the solution of a functional equation

Question

Suppose $f:\Bbb R \to \Bbb R$ is a bijective mapping such that $f(x+y)=f(x)+f(y)$,can we conclude that $f(x)=f(1)x$?

I know the fact that if $f$ is continuous,the above conclusion holds.

Answer 1

No, it is false for the same reason which makes it false without the hypothesis of bijectivity.

Consider a Hamel basis $\{\alpha_i\}_{ i\in\Bbb R}$ of $\Bbb R$ as a $\Bbb Q$-vector space with, say, $\alpha_0=1$. Then consider the $\Bbb Q$-linear map $T$ such that:

• $T(1)=T(\alpha_0)=\alpha_{j_0}$;
• $T(\alpha_1)=\alpha_{j_1}$ where $\alpha_{j_1}$ is some element of the Hamel basis which is neither $\alpha_{j_0}$ nor $\alpha_{j_0}\alpha_1$;
• $T(\alpha_i)=\alpha_{j_i}$, where the map $i\mapsto j_i$ from $\Bbb R\setminus\{0,1\}$ to $\Bbb R\setminus\{j_0,j_1\}$ is some bijection.

Since $T$ sends bijectively a basis to a basis, it is bijective. However, $T(\alpha_1)\ne T(1)\alpha_1$ by definition.