Discovering a function in two variables involved in a system of equations

Question

I have a problem in which I need to discover $f$ knowing that $$\left\{\begin{matrix}f(1,y)-f(0,y)=y\\f(x,1)-f(x,0)=x\end{matrix}\right.$$ Any hints to solve it?

Answer 1

We can't solve this equation with the amount of information given, but yes:

If we let $f(x,y) = xy$, then certainly $f(x,1)-f(x,0) = x-0=x$ and $f(1,y)-f(0,y) = y - 0 = y$, so this is one solution.

Answer 2

Hint. Assume $f:\mathbb{R}^2\to \mathbb{R}$. Take continuous $f$ (in fact you can prove it). Note that for all $\alpha,\beta \in\mathbb{R}$, $$ \left\{\begin{matrix} \beta f(1, y)-\beta f(0, y)=\beta y, \\ \alpha f( x,1)- \alpha f( x,0)=\alpha x \end{matrix}\right. \Leftrightarrow \left\{\begin{matrix} f(1, \beta y)- f(0, \beta y)=\beta y, \\ f( \alpha x,1)- f( \alpha x,0)=\alpha x \end{matrix}\right. $$ Ask yourself if the function $f$ is a quadratic form added to affine function: $$ f(x,y)= \begin{pmatrix}x & y\end{pmatrix}\cdot \begin{pmatrix} M_{11} & M_{12}\\ M_{21} & M_{22}\end{pmatrix} \begin{pmatrix}x \\ y\end{pmatrix} +\begin{pmatrix}V_1 & V_2\end{pmatrix}\cdot \begin{pmatrix}x \\ y\end{pmatrix}+C $$