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Suppose a function f is additive. Also $f(x)$ is positive whenever $x$ is positive and $f(x)$ is negative whenever $x$ is negative. My question is "Is it necessary that f is linear?" If yes then please give a proof and if no then I am looking for a counterexample.

I've heard that if an additive function is bounded, it must be linear. $f(x)\ge 0$ is a function bounded below. I search some previous asked questions here also in Google but couldn't find any satisfactory answer. Please help me. Thanks in advance

Sufaid Saleel
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1 Answers1

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Yes, this is true. The thing to remember about additive functions on $\mathbb R$ is that they are automatically linear on the rational numbers due to the equations

$$f(nx)=nf(x)$$ $$f(\frac1n)=\frac1nf(x)$$

which follow from additivity. So $f(q)=qf(1)$ for rational $q$, and if we can use any kind of continuity argument we can fill in the gaps to show $f$ is linear on $\mathbb R$.

Here, your condition guarantees monotonicity since if $x>y$ we can write $x=y+a$ for positive $a$, and then $f(x)=f(y)+f(a)>f(y)$. Monotonicity allows us to fill in the gaps since $f(x)$ must be between $a_nf(1)$ and $b_nf(1)$ for any two rational sequences $a_n$ and $b_n$ which approach $x$ respectively from below and from above, so $f(x)=xf(1)$.

Jack M
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