Minimization of a function on a convex subset of $\mathbb{R}^n$ Let $C \subset \mathbb{R}^{n}$ denote a non-compact convex subset of the Euclidean space $\mathbb{R}^n$ and $f : C \to \mathbb{R}^n$ a continuous function. What are the conditions to ensure that $f$ has a minimum on $C$ ?
If $C$ was compact, it would be OK.
If $f$ is convex, I guess something could be said.
Consider $C = ]0,+\infty[$ and $f(x) = x - \ln(x)$ for all $x \in C$. Then $f$ satisfies :
$$ \lim \limits_{\substack{x \to 0 \\ x > 0}} f(x) = \lim \limits_{x \to +\infty} f(x)  = +\infty \tag{$\star$} $$
$f$ is continuous (and convex) on $C$. The conditions $(\star)$ are enough to ensure that $f$ has a minimum on $C$. Indeed, by ($\star$), one can find $m,M \in C$ such that $f$ is arbitrarily large on $C \smallsetminus [m,M]$. By continuity, since $[m,M]$ is compact, $f$ has a minimum on $[m,M]$ and therefore on $C$. 
Does that follow from a theorem I don't know ? How would $(\star$) generalize in higher dimension (say, for $C = ]0,+\infty[^n \times \mathbb{R}^p$) ?

You need better control of the domain/function to say something. On $I=(0,+\infty)$ then function $x\mapsto 1/x-x$ is convex but has no lower bound. On a bounded domain a convex function will have a finite infimum but need not attain that inf in the interior (easy counter examples).
Concerning condition (*): If you happen to know that there is a compact subset $K\subset C$ so that $\inf_{C\setminus K} f > f(a)$ for some $a\in K$ then you are back in business (like in your example). $f$ need not be convex, nor does the domain.
