Can real linear map be defined using only additivity condition?

Question

Generally, linear map is defined to have the following two properties:

$f(x+y)=f(x)+f(y)$
$f(cx)=cf(x)$

If we restrict our discussion to real linear map ($c$ is real), will $f(x+y)=f(x)+f(y)$ imply $f(cx)=cf(x)$? It is easy to prove this if $c$ is rational. What about irrational numbers?

Not unless you add some kind of continuity assumption on $f$. — Dustan Levenstein, Aug 14 '16 at 03:55
@DustanLevenstein, I don't assume continuity. However, I cannot conjecture a counterexample myself even $f$ is not continuous. — atbug, Aug 14 '16 at 04:00

score 3 · Accepted Answer · answered Aug 14 '16 at 04:01

3

If you can check that $f$ is continuous, then the assertion is true since $\Bbb Q$ is dense in $\Bbb R$. I don't know if $f$ being measurable is already enough, maybe so.

If this is not the case, we can construct $f$ additive but not linear. The idea uses the existence of a Hamel basis of $\Bbb R$ as a $\Bbb Q-$vector space and you can read a sketch of it here, for example.

answered Aug 14 '16 at 04:01

Ivo Terek

77,665

2

+1. Note that this relies on the axiom of choice; for the OP, in ZF (set theory without choice) it is not provable that such an "alternate" solution exists. – Noah Schweber Aug 14 '16 at 04:20

Can real linear map be defined using only additivity condition?

1 Answers1