# Does strictly increasing functions $f_1$, $f_2$ ($f_1(0)=f_2(0)=0$) exist such that $f_1(x)\leq f(x)\leq f_2(x)$ for an increasing $f$ ($f(0)=0$)?

Let $f:[0, r] \to [0, \infty)$ be continuous, positive definite, and increasing function, i.e., it is continuous and

1) $f(0) = 0$ and $f(x) > 0$ for $x \in (0, r]$;

2) $f(x) \leq f(y)$ whenever $x < y$ for $x, y \in [0, r]$.

Then, I want to show that there exist continuous strictly increasing functions $f_1:[0,r] \to [0, \infty]$ and $f_2:[0,r] \to [0, \infty]$ such that

1) $f_1(0)=f_2(0) = 0$

2) $0 < f_1(x) \leq f(x) \leq f_2(x)$ for all $x \in (0, r]$.

I guess that such functions always exist for any function $f$ satisfying the properties, but stuck with the proof.

For $f_2$ you can chose:
$f_2(x)=2f(x)+x$
And for $f_1$ pick up: $$f_1(x) = \frac{1}{x}\int_0^x f(t) dt$$ for $x \neq 0$ and $f_1(0)=0$. $f_1$ is stricly increasing because $f$ is stricly positive on $(0, r]$ and less than $f$ because $f$ is increasing.
• The $2f(x)$ seems un-necessary, could just use $f(x)+x$. Nice example. – TravisJ Jun 6 '15 at 2:23
• if $f$ is not "strictly" increasing, then neither $f_1$ nor $f_2$ is strictly increasing. If the "strictly" condition wasn't there, one could simply use $f(x)$ for both $f_1$ and $f_2$. – TravisJ Jun 6 '15 at 2:15