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Function $f:[0,+\infty)\rightarrow\mathbb{R}$ satisfies $f(x+y)=f(x)+f(y)$ for every non-negative $x$ and $y$. It’s bounded from below with some non-positive constant $m$. Does it imply that $f$ has the form $f(x)=cx$ or is there another function satisfying these conditions?

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    $\begingroup$ Is $f$ continuous? $\endgroup$ – Thomas Andrews May 17 '13 at 15:50
  • $\begingroup$ Not necessarily. $\endgroup$ – Danylo Mysak May 17 '13 at 16:16
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Yes.

It follows easily from induction that $f(nx)=nf(x) \, \forall n \in \mathbb{Z}^+$. Thus $f(\frac{m}{n})=\frac{1}{n}f(m)=\frac{m}{n}f(1)$.

If $f(y)<0$ for some $y \in[0, +\infty)$, then $f(ny)=nf(y)<m$ for sufficiently large $n$. Thus $f(y) \geq 0 \forall y \in [0, +\infty)$. Now if $a \geq b$ then $f(b+(a-b))=f(b)+f(a-b) \geq f(b)$, so $f(x)$ is monotonic. This is sufficient to imply that $f(x)=xf(1)$ (where here we require $f(1) \geq 0$).

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