Existence of Optimal Solution

Question

Assume we are solving $\min\{f(x) \ | \ x \in S \}$.

If $f: \mathbb{R}^n \mapsto \mathbb{R}$ is a proper closed convex function, and $S$ is a non-empty closed convex set, does this imply that the above minimization problem has a non-empty solution set? Does an optimal solution always exist in such a setting?

What I know is that (by Wikipedia) a proper convex function is closed if and only if it is lower semi-continuous. Moreover, if I know $f$ is proper closed convex, then this implies the function is lower semi-continuous. By the extension of extreme value theorem to semi-continuous functions, we know that the above minimization has a non-empty solution set since $f$ is lower semi-continuous. So maybe this is it...

Edit: Based on the answer to this question below, I see that such an optimal value always exists. However, in Amir Beck's first-order methods book, there is the following set of assumptions:

Then don't (A,B,C) imply (D) anyways? Why are we also assuming (D) at all?

Answer 1

No, an optimal solution need not exist. Take $f: \mathbb{R} \to \mathbb{R}$ with $f(x) = e^{x}$.

However if you restrict $S$ to be compact instead of just closed, then you are guaranteed a solution. In fact, convexity is not required. For a simple proof, let $f(x_n) \to \inf_{x \in S} f(x)$. $x_n \in S$ has a convergent subsequence by compactness, and let $x \in S$ be its limit point. By lower semicontinuity $x$ minimizes $f$ on $S$. Lastly, note that the infimum above must be finite because $f$ is proper, hence $f(x) \neq \pm \infty$.

Answer 2

An optimal solution exists iff the KKT are satisfied.

As Mark Stone pointed out in the comments, the brief does not give us any guarantees on continuous differentiability, which is necessary for KKT. Without (D), there might or might not be KKT points, we can't tell. (D) imposes that we have at least one KKT point.