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Notation: Let $V$ be a normed vector space and $V^*$ to be its continuous dual space.

One of the consequences of the Hahn-Banach Theorem is the existence of non-trivial dual space.

Corollary 1: Let $z$ be any nonzero element in $V.$ Then there exists a continuous linear functional $\psi$ on $V$ such that $\|\psi\|\leq 1$ and $\psi(z)=\|z\|.$

Since $V^*$ is always a Banach space, we can apply corollary $1$ to obtain the following.

Corollary 2: For any nonzero element $\psi$ in $V^*,$ there exists a continuous linear functional $\mu$ on $V^*$ such that $\|\mu\|\leq 1$ and $\mu(\psi) = \|\psi\|.$

Now, recall James's Theorem.

James's Theorem: Let $V$ be a Banach space. Then $V$ is reflexive if and only if every continuous linear functional on $V$ attains its supremum on the closed unit ball $B_V,$ that is, for every $\psi\in V^*,$ there exists $z\in B_V$ such that $\|\psi\| = \psi(z).$

So here comes my question.

Question: Can we use Corollary $2$ to prove sufficiency of reflexivity in James's Theorem, that is, if a Banach space $V$ is reflexive, then every continuous linear functional on $V$ attains its supremum on $B_V?$

It seems to me that this is possible. Assume that $V$ is a reflexive Banach space. Pick any nonzero $\psi\in V^*.$ By Corollary $2$ and reflexivity of $V,$ there exists $z\in V$ such that $\|z\|\leq 1$ and $\psi(z) = \|\psi\|.$ (Since $V$ s reflexive, there exists some $z\in V$ such that $\mu = e_z$ where $e_z\in V^{**}$ is an evaluation functional). So sufficiency is proven.

Is there any mistake above?

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Your reasoning is correct. This direction of the theorem is easy; what is actually due to James is its converse.

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  • $\begingroup$ I see. Thanks for your reply. $\endgroup$ – Idonknow Apr 3 '18 at 11:27

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