A Banach space $X$ is said to have the Dunford-Pettis property if for any Banach space $Y$ every weakly compact operator $T:X\rightarrow Y$ is completely continuous. Recall that $T$ is completely continuous if it maps weakly convergent sequences to norm convergent sequences, or equivalently, if it maps weakly Cauchy sequences to norm Cauchy ones. There are many well-known characterizations of the Dunford-Pettis property (see J. Diestel, A survey of results related to the Dunford-Pettis property). In particular, it was known that $X$ has the Dunford-Pettis property if and only if $\langle x^{*}_{n},x_{n}\rangle \rightarrow 0$ for every weakly Cauchy sequence $(x_{n})_{n}$ in $X$ and every weakly null sequence $(x^{*}_{n})_{n}$ in $X^{*}$. Recently I am thinking about the following quantification of this characterization:

Let $X$ be a Banach space. The following are equivalent:

(1) $X$ has the Dunford-Pettis property.

(2) $\limsup\limits_{n}|\langle x^{*}_{n},x_{n}\rangle|\leq \delta((x_{n})_{n})$ whenever $(x_{n})_{n}$ is a bounded sequence in $X$ and $(x^{*}_{n})_{n}$ is a weakly null sequence in $B_{X^*}$, where $\delta((x_{n})_{n})=\sup\limits_{x^{*}\in B_{X^{*}}}\inf\limits_{n}\sup\limits_{k,l\geq n}|\langle x^{*},x_{k}-x_{l}\rangle|$. Clearly, $\delta((x_{n})_{n})=0$ if and only if $(x_{n})_{n}$ is weakly Cacuhy.

I am not sure if this characterization is true. Thank you!