Lemma 17.7.20702EBBE9C4C

Proof. (1) Suppose that $g(t_0)=-\infty $ for some $t_0\in I$, and let $t\in I\setminus \{ t_0\} $. Choose $s\in I$ such that $t$ lies strictly between $t_0$ and $s$, and write $t=(1-\lambda )t_0+\lambda s$ with $0<\lambda <1$. Since $g(s)<\infty $, convexity gives $g(t)\leq (1-\lambda )g(t_0)+\lambda g(s)=-\infty $. Hence either $g\equiv -\infty $ or $g$ is finite, and a finite convex function on $I$ is continuous by example 3.4.15.

(2) For $a<b<c$ in $I$ we have $\sigma (a,b)\leq \sigma (a,c)\leq \sigma (b,c)$, as shown in example 3.4.15; this gives the monotonicity of $\sigma $. Now let $t\in I$. For $r<t<r'$ in $I$ we have $\sigma (r,t)\leq \sigma (t,r')$, so $c=\sup _{r<t}\sigma (r,t)$ is finite and $c\leq \sigma (t,r')$ for every $r'>t$. These two inequalities say that $g(s)\geq g(t)+c(s-t)$ for $s<t$ and for $s>t$. The affine function $s\mapsto g(t)+c(s-t)$ is therefore dominated by $g$ and agrees with $g$ at $t$, which proves the remaining assertions.

(3) Fix $s\in (0,1)$. For $0<t<s$ we have $g(t)=g(s)-(s-t)\sigma (t,s)$. By (2), $\sigma (t,s)$ decreases as $t$ decreases, so it has a limit $\sigma _0\in [-\infty ,\infty )$ as $t\downarrow 0$. If $\sigma _0=-\infty $, then $g(t)\to +\infty $ as $t\downarrow 0$, which contradicts the upper bound. Hence $\sigma _0$ is finite and $g(t)\to g(s)-s\sigma _0$. The value at $0$ of the affine function that agrees with $g$ at $t$ and $s$ is $g(s)-s\sigma (t,s)$; it increases as $t$ decreases and tends to $g(s)-s\sigma _0$.

(4) Let $\delta \leq a<b\leq 1-\delta $. For $b<r<1$, (2) gives $\sigma (a,b)\leq \sigma (b,r)\leq 2C/(r-b)$, and letting $r\uparrow 1$ we get $\sigma (a,b)\leq 2C/(1-b)\leq 2C/\delta $. Similarly, for $0<r<a$ we have $\sigma (a,b)\geq \sigma (r,a)\geq -2C/(a-r)$, and letting $r\downarrow 0$ we get $\sigma (a,b)\geq -2C/a\geq -2C/\delta $. $\square$