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This question already has an answer here:

Find all functions $f:(0,\infty)\rightarrow(0,\infty)$ subject to the conditions:

$f(f(f(x)))+2x=f(3x)$ for all $x>0$ and $\displaystyle\lim_{x\to\infty}(f(x)-x)=0$

I tried as follows: Suppose $x_0=x,x_1=f(x_0)=f(x),...,x_n=f(x_{n-1})$. But I am facing problem due to $f(3x)$ term.

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marked as duplicate by Sil, Adrian Keister, Community May 30 at 18:41

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Note that $f(x)=x$ satisfies all three of your conditions, by inspection.

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  • $\begingroup$ Yes. But How to prove it mathematically. $\endgroup$ – J.Doe May 30 at 18:18
  • $\begingroup$ There's almost nothing to prove. $\displaystyle\lim_{x\to\infty}(f(x)-x)=\lim_{x\to\infty}(x-x)=\lim_{x\to\infty}0=0.$ Then $f(f(f(x)))+2x=f(f(x))+2x=f(x)+2x=x+2x=3x=f(3x).$ Finally, note that if $x>0,$ it follows that $f(x)=x>0,$ so that $f:(0,\infty)\to(0,\infty).$ Done. $\endgroup$ – Adrian Keister May 30 at 18:20
  • $\begingroup$ Oh, but you mean show that there aren't any other solutions. That's a lot harder. Not sure I know how to do that. Whatever function you had would certainly need to approach $f(x)=x$ asymptotically (the limit condition means you have a slant asymptote). $\endgroup$ – Adrian Keister May 30 at 18:21

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