How find the $f(x+f(y))+f(y+f(z))+f(z+f(x))=0,$

Question

:$f:R\longrightarrow R$ ,and is continuous such that $$f(x+f(y))+f(y+f(z))+f(z+f(x))=0,$$ find all $f$

Answer 1

There are two solutions: either $f(x) = c-x$ for some constant $c$, or $f(x) = 0$ for all $x$.

1. We claim that $f(x+f(y)) = f(x) - f(y)$.

   • Put $x = y = z = 0$, we get $f(f(0)) = 0$.
   • Put $x = y = f(0)$, we get $f(f(0) + f(z)) = -f(z)$.
   • Put $x = f(0)$, we get $-f(y) + f(y+f(z)) + f(z) = 0$, hence the claim.

2. Note that the image of $f$ is an additive subgroup of $\mathbb{R}$. There are two cases - either this subgroup is dense at 0 (hence in $\mathbb{R}$), or is not dense at 0 (hence discrete in $\mathbb{R}$)

   • For the first case, let $g(x) = x + f(x)$. Note that $$f(x+f(y)) = f(x) - f(y) \Rightarrow x + f(y) + f(x+f(y)) = x + f(x) \Rightarrow g(x+f(y)) = g(x)$$ for any $y$. Density of image of $f$ and continuity of $f$ then implies that $g$ is a constant, say $c$, then $f(x) = g(x) - x = c-x$.

   • For the second case, $f$ is a continuous map on a connected set. If the image is discrete, it must be a constant. It's easy to check that the only possible constant is 0.

Answer 2

1. Take $x=y=z=0$, we have $3f(f(0))=0$ so $f(f(0))=0$. Set $k=f(0)$; we have $f(k)=0$.

2. Take $y=0, x=z=k$, we have $f(k+k)+f(0)+f(f(0))=0$ so $f(2k)=-k$.

3. Take $y=k, z=0$, we have $f(x)+f(2k)+f(f(x))=0$. Hence $f(x)+f(f(x))=k$, for all $x$.

4. In particular, for all $t$ in the range of $f$, we have $f(t)=k-t$. The range, as a subset of $\mathbb{R}$, must therefore be symmetric about the point $\frac{k}{2}$.

If the range is $\mathbb{R}$, then $f(x)=k-x$ is the only family of solutions (one for each $k\in \mathbb{R}$). By inspection all of these work. There is likely some continuity argument to prove that this must be the case, but right now I can't come up with it.

Answer 3

Let me just point out that after you obtained $f(x+f(y))=f(x)-f(y)$ $(1)$ the solution can be finished as follows: as Vadim pointed out $f(x)=k-x$ on the image of $f.$ So if $f$ is not identical zero $f$ takes both negative and positive values (follows from the original equation). So $(1)$ implies that the image of $f$ is unbounded from below and from above. By the intermediate value theorem it is $\mathbb{R},$ and so $f(x)=k-x$ for all $x\in\mathbb{R}.$