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Here is the question: enter image description here

And here is the given answer: enter image description here

I would like to ask in the first line of the proof, why $\|x^*_k\|$ is less or equal than $1$? I guess zero belongs to $B(x^*_k,r)$ for any $k$, then $\|x^*_k - 0||<r<1$ but I don't know how to prove it. May I have some hints please?

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    $\begingroup$ $x_k^{*} \in B_E^{*}$ so $\|x_k^{*}\| \leq 1$. That is the definition of $B_E^{*}$. $\endgroup$
    – Kavi Rama Murthy
    May 26 at 5:48
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    $\begingroup$ $E^{*}$ is the set of all bounded linear maps from $E$ to $\mathbb K$. $B_E^{*}$ is the closed unit ball of $E^{*}$. $\endgroup$
    – Kavi Rama Murthy
    May 26 at 7:19
  • $\begingroup$ @Kavi Rama Murthy Thank you. I thought that $B^*_E$ is the set of all bounded linear map from $B_E$ to $\mathbb K$ rather than $B(E^∗)$. The notations are confusing me. $\endgroup$
    – math noob
    May 26 at 7:34

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