# Infimum of a continuous and convex functional in a convex subset of Hilbert space can be attained

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. $$F(x_n)\to \inf\{F(x):x\in K\}$$?

• Unless $K$ is closed, the statement is false. – Harald Hanche-Olsen May 8 '17 at 17:30
• @HaraldHanche-Olsen Yeah I forgot to mention K is closed – SHBaoS May 8 '17 at 17:35
• – nigel May 8 '17 at 17:38

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.