Set of poinwise convergence of a sequence of weak* continuous linear functionals is weak* closed?

Question

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?

Answer 1

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 $\ell^\infty$.