# Convex function with linear grow?

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?

• $f(x) = c\cdot x$ or what am I missing? May 12 '12 at 19:11
• Are you looking for a strictly convex function? May 12 '12 at 19:16
• Sorry, something is missing. Yes, I mean a strictly convex function. May 12 '12 at 19:23
• $f(x)=x+1-\sqrt{x+1}$.
– Did
May 12 '12 at 19:39

## 2 Answers

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 $f(x) = cx$.

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

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

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