Simple but entangled inequalities

Question

Do there exist functions $F,G$ on $[0,1]$ with $0\le F,G< 1$, such that for all $x, y\in [0,1]$ with $x+y\le 1$, the following hold?

1) $G(x)\le x$,

2) $G(1)<1$,

3) $F(x)>0$ if $x>0$,

4) $\min(y,F(x)) \le G(x+y)-G(x)$.

Answer 1

$\newcommand{\de}{\delta} \newcommand{\vp}{\varepsilon}$

No such functions $F,G$ exist.

Indeed, let $\vp_x:=F(x)$, so that, by property 3), $\vp_x>0$ for all $x\in(0,1]$. Take any $\de\in(0,1)$ and let \begin{equation} E:=E_\de:=\{x\in[\de,1]\colon\forall y\in[\de,x]\ \, G(y)\ge G(\de)+y-\de\}. \end{equation} Note that $\de\in E$. So, $E$ is a nonempty interval of the form $[\de,s]$ or $[\de,s)$ for some $s\in[\de,1]$.

If $E=[\de,s)$, then $s>\de$, because $E$ is nonempty. Also, by property 4), $G$ is nondecreasing. So, if $E=[\de,s)$, then \begin{equation} G(s)\ge G(s-)\ge\lim_{y\uparrow s}[G(\de)+y-\de]=G(\de)+s-\de, \end{equation} so that $s\in E$ and $E=[\de,s]$. Thus, in all cases, $E=[\de,s]$.

If $s\ne1$, then $s\in[\de,1)\subset(0,1)$. So, $\eta_s:=\vp_s\wedge(1-s)>0$ and, by property 4), for all $h\in(0,\eta_s]$ we have $G(s+h)\ge G(s)+h\ge G(\de)+s-\de+h$, so that $s+h\in E$ for all $h\in(0,\eta_s]$, which contradicts the fact that $E=[\de,s]$.

So, $s=1$, $E=[\de,1]$, and hence $G(1)\ge G(\de)+1-\de\ge1-\de$. Since $\de$ was an arbitrary number in $(0,1)$, we have $G(1)\ge1$, which contradicts property 2).

Thus, no such functions $F,G$ exist. One may also note that property 1) was not needed or used here.