# A functional inequality which calculates the limitation of human eyes

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.

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 $$[0,1]$$, $$0 \le x^2 \le x \le 2x-x^2 \le 1.$$ $$f_+(1-x) = 1-x^2 = 1-f_-(x).$$ $$f_-(x)f_-(y) = f_-(xy) = x^2y^2 \le f_-(x)f_+(y) \text{ since }f_-(y) \le f_+(y).$$ $$f_+(xy)-f_-(x)f_+(y) = 2xy - x^2y^2 - (2x-x^2)y^2 = 2xy(1-y) \ge 0.$$ $$f_+(x)f_+(y) - f_+(xy) = x(2-x)y(2-y) - xy(2-xy) = xy(2-2x-2y+2xy)= 2xy(1-x)(1-y) \ge 0.$$
• The trivial solution $$(f^-,f^+)(x)=(0,1)$$
• The quadratic solutions $$(f^-,f^+)(x)=(x(1-a+ax),x(1-ax+a))$$ (for any $$a\in[0,1]$$)