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Let $f:\mathbb R \to \mathbb R$ be a function such that $ f(x +f(y) + yf(x)) = y +f(x) + xf(y) ,\forall x,y \in \mathbb R$ , then is it true that $f(x)=x,\forall x \in \mathbb R$ ? If not true in general , then what if we also assume $f$ is bijective , or say continuous ?

Over $\mathbb C$ , $f(z)=\bar z$ is a bijective continuous , non-identity solution . Over $\mathbb R$ I can show that under the further assumption $\{f(x)/x : x\ne 0\}$ is countable, we must have $f$ is identity . With only the given functional equation , I can show that if $f(x_0)=0$ for some $x_0 $ then $x_0=0$ , and then $f(f(x))=x,\forall x \in \mathbb R$ i.e. $f$ is bijective . But I can't figure out anything else

Please help . Thanks in advance

user
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  • Just a thought: Once you have $f(f(x))=x$ you can substitute in $x\to f(x)$ in the original equation to get a symmetric form that holds for all real $x$. – Carl Schildkraut Jul 13 '17 at 16:22
  • @Carl Schildkraut : Ah uh ok . But you see, even $f(f(x))=x$ is got under the condition that $f$ has a root (in particular surjective) ... still I will try to see in the direction you pointed – user Jul 13 '17 at 16:24
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    You can get $f(f(x))=x$ by taking $x=a,y=0$ and $x=0,y=a$. This gives that $$f(a+f(0))=f(a)+af(0)$$ and $$f(f(a)+af(0))=a+f(0)$$ which implies that $$f(f(a+f(0)))=a+f(0)$$ whereupon substituting $x=a+f(0)$ yields $f(f(x))=x$. – Carl Schildkraut Jul 13 '17 at 16:26
  • @Carl Schildkraut : oh , thanks ! I didn't notice this – user Jul 13 '17 at 16:27
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    Another thing one can get (by taking $y=x$) is that $xf(x)+x+f(x)$ is a fixed point. This might be useful, but I'm not quite sure how. – Carl Schildkraut Jul 13 '17 at 16:27
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    With $x=y=0$, we get $f(f(0))=f(0)$. As $f$ is an involution [Carl Schildkraut], we find $f(0)=0$. If $x\ne 0$, we can let $y=-\frac x{f(x)}$ and get $ f(x+f(-\tfrac x{f(x)})-\tfrac x{f(x)}f(x)))=-\frac x{f(x)}+f(x)+xf(-\tfrac x{f(x)})$, which simplifies to $f(-\frac x{f(x)})=-\frac{f(x)}x$. In particular, $f(-1)=-1$. – Hagen von Eitzen Jul 13 '17 at 17:15
  • Hagen von Eitzen: It looks to me like your argument assumes that there exists a nonzero fixed point? – Harry Altman Jul 14 '17 at 00:05
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    Just let $ x = y = - 1 $ and you get $ f ( - 1 ) = - 1 $. – Mohsen Shahriari Aug 05 '17 at 15:19

1 Answers1

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If we assume continuity, then the identity function is the only possibility.

Edit: Here's a simpler argument based on Mohsen Shahriari's observation below that $f(-1)=-1$. I'll leave my original argument after.

By Carl Schildkraut's argument in the comments, $f$ is an involution and so in particular a bijection. So if we assume continuity, it must be either monotonically increasing or montonically decreasing.

Since $f(0)=0$ (as noted by Hagen von Eitzen above) and $f(-1)=-1$, it must be increasing. But this together with the fact that it's an involution forces it to be the identity (otherwise, picking some $x$ with $f(x)\ne x$, it would have to decrease between $x$ and $f(x)$).

Original more complicated argument for why it can't be monotonically decreasing follows:

If $f$ were monotonically decreasing, there could be at most one fixed point (or it would have to increase between the two fixed points.) We know zero is a fixed point. But also, for any $x$, $x+f(x)+xf(x)$ is a fixed point, and so $x+f(x)+xf(x)=0$. Solving, we get that, so long as $x\ne -1$, we have $f(x)=-\frac{x}{x+1}$.

Since $f$ is an involution on $\mathbb{R}$, whereas $x\mapsto -\frac{x}{x+1}$ is an involution on $\mathbb{R}\setminus\{-1\}$, we thus must have $f(-1)=-1$.

But this function isn't actually continuous, isn't actually monotonically decreasing, and doesn't satisfy the original functional equation.

Harry Altman
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