# Some hypothesis about the specific functions

Yesterday I found the question A question about the relation between two classes of functions which is strongly related to my former question Example of a special function. It's funny because Mark's question (unfortunately there weren't any answers for his question) is quite similar to the one I would like to ask now.
I strongly believe that for any continuous function $f$ which is:
1. positive and strictly increasing in $(0,a)$ for some $a>0$,
2. $f(0)=0$,
3. $\lim\limits_{x\rightarrow0^{+}}\frac{x}{f(x)}=0$
4. there doesn't exist an interval $(0,b)$, $b>0$, such that the map $x\rightarrow\frac{x}{f(x)}$ is strictly increasing for all $x\in(0,b)$,
one can always construct the continuous function $g$ satisfying $f>g$ in the interval $(0,a)$ for which $\lim\limits_{x\rightarrow0^{+}}\frac{f(x)}{g(x)}=1$, $g$ satisfies $1.$-$3.$ and there exists some interval $(0,\alpha)$, $\alpha>0$, such that the map $x\rightarrow\frac{x}{g(x)}$ is strictly increasing for all $x\in(0,\alpha)$.
I found the form of such functions $g$ for the functions $f$ given by Mr. Piau and Mr. Nicolas as the answers to my question Example of a special function but I couldn't find a general method to proof my hypothesis. Could someone give me some hints or maybe even counterexamples if my hypothesis is wrong?
I didn´t understand exactly the enviroment for the problem. But if I understood, the question is: "Given $f$ such that 1,2,3,4, I´m looking for an explicit form for a function $g$ such that bla bla bla". What about $g(x):= f(x) -x$? It seems to be an answer of the question I cited, but it doesn´t use a lot of the information you provide, so... – Giovanni De Gaetano Jan 11 '12 at 13:25
@Student73: You're absolutely right with your function $g$. But my mistake, I didn't add some conditions which the function $g$ have to satisfy in my hypothesis. Thanks for pointing me out this, even implicitly :) I'll fix it. – John Jan 12 '12 at 15:45