the solutions.

Blanco
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    You're not new here, you must be familiar with the policy and guidelines... what have you tried? – Clement C. Aug 20 '17 at 16:55
  • No, I mean "what have you tried to solve your question?" – Clement C. Aug 20 '17 at 17:00
  • Thanks, I will try@ClementC. – Blanco Aug 20 '17 at 17:06
  • Not sure if this will help: every monotonically increasing function is almost everywhere differentiable. – Batominovski Aug 20 '17 at 17:13
  • If this helps the only linear function that satisfies the equality is $f(x)=x$ – Marios Gretsas Aug 20 '17 at 17:17
  • @ClementC. Appears taken care of. – Simply Beautiful Art Aug 20 '17 at 17:21
  • Certainly $f(x)$ must be odd. – Simply Beautiful Art Aug 20 '17 at 17:22
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    @SimplyBeautifulArt That makes no sense. The function $f$ is defined only on $[1,+\infty)$. – Batominovski Aug 20 '17 at 17:24
  • @Batominovski Whoops! Completely missed that! If it were $f:\Bbb R\mapsto\Bbb R$, then it would have to be odd. – Simply Beautiful Art Aug 20 '17 at 17:46
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    Remark: The only Laurent polynomials $F(X)\in\mathbb{R}\left[X,\frac{1}{X}\right]$ such that $$F(X),\left(X^2+1\right)=X,\big(F\left(X^2\right)+1\big)$$ are of the form $$F_t(X):=t,X+\frac{1-t}{X},,$$ where $t\in\mathbb{R}$. In particular, if $t\geq \frac{1}{2}$, then the function $f:[+1,\infty)\to\mathbb{R}$ given by $f(x)=F_t(x)$ for all $x\geq 1$ is monotonically increasing. – Batominovski Aug 20 '17 at 18:03
  • @Batominovski Thanks, say, this problem may be more complicated than I thought... – Blanco Aug 20 '17 at 18:19
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    There are strange solutions like $$f(x) = x + \left(x-\frac{1}{x}\right)\left(A + B\sin\left(\frac{2\pi}{\log 2}\log\log x\right)\right) $$ to the functional equation. If $A > -\frac12$ and $\displaystyle;|B| < \frac{1+2A}{2\sqrt{1+\left(\frac{2\pi}{\log 2}\right)^2}};$, $f(x)$ will be increasing too. – achille hui Aug 20 '17 at 18:21
  • @achillehui, I am kind of interested in how you came up with this example! – Arash Aug 20 '17 at 18:47
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    @achillehui what on Earth or Hell did you have to summon to get access to that? – Clement C. Aug 20 '17 at 18:48
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    Perhaps the idea is that, ignoring the monotonicity, any solution is of the form $$f(x) = x + \left(x - \frac{1}{x}\right)\varphi(x) $$ for some $\varphi : [1, \infty) \to \mathbb{R}$ such that $\varphi(x^2) = \varphi(x)$, or equivalently, $x \mapsto \varphi(\exp(2^x))$ has period $1$. Now all we have to do is some engineering on $\varphi$. – Sangchul Lee Aug 20 '17 at 19:01
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    @SangchulLee Yup, you beat me in writing that up as a comment ;-p – achille hui Aug 20 '17 at 19:06

1 Answers1

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Clearly, $f(1)=1$. Now, the functional equation is equivalent to $$f(x^2)=\left(x+\frac1x\right)f(x)-1$$ This the same as $$f(x^2)-x^2=\left(x+\frac1x\right)(f(x)-x)$$ Or, if we define $g(x)=\dfrac{f(x)-x}{x-1/x}$ for $x>1$, we get $$g(x^2)=g(x)\tag1$$ Now the functions $g$ that satisfy $(1)$ are functions of form $g(x)=h(\ln(\ln x))$ where $ h:\mathbb{R}\to\mathbb{R}$ is an arbitrary periodic function having $\ln(2)$ as period.

So, the general solution of the proposed functional equation is $$f(x)=x+\left(x-\frac1x\right)h(\ln(\ln x))$$ Where $ h:\mathbb{R}\to\mathbb{R}$ is an arbitrary periodic function having $\ln(2)$ as period. Now, the monitonicity condition adds some conditions on our choices for $h$.

Felix Klein
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