Consider the space $\ell^p$ over the reals. Denote
$$
A = \lbrace x \in \ell^p; \lVert x \rVert_p \leq 1 \ \& \ \lVert x \rVert_\infty \leq \alpha \rbrace.
$$
For $p \in (1, \infty)$, $\alpha > 0$, find the set of extreme points of $A$.
I think the set of extreme points is
$$
B = \lbrace (x_n)_n \in S_{\ell^p}; \exists j: |x_j| = \alpha \rbrace
$$
where $S_{\ell^p}$ is the unit sphere. From $\lVert x \rVert \leq 1$ we have that $\alpha \leq 1$.
$\text{ext}A \supset B$
Pick $x \in B$, $y,z \in A$ and assume $x = \frac{1}{2}(y+z)$. Then
$$
1 = \lVert x \rVert = \frac{1}{2}\lVert y+z \rVert \leq \frac{1}{2} \left( \lVert y \rVert + \lVert z \rVert \right) \leq 1
$$
hence we have equalities.
For $j$ such that $|x_j| = \alpha$ we have
$$
\alpha = |x_j| = \frac{1}{2} |y_j + z_j| \leq \frac{1}{2}(|y_j| + |z_j|) \leq \alpha
$$
Again, we have equalities and $x=z=y$.
$\text{ext}A \subset B$
Pick $x \in A \setminus B$. Then $\forall j : |x_j| \neq \alpha$ and $ \lVert x \rVert_p < 1$, hence $\forall n: x_n \in (-1,1)$. We can find $\varepsilon > 0$ such that $[x_n – \varepsilon, x_n + \varepsilon] \subset [-1,1]$. Now define $y,z$ as $y_n = x_n – \varepsilon, z_n = x_n + \varepsilon$. Then $y,z \in A, y \neq z$ but $\frac{1}{2}(y+z) = x$. Hence $x \not\in \text{ext}A$.
Is this correct? Thank you.
Best Answer
The set of extreme points of $A$ is equal $$E:=\{x\in \ell^p\,:\, \|x\|_p=1, \ |x_n|\le a\}$$ Indeed every element $x$ in $E$ is an extreme point of the unit ball in $\ell^p.$ As $A$ is a subset of the unit ball, the element $x$ is an extreme point of $A.$
It remains to show that every extreme point of $A$ belongs to $E.$ To this end we will show that if $(x_n)$ belongs to $A$ and $\|x\|_p<1,$ then $x$ is not an extreme point of $A.$ Indeed, let $F=\{n\in \mathbb{N}\,:|x_n|=a\}.$ As $x_n\to 0$ we have $$m:=\inf_{n\notin F}(a-|x_n|) >0$$ Hence $$|x_n|\le a-m,\quad n\notin F$$ For $L\ge 1$ let $$y_n=\begin{cases}0 & n\in F \\ \displaystyle {m\over 2^{n/p}L} & n\notin F\end{cases}$$ Then $x+y$ and $x-y$ belong to $A.$ Indeed for $n\notin F$ we have $$|x_n\pm y_n|\le |x_n|+y_n\le (a-m)+m=a$$ $$\|x\pm y\|_p\le \|x\|_p+\|y\|_p\le \|x\|_p+\left (\sum_{n=1}^\infty {m^p\over 2^{n}L^p} \right )^{1/p}=\|x\|_p+{m\over L} $$ For $L$ large enough we obtain $\|x\pm y\|_p\le 1.$ Eventually we get $$x={1\over 2}(x+y)+{1\over 2}(x-y)$$ thus $x$ is not an extreme point of $A.$