Return to Question

1. $M(a,a)=a$

2. $M(a,b)=M(b,a)$.

6. $M(ac,bc)=cM(a,b)\qquad$ (homogeneousness)

7. $a < b \implies a\le M(a,b) \le b\qquad$ (order preservation)

8. continuity.

In general (3) does not imply (4) as can be seen in this example for $S=\{a,b,c,d,e\}$:

$$ \begin{array}{c|ccccc} M & a & b & c & d & e\\ \hline a & a & a & a & a & a\\ b & a & b & d & c & a\\ c & a & d & c & b & a\\ d & a & c & b & d & e\\ e & a & a & a & e & e\\ \end{array}$$

where $M(M(b,c),M(d,e)) \ne M(M(b,d),M(c,e))$.

UPDATE. If $M(x,y):=f^{-1}\big(\frac{f(x)+f(y)}{2}\big)$ then notice that we can translate and rescale $f$ and the equality still holds. Now we try to build $f$. To start, we are allowed to assume $f(1/2)=1/2$ and $f(2)=2$. The graph of $f$ can then be constructed in the following way, as per Eric Wofsey's comment below: for each 2 consecutive known points $(x_1, y_1=f(x_1))$ and $(x_2, y_2=f(x_2))$ build an intermediate point $\big(M_f(x_1,x_2), \frac{y_1+y_2}{2}\big)$. By density of the dyadics in $\mathbb{R}$ and properties (5), (7) and (8) of $M$, this procedure defines $f$ in the interval $I=[1/2,2]$. Associativity (hopefully in its weak form) should then be used to prove that the same procedure applied to 2 overlapping subintervals of $I$ yields identical results on the intersection. Finally, if that worked, conclude the proof by defining a second $f$ on $[1/4,4]$. This second $f$ can be translated and rescaled to satisfy $f(1/2)=1/2, f(2)=2$ and must then match the original $f$ in $[1/2,2]$. Repeating this step will extend $f$ to $\mathbb{R}^+$.

1. $M(a,a)=a\qquad$ (identity)

2. $M(a,b)=M(b,a)\qquad$ (commutativity).

6. $M(ac,bc)=cM(a,b)\qquad$ (homogeneousness - see UPDATE)

7. $a < b \implies a\le M(a,b) \le b\qquad$ (order preservation - see UPDATE)

8. continuity.

This counteraxample is wrong. After fixing my gawk code I found that there (3) always implies (4) in a set of 5. Thanks to Eric Wofsey for noticing.

In general (3) does not imply (4) as can be seen in this example for $S=\{a,b,c,d,e\}$:

$$ \begin{array}{c|ccccc} M & a & b & c & d & e\\ \hline a & a & a & a & a & a\\ b & a & b & d & c & a\\ c & a & d & c & b & a\\ d & a & c & b & d & e\\ e & a & a & a & e & e\\ \end{array}$$

where $M(M(b,c),M(d,e)) \ne M(M(b,d),M(c,e))$.

UPDATE. It turns out that neither (6) nor (7) are needed anywhere, at least for the questions explored in this thread.

This is superseded by Eric Wofsey's answer

UPDATE. If $M(x,y):=f^{-1}\big(\frac{f(x)+f(y)}{2}\big)$ then notice that we can translate and rescale $f$ and the equality still holds. Now we try to build $f$. To start, we are allowed to assume $f(1/2)=1/2$ and $f(2)=2$. The graph of $f$ can then be constructed in the following way, as per Eric Wofsey's comment below: for each 2 consecutive known points $(x_1, y_1=f(x_1))$ and $(x_2, y_2=f(x_2))$ build an intermediate point $\big(M_f(x_1,x_2), \frac{y_1+y_2}{2}\big)$. By density of the dyadics in $\mathbb{R}$ and properties (5), (7) and (8) of $M$, this procedure defines $f$ in the interval $I=[1/2,2]$. Associativity (hopefully in its weak form) should then be used to prove that the same procedure applied to 2 overlapping subintervals of $I$ yields identical results on the intersection. Finally, if that worked, conclude the proof by defining a second $f$ on $[1/4,4]$. This second $f$ can be translated and rescaled to satisfy $f(1/2)=1/2, f(2)=2$ and must then match the original $f$ in $[1/2,2]$. Repeating this step will extend $f$ to $\mathbb{R}^+$.

Given a set $S$, a function $M: S\times S \rightarrow S$ is a mean if it satisfies the properties:

1. $M(a,a)=a$

2. $M(a,b)=M(b,a)$.

and possibly

3. $M(M(a,b),M(a,c))=M(a,M(b,c))\qquad$ (weak associativity)

4. $M(M(a,b),M(c,d))=M(M(a,c),M(b,d))\qquad$ (strong associativity)

5. $a\ne b \implies a\ne M(a,b)\ne b\qquad$ (sharpness).

