Proving a function to be a difference

Question

I am concerned with a function $f:\mathbb{R}\times\mathbb{R}\rightarrow \mathbb{R}$ such that $$f(f(x_1,x_2),f(x_3,x_2))=f(x_1,x_3)$$ for any $x_1,x_2,x_3\in \mathbb{R}$, and $$f(x,0)=x$$ $$f(x,x)=0$$ for any $x\in\mathbb{R}$.

For example, such a function is $f(x_1,x_2)=x_1-x_2$. To what extent is it possible to reduce $f$ to the latter form?

Answer 1

It seems to me that there are a lot of functions that fulfill that requirement (if you don't include the assumption that they are linear) like $f(x_1,x_2) = \frac{x_1}{x_2}$.

If you assume it is linear then you can just plug in $f(x_1,x_2) = ax_1 + bx_2 + c$ and you get

$a^2x_1 +(ab+b^2)x_2 + abx_3 + (a+b+1)c = ax_1 + bx_3 + c$

From here you can see that there are many different choices for $a$,$b$,and $c$ that will give you the type of linear function you want.

Answer 2

Another solution is $$ f(x_1, x_2) = x_1 $$ then $$ f(f(x_1, x_2), f(x_3, x_2)) = f(x_1, x_3) $$

Answer 3

Carrying on from @AWashburn’s post:

$$f(x_1,x_2)=ax_1+bx_2+c\ \implies \\ a^2x_1+(ab+b^2)x_2+abx_3+(a+b+1)c\ =\ ax_1+bx_3+c$$

Comparing coefficients gives

$$\begin{array}{rcl} a^2 &=& a \\ ab+b^2 &=& 0 \\ ab &=& b \\ (a+b+1)c &=& c \end{array}$$

The first equation gives $a=0$ or $a=1$; the second gives $b=0$ or $a+b=0$ $\implies$ $b=-a=0\ \text{or}\ -\!1$. We have the following:

1. If $b=0$ then $(a+b+1)c=(a+1)c=c$ $\implies$ $a=1,c=0$ or $a=0,c=\ \text{any value}$. Hence $f(x_1,x_2)=x_1$ or $f(x_1,x_2)=c$ (constant).

2. If $b=-1$ then $(a+b+1)c=ac=c$ $\implies$ $a=c=0$ (impossible since $ab=b\ne0$) or $a=1,c=\ \text{any value}$. Hence $f(x_1,x_2)=x_1-x_2+c$.

Hence, if $f$ is linear, the only possibilities are $$\boxed{f(x_1,x_2)=c}$$ or $$\boxed{f(x_1,x_2)=x_1}$$ or $$\boxed{f(x_1,x_2)=x_1-x_2+c}$$ Of course if $f$ is not linear there may be other solutions. Since this question is tagged as “linear-transformations” it is presumed that a linear solution is required.