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I was reading Haim Brezis Weak* topology section on page 6 4enter image description here

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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

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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.

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  • $\begingroup$ Thanks a lot Sir $\endgroup$ – MathLover Oct 11 '19 at 11:08

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