0
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ @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
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ $f(x)≥2/3x$ 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
  • 1
    $\begingroup$ 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

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.