Let $\mathrm{epi}(f)=\{(x,\alpha)\in \mathbb{R}^{n+1}| f(x)\le\alpha\}$
Prove or disprove:
If $f$ is coercive and $\mathrm{epi}(f)$ closed, the problem
$\min f(x), x\in \mathbb{R}^n$
has at least one solution.
If $f$ is coercive, then $\lim_{\|x\| \rightarrow \infty} f(x)=\infty$. But I don't know how this helps me. Thanks for any hint.
Hint:
This is the corollary of Bolzano–Weierstrass theorem.
If $f$ is optimized over an unbounded set, then for a coercive function $f$, we can take a bounded interval say $\mathcal{B}$, such that $\lvert f(x) \rvert \leq a \ \forall x \in \mathcal{B}$. Now, we can use the Bolzano–Weierstrass theorem to prove the existence of at least one infimum.
To prove the attainability of the infimum, $f$ has to be lower semi-continuous. If the epigraph of $f$ is closed, then $f$ is lower semi-continuous. A deligant prove can be found here.