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how to prove:

If $X$ is a separable Banach space, then there exists a sequence $\{f_n\}_{n\ge 0}\subset X^*$ such that $$\|x\|=\sup_n |f_n(x)|$$ for all $x\in X$

I don't know how to deal with "separable" since it is a property of $X$, not $X^*$, any hints would be appreciate!

Thanks to Daniel's comment:

Let $\Lambda$ be the unit sphere of $X^*$, then $\forall f\in\Lambda$ $$|f(x)|\le \|f\|_{X^*}\|x\|=\|x\|$$ hence $\sup_{\alpha\in\Lambda}|f_\alpha(x)|\le \|x\|$ consider $f(x)=\|x\|$, we have $\sup_{\alpha\in\Lambda}|f_\alpha(x)|= \|x\|$.

finally by the weak* separability of $\Lambda$, we can find a sequence s.t. $$\sup_{\alpha\in\Lambda}|f_\alpha(x)|=\sup_n|f_n(x)|$$

Lookout
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  • The separability of $X$ implies the weak$^{\ast}$-separability of the closed unit ball of $X^{\ast}$. – Daniel Fischer Dec 02 '16 at 14:49
  • @DanielFischer Thanks a lot. – Lookout Dec 02 '16 at 15:01
  • $X$ is separable : it has a basis ${e_k}$. For each $e_k$, find a sequence ${g_{m}(k)}m \subset X^*,|g{m}(k)| = 1$ such that $\sup_m |g_{m}(k)(e_k)| = |e_k|$, and build $f_n$ from this. – reuns Dec 02 '16 at 15:42

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