Is there any guarantee that a norm one operator, $T$ will attain its norm at some extreme point?

Question

I am studying about extreme points and extreme operators. I have a question regarding norm one operators. For a finite dimensional Banach space $X$, using Hahn-Banach theorem we know that the set $$M_T=\{x\in S_X:\|Tx\|=\|T\|\}, T\in B(X)$$ is non-empty. Can we gurantee that for a norm one operator, $T$, $M_T$ will contain at least one extreme point of the unit ball, $B_X$? A detail explanation will be very much helpful. Thank you in advance.

Answer 1

You need to assume that $X\ne \{0\}$.

Take $x$ with $\|x\|\le1$ such that $\|Tx\|= \|T\|$. Then there is $f\in X^*$ with $\|f\|\le1$ and $f(Tx)= \|Tx\|$.

In addition, if $\|y\|\le1$ then $$ f(Ty) \le \|f\| \cdot \|T\| \cdot \|y\| \le \|T\|. $$ Define the hyperplane $H$ by $$ H = \{y \in X: \ f(Ty) = \|T\|\}. $$ Then the unit ball $B_X$ is on one side of the hyperplane $H$ (due to the above inequality). In addition $x\in H \cap B_X$. Then it is easy to check that $H \cap B_X$ is a closed face of $B_X$, hence it is compact and has an extreme point. This extreme point is also an extreme point of $B_X$. (This argument is part of the proof of the Krein-Milman theorem.)

The finite-dimensionality is only used to argue compactness of $H \cap B_X$. I think the argument also works in reflexive Banach spaces (where one uses weak compactness).