When $S$ is an abelian groupoid or an ordered set or a topological space, $M$ can have additional specific requirements, such as:

6. $M(ac,bc)=cM(a,b)\qquad$ (homogeneousness)

7. $a < b \implies a\le M(a,b) \le b\qquad$ (order preservation)

8. continuity.

In general (3) does not imply (4) as can be seen in this example for $S=\{a,b,c,d,e\}$:

$$ \begin{array}{c|ccccc} M & a & b & c & d & e\\ \hline a & a & a & a & a & a\\ b & a & b & d & c & a\\ c & a & d & c & b & a\\ d & a & c & b & d & e\\ e & a & a & a & e & e\\ \end{array}$$

where $M(M(b,c),M(d,e)) \ne M(M(b,d),M(c,e))$.

Here are some of the questions that come to mind.

Q1. Is there a finite example where (3) and (5) hold, but not (4)? I know that $S$ will need to have at least 6 elements.

Q2. Does $M$ in the above example naturally extend to a mean in $\mathbb{R}[a,b,c,d,e]$ where both (3) and (6) hold?

Another example: if $A$ and $G$ are the arithmetic and geometric means on $\mathbb{R}^+$, it's easy to check that the mean function $M(x,y)=G(A(x,y),G(x,y))$ satisfies all the properties except (3) and (4).

Q3. Assuming all of the above properties except (4) hold for $M$ on $\mathbb{R}^+$, does (4) follow?

Q4. My starting point leading to this post: if all the above properties, including (4), hold for $M$ on $\mathbb{R}^+$, does it follow that $M$ is equivalent to the arithmetic mean, in the sense that $M(x,y)=f^{-1}\big(\frac{f(x)+f(y)}{2}\big)$ for some continous strictly monotonic function $f: \mathbb{R}^+ \to \mathbb{R}$?

I welcome suggestions for improvements to this post and references to relevant work.

UPDATE. If $M(x,y):=f^{-1}\big(\frac{f(x)+f(y)}{2}\big)$ then notice that we can translate and rescale $f$ and the equality still holds. Now we try to build $f$. To start, we are allowed to assume $f(1/2)=1/2$ and $f(2)=2$. The graph of $f$ can then be constructed in the following way, as per Eric Wofsey's comment below: for each 2 consecutive known points $(x_1, y_1=f(x_1))$ and $(x_2, y_2=f(x_2))$ build an intermediate point $\big(M_f(x_1,x_2), \frac{f(x_1)+f(x_2)}{2}\big)$. By density of the dyadics in $\mathbb{R}$ and properties (5), (7) and (8) of $M$, this procedure defines $f$ in the interval $I=[1/2,2]$. Associativity (hopefully in its weak form) should then be used to prove that the same procedure applied to 2 overlapping subintervals of $I$ yields identical results on the intersection. Finally, if that worked, conclude the proof by defining a second $f$ on $[1/4,4]$. This second $f$ can be translated and rescaled to satisfy $f(1/2)=1/2, f(2)=2$ and must then match the original $f$ in $[1/2,2]$. Repeating this step will extend $f$ to $\mathbb{R}^+$.

Given a set $S$, a function $M: S\times S \rightarrow S$ is a mean if it satisfies the properties:

1. $M(a,a)=a$

2. $M(a,b)=M(b,a)$.

and possibly

3. $M(M(a,b),M(a,c))=M(a,M(b,c))\qquad$ (weak associativity)

4. $M(M(a,b),M(c,d))=M(M(a,c),M(b,d))\qquad$ (strong associativity)

5. $a\ne b \implies a\ne M(a,b)\ne b\qquad$ (sharpness).

When $S$ is an abelian groupoid or an ordered set or a topological space, $M$ can have additional specific requirements, such as:

6. $M(ac,bc)=cM(a,b)\qquad$ (homogeneousness)

7. $a < b \implies a\le M(a,b) \le b\qquad$ (order preservation)

8. continuity.

In general (3) does not imply (4) as can be seen in this example for $S=\{a,b,c,d,e\}$:

$$ \begin{array}{c|ccccc} M & a & b & c & d & e\\ \hline a & a & a & a & a & a\\ b & a & b & d & c & a\\ c & a & d & c & b & a\\ d & a & c & b & d & e\\ e & a & a & a & e & e\\ \end{array}$$

where $M(M(b,c),M(d,e)) \ne M(M(b,d),M(c,e))$.

Here are some of the questions that come to mind.

Q1. Is there a finite example where (3) and (5) hold, but not (4)? I know that $S$ will need to have at least 6 elements.

