Find all pairs of functions $f^-,f^+:[0,1]\rightarrow[0,1]$ 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:
-Given 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 not use something 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 be chosen 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 $f^-(x),f^+(x)$ be a lower bound and upper bound of $\frac{AC}{AB}$ with $C$ is the point P chooses. 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 figures (Weber-Fechner Laws), so $f^-,f^+$ is the same for all line segments, and so is defined. So find $C$ such that $\frac{BC}{BA}=1-x$ is the same of 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 choosing by bare eye (that means 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 choosing $D$ 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 $\frac{f^-(xy)}{f^+(y)}\leq f^-(x),\frac{f^+(xy)}{f^-(y)}\geq f^+(x)$.

Combining those results, 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 with 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: Does that functional inequation or that idea or something similar already appear in the literature?
Question 2: Is there any non-trivial solution and for each real number $F\in(0,\frac{1}{2}]$, is 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?
Question 3: Find the method to solve the general case, like computing the error when finding the center of a triangle by bare eye, using the two assumptions above.

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.

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 of 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 with 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?
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.

Find all pairs of functions $f^-,f^+:[0,1]\rightarrow[0,1]$ 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:
-Given 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 not use something 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 be chosen 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 $f^-(x),f^+(x)$ be a lower bound and upper bound of $\frac{AC}{AB}$ with $C$ is the point P chooses. 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 figures (Weber-Fechner Laws), so $f^-,f^+$ is the same for all line segments, and so is defined. So find $C$ such that $\frac{BC}{BA}=1-x$ is the same of 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 choosing by bare eye (that means 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 choosing $D$ 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 $\frac{f^-(xy)}{f^+(y)}\leq f^-(x),\frac{f^+(xy)}{f^-(y)}\geq f^+(x)$.

Combining those results, 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 with 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: Does that functional inequation or that idea or something similar already appear in the literature?
Question 2: Is there any non-trivial solution and for each real number $F\in(0,\frac{1}{2}]$, is 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?
Question 3: Find the method to solve the general case, like computing the error when finding the center of a triangle by bare eye, using the two assumptions above.

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 of 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 two trivial solutions: for $F=0$ we have $f^-(x)=0,f^+(x)=1$ 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?
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.

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 of 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 with 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?
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.