$f(x)>0$ and $f(y)>0$ implies $f(x+y)>0$, then there must exist an linear function $g$ such that $g(x)>0$ iff $f(x)>0$?

Question

Background: Let $x,y\in\mathbb (0,+\infty)^n$. $f$ is a continuous function on $\mathbb R^n_+=(0,+\infty)^n$. Consider the following condition (1), the sign of $f(x+y)$ is dependent on the sign of $f(x)$ and $f(y)$, formally:

$f(x)\geq0$ and $f(y)\geq\mathrel{(>)}0$ implies that $f(x+y)\geq\mathrel{(>)}0$,

$f(x)\leq0$ and $f(y)\leq\mathrel{(<)}0$ implies that $f(x+y)\leq\mathrel{(<)}0$.

Hypothesis: $f$ satisfies condition (1) if and only if there exists a linear function $g(x)=c\cdot x$ such that: $g(x)\geq0 \iff f(x)\geq 0$.

Motivation: It is known from Riesz's representation theorem that if $f(x)+f(y)=f(x+y)$, then $f(x)=c\cdot x$. Here is an effort to generalize the theorem by weakening the prerequisites.

Riesz's theorem require that the value of $f(x+y)$ is fully determined by $f(x)$ and $f(y)$. I only require that the sign of $f(x+y)$ is fully determined by $f(x)$ and $f(y)$. Although this hypothesis seems "exaggeratedly wrong" to me, I have yet to find a counterexample after thinking for a few monthes.

I also believe that such a simple idea has been thought before. Any ideas or references will help.

One might start with the facts that:

1. If $f(x)=f(y)=0$, then $f(x+y)=f(x)+f(y)$, just like Riesz's theorem.
2. If $f(x)>0$, then $f(\frac{x}{2})>0$.
3. If $f(x)>0$ and $f(y)>0$, then $f(ax+by)>0$ for any $a>0$ and $b>0$.