Doubt in proof of Milman-Pettis Theorem in Brezis

Question

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

Answer 1

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.

Answer 2

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 $$ \langle \xi,\tilde{f}\rangle_{E^{**},E^*} \geq 1-\frac{\delta}{2}. $$