# Simple but entangled inequalities

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

• The comments above are superseded by edits. – Yaakov Baruch Nov 22 '18 at 23:32
• Is $F$ allowed to be discontinuous? – Reid Barton Nov 23 '18 at 0:14
• What would obstruct G from being x up to 1/2, and then (x+3/2)/4 afterwards? (Answer: slope of 1/4 is too small.).Gerhard "There Should Be Enough F" Paseman, 2018.11.22. – Gerhard Paseman Nov 23 '18 at 0:24
• Yes, $F$ (and $G$) can be discontinuous. – Yaakov Baruch Nov 23 '18 at 0:26

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

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.

• For those, like me, not used to the notation, notice that $a\wedge b$ above means $\min(a,b)$. – Yaakov Baruch Nov 23 '18 at 8:52
• In lattice theory en.wikipedia.org/wiki/Lattice_(order), the notation $a\wedge b$ is standard. It has twice as few characters (counting $\min$ as one character) as $\min(a,b)$ does. When one writes by hand, its $3$ versus $8$ characters. – Iosif Pinelis Nov 25 '18 at 21:49