1
$\begingroup$

If $f$ and $g$ are commutative operations $$\mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R},$$ then for any constants $a,b \in \mathbb{R}$, the system of equations $$f(x,y) = f(a,b), \qquad g(x,y) = g(a,b)$$ must include $(x,y) = (a,b)$ and $(x,y) = (b,a)$ among its solutions.

I conjecture that if $f$ is given by $f(x,y) = xy$, and if $g$ can be expressed as $g(x,y) = G(x) + G(y)$ for some $G : \mathbb{R} \rightarrow \mathbb{R}$ that's continuous and injective, then the above system has only these solutions.

The injectivity requirement in particular is designed in particular to block $G(x) = x^2,$ which would break the theorem if it were allowed, while allowing the choices $G(x) = x^3$ and $G(x) = \mathrm{tan}^{-1}(x),$ which seem to be consistent with the theorem (based on graphical considerations).

Here's a graphical example illustrating the case $a = 2, b = 3, G(x) = x^3$:

enter image description here

Question. Is this true? If not, what would be a counterexample? If so, how might we prove it?

$\endgroup$
  • $\begingroup$ The term "commutative function" is rather unusual : Do you mean $f(x,y)=f(y,x)$ for all $x,y$ ? $\endgroup$ – Jean Marie May 9 at 11:57
  • $\begingroup$ @JeanMarie, yep. I've just changed it. $\endgroup$ – goblin May 9 at 11:59
1
$\begingroup$

It's not true.

Define

$$G(x)= \begin{cases} x, & \text{for } x \le 4 \\ \frac12x+2, & \text{for } x > 4. \\ \end{cases} $$

This is a monotonically increasing, continous (check $x=4$ on both 'branches') function, thus injective.

Especially note that

$$G(2)=2, G(3)=3, G(4)=4, G(6)=5$$

and thus $g(2,6)=7=g(3,4)$, and of course $2\times6=3\times4$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.