Extreme Points of the Closed Unit Ball of a Normed Linear Space

# Extreme Points of the Closed Unit Ball of a Normed Linear Space

Recall from the Extreme Subsets and Extreme Points of a Set in a LCTVS page that if $X$ is a locally convex topological vector space and $K$ is a nonempty convex subset of $X$ then $E \subseteq K$ is said to be an extreme subset of $K$ if:

- 1. $E$ is nonempty.

- 2. $E$ is closed.

- 3. $E$ is convex.

- 4. Whenever $x = \lambda u + (1 - \lambda )v$ where $u, v \in K$ and $\lambda \in [0, 1]$ then $u, v \in E$.

Furthermore, a point in $x \in K$ is said to be an extreme point of $K$ if $\{ x \}$ is an extreme subset of $K$.

The following proposition tells us that if $X$ is a normed linear space then the extreme points of the closed unit ball $x \in B_X$ of $X$ are such that $\| x \| = 1$.

Proposition 1: Let $X$ be a normed linear space and let $B_X$ denote the closed unit ball of $X$. If $x \in B_X$ is an extreme point of $B_X$ then $\| x \| = 1$. |

**Proof:**Let $x \in B_X$ be an extreme point of $B_X$. Since $x \in B_X$ we have that $\| x \| \leq 1$.

- Suppose that $\| x \| = 0$. Then $x = 0$. By taking $y \in B_X$ with $\| y \| = 1$, we can consider the line segment joining $y$ and $-y$ given by the equation:

\begin{align} \quad \tilde{x} = \lambda y + (1 - \lambda) (-y) \:, \quad \lambda \in [0, 1] \end{align}

- Observe that when $\lambda = 1/2$ we have that $\tilde{x} = 0 = x$. But clearly $y, -y \not \in \{ x \}$ and so $\{ x \}$ is not an extreme subset of $B_X$, i.e., $x$ is not an extreme point of $B_X$ - a contradiction. So the assumption that $\| x \| = 0$ is false. Thus $0 < \| x \| \leq 1$.

- Now suppose that $0 < \| x \| < 1$. We can consider the line segment joining $0$ and $\displaystyle{\frac{x}{\| x \|}}$ given by the equation:

\begin{align} \quad \tilde{x} = \lambda \frac{x}{\| x \|} \: , \quad \lambda \in [0, 1] \end{align}

- Observe that when $\lambda = \| x \| \in (0, 1)$ we have that $\tilde{x} = x$. But $0, \frac{x}{\| x \|} \not \in \{ x \}$, again contradicting $x$ being an extreme point of $B_X$. Thus the assumption that $0 < \| x \| < 1$ is false.

- Therefore, if $x$ is an extreme point of $B_X$ then $\| x \| = 1$. $\blacksquare$