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?

  • $\begingroup$ You should be more careful when using the words "easy" and "clearly". You can't deduce $f(0)=0$ from what you observed, for instance. $\endgroup$ – Julien Apr 9 '13 at 23:50

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

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



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


$$ f(2f(x)+f(y)) = 2x+f(y) $$ Now consider values of $x$ and $y$ such that $2f(x)=-f(y)$. Now, we can write that

$$ f(0) = 2(x-f(x)) $$ Therefore, we may write that $$ f(x) = x-\frac{f(0)}{2} $$ Substituting $x=0$, we get $$ f(0) = -\frac{f(0)}{2} $$ And therefore $f(0)=0$, thereby giving $f(x)=x$.

This analysis required an assumption: that values of $x$ and $y$ could be found such that $2f(x)=-f(y)$. This can be confirmed - for any value of $y$, we have that $$ f(2f(x)+f(y))=2x+f(y) $$ Therefore, for any given real number, $a$, we may choose $x_a=\frac{f(y)-a}{2}$, and thus we have $f(2f(x_a)+f(y))=a$, and thus the function may take the value $a$. As this must be true of any $a$, we know that the function must take all values, and thus there must exist a solution to $2f(x)=-f(y)$ for any $x$ or for any $y$.


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