# Doubt in proof of Milman-Pettis Theorem in Brezis

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

Why such f exist such that $$<\psi,f>>1-\delta/2$$  Any Help will be appreciated

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.
• @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$is the one that works. – Kavi Rama Murthy Oct 27 '19 at 11:24 • No.$\xi$is from$E^{\ast\ast}$, not from$E$, so weak$^\ast$convergence is not enough. For$\xi\in E$, this stronger statement follows already from Hahn-Banach. – MaoWao Oct 28 '19 at 7:00 • @MaoWao You are right. I was thinking of$\xi \in E\$ when I wrote the stronger statement. – Kavi Rama Murthy Oct 28 '19 at 7:15
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 $$\langle \xi,\tilde{f}\rangle_{E^{**},E^*} \geq 1-\frac{\delta}{2}.$$