Q2. Does $M$ in the above example naturally extend to a mean in $\mathbb{R}[a,b,c,d,e]$ where both (3) and (6) hold?

Another example: if $A$ and $G$ are the arithmetic and geometric means on $\mathbb{R}^+$, it's easy to check that the mean function $M(x,y)=G(A(x,y),G(x,y))$ satisfies all the properties except (3) and (4).

Q3. Assuming all of the above properties except (4) hold for $M$ on $\mathbb{R}^+$, does (4) follow?

Q4. My starting point leading to this post: if all the above properties, including (4), hold for $M$ on $\mathbb{R}^+$, does it follow that $M$ is equivalent to the arithmetic mean, in the sense that $M(x,y)=f^{-1}\big(\frac{f(x)+f(y)}{2}\big)$ for some continous strictly monotonic function $f: \mathbb{R}^+ \to \mathbb{R}$?

I welcome suggestions for improvements to this post and references to relevant work.

UPDATE. If $M(x,y):=f^{-1}\big(\frac{f(x)+f(y)}{2}\big)$ then notice that we can translate and rescale $f$ and the equality still holds. Now we try to build $f$. To start, we are allowed to assume $f(1/2)=1/2$ and $f(2)=2$. The graph of $f$ can then be constructed in the following way, as per Eric Wofsey's comment below: for each 2 consecutive known points $(x_1, y_1=f(x_1))$ and $(x_2, y_2=f(x_2))$ build an intermediate point $\big(M_f(x_1,x_2), \frac{y_1+y_2}{2}\big)$. By density of the dyadics in $\mathbb{R}$ and properties (5), (7) and (8) of $M$, this procedure defines $f$ in the interval $I=[1/2,2]$. Associativity (hopefully in its weak form) should then be used to prove that the same procedure applied to 2 overlapping subintervals of $I$ yields identical results on the intersection. Finally, if that worked, conclude the proof by defining a second $f$ on $[1/4,4]$. This second $f$ can be translated and rescaled to satisfy $f(1/2)=1/2, f(2)=2$ and must then match the original $f$ in $[1/2,2]$. Repeating this step will extend $f$ to $\mathbb{R}^+$.

Given a set $S$, a function $M: S\times S \rightarrow S$ is a mean if it satisfies the properties:

1. $M(a,a)=a$

2. $M(a,b)=M(b,a)$.

and possibly

3. $M(M(a,b),M(a,c))=M(a,M(b,c))\qquad$ (weak associativity)

4. $M(M(a,b),M(c,d))=M(M(a,c),M(b,d))\qquad$ (strong associativity)

5. $a\ne b \implies a\ne M(a,b)\ne b\qquad$ (sharpness).

When $S$ is an abelian groupoid or an ordered set or a topological space, $M$ can have additional specific requirements, such as:

6. $M(ac,bc)=cM(a,b)\qquad$ (homogeneousness)

7. $a < b \implies a\le M(a,b) \le b\qquad$ (order preservation)

8. continuity.

In general (3) does not imply (4) as can be seen in this example for $S=\{a,b,c,d,e\}$:

$$ \begin{array}{c|ccccc} M & a & b & c & d & e\\ \hline a & a & a & a & a & a\\ b & a & b & d & c & a\\ c & a & d & c & b & a\\ d & a & c & b & d & e\\ e & a & a & a & e & e\\ \end{array}$$

where $M(M(b,c),M(d,e)) \ne M(M(b,d),M(c,e))$.

Here are some of the questions that come to mind.

Q1. Is there a finite example where (3) and (5) hold, but not (4)? I know that $S$ will need to have at least 6 elements.

Q2. Does $M$ in the above example naturally extend to a mean in $\mathbb{R}[a,b,c,d,e]$ where both (3) and (6) hold?

Another example: if $A$ and $G$ are the arithmetic and geometric means on $\mathbb{R}^+$, it's easy to check that the mean function $M(x,y)=G(A(x,y),G(x,y))$ satisfies all the properties except (3) and (4).

Q3. Assuming all of the above properties except (4) hold for $M$ on $\mathbb{R}^+$, does (4) follow?

Q4. My starting point leading to this post: if all the above properties, including (4), hold for $M$ on $\mathbb{R}^+$, does it follow that $M$ is equivalent to the arithmetic mean, in the sense that $M(x,y)=f^{-1}\big(\frac{f(x)+f(y)}{2}\big)$ for some continous strictly monotonic function $f: \mathbb{R}^+ \to \mathbb{R}$?

I welcome suggestions for improvements to this post and references to relevant work.

