Boundary limits of a normalized finite convex generator exist

Fix $c>0$ and let $u=c\mathrm{/}2$, $a=3c\mathrm{/}4$, $b=5c\mathrm{/}4$, and $v=3c\mathrm{/}2$. For $r<s$, write $m(r,s)=(f(s)-f(r))\mathrm{/}(s-r)$. If $a\le x<y\le b$, convexity at $a$ between $u,x$, at $x$ between $u,y$, at $y$ between $x,v$, and at $b$ between $y,v$ gives $m(u,a)\le m(u,x)\le m(x,y)\le m(y,v)\le m(b,v)$ by rearranging the four two-point inequalities in [F1]. All these endpoint slopes are finite. Thus with $M=\max \{\mathrm{\mid }m(u,a)\mathrm{\mid },\mathrm{\mid }m(b,v)\mathrm{\mid }\}$, one has $\mathrm{\mid }f(y)-f(x)\mathrm{\mid }\le M\mathrm{\mid }y-x\mathrm{\mid }$ for $x,y\in [a,b]$, including $x=y$.

Fix $a=1\mathrm{/}2$ and set $n(x)=(f(a)-f(x))\mathrm{/}(a-x)$ for $0<x<a$. If $0<x<y<a$, then $y$ is a convex combination of $x$ and $a$; [F1] gives $f(y)\le \frac{a-y}{a-x}f(x)+\frac{y-x}{a-x}f(a)$. Rearranging yields $n(x)\le n(y)$. Also $a$ is a convex combination of $x$ and $1$, so another application of [F1] gives $n(x)\le (f(1)-f(a))\mathrm{/}(1-a)$, a finite upper bound. Therefore $L:={\inf }_{0<x<a}n(x)$ is either real or $-\mathrm{\infty }$; the infimum property from [F2] gives $n(x)\to L$ as $x\downarrow 0$ (if $L$ is finite, approximate it within any $\epsilon >0$; if $L=-\mathrm{\infty }$, use any real lower target). Since $f(x)=f(a)-(a-x)n(x)$, its limit is $f(a)-aL$ if $L$ is finite and $+\mathrm{\infty }$ if $L=-\mathrm{\infty }$. Thus $f(0)$ exists and is not $-\mathrm{\infty }$.

Given $\epsilon >0$, choose $\delta =\min \{c\mathrm{/}4,\epsilon \mathrm{/}(M+1)\}$. If $\mathrm{\mid }x-c\mathrm{\mid }<\delta$, then $x,c\in [a,b]$, and step 1.1 gives $\mathrm{\mid }f(x)-f(c)\mathrm{\mid }\le M\mathrm{\mid }x-c\mathrm{\mid }<\epsilon$. This is continuity at the arbitrary $c>0$, so $f$ is continuous on $(0,\mathrm{\infty })$.

For $t>1$, put $r(t)=f(t)\mathrm{/}(t-1)$. If $1<s<t$, then $s$ is a convex combination of $1$ and $t$. By [F1] and $f(1)=0$, this gives $r(s)\le r(t)$. Hence $r$ is nondecreasing on $(1,\mathrm{\infty })$ and bounded below on $[2,\mathrm{\infty })$ by the finite value $r(2)$. By the supremum property in [F2], its limit is the real supremum $R$ if it is bounded above, and is $+\mathrm{\infty }$ otherwise: for finite $R$, values eventually lie between $R-\epsilon$ and $R$; when unbounded, they eventually exceed any prescribed real number. Since $f(t)\mathrm{/}t=((t-1)\mathrm{/}t)r(t)$ and $(t-1)\mathrm{/}t\to 1$, this quotient tends to $R$ if $R$ is finite; if $R=+\mathrm{\infty }$, the factor is at least $1\mathrm{/}2$ for $t\ge 2$ and the quotient tends to $+\mathrm{\infty }$. Thus $f_{\mathrm{\infty }}$ exists and is not $-\mathrm{\infty }$.