Lately, I've been studying optimization and came across with the following problem formulation:
$$ \text{min} \;f_0(x) \\ \text{subject to} \begin{align}\;\;f_i(x) &≤ 0, \;\; i = 1, ..., m \\ \;\; h_i(x) &= 0, \;\;i = 1, ..., p\\ \end{align} $$ and the domain of the problem is defined to be
$$ D = \bigcap^m_{i=0} D_{f_i} \cap \bigcap^p_{i=0} D_{h_i} $$
Now, for a convex optimization problem, it is said that all $f_i$ must be convex and $h_i$ must be affine transformations. Then it suggests that the domain of the problem is convex since convex sets are closed under intersection.
If $f_i$ is convex, then $D_{f_i}$ is a convex set by definition. But what about affine transformations? Are the domains of affine transformations also convex by definition? Let's say I have an affine function $y = 3x + 2$ defined on $[-1, 2) \cup (3, 5]$. Domain is not convex, does it suggest that my function is not affine? If this is the case, can you please provide a reference to a reliable definition of affine functions?
For example for linear transformations, we can say that the definition implies convexity of the domain. Otherwise $T(\lambda u + (1 - \lambda) v) = \lambda T(u) + (1 - \lambda) T(v)$ doesn't hold for all $u, v \in D_T$. Is there such a property of affine transformations we can use to achieve convexity?
Thanks
The function $f$ is affine if and only if $$ f(\lambda x + (1-\lambda)y) = \lambda f(x) + (1-\lambda) f(y) $$ for all $x,y$ in the domain of $f$ and $\lambda\in (0,1)$.
The domain of an affine function has to be a affine subspace, otherwise the definition makes no sense. Affine subspaces are convex.
You can view it also differently: a function $f$ is affine iff $f$ and $-f$ are convex.