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My task is the following: Let $X$ be a nonreflexive Banach space. Prove that there exists a closed vector subspace $N \subset X^*$ such that $N \neq (^{\bot} N)^{\bot}$. As usual here $(^{\bot} N)$ denotes the preannhilator of a subset $N \subset X^*$ and $ M ^{\bot}$ denotes the annhilator of a subset $M \subset X$.

I definetively need to use the non reflexiveness of the space $X$. So maybe we can work with some $h \in X^{**}$ such that $i_x \neq h$ for any $x \in X$. Then we can perhaps consider its Kernel, which we know is a closed subsapce of $X^*$. Other than that I am not sure how to proceed.

Maybe we could somehow conclude that $^{\bot} Ker(h)$ is trivial, which will then show that the required equality does not hold. But this is just speculation. Any help is appreciated!

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You have the right idea: Since $X$ is not reflexive, there exists $x^{**} \in X^{**}$ such that $x^{**} \notin i(X)$. [Here $i \colon X\to X^{**}$ is the natural embedding.] Let $N= \ker x^{**}$. Then, $${}^\bot N = \{0\}. $$ Hint: Suppose there exists nonzero $x \in X$ such that $x^*(x)=0$, for every $x^* \in N$. Consider the functional $i(x) \in i(X)$. Show that $$ \ker i(x) \supseteq \ker x^{**} $$ and so, from a standard linear algebra Lemma, there exists $λ\in \mathbb R$ such that $x^{**}=λi(x) = i(λx) \in i(X)$.

Since ${}^\bot N=\{0\} , $ we must have that $({}^\bot N)^\bot =X^* $.

Evangelopoulos Foivos
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