Prove that isometries are extreme points of the unit ball of the space of bounded operators I have a request for any ideas to prove:


*

*If $H$ is a Hilbert space, then any unit vector is an extreme point of the unit ball of $H$.

*Every isometry is an extreme point of the unit ball of the Banach space of bounded operators on $H$.


I suspect that 2. is an application of 1., but I've tried inequalities without success.
Thanks.
 A: Assume that $p\in\operatorname{Sphere}_H(0,1)$ is not an extreme point, then there exist $x,y\in\operatorname{Sphere}_H(0,1)$ such that $x\neq y$ and $p=tx+(1-t)y$ for some $t\in(0,1)$. From this answer we infer that 
$$
\operatorname{Re}\langle x,y\rangle\leq\langle x,y\rangle<\Vert x\Vert\Vert y\Vert=1
$$
$$
1=\Vert p\Vert^2=
\Vert tx+(1-t)y\Vert^2=t^2\Vert x\Vert^2+(1-t)^2\Vert y\Vert+2t(1-t)\operatorname{Re}\langle x,y\rangle\\
<t^2+(1-t)^2+2t(1-t)=1
$$
Contradiction, hence $p$ is an extreme point.
Assume an isometry $I\in\operatorname{Sphere}_{\mathcal{B}(H)}(0,1)$ is not an extreme point, then there exist $S,T\in \operatorname{Sphere}_{\mathcal{B}(H)}(0,1)$ such that $S\neq T$ and $I=tT+(1-t)S$ for some $t\in(0,1)$. Consider arbitrary $p\in\operatorname{Sphere}_H(0,1)$ and define $x=S(p)$, $y=T(p)$, then
$$
I(p)=tx+(1-t)y
$$
By definition of $S$ and $T$ we have $\Vert x\Vert\leq\Vert S\Vert\Vert p\Vert\leq1$ and $\Vert y\Vert\leq\Vert T\Vert\Vert p\Vert\leq 1$. Assume that $\Vert x\Vert< 1$ or $\Vert y\Vert<1$, then
$$
1=\Vert p\Vert=\Vert I(p)\Vert=\Vert t x+(1-t) y\Vert\leq t\Vert x\Vert+(1-t)\Vert y\Vert<t+(1-t)=1
$$
Contradiction, so $\Vert x\Vert=1$, $\Vert y\Vert=1$. Since $I(p)=tx+(1-t)y$ and $\Vert x\Vert=\Vert y\Vert=\Vert I(p)\Vert=1$, from result of previous paragraph we conclude that $x=y$, i.e. $S(p)=T(p)$. Since $p\in\operatorname{Sphere}_H(0,1)$ is arbitrary, then $S(q)=T(q)$ for all $q\in H$. In other words $S=T$, contradiction. Hence $I$ is an extreme point. 
