Extreme points of the ell-infinity unit ball

Suppose $x$ lies in the closed unit ball and $\mathrm{\mid }{x}_N\mathrm{\mid }<1$ for some $N$. Over $\mathbb{R}$, choose $0<\epsilon \le 1-\mathrm{\mid }{x}_N\mathrm{\mid }$ and put $u_N=1$; over $\mathbb{C}$, put $u_N=1$ if $x_N=0$ and $u_N=i{x}_N\mathrm{/}\mathrm{\mid }{x}_N\mathrm{\mid }$ otherwise, and choose $0<\epsilon \le \sqrt{1-\mathrm{\mid }{x}_N{\mathrm{\mid }}^2}$. With $e$ supported at $N$ and equal there to $u_N$, both $x+\epsilon e$ and $x-\epsilon e$ have sup norm at most one, are distinct, and have midpoint $x$. Thus $x$ is not extreme by [F1].

Conversely suppose $\mathrm{\mid }{x}_n\mathrm{\mid }=1$ for all $n$ and $x=(1-t)y+tz$ with $y,z$ in the unit ball and $0<t<1$. For each $n$, the scalar identity $\mathrm{\mid }(1-t)y_n+t{z}_n{\mathrm{\mid }}^2=(1-t)\mathrm{\mid }{y}_n{\mathrm{\mid }}^2+t\mathrm{\mid }{z}_n{\mathrm{\mid }}^2-t(1-t)\mathrm{\mid }{y}_n-z_n{\mathrm{\mid }}^2$ gives $1\le 1-t(1-t)\mathrm{\mid }{y}_n-z_n{\mathrm{\mid }}^2$ by [F2]. Hence $y_n=z_n$, and their convex combination equals $x_n$, so $y_n=z_n=x_n$ for every $n$.

Step 1.1 excludes exactly the sequences with an interior coordinate, while step 1.2 and [F1] prove every sequence with all coordinates on the scalar unit circle is extreme.