0
$\begingroup$

I'm looking for a continuous, strictly increasing, strictly convex function $f: \mathbb{R}_{\geq 0} \rightarrow \mathbb{R}_{\geq 0}$, with $f(0)=0$, and such that

$ \lim_{x \rightarrow\infty} \frac{f(x)}{x} \leq c $

for some $c \in \mathbb{R}_\geq 0$.

Suggestions?

  • 1
    $f(x)=x+1-\sqrt{x+1}$.2012-05-12

2 Answers 2

2

Let $g\colon \mathbb{R}_{\geq 0}\to \mathbb{R}_{>0}$ be any monotonically increasing function such that $\lim_{x\to\infty} g(x) = c$. Then the function $f(x) = \int_0^x g(t)\,dt$ will work. In the case $g\equiv c$, you of course get martini's suggestion of $f(x) = cx$.

1

$ x - 1 \; + \; \; \frac{1}{x+1} $

If you really want strict increasing and strict convex to include $0$ and slightly negative numbers,

$ x - \frac{1}{2} \; + \; \; \frac{1}{x+2} $