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 $\lambda \geq 1$.
Answer
No, this inequality is guaranteed exclusively for integer $\lambda$.
Take e.g. $f(x) = x + |\sin x|$. Then $$
f(x + y) = x+ y + |\sin x\cos y + \sin y \cos x|\\\le x+ y + |\sin x\cos y| + |\sin y \cos x|\\
\le x+ y + |\sin x| + |\sin y| = f(x) + f(y).$$
It also satisfies all other assumptions (note that the last one follows from the Cauchy functional inequality).
Then for any $\lambda>1$, $f(\lambda \pi) - \lambda f(\pi) = |\sin\lambda \pi| - \lambda |\sin \pi| = |\sin\lambda \pi|$ is positive unless $\lambda\in \mathbb Z$.
No comments:
Post a Comment