Given an increasing function $f:[0,\infty)\to[0,\infty)$, we can define $$F(x)=\int\_x^{x+1} f(t)dt,$$
which is continuous, increasing function satisfying $$f(x)\leq F(x)\leq f(x+1).$$


Question) For such $f$ can we construct a continuous increasing function $g:[0,\infty)\to[0,\infty)$ such that for all $x\in [0,\infty),$
$$c\_1 f(x)\leq g(x) \leq c\_2 f(x),$$
for some positive constant $c\_1,c\_2?$



Given an increasing function, need to construct a continuous increasing function equivalent to given function
Using the integer-part function, let
$$f(x) = x ^ {\lfloor x \rfloor +1}$$


Then there is no $g$ satisfying the given condition for this $f$.


For any such $g$, let $b = \lfloor c\_2/c\_1 \rfloor +2$. Then
\begin{align}
b-1<x<b \implies & g(x) \le c\_2 x^b\\
b<x<b+1 \implies & g(x) \ge c\_1x^{b+1}
\end{align}


By the continuity of $g$, $g(b)$ must satisfy both of those inequalities, so
$$c\_2 b^b \ge g(b) 
\ge c\_1 b^{b+1}$$
$$\frac{c\_2}{c\_1}\ge b$$
which is impossible.


