every denting point and strongly exposed point is extreme point

Question

If $X$ be a Banach space and $K$ is a subset of $X$, then I want to prove

1. Every denting point of $K$ is extreme point

2. Every strongly exposed point of $K$ is extreme point

3. $K$ is the closed convex hull of its denting points and inparticular $K$ is is the closed convex hull of its strongly points

please help me to answering above problem.

The point $x_{0} \in K$ is called exposed point, if there exists a linear functional $x^{*}\in X^{*}$ s.t $x^{*}(x_{0})>x^{*}(x)$, for any $x \in K \setminus \{x_{0}\}$.

The point $x_{0} \in K$ is called strongly exposed point, if there exists a linear functional $x^{*}\in X^{*}$ s.t for every $\epsilon>0$, there is $\delta>0$ s.t the slice $S(x^{*},\delta)$ contains $x$ and has diameter less than $\epsilon$.

The point $x_{0} \in K$ is called denting point if for every $\epsilon>0$, $$x \notin \overline{con}(K \setminus B_{\epsilon}(x)) $$.

Answer 1

What are the hypothesis on $K$?

Observe that a strongly exposed point is a denting point and a denting point is an extreme point. So, for instance take the set $K$ as the unit ball of $c_0$. This set is convex and closed but it has no extreme point. Hence, $K$ can not be the closed convex hull of any kind of extreme points of $K$.

Items 1 and 2 can be shown by "reductio"