0

Consider functions through the origin $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $f(x+y) = f(x) + f(y)$ for all $x,y \in \mathbb{R}.$

I want to try to figure out if these functions are linear in scalar multiplication, i.e. $\forall \lambda \in \mathbb{R}. f(\lambda x) = \lambda f(x).$

I have an intuition that this should work, since $f(x+x) = f(x) + f(x) = 2f(x) \implies f(2x) = 2f(x).$ This would imply that $f(3x) = f(2x + x) = f(2x) + f(x) = 2f(x) + f(x) = 3f(x).$ I'm wondering if someone could help me the induction step of this problem.

  • 1
    The strongest identity one can prove assuming only that the Cauchy equation is satisfied is $f(rx)=rf(x)$ for all $r\in\mathbb{Q}$. There exist in fact non trivial Cauchy-functions for which $f(\lambda x)=\lamda f(x)$ isn't satisfied for all $x$. For this, see https://en.wikipedia.org/wiki/Cauchy%27s_functional_equation. – Redundant Aunt Dec 30 '15 at 18:12

1 Answers1

1

Induction step: assume the claim holds for $n\in \mathbb{N}$. $$f(nx) = f((n-1)x+x) = f((n-1)x) + f(x) = (n-1)f(x) + f(x) = nf(x)$$ Note you can also show $f(-x) = -f(x)$ and $f(x/m) = f(x)/m$ for all $m \in \mathbb{N}$. This proves $f(\lambda x) = \lambda f(x)$ for all $\lambda \in \mathbb{Q}$. Unfortunately, this is the best we can do. If you assume continuity you can prove the claim for all real $\lambda$. To do so, use density of $\mathbb{Q}$ in $\mathbb{R}$.

GiantTortoise1729
  • 3,048