Concerning this draft here: http://hajek.ece.illinois.edu/Papers/randomprocJuly14.pdf on page 71 in the proof of Proposition 2.17, the inequalities (2.8) and (2.9) I don't understand how to derive these two inequalities.
For the first inequality it should follow from the convexity of $f$, but then I need to find a suitable form for the definition of convexity. But I don't see how exactly.
Can you help here?
Thanks!
P.S As I was requested in the comments. We assume that $f$ is convex, that $h>0$ is small enough such that: $s<s+h<t<t+h$, and the inequalities: $$(2.8)\ \ f(s+h)(t-s+h)\le (t-s)f(s)+hf(t+h)$$ $$(2.9)\ \ f(t)(t-s+h)\le hf(s)+(t-s)f(t+h)$$
I hope you agree that if $a<b<c$ (that is, $b$ is between $a$ and $c$) then $$b=\lambda a+\left(1-\lambda\right)c$$ for some $\lambda\in(0,1)$. $\lambda$ is the coefficient of betweenness that we can compute explicitly: since \begin{align} b&=\lambda a+\left(1-\lambda\right)c\\&=\left(a-c\right)\lambda+c \end{align} it follows that \begin{align} \lambda&=\frac{b-c}{a-c}=\frac{c-b}{c-a}\\1-\lambda&=\frac{b-a}{c-a} \end{align}
Then we can just substitute and apply the definition of convexity:
\begin{align} f(b)&=f\left(\lambda a+\left(1-\lambda\right)c\right)\\&\le\lambda f(a)+\left(1-\lambda\right)f(c)\\&=\frac{c-b}{c-a}f(a)+\frac{b-a}{c-a}f(c) \end{align}
Multiplying by the (positive) denominator $c-a$, we get $$\left(c-a\right)f(b)\le(c-b)f(a)+(b-a)f(c).$$
Take $s < s+h < t+h$ for $a<b<c$ and you'll get 2.8.
Take $s < t < t+h$ for $a<b<c$ and you'll get 2.9.