I have one question regarding the derivative of a convex function. Let's say we have two functions (one convex, one linear) with the inequality sign as follows:
$ s^{1-k} > \frac{1-(1-p)s}{p}, \forall s \in (0, 1), p \in (0, 1) $ If this is OK to say, then what is the reason behind this?
Thanks you!
Let $f(s)=s^{1-k}-\frac {1-(1-p)s} p$. Suppose $k \leq \frac 1 p$. Then $f'(s)=(1-k)s^{-k}+\frac {1-p} p$. For $s$ near $1$ we can see that $f'(s) >0$ (because $p<1$). Hence $f$ is increasing in $(1-\delta ,1)$ for some $\delta >0$. Since $f(1-)=0$ it follows that $f(s) <0$ in $(1-\delta ,1)$ but this contradicts the given inequality. Hence we must have $k >\frac 1 p$.