UPDATE. If $M(x,y):=f^{-1}\big(\frac{f(x)+f(y)}{2}\big)$ then notice that we can translate and rescale $f$ and the equality still holds. Now we try to build $f$. To start, we are allowed to assume $f(1/2)=1/2$ and $f(2)=2$. The graph of $f$ can then be constructed in the following way, as per Eric Wofsey's comment below: for each 2 consecutive known points $(x1, y1=f(x1))$ and $(x2, y2=f(x2))$ build an intermediate point $(M_f(x1,x2), \frac{f(x1)+f(x2)}{2})$. By density of the dyadics in $\mathbb{R}$ and properties (5), (7) and (8) of $M$, this procedure defines $f$ in the interval $I=[1/2,2]$. Associativity (hopefully in its weak form) should then be used to prove that the same procedure applied to 2 overlapping subintervals of $I$ yields identical results on the intersection. Finally, if that worked, conclude the proof by defining a second $f$ on $[1/4,4]$. This second $f$ can be translated and rescaled to satisfy $f(1/2)=1/2, f(2)=2$ and must then match the original $f$ in $[1/2,2]$. Repeating this step will extend $f$ to $\mathbb{R}^+$.

Given a set $S$, a function $M: S\times S \rightarrow S$ is a mean if it satisfies the properties:

1. $M(a,a)=a$

2. $M(a,b)=M(b,a)$.

and possibly

3. $M(M(a,b),M(a,c))=M(a,M(b,c))\qquad$ (weak associativity)

4. $M(M(a,b),M(c,d))=M(M(a,c),M(b,d))\qquad$ (strong associativity)

5. $a\ne b \implies a\ne M(a,b)\ne b\qquad$ (sharpness).

When $S$ is an abelian groupoid or an ordered set or a topological space, $M$ can have additional specific requirements, such as:

6. $M(ac,bc)=cM(a,b)\qquad$ (homogeneousness)

7. $a < b \implies a\le M(a,b) \le b\qquad$ (order preservation)

8. continuity.

In general (3) does not imply (4) as can be seen in this example for $S=\{a,b,c,d,e\}$:

$$ \begin{array}{c|ccccc} M & a & b & c & d & e\\ \hline a & a & a & a & a & a\\ b & a & b & d & c & a\\ c & a & d & c & b & a\\ d & a & c & b & d & e\\ e & a & a & a & e & e\\ \end{array}$$

where $M(M(b,c),M(d,e)) \ne M(M(b,d),M(c,e))$.

Here are some of the questions that come to mind.

Q1. Is there a finite example where (3) and (5) hold, but not (4)? I know that $S$ will need to have at least 6 elements.

Q2. Does $M$ in the above example naturally extend to a mean in $\mathbb{R}[a,b,c,d,e]$ where both (3) and (6) hold?

Another example: if $A$ and $G$ are the arithmetic and geometric means on $\mathbb{R}^+$, it's easy to check that the mean function $M(x,y)=G(A(x,y),G(x,y))$ satisfies all the properties except (3) and (4).

Q3. Assuming all of the above properties except (4) hold for $M$ on $\mathbb{R}^+$, does (4) follow?

Q4. My starting point leading to this post: if all the above properties, including (4), hold for $M$ on $\mathbb{R}^+$, does it follow that $M$ is equivalent to the arithmetic mean, in the sense that $M(x,y)=f^{-1}\big(\frac{f(x)+f(y)}{2}\big)$ for some continous strictly monotonic function $f: \mathbb{R}^+ \to \mathbb{R}$?

I welcome suggestions for improvements to this post and references to relevant work.

UPDATE. If $M(x,y):=f^{-1}\big(\frac{f(x)+f(y)}{2}\big)$ then notice that we can translate and rescale $f$ and the equality still holds. Now we try to build $f$. To start, we are allowed to assume $f(1/2)=1/2$ and $f(2)=2$. The graph of $f$ can then be constructed in the following way, as per Eric Wofsey's comment below: for each 2 consecutive known points $(x_1, y_1=f(x_1))$ and $(x_2, y_2=f(x_2))$ build an intermediate point $\big(M_f(x_1,x_2), \frac{f(x_1)+f(x_2)}{2}\big)$. By density of the dyadics in $\mathbb{R}$ and properties (5), (7) and (8) of $M$, this procedure defines $f$ in the interval $I=[1/2,2]$. Associativity (hopefully in its weak form) should then be used to prove that the same procedure applied to 2 overlapping subintervals of $I$ yields identical results on the intersection. Finally, if that worked, conclude the proof by defining a second $f$ on $[1/4,4]$. This second $f$ can be translated and rescaled to satisfy $f(1/2)=1/2, f(2)=2$ and must then match the original $f$ in $[1/2,2]$. Repeating this step will extend $f$ to $\mathbb{R}^+$.