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I'm aware of this post, but I have not yet been granted permission to comment on these other posts, so I hope you'll excuse asking for a clarification here.

For clarity, I'll restate the question. If $X$ is a Banach space and $X^*$ (the space of continuous linear functionals) is separable, then $X$ is separable.

In this answer, the chosen answer provides the following argument: Choose a countable dense subset of $X^*$, say $\{f_n\}$. Choose $x_n\in X$ so that $|f_n(x_n)|\ge 1/2||f_n||_{op}$. If $Y$, the rational span of $\{x_n\}$, is not dense, then Hahn-Banach guarantees the existence of a nonzero continuous functional $f$ on $X$ so that $f$ vanishes on $\overline{Y}$. Find an $n$ for which $||f-f_n||_{op}<1/4$, then since $f(x_n)=0$, we have that

$1/4>||f-f_n||_{op}\ge |f(x_n)-f_n(x_n)|=|f_n(x_n)|$

Everything so far is elementary. Now, excuse me if I'm missing something simple, but he/she concludes that $|f_n(x_n)|>1/2$, and immediately reaches the contradiction. But we only know that $|f_n(x_n)|>1/2||f_n||_{op}$, no?

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up vote 3 down vote accepted

You can rescale $f$ so that it has unit norm. Then $\|f-f_n\|_{op}<1/4$ implies $\|f_n\|_{op}>3/4$. Now you can finish the chain of inequalities with $|f_n(x_n)|>(1/2)\|f_n\|_{op}>3/8$, which is still a contradiction.

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Ah, of course! Thank you. –  user39992 Dec 27 '12 at 22:00

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