Why must a function be closed for descent methods to work?

Question

At the beginning of chapter 9 of Boyd & Vandenberghe's Convex Optimization, about unconstrained minimization, it is said:

The starting point $x^{(0)}$ for a method must lie in $\mathbf{dom} f$, and in addition the sublevel set $$S = \{x \in \mbox{dom} \, f \mid f(x) \leq f(x^{(0)}) \}$$ must be closed. This is satisfied for all $x \in \mathbf{dom} f$ if the function is closed.

In other paper I've found an assumption, that a function $f$ must be convex, proper and closed in order that: minimize f is solvable. I understand why are the first two assumptions needed, but why is it important for the function to be closed so that problem is solvable? I would appreciate any explanation for the intuition (and what does this mean for the epigrpah for the objective function).

Answer 1

Here's an example of what can go wrong. Consider the domain

$D=\left\{x \in R^{2} | x_{2} > 0 \right\} \cup (0,0)$

Note that $D$ is convex.

Let

$f(x)=x_{2}$

Here, $f$ is convex on $D$, and the unique minimum of $f$ over $D$ is at the origin. However, the level sets of $f$ are not closed.

Now, apply the method of gradient descent to $f$, starting at the point $(1,1)$. With appropriate step lengths, you can get the method to converge to $(1,0)$, which is outside of $D$ and nowhere near the minimizer at the origin.

Answer 2

Another example is minimizing $f(x) = 1/x$. The minimum exists, and is 0, but the minimizer does not exist, and most standard methods behave oddly on this problem. The function's epigraph is

$$ \mathbf{epi}~f = \{(x,f(x)): x\in\mathbb R\} $$ and does not include the point $(x,0)$ for any value of $x$ (but the sequence $(n,f(n))$ is in the epigraph and approaches the x axis). Therefore the epigraph is not closed, and the function is not closed.