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The problem I am trying to solve now is to find all continous functions satistying

$f(x) = f(x^{2} + \frac{x}{3} + \frac{1}{9})$ for all $x \in \mathbb{R}$

It is the first time for me to face this sort of problem, so it is tricky for me to come up with the idea.

The idea I am taking into consideration is to find the sequence $\{ x_{n} \}$ such that

$x_{n+1} = x_{n}^{2} + \frac{x_{n}}{3} + \frac{1}{9}$

But from here I don't know how to proceed. Is it a right approach? Any other ideas? Please help me :)

Thanks in advance

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1 Answer 1

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The first thing to do is to draw a sketch of the graph of the function $g:x\mapsto x^2+\frac13x+\frac19$ and of the line $y=x$. Let $z=\frac13$ and $t=-\frac23$. Three regions, separated by two special points, appear:

  • if $x\lt t$, then $g(x)\gt z$;
  • if $x=t$, then $g(x)=z$;
  • if $t\lt x\lt z$, then $t\lt g(x)\lt z$;
  • if $x=z$, then $g(x)=z$;
  • if $x\gt z$, then $g(x)\gt z$.

As a consequence, if $t\leqslant x\leqslant z$, then $g(x)=z$ or $g^n(x)\to z$ when $n\to\infty$, where $g^n(x)$ is the $n$th iterate of $x$ by the function $g$. Hence, by continuity, $f(x)=f(z)$.

If $x\lt t$, then $f(x)=f(g(x))$ with $g(x)\gt z$, hence it remains to study $f$ on $(z,\infty)$.

If $x\gt z$, then $f(x)=f(h(x))$ where $h(x)$ is the unique antecedent of $x$ by $g$ in $(z,\infty)$, that is, $h(x)=-\frac16+\sqrt{x-\frac1{12}}$ for every $x\gt z$. Iterating, one sees that $f(x)=f(h^n(x))$ for every nonnegative $n$. Since $h^n(x)\to z$ when $n\to\infty$, by continuity, $f(x)=f(z)$.

Finally, the only solutions $f$ are the constant functions.

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