I have reading an answer post before , and there's something I do not understand .
For each $f_n$ in the unit ball of $X^*$ , why there exist $x_n \in X$ with $\|x_n \| \le1$ such that $f_n(x_n) \ge \frac{1}{2}$ .
I can show that for each $f_n$ , there exist $x \in X$ , $f_n(x)=a \neq 0$ . So let $x_n = \frac{x}{a}$ , we get the inequality . But how to get the desired $x_n$ with $x_n$ also in the unit ball of $X$ ?
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J.Guo
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1I think that both comments and the answer there explain that the proof as written there assumes that $f_n$ belong to the unit sphere, i.e., $|f_n|=1$. Using that, you can get $x_n$ such that $f_n(x_n)\ge\frac12$. – Martin Sleziak Dec 08 '18 at 13:43
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That is false. It should read, the unit sphere, not the unit ball.
Ben W
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Is it right that : $f $ belongs to the unit sphere means $|f | =1 $ and $f $ belongs to the unit ball means $| f | \le 1$ ? – J.Guo Dec 08 '18 at 13:48
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