Yesterday, I thought of an interesting open-ended question.
Given a function $f:D \to\mathbb{R}$, (for a suitable $D\subset\mathbb{R^2}$) when can we define an invertible function $g$ such that:
$$f(x,y)=g\Big(g^{-1}(x) +g^{-1}(y)\Big)$$
where $g:\mathbb{R}\to Im(f)?$
From this definition, we immediately obtain some conditions that must be met. For instance, $f$ must be symmetric in $x,y$.* Secondly, the image of $f$ must not be countable (or $g$ is not invertible), but $f$ need not be injective, as I can show below with some examples.
- Suppose $f(x,y) = x+y.$ Then $g(t) = t$ is sufficient (but not the only solution).
- Suppose $f(x,y) = xy.$ Then $g(t) = e^t$ is sufficient. (In this case, $D = \mathbb{R^+ \cup \{0\}}.$)
Suppose that $f(x,y) =$
$$\frac{x+y}{1-xy}$$
Then $g(t) = \tan(t)$ is sufficient.
However, I also found an instance in which finding such a function $g$ would be impossible. Consider $f(x,y)=x^2+y^2.$ In this case, it is easy to work out that for any satisfactory $g$ we must have $ g^{-1}(0) = 0 \implies \forall x, g^{-1}(x) = g^{-1}(x^2)$. Then $g^{-1}$ clearly cannot be invertible, and we have a contradiction; thus, $f$ cannot be written in this form.
In summary, $f$ must be symmetric and there must exist an injection from $\mathbb{R}$ to $Im f$, but we do not require that $f$ is linear. What exactly is this criterion? Any progress, even if it is specific to just one function, or any other food for thought would be appreciated.
*Although, this could easily be generalised e.g. we could have that $f$ is anti-symmetric i.e. that $f$ is symmetric in $x,-y$.
Note: I apologise if the title was not clear. Please feel free to rephrase it if there is better terminology for what I described.