A functional inequality which calculates the limitation of human eyes

Question

Find all pair of function $f^-,f^+:[0,1]\rightarrow[0,1]$ such that:
(1)$f^-(x)\leq x\leq f^+(x)$.
(2)$f^-(x)+f^+(1-x)=1$.
(3)$f^-(x)f^-(y)\leq f^-(xy)\leq f^-(x)f^+(y)$.
(4)$f^+(x)f^-(y)\leq f^+(xy)\leq f^+(x)f^+(y)$.
for all $x,y\in [0,1].$

Motivation: I want to compute explicitly the relative error of estimating distance by bare eye. Consider this test:
-Give a line segment $AB$ and a real number $x\in [0,1]$. We ask a person P to choose a point $C$ on $AB$ such that $\frac{AC}{AB}=x$, P must choose $C$ by their bare eye and don't use thing such as a finger to estimate, but P can do some algorithm such as choose a real number $y\geq x$, find $D$ such that $\frac{AD}{AB}=y$, then find $C$ such that $\frac{AC}{AD}=\frac{x}{y}$, but every point in the process must choose by bare eye. (sometime like $x$ or $1-x$ is too small and P can't image where $C$ should be, we can show them a line segment $XY$ which $XY\neq AB$ and a point $Z$ such that $\frac{XZ}{XY}=x$).

Let $f^-(x),f^+(x)$ is lower bound and upper bound of $\frac{AC}{AB}$ with $C$ is the point P choosing. We should have (1) because P can't always choose $C$ on the left (or right) of where $C$ should be. To have more information of $f^-,f^+$, I have two assumptions:
-The relative error should be the same for all similar figure (Weber-Fechner Laws), so $f^-,f^+$ is the same for all line segment, and so is defined. So find $C$ such that $\frac{BC}{BA}=1-x$ is the same as the above test. So we have (2).
-The best way for P to choose should be just choose $C$ by their bare eye, so any algorithm (which can choose $C$ if don't count error) can't be more exact than just choose by bare eye (that mean our brain is quite perfect). Now in the test we give other real number $y\in [0,1]$ and ask P to choose another point $D$ such that $\frac{AD}{AB}=xy$. Consider three following ways:
+Just choose $C,D$ normally. This should be the best approach.
+Choose $C$ first, then choose $D$ such that $\frac{AD}{AC}=y$, P can't choose $D$ this way more exactly than choose $D$ normally, so we have $f^-(x)f^-(y)\leq f^-(xy),f^+(x)f^+(y)\geq f^+(xy)$.
+Choose $D$ first, then choose $C$ such that $\frac{AD}{AC}=y$, similarly for the point $C$, we have $\frac{f^-(xy)}{f^+(y)}\leq f^-(x),\frac{f^+(xy)}{f^-(y)}\geq f^+(x)$.

Combine of those result, we have (3),(4) and so we have the above functional inequation. It seems like for each real number $F\in(0,\frac{1}{2}] $, there is a unique solution $f^-,f^+$ such that $f^-(\frac{1}{2})=F$. We have three trivial solutions: for $F=0$ we have $f^-(x)=0,f^+(x)=1$ or the same for all $x\in (0,1)$ but $f^+(0)=0,f^-(1)=1$ and for $F=\frac{1}{2}$ we have $f^-(x)=f^+(x)=x$.

Question 1: Is that functional inequation or that idea or something similar already appear in the literature?
Question 2: Is there any non-trivial solution and is for each real number $F\in(0,\frac{1}{2}]$, there is a unique solution $f^-,f^+$ such that $f^-(\frac{1}{2})=F$. If the answer is yes, how to compute $f^-(x),f^+(x)$? Or can you suggest some method or idea to solve this question?
Update: thank Christophe Leuridan answer, it seems like the solution is $f^-(x)=x^a,f^+(x)=1-(1-x)^a$ with $a=-log_2(F)\geq 1$ and the question now is there another solution?
Question 3: Finding the method to solve the general case, like computing the error when finding the center of a triangle by bare eye, using two assumption above.

Answer 1

Nice questions, with nice motivation.

I think that $f_-(x) = x^2$ and $f_+(x) = 2x-x^2 = 1-(1-x)^2$ is a non-trivial solution. Indeed, for every $x$ and $y$ in $[0,1]$, $$0 \le x^2 \le x \le 2x-x^2 \le 1.$$ $$f_+(1-x) = 1-x^2 = 1-f_-(x).$$ $$f_-(x)f_-(y) = f_-(xy) = x^2y^2 \le f_-(x)f_+(y) \text{ since }f_-(y) \le f_+(y).$$ $$f_+(xy)-f_-(x)f_+(y) = 2xy - x^2y^2 - (2x-x^2)y^2 = 2xy(1-y) \ge 0.$$ $$f_+(x)f_+(y) - f_+(xy) = x(2-x)y(2-y) - xy(2-xy) = xy(2-2x-2y+2xy)= 2xy(1-x)(1-y) \ge 0.$$

Answer 2

Just a few more solutions I found before Mathematica locked up:

• The trivial solution $$(f^-,f^+)(x)=(0,1)$$
• The quadratic solutions $$(f^-,f^+)(x)=(x(1-a+ax),x(1-ax+a))$$ (for any $a\in[0,1]$)

I'm sure there are more.