On weak convergence

Question

I have the following Statement to prove

Let $C$ be a closed, bounded and convex subset of a $\mathbb{K}$-Vectorspace $X$. Define a Support function $S_C:X^*\rightarrow\mathbb{R}$, $f\rightarrow sup_{x\in C}f(x)$

To show is now:

i) $\forall y\in X:y\in C\iff \forall f\in X^*: f(y)\leq S_C(f)$

ii) $(x_n)_n\subset C$, $(x_n)_n\rightarrow x\in X$ weakly $\Rightarrow x\in C$

iii) Let detine the function $d_C: X\rightarrow \mathbb{R_\geq 0}, y\rightarrow d_C(y)= inf_{x\in C}||x-y||$. If X is reflexive $\Rightarrow \forall y\in X$ there exists the infimum.

I started the fist point proving this direction "$\Rightarrow$" but i've doubts obout the other one. I dont'see the argument that gives the conclusion.

At the second point I was thinking on a contradiction-proof trying to assume $\notin C$. Then since the sequence is weak convergent, I know that $f(x_n)\rightarrow f(x) \forall f\in X^*$ but then I don't know how to procede..

Any tips would be helpful. thanks

Answer 1

1. Obviously, $y \in C$ implies $f(y) \leq S_C(f)$ for any $f \in X^{\ast}$. Now we prove that whenever $y \notin C$ we can define $g \in X^{\ast}$ such that $g(y)> S_C(g)$. Without loss of generality, we may assume that $0 \in C$. As $C$ is closed, we know that $d:=d(y,C)>0$. Now we define the open blow-up of $C$ by $$C_d := C+B(0,d/2) := \{x; x=c+b, c \in C, b \in B(0,d/2)\}.$$ Note that $C_d$ is a convex and absorbing set such that $d(x,C_d) \geq d/2$. In particular, the Minkowski functional $$p_{C_d}(x) := \inf\{\lambda \in (0,\infty); x \in \lambda C_d\}$$ is a sublinear function satisfying $p_{C_d}(y)>1$. Moreover, as $B(0,d/2) \subseteq C_d$, it is not difficult to see that $$p_{C_d}(x) \leq 4\frac{\|x\|}{d}, \qquad x \in X \tag{1} $$ Now we set $f: \text{lin}\{y\} \to \mathbb{R}$, $f(\lambda y) := \lambda \cdot p_{C_d}(y)$. By the Hahn-Banach theorem, there exists $g:X \to \mathbb{R}$ linear such that $$g(x) \leq p_{C_d}(x), \qquad x \in X \tag{2}$$ and $g(y)=p_{C_d}(y)$. In particular, we see that $$g(x) \leq p_{C_d}(x) \leq 1 < p_{C_d}(y)=g(y)$$ for any $x \in C$ and $g \in X^{\ast}$ by $(1)$ and $(2)$. This finishes the proof. (If you know the supporting hyperplane theorem, the proof is much more easier.)
2. Let $(x_n)_{n \in \mathbb{N}} \subseteq C$ such that $x_n \to x$ weakly. By (i), we have $$x \in C \Leftrightarrow \forall f \in X^{\ast}: f(x) \leq S_C(f).$$ Now, as $x_n \to x$ weakly, we know that $f(x_n) \to f(x)$ as $n \to \infty$. Hence, $$f(x) = \lim_{n \to \infty} f(x_n) \leq S_C(f).$$
3. Since $X$ is reflexive, the unit ball in $X$ is weakly compact. By definition, we can choose a sequence $(x_n)_n \subseteq C$ such that $\|y-x_n\| \to d_C(y)$. Show that $(x_n)$ admits a weakly convergent subsequence and that the limit $x \in C$ satisfies $d(y,C) = \|x-y\|$.