Lemma 5.41.5742009E66DA3

Proof. Since $g$ is decreasing and finite somewhere, it is finite on a half-line $(-\infty ,\tau _1)$. Hence the defining supremum is finite for $t>0$: it is bounded above by $ct$ and below by any one of its finite terms. Concavity and upper semicontinuity give $g(c)=\lim _{\tau \uparrow c}g(\tau )$, so including $\tau =c$ does not change the supremum. The Legendre transform $t\mapsto \sup _\tau (g(\tau )+t\tau )$ of $-g$ is infinite for $t<0$. At $t=0$ it equals $m=\sup _{\tau <c}g(\tau )$, and the bounds

show that $f(t)\to m$ as $t\downarrow 0$. The biconjugacy theorem for the closed convex function $-g$ therefore gives the inverse formula; the value at $t=0$ may be omitted without changing the infimum.

Conversely, start with $f$. At each $t>0$, choose the slope $p$ of a supporting line of $f$. The upper bound $f(s)\leq cs$ as $s\to \infty $ forces $p\leq c$. If $p<c$, the supporting line gives $g(p)=f(t)-tp$, so the first formula recovers $f(t)$. If $p=c$, then $f(s)-cs$ is constant for $s\geq t$, because this function is convex and bounded above, hence decreasing, while its supporting line at $t$ is horizontal. Consequently, for every $\tau <c$,

and letting $\tau \uparrow c$ again recovers $f(t)$. In particular, $g$ is finite somewhere. As an infimum of affine functions with slopes $-s<0$, it is upper semicontinuous, concave and decreasing. Letting $s\downarrow 0$ in the bound $f(s)-s\tau \leq s(c-\tau )$ gives $g\leq 0$, and letting $s\to \infty $ gives $g=-\infty $ for $\tau >c$. The endpoint identity now follows from the first part. The case of functions identically equal to $-\infty $ follows directly from the formulas. $\square$