Map that satisfies $f(\lambda x) = \lambda f(x)$ but not $f(x+y) = f(x)+f(y)$

Question

Could you give me example of maps $f:\mathbb R \to \mathbb R$ that satisfy $$ f(\lambda x) = \lambda f(x) \quad \forall x,\lambda \in \mathbb R $$ but not $ f(x+y) = f(x)+f(y) $? Thanks in advance.

Answer 1

Let $f(1)=a$. Then $f(\lambda\cdot 1)=\lambda f(1)=\lambda a$. Or, to put it in more familiar letters, $f(x)=ax$. So $f(x+y)=f(x)+f(y)$.

Remarks: Things get more interesting if, for example, we restrict $\lambda$ to rational values.

Things also get more interesting if we ask about the converse problem. It turns out that there are exotic solutions to the Cauchy functional equation $f(x+y)=f(x)+f(y)$.

Things also get more interesting if replace $\mathbb{R}$ by $\mathbb{R}^2$. Then it is perfectly possible to have $f(\lambda x)=\lambda f(x)$ for all vectors $x$ and scalars $\lambda$, without having $f(x+y)=f(x)+f(y)$. For example, map all points on the $x$-axis to $(0,0)$, and map off-axis points to themselves.

Answer 2

No. Choose $z$ with $f(z)\ne 0$. Then $f(x+y)=\frac{x+y} zf(z)=\frac xzf( z)+\frac yz f(z)=f(x)+f(y)$