I've come across the following functional equation:

Determine all surjective functions $f:\mathbb{R_{>0}}\to\mathbb{R_{>0}}$ which satisfy for all $x\in\mathbb{R_{>0}}$: $$ 2xf(f(x))=f(x)\left(x+f(f(x))\right) $$ My approach so far:

The equation is equivalent to: $$ x=\frac{f(x)f(f(x))}{2f(f(x))-f(x)} $$ Which implies injectivity. Thus, $f$ is bijective and therefore, there exists a bijective function $f^{-1}:\mathbb{R_{>0}}\to\mathbb{R_{>0}}$ such that $f^{-1}(f(x))=x$. This allows the, in my opinion, very beautiful reformulation:

Find all (bijective) functions such that for all $x\in\mathbb{R_{>0}}$: $$ \frac{f^{-1}(x)f(x)}{x}=\frac{f^{-1}(x)+f(x)}{2} $$ But now I don't know how to proceed further. So any help will be appreciated.

  • 1
    $\begingroup$ By inspection, $f(x)=x$ is at least one solution... $\endgroup$
    – user170231
    Sep 10 '15 at 17:44
  • $\begingroup$ Is it an open problem or you know that there exists a (nice) solution to it? $\endgroup$
    – Tintarn
    Sep 10 '15 at 17:45
  • 1
    $\begingroup$ The standard approach would be to fix some $x_0$ and then define recursively $x_{n+1}=f(x_n)$. Then, hopefully, you can find a closed form for this recursion and in some way use $x_k>0$ to conclude that $x_k=x_0$ for all $k$... $\endgroup$
    – Tintarn
    Sep 10 '15 at 17:48
  • 1
    $\begingroup$ For any $f(x)$ which solves it, $f^{-1}(x)$ must also be a solution because of symmetry. $\endgroup$ Sep 10 '15 at 17:52

I will write a sketch of a solution: As suggested in my comment above, we fix some $x_0>0$. Define $x_{n+1}=f(x_n)$ recursively.

Note that your equation implies $x_k>0$ and $2x_kx_{k+2}=x_kx_{k+1}+x_{k+1}x_{k+2}$ for all $k$. Let $y_k=\frac{1}{x_k}$.

Then this equation is equivalent to $y_{n+2}+y_n=2y_{n+1}$ i.e. $y_n=an+b$ for some $a,b$. Since $y_n>0$ for all $n$ we conclude $a \ge 0$.

Also, by your ideas $f$ must be bijective i.e. $f^{-1}$ exists and so we can define $y_{-1},y_{-2},\dotsc$ which satisfy the same equation and are also positive.

Now, similar to above we can conclude $a \le 0$ because otherwise $y_k$ would be negative for $k \to -\infty$. So $a=0$ and hence $y_n$ is constant.

In particular, we find $f(x_0)=x_1=\frac{1}{y_1}=\frac{1}{y_0}=x_0$.

Since $x_0$ was chosen arbitrarily, we find $f(x)=x$ for all $x$ which is indeed a solution.

Remark: This really should be the standard approach when you see problems that only involve $x,f(x),f(f(x)),\dotsc$ in some combination. But often for these methods to work, you need some additional constraint such as continuity or, in this case, that $f(x)>0$ for all $x$.


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