# How to solve the functional equation $f(x + f(x +y ) ) = f(2x) + y$?

Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ that satisfy the following equation:

$$f(x + f(x +y ) ) = f(2x) + y,\quad \forall x,y\in\mathbb{R}$$

The only function I have found is $f(x) = x$, but I think there are more.

If we put $x=0$ we get $$f(f(y)) = f(0)+y$$ so $f$ is injective. Now let $y=0$, then $$f(x+f(x)) = f(2x)\Longrightarrow x+f(x) = 2x$$ so $f(x)=x$ for all $x$.

• Cool. Kind of amazing that there are these two very disparate proofs. – Thomas Andrews Jan 22 '18 at 19:20

Given $z$, let $x=f(z)$. and $y=z-x.$ Then you get:

$$f(x+f(x+y))=f(2f(z))$$ and $$f(2x)+y=f(2f(z))+z-f(z)$$

From this you get $z=f(z).$

A cute variation of Christian's very nice answer:

\begin{align} 2z+f(0)&=f(f(2z))&[x=0,y=2z]\\ &=f(f(z+f(z)))&[x=z,y=0]\\ &=z+f(z)+f(0)&[x=0,y=z+f(z)] \end{align}

So $f(z)=z.$

This is avoiding the reference to being an injection, by implicitly using the right inverse $g(z)=f(z)-f(0).$

• Very elegant, lovely. – Patrick Stevens Jan 22 '18 at 19:15