---
tag: W35ZP39S4R
kind: definition
name: Dunford–Pettis property
msc: [46B10, 46B20]
defines: QQSNZX1QMM
uses: [06GZ956657, 4A82S5FVRT, P6KHVSTPZV]
sources:
  - url: https://en.wikipedia.org/wiki/Dunford%E2%80%93Pettis_property
---

## Definition

Let $X$ be a real or complex [[Banach space|06GZ956657]] with [[continuous dual|4A82S5FVRT]] $X^*$.
The space $X$ has the *Dunford–Pettis property* if the following holds.
Let $(x_n)$ be a sequence in $X$ that converges to $0$ in the [[weak topology|P6KHVSTPZV]] $\sigma(X, X^*)$.
Let $(f_n)$ be a sequence in $X^*$ that converges to $0$ in the weak topology $\sigma(X^*, X^{**})$.
Then $f_n(x_n) \to 0$.

## 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.