Consider Banach space $X$ with unit ball $B_X$. There is the cannonical embeding $\iota:X\to X^{**}$. Does this always preserve extreme points of $B_X$? I.e., is it always true that $\iota(\partial_e(B_X))\subset \partial_e(B_{X^{**}})$?
I tried to prove it by having $x\in\partial_eB_X$ and $y,z\in B_{X^{**}}\setminus\{x\}$ such that $\frac{y+z}2=x$ and looking at the sequances $(y_k),(z_k)\subset B_X$ weakly-* converging to $y$ and $z$, but I can't seem to squeeze the contradiction out of it.
Also, by Milman's theorem, it is obvious for isolated points of $\partial_eB_X$, but that really don't need to be all of them…
Best Answer
Here is a short description of the idea in the paper of Morris: Disappearance of extreme points. I didn't work up all details. If anyone would like to elaborate further (or finds a mistake in my interpretation), feel free to edit my answer.
Consider $X=c_0$. The main effort is to construct an injective continuous linear operator $T:c_0 \to l^2$ such that $T^{\ast \ast}: l^\infty \to l^2$ has an infinite dimensional kernel $N(T^{\ast \ast})$. Then $c_0$ is equivalently renormed by $\|x\|_T:=\|x\|_\infty + \|Tx\|_2$. Now $(c_0,\|\cdot\|_T)$ is strictly convex (hence each boundary point of the unit ball is an extreme point, whereas the unit ball wrt $\|\cdot\|_\infty$ has no extreme points). The norm in $(c_0,\|\cdot\|_T)^{\ast \ast}$ is then given by $x \mapsto \|x\|_\infty + \|T^{\ast \ast}x\|_2$ $(x \in l^\infty)$, and is also denoted by $\|\cdot\|_T$.
Now let $x =(x_n)_{n=1}^\infty\in c_0$ with $\|x\|_T=1$. We show that $x$ is not an extreme point of the unit ball in $(c_0,\|\cdot\|_T)^{\ast \ast}$. Let $m \in \mathbb{N}$ be such that $|x_n| \le \|x\|_\infty/2$ $(n \ge m)$. Note that $m >1$. Since $N(T^{\ast \ast})$ is infinite dimensional there is a nonzero element $y =(y_n)_{n=1}^\infty\in N(T^{\ast \ast})$ with $y_n =0$ $(n < m)$. Scale $y$ such that in addition $\|y\|_\infty \le \|x\|_\infty/2$. Then $$ \|x \pm y\|_\infty = \|x\|_\infty, \quad \|T^{\ast \ast}(x \pm y)\|_2 = \|T^{\ast \ast}x\|_2 = \|Tx\|_2. $$ Hence $\|x \pm y\|_T = \|x\|_T=1$, and so $x$ is not an extreme point.