Functional equation: $f(f(x,y), z) = f(x, f(y,z))$

Question

I am curious to know about the functions $f \colon \mathbb{R}^2_{\geq 0} \to \mathbb{R}_{\geq 0}$ that satisfy the following equality. For each $\{x,y,z\} \subseteq \mathbb{R}_{\geq 0}$,$$f( f(x,y), z) = f(x, f(y,z)).$$ Examples of such functions include

$f_1(x,y) = 1$
$f_2(x,y) = x$
$f_3(x,y) = x + y + 1$
$f_4(x,y) = xy$
$f_5(x,y) = \max\{x,y\}$
$f_6(x,y) = \sqrt{x^2 + y^2}$

In addition, I require $f(x,y) = f(y,x)$ (disqualifies $f_2$) and $f(x,0) = x$ (disqualifies $f_1$, $f_3$, $f_4$).

From the examples, my wild conjecture is that $f$ has to be homogeneous of degree 1, but I cannot prove this. Any pointers on how to proceed are much appreciated!

Edit: I was also considering the following related question.

Let $f$ be homogeneous of degree 1 and satisfy the additional requirements, so $f(x,y) = f(y,x)$ and $f(x,0) = x$. Does $f$ then satisfy the main equality, $f( f(x,y), z) = f(x, f(y,z))$?

This also turns out to be false; a counter-example is $f(x,y) = \sqrt{xy} + x + y$.

In fact, the second criterion doesn't make sense, as $(x, 0)$ isn't even in the domain of a function $\Bbb R_+^2 \to \Bbb R_+$. — Travis Willse, Feb 28 '17 at 15:27
Thank you for all the comments. Zelos: As $x \geq 0$, $\max{x,0} = x$, right? Travis: You are right. I have edited to include 0. Nate and Jonas: Yes, $f_3(x,y) = xy$ was the one I should have included and now do. — user420879, Feb 28 '17 at 15:35
Egreg: By that I mean the non-negative real numbers, so including 0. — user420879, Feb 28 '17 at 15:38
Unless you restrict it somewhat there are going to be an awful lot of these. Every time you have solution $f$, there is a solution $\alpha^{-1}\circ f\circ(\alpha,\alpha)$ for each $\alpha$ which fixes $0$ and permutes $\mathbb{R}^{>0}$ as you please. — ancient mathematician, Feb 28 '17 at 15:45
Thanks, ancient mathematician, but I don't quite follow. What does this mean for, say, $f(x,y)=x+y$? Which are the related functions? — user420879, Feb 28 '17 at 15:55
J G, for any bijection $\alpha\colon \Bbb R_{++}\to \Bbb R_{++}$ such that $\alpha(0) = 0$, function $g=\alpha^{-1} \circ f\circ(\alpha,\alpha)$ is associative, commutative and has identity $0$. In particular, for $f(x,y) = x + y$, $g(x,y) = \alpha^{-1}(\alpha(x)+\alpha(y))$. — Ennar, Feb 28 '17 at 16:10
Boba: Thanks for the even simpler example. Ennar: Thanks for the clarification, I see what you mean. — user420879, Feb 28 '17 at 16:33

score 4 · Answer 1 · answered Feb 28 '17 at 15:51

The property that $f(x,f(y,z)) = f(f(x,y),z)$ means that $f$ is an associative product.

$f(x,y) = f(y,x)$ means $f$ is a commutative product.

$f(x,0) = f(0,x) = x$ means $0$ is the unit of this product.

The triple $(\mathbb{R}_{\ge0},f,0)$ is called a commutative monoid, and your question is essentially about characterizing commutative monoid structures on $\mathbb{R}_{\ge0}$.

Unfortunately without more constraints there are a huge number of such structures that can behave pretty wildly. Here is an example of such a wild example:

Fix an arbitrary bijection $g: \mathbb{R}_{\ge0} \to \mathcal{P}(\mathbb{N})$ where $\mathcal{P}$ denotes the power set, which sends $0$ to the empty set. We can define a $f$ via the formula $f(x,y) = g^{-1}(g(x) \cup g(y))$. This will have the property that whatever $x$ gets sent to the whole set $\mathbb{N}$ satisfies $f(x,y)=x$ for all $y$, and this is definitely not homogeneous.

Thanks a lot for the explanation, Nate! The additional constraints that I can think of is to assume that $f$ is strictly increasing and continuous in both of its arguments. This disqualifies $f(x,y) = \max{x,y}$ at least. But there probably still are lots of wild functions left, aren't there? — user420879, Feb 28 '17 at 16:53

score 2 · Answer 2 · answered Feb 28 '17 at 15:43

2

Try $f(x,y) = xy+x+y = (x+1)(y+1)-1$. It satisfies all of the conditions and is not homogeneous.

answered Feb 28 '17 at 15:43

Jonas Meyer

53,602

Functional equation: $f(f(x,y), z) = f(x, f(y,z))$

2 Answers2