Dunford–Pettis property

Remarks

Example: $\ell^1$ has the Dunford–Pettis property. By the Schur property of $\ell^1$, a weakly null sequence $(x_n)$ in $\ell^1$ tends to $0$ in norm. A weakly convergent sequence $(f_n)$ is bounded, by the uniform boundedness principle. So $\abs{f_n(x_n)} \le \norm{f_n} \, \norm{x_n} \to 0$.

Non-example: $\ell^2$ does not have it. The unit vectors $e_n$ tend weakly to $0$, and so do the functionals $f_n = \ip{\cdot}{e_n}$ in the dual of $\ell^2$. But $f_n(e_n) = 1$ for all $n$. More generally, no infinite-dimensional reflexive space has the property.

An equivalent form: every weakly compact operator from $X$ to a Banach space $Y$ maps weakly convergent sequences to norm convergent sequences. The spaces $L^1(\mu)$, $C(K)$ for compact Hausdorff $K$, and $c_0$ have the property.