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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.

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  • $\begingroup$ Why do you want to consider unbounded operators on Banach spaces? (Reflexive spaces are complete.) They exist only by the Axiom of Choice. However, if you want to play this game, you will quickly find silly counterexamples. Take a bounded operator, restrict it to the span of $E$ which is not everything and then extend to an unbounded operator... $\endgroup$ – Tomek Kania Mar 12 '17 at 16:22
  • $\begingroup$ It was just curiosity whether this condition is sufficient to be bounded. I am afraid your answer only solves the second part of my (rephrased) question, doesn't it? $\endgroup$ – user342207 Mar 12 '17 at 16:25
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    $\begingroup$ I have included my above comment which you have probably missed. $\endgroup$ – Tomek Kania Mar 12 '17 at 16:34
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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...

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