Easy functional equation

Find all functions $f:\mathbb{R} \rightarrow \mathbb{R}$ such that:

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

If you put $x=y=0$, you get $f(3f(0))=f(0)$. What deductions about $f(0)$ can you then make?

Clearly from above $f(0)=0$ is a solution . . . so,

Putting $x=0$ gives $f(2f(0)+f(y))=f(y)$

$\rightarrow$ $f(f(y))=f(y)$

So $f(x)=x$ is a solution, but is it the only one?

I think it probably is, but how to prove?

Answer

$$f(2f(x)+f(y))=2x+f(y)\qquad \forall x,y \in \mathbb{R}.$$
Interchaning $x$ and $y$ you get

$$f(f(x)+2f(y))=f(x)+2y \,.$$

Claim 1: $f(x)$ is 1 to 1.

Indeed, if $f(x)=f(y)$ then

$$2x+f(x)=2x+f(y)=f(2f(x)+f(y))=f(f(x)+2f(y))=f(x)+2y $$

This implies that $x=y$.

Now, you can do part of what you did:

$$f(2f(0)+f(y))=f(y)\qquad \forall y \in \mathbb{R}.$$

Since $f$ is 1 to 1 you get

$$2f(0)+f(y)=y \,.$$

Thus

$$f(y)=y-2f(0)\,.$$

Setting $y=0$ you get $f(0)=0$ and thus $f(x)=x$ is the only solution.