# Cauchy functional inequality

Given a function on a closed interval $f\colon I\subset \mathbb{R}\to \mathbb{R}$ with $$f(x+y) \leq f(x) + f(y).$$ Moreover, I know that $f$ is

• monotonic increasing
• continuous on all points except countably many
• continuous from the right
• such that the limit from the left at each point exists.
• $f(nx) \leq nf(x)$ for all $n\in\mathbb{N}$.

Can I say more on this function? In particular, I would like to show that $f(\lambda x) \leq \lambda f(x)$ for all $\lambda \geq 1$ or at least for all rational $\lambda \geq 1$.

• Maybe you should also require that $f$ is strictly positive. For instance, consider $f(x)=x-2$ on the interval $I=[0,2]$. This function satisfies all of the hypotheses listed, but taking $\lambda =2$ and $x=1$, $$0=f(2) = f(2\cdot 1) \nleq 2\cdot f(1) =2(-1)=-2.$$ I am not sure if that suffices to prove the desired statement though. – Alberto Debernardi May 7 '16 at 9:17
• Note that your $f$ does not satisfy the subadditivity. Your example shows $f(x+x) \nleq f(x) + f(x)$ – Peter May 7 '16 at 10:31
• If $f(lx)\leq l f(x)$ for positive rational $l$ then for $x , x/n \in I$ we have $f(x)\leq n f(x/n)\leq n\cdot n^{-1} f(x)=f(x)$, implying $f(x)/n=f(x/n).$ – DanielWainfleet May 7 '16 at 12:48
• @user254665, the question is whether this holds for some $\lambda \ge 1$; you try to apply this for any $\lambda>0$, even for $\lambda<1$. – zhoraster May 7 '16 at 12:54
• @zhoraster. The Q says "for all" as of May 7, 11:00 EDT – DanielWainfleet May 7 '16 at 15:10

No, this inequality is guaranteed exclusively for integer $\lambda$.
Take e.g. $f(x) = x + |\sin x|$. Then $$f(x + y) = x+ y + |\sin x\cos y + \sin y \cos x|\\\le x+ y + |\sin x\cos y| + |\sin y \cos x|\\ \le x+ y + |\sin x| + |\sin y| = f(x) + f(y).$$
Then for any $\lambda>1$, $f(\lambda \pi) - \lambda f(\pi) = |\sin\lambda \pi| - \lambda |\sin \pi| = |\sin\lambda \pi|$ is positive unless $\lambda\in \mathbb Z$.