I am not an expert in operator algebras, so if the answer to this question might be trivial, that might be one reason for that:
Let $\mathcal{A}$ be a $C^\ast$-algebra. Then $\mathcal{A}^{\ast \ast}$ is a $W^\ast$-algebra. Consider the multiplier algebra $$M(\mathcal A) = \lbrace x \in \mathcal{A}^{\ast \ast} : \forall a \in \mathcal A \colon ax, xa \in \mathcal{A} \rbrace.$$ Suppose that $\mathcal A^\ast$ has the Dunford-Pettis property. Then one knows that if $(y_n)_n$ and $(x_n)_n$ are sequences in $\mathcal A^\ast$ and $\mathcal A^{\ast \ast}$ converging to $0$ with respect to $\sigma(\mathcal A^\ast, \mathcal A^{\ast \ast})$ and $\sigma(\mathcal A^{\ast \ast}, \mathcal A^{\ast \ast \ast})$, respectively, then $\langle x_n, y_n \rangle \to 0$ as $n \to \infty$.
Is it possible to show from that if $(y_n)_n$ and $(x_n)_n$ are sequences in $\mathcal A^\ast$ and $M(\mathcal A)$ converging to $0$ with respect to $\sigma(\mathcal A^\ast, \mathcal A^{\ast \ast})$ and $\sigma(M(\mathcal A), \mathcal A^\ast)$, respectively, then $\langle x_n, y_n \rangle \to 0$ as $n \to \infty$?
Indeed, in a commutative setting, Gelfand's theorem yields that $\mathcal A$ is isometrically isomorphic to $C_0(L)$, i.e., the space of continuous functions that vanish at infinity, for some locally compact space $L$. Then $\mathcal A^\ast$ is isomorphic to $\mathcal M(L)$ (the space of measures of bounded variations). Moreover, $M(\mathcal A)$ is isomorphic to $C_b(L)$, the space of continuous bounded functions on $L$. In this setting, it is possible to use arguments from the theory of Banach lattices to show that if $(\mu_n)_n$ and $(f_n)_n$ are sequences in $\mathcal M(L)$ and $C_b(L)$ converging to $0$ with respect to $\sigma(\mathcal M(L), \mathcal M(L)^{\ast})$ and $\sigma(C_b(L), \mathcal M(L))$, respectively, then $\langle f_n, \mu_n \rangle \to 0$ as $n \to \infty$. The main idea one needs is that weakly compact sets in $L^1$-type spaces (like the spaces of measures) are almost order bounded. Note here that $\mathcal M(L)$ has the Dunford-Pettis property.
So, my question is, basically, if the above statement still holds in a commutative setting. The real problem is that in a non-commutative setting, one cannot employ lattice arguments. One still has ordered Banach spaces in this case, but it is very unclear to me, if that structure is helpful in the context. However, I am quite unaware of deeper techniques from the theory of operator algebras and wonder if they might be useful for proving the statement. I am also completely unaware if there might be easy counter examples in the non-commutative setting.
