Is $\frac{\phi(r)}{r}$ dominated by $\phi'(r)$?

Question

Let $\phi:[0,1] \to \mathbb R$ be a smooth strictly increasing function satisfying $\phi(0)=0, \phi'(r)>0$ for every $r \in [0,1]$.

Is it true that $\frac{\phi}{r} \le \phi'(r)$ for every $r \in (0,1]$?

If not, is it true that $\frac{\phi}{r} \ge \phi'(r)$ for every $r \in (0,1]$? Or can $\frac{\phi}{r},\phi'(r)$ "cross" each other?

Note that $\phi(r)=\phi(r)-\phi(0)=\int_0^r \phi'(t)dt$.

Thus, if $\phi'$ is non-decreasing, we $$\phi(r) \le\int_0^r \phi'(r)dt=r\phi'(r) \Rightarrow \frac{\phi}{r} \le \phi'(r)$$

as required.

Answer 1

Not true in general. For example, take $\phi$ to be almost constant near $r=0$, almost constant and equal to 1 near $r=1$, and having a sharp transition from 0 to 1 near $r=\frac12$. Then for $r\approx1$ we have $\phi(r)/r\approx1$ but $\phi'(r)\approx0$, and for $r=\frac12$ we have $\phi(r)/r\leq2$ but $\phi'(r)$ very large.