Consider the space $\ell^p$ over the reals. Consider a subset of the unit ball
$$
A = \lbrace (x_n)_n^\infty \in \ell^p; \lVert x \rVert_p \leq 1 \ \& \ \lVert x \rVert_\infty \leq a \rbrace
$$
for $a > 0$, $p \in (1,\infty)$. Find the convex hull of extreme points of $A$.
WLOG $a\leq 1$.
I already know that the set of extreme points of $A$ is
$$
\text{ext} A = \lbrace (x_n)_n^\infty \in \ell^p; \lVert x \rVert_p = 1 \ \& \ \forall n: |x_n| \leq a \rbrace.
$$
I believe $\text{conv} \ \text{ext} A = A$.
Firstly $\text{conv} \ \text{ext} A \subset A$:
Pick $y^1, y^2, \ldots, y^j \in \text{ext} A$, $t_1, \ldots, t_j \geq 0$, $\sum_{n=1}^j t_n = 1$ and write $x$ as the convex combination of $y^1, \ldots, y^j$, i.e. $x = \sum_{n=1}^j t_n y^n$.
Then
$$\lVert x \rVert_p \leq t_1 \overbrace{\lVert y^1 \rVert_p}^{=1} + \ldots + t_j \overbrace{\lVert y^j \rVert_p}^{=1} \leq \sum_{n=1}^j t_n = 1.$$
It also holds
$$|x_n| \leq t_1 \overbrace{|y^1_n|}^{\leq a} + \ldots + t_j \overbrace{|y^j_n|}^{\leq a} \leq a \sum_{n=1}^j t_n = a.$$ Hence $x \in A$.
I do not know how to show $\text{conv} \ \text{ext} A \supset A$. The hint is to first assume $x$ with finite support (that is $ x = (x_n)_n, \lbrace n; x_n \neq 0 \rbrace$ is finite) and then the general case. I am not sure how to utilize this, since I can't think of an element which belong to $\text{ext} A$ for $a \leq 1$ arbitrary.
How could I proceed? Thank you.
Best Answer
We will show the remaining inclusion.
Pick $x \in A$, we will show that $ x \in \text{conv} \ \text{ext} \ A$.
We'll begin with the assumption that $x$ has finitely many non-zero coordinates, that is $x = (x_1, x_2, \ldots, x_m, 0 ,0 , \ldots)$. Since $x \in A$ we know that $\lVert x \rVert_p \leq 1$ and $\forall j: |x_j| \leq a$. If $\lVert x \rVert_p = 1$, then $x \in \text{ext} \ A \subset \text{conv} \ \text{ext} \ A$.
If $\lVert x \rVert_p < 1$, we can find $k \in \mathbb{N}_0$ and $\beta \in [0,a)$ such that $\lVert x \rVert_p^p +k a^p + \beta^p = 1$. Then we define $$ y= (x_1, x_2, \ldots, x_m, \underbrace{a, a, \ldots, a}_{k\text{-times}}, \beta, 0, 0, \ldots), $$ $$ w= (x_1, x_2, \ldots, x_m, \underbrace{-a, -a, \ldots, -a}_{k\text{-times}}, -\beta, 0, 0, \ldots). $$ Then $ \lVert y \rVert_p^p = \lVert x \rVert_p^p +k a^p + \beta^p = 1 = \lVert w \rVert_p^p $ and $\forall j: |y_j|, |w_j| \leq a$. Hence $y,w \in \text{ext} \ A$. At the same time $$ \frac{1}{2}y + \frac{1}{2}w = x. $$ This means $x \in \text{conv} \ \text{ext} \ A$.
Now we will assume $x \in A$ is general. Again, if $\lVert x \rVert_p = 1$, $x$ belongs to the convex hull of $\text{ext} \ A$. If $\lVert x \rVert_p <1$, we can find $m \in \mathbb{N}_0$ and $\beta \in \mathbb{R}$ such that $|x_{m+1}| \leq \beta < a$ (since $(x_n)$ converges to zero) and $$ m a^p + \beta^p + \sum_{j=m+2}^\infty |x_j|^p = 1. $$
Now denote $K = \overbrace{[-a,a] \times [-a,a] \times \ldots \times [-a,a]}^{m\text{-times}} \times [-\beta,\beta] \subset \mathbb{R}^{m+1}.$ Then $K$ is convex and compact (compact in usual norm topology, hence compact also in $\ell^p$ topology, since the dimension is finite). The set of extreme points is the set of vertices of this "cube", that is $$ \text{ext} \ A = \lbrace (\pm a, \pm a, \ldots, \pm a, \pm \beta) \rbrace, $$ where we consider all possible combinations of signs of $a$ and $\beta$. From Carathéodory's theorem $\text{conv} \ \text{ext} \ K = K$.
Since $x = (x_1, \ldots, x_{m+1}) \in K$, we can find $t_1, \ldots, t_{m+1} \geq 0$, $\sum t_j = 1$ and $y^1, \ldots, y^{m+1} \in \text{ext} \ K$ such that $$ x = \sum_{n=1}^m t_n y^n. $$
Now for $j \in \lbrace 1, \ldots, m+1 \rbrace $ denote $w^j = (y_1^j, y_2^j, \ldots, y_{m+1}^j, x_{m+2}, x_{m+3}, \ldots)$. Then $\forall n: |w_n^j| \leq a$ and $\lVert w^j \rVert_p^p = m |a|^p + |\beta|^p + \sum_{n=m+2}^\infty |x_n|^p = 1$. Thus $w^j \in \text{ext} \ A$ for all $j$.
At the same time \begin{align} t_1 w^1 + \ldots + t_{m+1} w^{m+1} &= \left( \sum_{j=1}^{m+1} t_j y_1^j, \ldots, \sum_{j=1}^{m+1} t_j y_{m+1}^j, \sum_{j=1}^{m+1} t_j x_{m+2}, \ldots \right) \\ &= (x_1, x_2, \ldots, x_{m+1},x_{m+2}, \ldots) \\ &= x. \end{align} Thus $x \in \text{conv} \ \text{ext} \ A$ which means $\text{conv} \ \text{ext} \ A = A$.