TITLE: Existence of minimizer for strongly convex function on closed, convex set
QUESTION [8 upvotes]: Let $f : \mathbb{R}^n \to \mathbb{R}$ be strongly convex. Consider the problem
$$\min_{x \in A} f(x)$$
where $A \subseteq \mathbb{R}^n$ is nonempty, convex, and closed. Also assume $\inf_{x \in A} f(x) < \infty$ (if this is not the case, then $f(x) = \infty$ for all $x \in A$, which is not a particularly interesting scenario).
I know that if $f$ has a minimizer, on $A$, then it is unique (see this thread for an explanation of why). My question is are these conditions sufficient to guarantee the existence of such a minimizer? If not, what else would need to be assumed to imply this?
EDIT: Here is the definition of strongly convex (from Wikipedia): $f$ is strongly convex if there exists a constant $m > 0$ such that for all $x$ and $y$ in the domain and $\lambda \in [0,1]$ the following inequality holds:
$$f(\lambda x + (1-\lambda)y) \leq \lambda f(x) + (1-\lambda) f(y) - \frac{1}{2} m\lambda (1-\lambda) \| x - y \|^2$$
This is equivalent to the following:
$$g := f - \frac{m}{2}\|\cdot\|^2$$
is convex, where $\|\cdot\|$ is induced by an inner product space. See this paper for more on the equivalence of these definitions.

REPLY [2 votes]: Strong convexity alone is not enough to guarantee existence of a minizer. To see this consider the function $f:[0,1] \to \mathbb{R}$,
$$
f(x) =
\begin{cases}
x^2 & \text{if $x > 0$,} \\
1 & \text{if $x=0$.}
\end{cases}
$$
This function is $2$-strongly convex, i.e. it satisfies
$$
f(tx+(1−t)y) \le tf(x)+(1−t)f(y) − t(1−t)(x−y)^2
$$
for all $x,y,t \in [0,1]$. However, $f(x)$ doesn't have a minimizer.
This means that an extra assumption on $f$ is required in order to guarantee the existence of a minimizer.
Edit: The standard extra assumption that is often made is that the function is lower semi-continuous. In that case, the minimizer exists and is unique. The example above is NOT lower semi-continuous.
Edit2: Convex function $f:\mathbb{R}^n \to \mathbb{R}$ is necessarily continuous on the whole domain $\mathbb{R}^n$. So my counter-example works only if the domain of the function is a strict subset of $\mathbb{R}^n$.