# Can a multivariable mapping that is linear in each variable separately have a local extrema?

Let $$f:\mathbb{R}^n\rightarrow \mathbb{R}^m$$, $$m be a mapping that is linear in each variable separately (i.e., in each of the functions $$f_i(x_1,\cdots,x_n)$$, $$1\leq i\leq m$$, the degree of each variable $$x_j$$ is at most 1, but there can be a product of several variables, e.g., $$7x_1x_3x_{n-1}$$).

I define a local boundary point as a point $$\boldsymbol{x}\in \mathbb{R}^n$$ with $$f(\boldsymbol{x})$$ not on the global boundary of $$f$$, that for some $$\delta>0$$ there exist a ball $$\mathcal{B}_\delta(\boldsymbol{x})$$ of radius $$\delta$$ around $$\boldsymbol{x}$$ such that there is no $$\epsilon>0$$ for which $$f(\mathcal{B}_\delta(\boldsymbol{x}))$$ will contain a ball of radius $$\epsilon$$ around $$f(\boldsymbol{x})$$. Or in simple words: there is some direction in the $$f$$-plane I can't step to from $$\boldsymbol{x}$$.

Can this mapping have a local boundary point? The Jacobian can definitely be rank deficient, but it is counterintuitive for such mapping to have a local boundary point.