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I was reading Proof of Milman-Pettis Theorem. In that I do not understand highlighted text

Why such f exist such that $<\psi,f>>1-\delta/2$ enter image description here

enter image description here

Any Help will be appreciated

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In any normed linear space (not equal to $\{0\}$) $\|x\|=\sup \{|f(x)|: \|f\|= 1\}$. In this case $\sup \{|f(\xi)|: \|f\|= 1\}=1$ and the result follow by definition of supremum.

An even stronger statement is holds if $\xi \in E$. We can find $f$ such that $\|f\|=1$ and $f(\xi)=1$. This follows from Banach -Alaoglu Theorem.

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  • $\begingroup$ Can you elaborate how this stronger statement follows from Banach-Alaoglu? $\endgroup$ – MaoWao Oct 27 '19 at 9:12
  • $\begingroup$ @MaoWao The closed unit ball of $E^{*}$ is compact in weak* topology. We can take $(f_i)$ in the ball such that $f_i(\xi) \to 1$ and then extract a subnet converging to some f$. This $f$ is the one that works. $\endgroup$ – Kavi Rama Murthy Oct 27 '19 at 11:24
  • $\begingroup$ No. $\xi$ is from $E^{\ast\ast}$, not from $E$, so weak$^\ast$ convergence is not enough. For $\xi\in E$, this stronger statement follows already from Hahn-Banach. $\endgroup$ – MaoWao Oct 28 '19 at 7:00
  • $\begingroup$ @MaoWao You are right. I was thinking of $\xi \in E$ when I wrote the stronger statement. $\endgroup$ – Kavi Rama Murthy Oct 28 '19 at 7:15
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It comes from the definition of norm: $$ \|\xi\|_{E^{**}}=1=\sup_{\|f\|_{E^*=1}} \langle \xi,f\rangle_{E^{**},E^*}. $$ Then for each $\frac{\delta}{2}>0$ there exists $\tilde{f}\in E^*, \|f\|_{E^*}=1$ so that $$ \langle \xi,\tilde{f}\rangle_{E^{**},E^*} \geq 1-\frac{\delta}{2}. $$

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