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Is it always true that a weakly null sequence $(x_n)_{n \in \mathbb{N}}$ implies the inequality $$|x^*(x_n)|<\epsilon?$$

Recall that a weakly null sequence $(x_n)_{n \in \mathbb{N}}$ is a sequence converges to $0$ weakly. So for all linear functionals $x^* \in X^*$, we have
$$|x^*(x_n) - x^*(0)|<\epsilon$$

My guess is that $x^*(0)=0$ but it seems not always the case to me. Can anyone enlighten me?

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    $\begingroup$ Any linear map takes 0 to 0 $\endgroup$ Commented Oct 22, 2015 at 4:07

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As pointed out by Prahlad Vaidyanathan, you always have $x^*(0) = 0$. Indeed, by linearity you have $x^*(0) = x^*(0+0) = x^*(0) + x^*(0)$ or $x^*(0) = x^*(0\cdot 0) = 0 \cdot x^*(0) = 0$.

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