Let $X$ be a separable space and let $x^{**}\in X^{**}$. If $x^{**}(x^{*}_{n})\rightarrow 0$ for each weak*-null sequence $(x^{*}_{n})_{n}$ in $X^{*}$, is $x^{**}$ in $X$ ?
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asked Aug 15, 2021 at 3:08
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$\begingroup$ A Banach space $X$ has Mazur property if every weak* sequentially continuous linear $\mu:X^{\ast} \to\mathbb{C}$ is (weak* continuous, so) in $X$. In this language, separable Banach spaces have Mazur property. $\endgroup$– Onur OktayCommented Aug 15, 2021 at 6:48
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Yes, because $(B_{X^*},w^*)$ is compact metrizable. So $x^{**}$ is $w^*$-continuous at any $x^*\in B_{X^*}$.
