Dunford--Pettis type properties and the Grothendieck property for function spaces

TL;DR

This paper investigates Dunford-Pettis and Grothendieck properties in function spaces $C_k(X)$ and $C_p(X)$, extending classical results to broader classes of spaces and establishing conditions for these properties.

Contribution

It extends Grothendieck's classical results by identifying new conditions under which $C_k(X)$ has Dunford-Pettis and Grothendieck properties, and analyzes these properties for $C_p(X)$.

Findings

$C_k(X)$ has Dunford-Pettis properties for hemicompact, cosmic, ordinal, and locally compact paracompact spaces.

$C_k(X)$ has the Grothendieck property iff every functionally bounded subset of $X$ is finite for cosmic spaces.

$C_p(X)$ always has Dunford-Pettis properties, and has the Grothendieck property iff every functionally bounded subset of $X$ is finite.

Abstract

For a Tychonoff space $X$, let $C_k(X)$ and $C_p(X)$ be the spaces of real-valued continuous functions $C(X)$ on $X$ endowed with the compact-open topology and the pointwise topology, respectively. If $X$ is compact, the classic result of A.~Grothendieck states that $C_k(X)$ has the Dunford-Pettis property and the sequential Dunford--Pettis property. We extend Grothendieck's result by showing that $C_k(X)$ has both the Dunford-Pettis property and the sequential Dunford-Pettis property if $X$ satisfies one of the following conditions: (i) $X$ is a hemicompact space, (ii) $X$ is a cosmic space (=a continuous image of a separable metrizable space), (iii) $X$ is the ordinal space $[0,\kappa)$ for some ordinal $\kappa$, or (vi) $X$ is a locally compact paracompact space. We show that if $X$ is a cosmic space, then $C_k(X)$ has the Grothendieck property if and only if every functionally bounded subset of $X$ is finite. We prove that $C_p(X)$ has the Dunford--Pettis property and the sequential Dunford-Pettis property for every Tychonoff space $X$, and $C_p(X) $ has the Grothendieck property if and only if every functionally bounded subset of $X$ is finite.