Set of extremal points of closed unit ball in $\mathbb{C}^n$

Question

As an exercise to familiarise myself with the concept of extremal points, I'm trying to determine the set $S$ of extremal points for the closed unit ball with respect to the $p$-norm, $p \in \mathbb{N}$: $$ \overline{\mathbb{B}} = \left\{z \in \mathbb{C}^n : \Vert z \Vert_p = \left(\sum_{j = 1}^n |z_j|^p\right)^{\frac{1}{p}} \leq 1 \right\}.$$ I already know that $S \subset \partial\mathbb{B}$ by a general fact for extremal points, and I think I know what to do for the case of $p = 2$. In this case, $S = \partial\mathbb{B}$ because the $2$-norm is given by an inner product which allows for the Cauchy-Schwarz inequality to hold.

However, for $p \neq 2$, I'm not sure what to do. I think for $p = 1$, there is a strict inclusion and that the extremal points are exactly of the form $a_j = \exp(i\theta_j)e_j$, where $\theta_j \in \mathbb{R}$ and $e_j$ is the $j$-th standard basis vector in $\mathbb{C}^n$, but I'm stuck verifying this. I also think that $S = \partial \mathbb{B}$ for all $p \geq 2$, because my intuition says that it is not possible to have a non-degenerate straight line segment that will solve the equation $$\sum_{j = 1}^n |z_j|^p - 1 = 0.$$ Are my guesses right and could I get some suggestions on what to try? Thanks!

Edit: Thanks to Robert's hint in the comments, I managed to figure out the case of $p > 1$. Let $c \in \partial\mathbb{B}$ and suppose there are $a, b \in \overline{\mathbb{B}}$ such that $c = (1 - t)a + tb$ for some $t \in (0, 1)$. Then the triangle inequality implies that we necessarily have $a, b \in \partial\mathbb{B}$. In particular, $$1 = \Vert (1 - t)a + tb \Vert_p \leq (1 - t)\Vert a \Vert_p + t\Vert b \Vert_p = 1$$ so that the inequality turns into an equality. Minkowski's inequality then implies that there is $\lambda > 0$ such that $(1 - t)a = \lambda tb$. (See this post.) Taking norms on both sides implies $a = b$ and hence that $c = a = b$. So $c \in S$, which shows $\partial \overline{\mathbb{B}} \subset S$.

However, I'm still not sure about the case for $p = 1$. Again, any suggestions would be greatly appreciated!

Answer 1

I believe I have managed to figure out the case of $p = 1$, but my solution ended up being tediously long. However, I still think it is worth at least putting a sketch solution just so this question sits on the answered list.

The extremal points of $\overline{\mathbb{B}}$ with respect to the $1$-norm is indeed the set $\{\exp(i\theta)e_j : j = 1, \dots, n, \theta \in \mathbb{R}\}$. To see this:

1. We may assume that $n \geq 2$, as for $n = 1$, all the $p$-norms coincide on $\mathbb{C}$. Supposing that $z \in \partial\mathbb{B}$ has at least two non-zero coordinates, say $z_1$ and $z_2$, then we set $$a = \frac{z_1}{|z_1|}e_1 \quad \text{and} \quad b = \frac{1}{1 - |z_1|}(z - z_1e_1).$$ Here, $0 < 1 - |z_1| < 1$ and $a, b \in \partial\mathbb{B}$, so that by choosing $t = 1 - |z_1|$, we get $z = (1 - t)a + tb$. This says that any point on the boundary of $\mathbb{B}$ that has at least two non-zero coordinates cannot be an extremal point.
2. We still need to verify that $\{\exp(i\theta)e_j : j = 1, \dots, n, \theta \in \mathbb{R}\}$ are extremal points. To do this, one assumes that if any of these points can be written in the form $(1 - t)a + tb$ for some $a, b \in \overline{\mathbb{B}}$ and $0 < t < 1$, then the triangle inequality implies that $a$ and $b$ must lie on the boundary of $\mathbb{B}$. In fact, the triangle inequality implies that $a$ and $b$ necessarily have one non-zero coordinate, and that this coordinate must be the same coordinate as the non-zero coordinate of $(1 - t)a + tb$. By passing through the $1$-norm again, one deduces that the non-zero coordinates of $a$ and $b$ must be positively linearly dependent (in a similar argument for $p > 1$). This will imply that $a = b$, and hence that the set of extremal points for $\overline{B}$ in the $1$-norm is exactly $\{\exp(i\theta)e_j : j = 1, \dots, n, \theta \in \mathbb{R}\}$.