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I am trying to solve the following functional equation: $f(x)$ is a continuous function, satisfying (1) $f(f(x))=x$
(2) $ 3f(-3x)-f(x)= 3x^2 $ for $x>0$.

From (1) & (2), I found that $f(x)$ should be decreasing, and $f(x) = f^{-1} (x)$.

But, I cannot figure out how to use (2).

How can I use (2) to solve this?

Thanks!

five manifold
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    "From (1), I found that $f(x)$ should be decreasing" What about $f(x)=x$? It satisfies (1) and is increasing. – Arthur Jul 17 '18 at 04:36
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    Yes, only from (1), $f(x)$ can be the solution. But from (2), if $f(x)=x$ for $x>0$, $f(x)$ is not one-to-one. – five manifold Jul 17 '18 at 04:39
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    Sure. But it means "From (1), I found that $f(x)$ should be decreasing" is wrong. – Arthur Jul 17 '18 at 04:40
  • For (2) you have condition $x>0$. Is there any condition for (1)? – Saša Jul 17 '18 at 10:42
  • (1) is Babbage's equation with prodigious orbits of solutions. (2) has the particular solution $(3/26)x^2 +a/|x|$, but it may not be easy to fit onto one of those orbits.... – Cosmas Zachos Jul 17 '18 at 15:57

1 Answers1

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There are no solutions.

(The question already notes $f$ is decreasing; I include a proof for completeness).

By (1), $f$ is bijective.

Since $f$ is assumed continuous, this implies $f$ is strictly monotone by this result. That is, $f$ is either strictly increasing or strictly decreasing.

Taking the limit as $x\to0^+$ in (2) gives $f(0)=0$. Setting $x=1$ gives $$ 3f(-3)-f(1)=3.\tag{3} $$ If $f$ is increasing, then $f(-3)<0$ and $f(1)>0$, contradicting (3). Thus $f$ is decreasing. In particular $f(-3)>0$, so (3) gives $$ f(1)=3f(-3)-3>-3. $$ Thus $1=f(f(1))<f(-3)$. Now (3) gives $$ f(1)=3f(-3)-3>0=f(0), $$ a contradiction.

stewbasic
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