In a Reflexive banach space, given a closed convex set $C$ and some point $y$, there is a point in $C$, of minimal distance to $y$ In a Reflexive space, given a closed convex set $C$ and some point $y$, there is a point in $C$, of minimal distance to $y$
All i could figure out is that there is a sequence $z_n \in C$ z.t $\|y-z_n\|\to min distance$ and that $z_n$ converges weakly as it is a bounded sequence. I am not sure where to go from here
 A: You are on the right track. Since $C$ is not only closed but also convex, it is weakly closed. The weak limit of the sequence $(z_n)_{n \in \mathbb{N}}$, lets call it $z$, therefore satisfies $z \in C$. Can you show that $z$ indeed minimizes the distance using that the norm is (sequentially) weakly lower semi-continuous?
A: A subset $C$ of X which  has the property that for every $x \in X$, the distance $\operatorname{dist}(x,C)$ is attained (i.e there exists a $c \in C$ s.t $||x-c||=\operatorname{dist}(x,C)$) is called proximinal.
We shall show that if $C$ is a closed and bounded subset of a reflexive space $X$, then $C$ is proximinal. Let $x \in X$ and  $d=\operatorname{dist(x,C}):= \inf \{ ||{x-c}|| \colon c \in C\}$. By the definition of the infimum, for all $n \in \mathbb N$, we can find a $ z_n \in B_{d+1/n}(x) \cap C := 
\{ c \in C \colon ||x-c|| \leq d+1/n\}$ (so the intersection is not empty). Define the sequence of subsets  $(A_n)_n \subset 2^X$, by $A_n = B_{d+1/n}(x) \cap C$. Notice that $B_{d+1} \supset A_1 \supset A_2 \supset \dots \ $ and $B_{d+1}$ is weakly compact (since X is reflexive). Furthermore, one can easily check that the famlily $(A_n)_n$ has the finite intersection property and so, we can find a $z \in \bigcap_{n \in \mathbb N} B_{d+1/n}\cap C$. The point $z$ is the required.
In fact, it can be shown that a normed space X is reflexive if and only if every closed and convex subset of X is proximinal. (The opposite direction is due to James' theorem.)
