Let's look at the constrained optimization problem:
$ \displaystyle \min_{\mathbf{x} \in S} f(\mathbf{x})$
I have looked through several optimization textbooks and in all of them the relationship between the feasible set $S \subset \mathbb{R}^n$ and the domain of the objective function $f(\mathbf{x})$ is described rather poorly. These books just say that $f$ should be defined on $S$ or on some set which contains $S$ (i.e. $S \subseteq \operatorname{dom} f \subset \mathbb{R}^n$). The problem with these textbooks is that they have several theorems which look like
If the objective function $f(\mathbf{x})$ is continuously differentiable on the feasible set $S$, then ... (some statement)
Sometimes these theorems have some additional requirements on the feasible set $S$ (convex set/close set/open set, etc.). And sometimes they require $f$ to be convex or to be twice continuously differentiable on $S$.
I want to know what does the following statement usually mean in optimization textbooks: "the objective function $f(\mathbf{x})$ is continuously differentiable on the feasible set $S$" ?
The problem is that the set $S$ is not necessarily open, and in my calculus textbooks there is a definition of continuous differentiability only for the case of an open set. So I am highly confused now.
My guess is that optimization textbooks (at least for undergraduate students) implicitly assume two things:
- $S \subseteq \operatorname{int} \operatorname{dom} f \subset \mathbb{R}^n$
- The statement "the objective function $f(\mathbf{x})$ is continuously differentiable on the feasible set $S$" means that $f(\mathbf{x})$ is continuously differentiable in every point of $S$ (i.e. for each point $\mathbf{x} \in S$ we have that all first-order partial derivatives of $f$ are defined in some its neighborhood $U(\mathbf{x})$ and are continuous in $\mathbf{x}$).
Is my guess correct?