Existence of a minimum for a convex function

Question

Let $f$ be a convex function on $\mathbb{R}^n$ and assume that for all $s$ on the unit sphere, $$\lim_{c \to \infty} f(c s)=\infty.$$ Does this imply that $f$ admits a minimum? Note that the condition above is weaker than coercivity, which does imply the existence of a minimum.

Answer 1

A friend gave me the solution. It suffices to prove that $f$ is coercive, namely that for all $A>0$, there exists $B>0$ such that $\|x\|>B$ implies $f(x)>A$. Suppose $f$ is not coercive. Then, there exists $A>0$ and a sequence $(s_n)_{n\ge 1}$ on the unit sphere $\mathbb{S}$ such that $f(ns_n)< A$. Without loss of generality, assume $A>f(0)$. Since $(s_n)_{n\ge 1}$ belongs to a compact set, it admits a converging subsequence; let $s_\infty$ denote its limit. By assumption, there exists $n_0\ge 1$ such that $f(n_0 s_\infty)>A+2$. Since $f$ is convex, it is continuous and thus $s\mapsto f(n_0s)$ is also continuous on $\mathbb{S}$. As a result, there exists a neighborhood $V\subset \mathbb{S}$ of $s_\infty$ such that for all $s\in V$, $f(n_0 s)>A+1$. By definition of $s_\infty$, there also exists $n>n_0$ such that $s_n\in V$. Then, $f(n_0 s_n)>A+1$. Now, because $n_0s_n = (n_0/n) \times (n s_n) + (1-n_0/n)\times 0$, we have, by convexity, $A +1 < f(n_0 s_n) \le \max(f(n s_n), f(0)) < A$, a contradiction.