I was reading Haim Brezis Weak* topology section on page 6 4
I understand everything except fact that
why in particular $|<f,x_i>|=0\implies \phi(f)=0$
Any Help in this regard will be appreciated Thanks you
If $\langle f,x_i\rangle=0$ for all $i$, then $tf\in V$ for all $t\in \mathbb R$. This implies $$ \phi(tf) = t \phi(f) <1 \quad \forall t\in \mathbb R, $$ and $\phi(f)$ has to be zero.
Copyright © 2021 JogjaFile Inc.
If $\langle f,x_i\rangle=0$ for all $i$, then $tf\in V$ for all $t\in \mathbb R$. This implies $$ \phi(tf) = t \phi(f) <1 \quad \forall t\in \mathbb R, $$ and $\phi(f)$ has to be zero.