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How to prove that the set of extreme points of $B_{\ell^1} = \{v \in \ell^1 : \| v \| \le 1\}$ is $\{ +e^N, -e^N : N=1,2,3,\ldots \}$, where $e^N$ denotes the Nth standard basis element in $\ell_1$: $e_n^N=0$ if $n\neq N$ and $e_N^N =1$.

Also, is $B_{\ell^1}$ the norm closure of the convex hull of its set of extreme points? How to prove this? I don't know how to do it. Please help.

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First, if $e^N = \frac 12(x+y)$ with $x,y \in B_{\ell^1}$, then $1 = e^N_N = \frac 12(x_N + y_N)$ with $|x_N|, |y_N| \le 1$. Therefore $x_N = y_N = 1$ and since $\|x\|_1, \|y\|_1 \le 1$ we have $x=y= e^N$. So all $e^N$ are extreme points of $B_{\ell^1}$, so are $-e^N$.

Otherwise, if $x \in S_{\ell^1}$ with $x \ne \pm e^N$ for all $N$, there are $n \ne m$ with $x_n, x_m \ne 0$ and therefore $0 < |x_n|, |x_m| < 1$, let $\eta := \min\{|x_n|, |x_m|, 1 - |x_n|, 1-|x_m|\}$ and set $y = x + \eta e^n - \eta e^m$ and $z = x - \eta e^n + \eta e^m$. Then $y,z \in B_{\ell^1}$, $x = \frac 12(y+z)$ and $y\ne z$. So $x$ is not extreme.

No let $x \in B_{\ell^1}$. We have \[ \sum_{i=1}^n x_i e^i = \sum_{i=1}^n |x_i|\cdot \mathrm{sgn} x_i\cdot e_i + \frac 12\left(1 - \sum_{i=1}^n |x_i|\right)e^{1} + \frac 12 \left(1 - \sum_{i=1}^n |x_i|\right)(-e^{1}) \] so $\sum_{i=1}^n x_i e^i \in\mathrm{conv}\mathrm{ext} B_{\ell^1}$ for each $n$. But $\sum_{i=1}^n x_ie^i \to x$ in norm, so we are done.

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