Let $\{\Lambda_n\}$ be a sequence of weak* cont. linear functionals on $X^{*},$ the dual of a Banach space $X.$ As every weak* cont. functinal is also norm continuous on $X^{*},$ assume that the family is also uniformly bounded that is $||\Lambda_n||_{X^{*}}\leq1\forall n\in\mathbb{N}.$ Consider the set $$A:=\{\varphi\in X^{*}:\lim\Lambda_n\varphi\,\,\text{exists}\}.$$

The set $A$ is norm closed in $X^{*}.$ I have the following question;

Is $A$ weak* closed also?

  • 2
    $\begingroup$ A weak star continuous linear functional on $X^*$ can be seen as an element of $X$. You need to rewrite the details of your question. $\endgroup$ – Neutral Element Nov 22 '15 at 13:16
  • $\begingroup$ Yes, I did try that although I'm not sure how it's going to help. $\endgroup$ – Abelvikram Nov 22 '15 at 13:36

This is not true. For a counterexample, take $X = \ell^1$, $X^* = \ell^\infty$. Define $\Lambda_n \in X^{**}$ by $$\Lambda_n(x) = x_n$$ for $x \in \ell^\infty$. Then, it is easy to see that this sequence $\{\lambda_n\}$ satisfies your assumptions. Moreover, $$A = \{x \in \ell^\infty: \lim_{n \to \infty} x_n \text{ exists}\}.$$ This set is not weak-$*$ closed. In fact, the set $$c_0 = \{x \in \ell^\infty : \lim_{n \to \infty} x_n = 0\}$$ is already weak-$*$ dense in $\ell^\infty$.

  • $\begingroup$ But $A$ contains $c_0,$ so it is also weak* dense in $l^{\infty}?$ $\endgroup$ – Abelvikram Nov 23 '15 at 1:06
  • $\begingroup$ Yes. And, obviously, $A \ne \ell^\infty$. Hence, it cannot be weak-$*$ closed. $\endgroup$ – gerw Nov 23 '15 at 7:20

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