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.
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.
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.
I'm not sure if we can derive all analytic solutions to this functional equation outside of polynomials, but we can prove that in polynomial space the only possibilities are $f(x) = x$ and $f(x) = -x$
First, let's assume that $f(x) = ax + b$. Then, we have:
$f(x + f(x)) = a(x + f(x)) + b = ax + b + af(x) = f(x) + af(x) = (a+1)f(x)$
Therefore, $f(x + f(x)) = x + f(x) = (a + 1)f(x)$, which means that $x = a \cdot f(x)$
Then, $x = a(ax + b)$, and thus $(a^2-1)x + ab = 0$
This gives us a system of equations: $a^2 - 1 =0$ and $ab=0$
From the first one, $a = \pm 1$, which means $a \neq 0$ and this forces $b = 0$; therefore the only linear solutions to the equation are $f(x) = x$ and $f(x) = -x$.
Now, suppose that there are polynomial solutions of degree higher than $1$. This means we can write $f(x) = a_0 + a_1x + a_2x^2 + a_3x^3 + ... a_nx^n$.
Then, $f(x + f(x))$ can be written as $a_0 + a_1(x + f(x)) + a_2(x + f(x))^2 + a_3(x + f(x))^3 + ... a_n(x + f(x))^n$
Now, focus on that last term: $a_n(x + f(x))^n$. The largest term in the polynomial expansion of this term has a degree of $n^2$. However, the degree of $x + f(x)$ is just the degree of $f(x)$, which is $n$; therefore, the only possible solutions happen when these degrees are equal, or simply: $n^2 = n$, which gives the solutions $n = 0$ and $n = 1$, covered by the prior part of this solution. Any degree higher than 1 and we don't have any solutions.
We cannot, unfortunately, expand this idea to cover infinite power series, so I can't provide any other obvious solutions to the equation.
