Is it true for a bilinear function function that $\phi(x,y)=\phi (x+x+x+\dots,y)$? If so, how can we construct a bilinear function with this property?

Question

Let $E$ and $F$ be vector spaces. The a mapping $\phi : E\times F \to \Gamma $ satisfying $$\phi(\lambda x_1 + \mu x_2, y) = \lambda \phi (x_1, y) + \mu \phi (x_2,y),$$ $$\phi(x,\lambda y_1 + \mu y_2) = \lambda \phi (x,y_1) + \mu \phi (x,y_2),$$ is called a bilinear function in $E \times F$.

From this definition of bilinear function, one can conclude that \begin{align} \phi(x,y) &= \phi(x+x-x, y)\\ & = \phi (x,y) + \phi (x,y) + \phi (-x,y)\\ & = \phi (x,y) + \phi (x,y) + (-1) \phi (x,y)\\ & = \phi (x,y) + \phi (x,y) + \phi (x,-y)\\ & = \phi (x+x+x,y). \end{align}

Since I can also write $x = x + x - x + x - x\dotsb $ infinitely many times, a bilinear function has to satisfy $\phi (x,y) = \phi (x+x+x+\dotsb, y)$ regardless of how I expressed $x$, and this really made me uncomfortable.

My question is, first of all, is this really the case with bilinear functions, i.e is my conclusion correct. Secondly, isn't this causing any problem in defining a map with the given condition. I mean, as I've said, this kind of freedom really bothers me, so I'm looking for an explanation to satisfy this conditions while defining a bilinear function.

Answer 1

The error is in the step $\phi(x, y) + \phi(x,y) + \phi(x,-y)$; $\phi$ is a linear map in the second argument when holding the first fixed, so this actually sums to $\phi(x, y+y-y) = \phi(x,y)$.