Find all solutions to the functional equation $f:\mathbb{R}\rightarrow \mathbb{R}$ $$f(x+f(x))=f(x)+x$$ I have no idea how to solve this. I can substitute $x=0$ to obtain $f(f(0))=f(0)$. But other than that I can't make any progress. There are two obvious solutions, namely $f(x)=x$ and $f(x)=-x$. But are there any other solutions?

Thanks in advance.

  • $\begingroup$ You sure you aren't missing something here? Does $ f $ have to be continuous? Functional equations like this can be really hairy when you consider logic-monsters, like Dirichlet's function. $\endgroup$ – theREALyumdub May 2 '18 at 1:59
  • $\begingroup$ @Winther in this question we are working in reals, not integers. $\endgroup$ – abc... May 2 '18 at 2:13

There are many many solutions. Let $S\subseteq\mathbb R$ be any nonempty subset satisfying $2S\subseteq S$. Pick any function $g:\mathbb R\setminus S\to S$. Let $$ f(x)=\begin{cases} x&\text{if }x\in S\\ g(x)-x&\text{otherwise.} \end{cases} $$ For each $x\in\mathbb R$ we have $f(x)+x\in S$, so $f(f(x)+x)=f(x)+x$.

Even if we require $f$ to be continuous, there are as many solutions as continuous functions on $\mathbb R$. Indeed set $S=\{x\in\mathbb R\mid x\geq0\}$ and pick any continuous function $g:\mathbb R\setminus S\to S$ satisfying $\lim_{x\to0^-}g(x)=0$; then construct $f$ as above.

  • $\begingroup$ Thanks! That solves the problem. $\endgroup$ – abc... May 2 '18 at 2:22
  • $\begingroup$ This seems right, but the post linked also says there are only two solutions there, $ f(n) = n $ and $ f(n) = -n $. What's the catch here? Just because we have the continuous case? $\endgroup$ – theREALyumdub May 2 '18 at 2:36
  • $\begingroup$ @theREALyumdub The question linked by Winther is not the same (it has two free parameters). Even if you require $f$ to be continuous, you can still have $f(x)=x$ for $x\geq0$ and choose $f$ however you want for $x<0$ as long as $f(x)\geq-x$. $\endgroup$ – stewbasic May 2 '18 at 3:04

I want to qualify this by saying I know almost nothing about functional equations, but here's a little gimmick I see here:

$ f : \mathbb{R} \mapsto \mathbb{R} $ defined by

$$ f( f(x) + x) = f(x) + x $$

could be thought of a lot more succinctly as a relation showing that

$ f(c) = c $, for all $ c \in \mathbb{R} $, with $ c = f(x) + x $.

If you consider the entire real line as your domain, then $ f(x) $ has to have a "fixed point" at every value $ f(x) + x $ attains. I think it shouldn't be hard to go on and show that there are only two solutions (so I rescind my comment on your original post).

That's also just my take on what the linked post says, but it's a bit more technical and general.

  • 1
    $\begingroup$ But f(x)=-x is also a solution $\endgroup$ – abc... May 2 '18 at 2:12

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