Let $X, Y$ be normed spaces and define $B_1 := \{x \in X \;: \; \|x\|_X \leq 1\}$. The operator norm of a linear operator $T : X \to Y$ is in general defined as $ \|T\|_{op} := \sup \{ \| T x \| \; : \; x \in B_1 \}.$ If $X$ is assumed to be reflexive, the convex hull of the extreme points $E$ on $B_1$ is norm dense in $B_1$. Suppose now that $T$ satisfies $K := \sup \{ \| Tx \| \; : \; x \in E \} < \infty$. Is this condition sufficient for the operator $T$ to be bounded and $K = \| T \|_{op}$?
Any comment is appreciated.
This is an answer to the previous version of this question:
Let $T\colon X\to Y$ be a bounded linear operator. By reflexivity of $X$ and the Krein-Milman theorem, the set $E$ is norming for $X^*$, i.e., for any $f\in X^*$ we have $\|f\|=\sup_{x\in E}|\langle f,x\rangle|$. However, $$\begin{array}{lcl}\sup_{x\in E} \|Tx\|&= &\sup_{x\in E}\sup_{g\in B_{Y^*}} |\langle g, Tx\rangle| \\ &= &\sup_{x\in E}\sup_{g\in B_{Y^*}} |\langle T^*g, x\rangle|\\ &= &\sup_{g\in B_{Y^*}} \sup_{x\in E}|\langle T^*g, x\rangle|\\ &=&\sup_{g\in B_{Y^*}} \|T^*g\|\\ &=&\|T^*\|\\ &=&\|T\|.\end{array}$$
Here is an answer to the new question.
If we care about pathological examples of unbounded linear operator defined on Banach spaces, we may proceed as follows. Let $X$ be an infinie-dimensional reflexive space. Then ${\rm span}\, E\neq X$. Let $T$ be any extension of the identity operator on ${\rm span}\, E$ to an unbounded operator, then...