Let $K$ be a non-empty closed convex subset of a real Hilbert space $X$. Let $F:K \to \mathbb{R}$ be a continuous and convex linear functional s.t. if $K$ is unbounded, then $\lVert x_n \rVert\to \infty$ implies $F(x_n)\to \infty$.

Prove there is a $x_0 \in K$ s.t. $F(x_0)\leq F(x)$ for all $x\in K$

I have no intuitive if $K$ is bounded. Can anyone give me some hints?

Maybe construct a sequence $x_n$ in $K$ s.t. $F(x_n)\to \inf\{F(x):x\in K\}$?


For any constant $c$, $F^{-1}((-\infty,c]) = \{x \in K: F(x) \le c \}$ is closed, convex and bounded. It is weakly closed by the Hahn-Banach separation theorem. By Banach-Alaoglu, it is weakly compact. For a sequence $c_n$ decreasing to $\inf_K F$, $F^{-1}((-\infty, c_n])$ are nested weakly compact sets, and by the Finite Intersection Property they have a nonempty intersection.


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