# Additive and Bijective function on the real line

I was trying to solve the following functional equation:

$$f(f(x-y))=f(x)-f(y)$$.

And I concluded that $$f$$ must be additive and bijective. The question is:

Let $$f:\mathbb{R} \to \mathbb{R}$$ be an additive bijective function. Is it true that $$f$$ has to be of the form $$f(x)=cx$$, $$c\in\mathbb{R}-\{0\}$$?

What if we impose the condition that $$f(f(x))=f(x)$$. Is the identity map the only solution?

Edit: if $$f$$ is surjective and $$f(f(x))=f(x)$$, then $$f$$ is the identity map (almost by definition)

• Constant functions $F$ also satisfy $F\circ F=F$. – lulu Apr 24 at 20:00
• For what concerns the additive and bijective function, you'd like to read this: en.m.wikipedia.org/wiki/Cauchy%27s_functional_equation. For it to be bijective, just do this: take a basis of $\Bbb R$ over $\Bbb Q$ and make the function swap two elements of the basis, while being the identity on the others. – Marco Vergamini Apr 24 at 20:05