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Does there exist real number $a, b$ and onto function $f : \mathbb{R} \to \mathbb{R}$ satisfying $$ f(f(x))=bxf(x)+a$$ for all real numbes $x$ ?

My attempt :

Let $f(x_1)=f(x_2)$, so $f(f(x_1))=f(f(x_2))$

so $bx_1f(x_1)=bx_2f(x_2)$

then $bx_1f(x_2)=bx_2f(x_2)$

For $b \not= 0$, we have $x_1=x_2$

Hence $f$ is one-to-one function.

Please suggest how to proceed.

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    $\begingroup$ Would $f(x) = x^{(\sqrt{5}+1)/2}$ be one? Looks surjective to me. $\endgroup$ – cr001 Sep 16 '17 at 6:27
  • $\begingroup$ @cr001. How do you choose the values of $a, b$ ? $\endgroup$ – carat Sep 16 '17 at 6:33
  • $\begingroup$ @cr001: That has trouble if $x$ is negative... $\endgroup$ – Eric Wofsey Sep 16 '17 at 6:35
  • $\begingroup$ $b=1, a=0$ for this specific function. $\endgroup$ – cr001 Sep 16 '17 at 6:35
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    $\begingroup$ As long as $b \ne 0$, we have $g(x) = bf(b^{-1}x)$ satisfies $g(g(x)) = xg(x) + ab$. We only need to consider the case $b = 1$ (or $0$). $\endgroup$ – achille hui Sep 16 '17 at 7:07
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Hint: First, consider $w$ such that $f(w)=0$ and see what happens when you iterate $f$ on $w$. Then consider $v$ such that $f(v)=-1/b$ and see what happens when you iterate $f$ on $v$.

A full solution is hidden below.

Note that $b\neq 0$, since otherwise we would have $f(f(x))=a$ for all $x$ and $f$ could not be surjective. Now let $w$ be such that $f(w)=0$. We then have $$f(0)=f(f(w))=bwf(w)+a=a$$ and therefore $$f(a)=f(f(0))=b\cdot 0\cdot f(0)+a=a$$ and therefore $$a=f(f(a))=baf(a)+a=ba^2+a.$$ Since $b\neq 0$, this implies $a=0$.

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Now let $v$ be such that $f(v)=-1/b$. We then have $$f(-1/b)=f(f(v))=bvf(v)=-v$$ and therefore $$f(-v)=f(f(-1/b))=b\cdot(-1/b)\cdot f(-1/b)=v$$ and therefore $$f(v)=f(f(-v))=b\cdot(-v)\cdot f(-v)=-bv^2.$$ But $f(v)=-1/b$ as well, so $v=\pm 1/b$. If $v=-1/b$ we have both $f(v)=v$ and $f(v)=-v$, which is a contradiction. If $v=1/b$ we have both $f(-v)=-v$ and $f(-v)=v$, again a contradiction. Thus no such $f$ exists.

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    $\begingroup$ On the other hand, if you allow just a single number to be omitted from the image of $f$, then it is possible. For instance, with $a=0$ and $b=1$, you can get an $f$ whose image is $\mathbb{R}\setminus\{-1\}$. Here's a sketch of the construction. Start by defining $f(0)=0$, $f(1)=1$, and the orbit of $-1$. Then by transfinite induction define $f$ to be a bijection from the rest of $\mathbb{R}$ to itself, choosing one new orbit of $f$ and $f^{-1}$ at a time. $\endgroup$ – Eric Wofsey Sep 16 '17 at 8:39
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    $\begingroup$ (You must omit $-1$ from the image since the induction step of choosing one new orbit of $f$ and $f^{-1}$ fails if your new orbit is supposed to contain $-1$. Specifically, if you pick a number $x$ that you want to be in the new orbit and want to choose a value $y$ for $f(x)$, all the other values in the orbit will have the form $x^ky^\ell$ where $k$ and $\ell$ are certain Fibonacci numbers. For the construction to be well-defined, you need all these values to be distinct, which fails if $x=-1$.) $\endgroup$ – Eric Wofsey Sep 16 '17 at 8:43
  • $\begingroup$ Thank you for your thorough explanation, Eric Wolsey :) $\endgroup$ – carat Sep 16 '17 at 9:47

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