Be $f:\mathbb R^ +\mapsto\mathbb R$ a function that satisfies the following conditions:

a)$ f(f(f(x)))+2x=f(3x)$ for every $x\gt 0$;

b) $\lim_{x \to \infty} (f(x)-x)=0$.

This was proposed by Gabriel Dospinescu at a romanian competitions about a decade ago. No idea how to approach it. Thank you!

  • $\begingroup$ What is the question? To determine every possible $f$ ? $\endgroup$ – Pedro M. Mar 28 '15 at 18:29
  • $\begingroup$ yes. i forgot to mention:D thanks. $\endgroup$ – Dan Leonte Mar 28 '15 at 18:45
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    $\begingroup$ @DanLeonte Just as a matter of notation, the symbol $\mapsto$ is usually (as a standard) used for discussing where an element is sent. For example, we might say the element $x$ is sent to $x^{2}$ by writing $x \mapsto x^{2}$. But when writing a function's domain and codomain, we use a regular arrow (code: $\text{ \to }$). So, it would look like $f: \Bbb R^{+} \to \Bbb R^{+}$, with code $\text{f: \Bbb R^{+} \to \Bbb R^{+}}$. $\endgroup$ – layman Mar 28 '15 at 18:49

The function must be defined as follow $f:\mathbb{R^+}\to\mathbb{R^+}$,otherwise the functional equation can not be true because it involves $f(f(x))$

we know that if replacing $x$ by $\frac{x}{3}$ we obtain : $$f(x)=\frac{2}{3}x+f(f(f(\frac{x}{3})))\geq\frac{2}{3}x$$

now let $u_0=\frac{2}{3}$ assuming that $f(x)\geq u_nx$ one can prove that: $$f(x)\geq u_{n+1}x$$ with $u_{n+1}=\frac{u_n^3+2}{3}$ and because $u_n \rightarrow 1$ when $n$ tends to infinity we conclude that $$f(x)\geq x$$ for evry positive real $x$. Now for any real $a$ let $f(a)=a+t$ then $(3a)\geq 3a+t$ and by induction $f(3^na)\geq 3^na+t$ and using the second condition we have $t=0$

Conclusion The only function verifying the two conditions is the identity.

  • $\begingroup$ $f(x)≥2/3x$ is true only for high-enough x! $\endgroup$ – Dan Leonte Mar 28 '15 at 19:09
  • $\begingroup$ for every positive $ x>0$We have $f(x)=\frac{2}{3}x+f(f(f(\frac{x}{3})))\geq \frac{2}{3}x$ so why for hight enought x?, hmmmm you changed the definition of $f$!!!! $\endgroup$ – Elaqqad Mar 28 '15 at 19:15
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    $\begingroup$ in fact $f(x)$ must be positive for every $x$ otherwise the functional equation will not have any sense because there is $f(f(x))$ $\endgroup$ – Elaqqad Mar 28 '15 at 19:21
  • $\begingroup$ yes Elaqqad sory for editing the question so late and thankyou so much for your effort:D $\endgroup$ – Dan Leonte Mar 28 '15 at 19:26
  • $\begingroup$ @DanLeonte my answer is correct and because you can not edit the question like you did in fact the function must be defind as $f:\mathbb{R}^+\to\mathbb{R}^+$ , otherwise the term $f(f(f(x)))$ will have nosense!!!!!!!!, if you're interested in the answer you can read mine if you're interested in something else $\cdots$ $\endgroup$ – Elaqqad Mar 28 '15 at 19:29

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