For an extreme contraction, $T$, $\|T\pm U\|\leq 1$ implies $U=0$. How?

Question

$\textbf{Extreme Contraction:}$ An extreme point of the set of all operators $T$ of norm less than or equal to one.

I was reading a proof of a result on extreme points and extreme contractions from a paper entitled as "ON EXTREME OPERATORS IN FINITE-DIMENSIONAL SPACES" written by "J. LINDENSTRAUSS AND M. A. PERLES", there I met with a statement which says if for an extreme contraction, $T$, $\|T\pm U\|\leq 1$, for some operator $U$ then $U=0$.

I was going to prove this result by a contradiction for a better understanding. By assuming $\|U\|>0$ I tried to show $\|T\|\ne 1$ but do not able to conclude. If anyone can explain the above statement it will be very much helpful for me. Thank you in advance.

Answer 1

This is true in any normed vector space, not only for operator norms. If $\|T\pm U\|\leq 1$ with $U \ne 0$ then $$ T = p T_1 + (1-p) T_2 $$ where $T_1 = T+U$, $T_2 = T-U$ are distinct elements of the unit ball, and $p = 1/2$. So $T$ is not an extreme point of the unit ball.