Find all pair of function $f^,f^+:[0,1]\rightarrow[0,1]$ such that:
(1)$f^(x)\leq x\leq f^+(x)$.
(2)$f^(x)+f^+(1x)=1$.
(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 $1x$ 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 (WeberFechner Laws), so $f^,f^+$ is the same for all line segment, and so is defined. So find $C$ such that $\frac{BC}{BA}=1x$ 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 nontrivial 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 $f^(x)=x^a,f^+(x)=1(1x)^a$ 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.
2 Answers
Nice questions, with nice motivation.
I think that $f_(x) = x^2$ and $f_+(x) = 2xx^2 = 1(1x)^2$ is a nontrivial solution. Indeed, for every $x$ and $y$ in $[0,1]$, $$0 \le x^2 \le x \le 2xx^2 \le 1.$$ $$f_+(1x) = 1x^2 = 1f_(x).$$ $$f_(x)f_(y) = f_(xy) = x^2y^2 \le f_(x)f_+(y) \text{ since }f_(y) \le f_+(y).$$ $$f_+(xy)f_(x)f_+(y) = 2xy  x^2y^2  (2xx^2)y^2 = 2xy(1y) \ge 0.$$ $$f_+(x)f_+(y)  f_+(xy) = x(2x)y(2y)  xy(2xy) = xy(22x2y+2xy)= 2xy(1x)(1y) \ge 0.$$
Just a few more solutions I found before Mathematica locked up:
 The trivial solution $$(f^,f^+)(x)=(0,1)$$
 The quadratic solutions $$(f^,f^+)(x)=(x(1a+ax),x(1ax+a))$$ (for any $a\in[0,1]$)
I'm sure there are more.

$\begingroup$ Thank you. I've known that so here the update mathoverflow.net/questions/425891/… $\endgroup$ Commented Jul 26, 2022 at 11:30