4
$\begingroup$

Recall that an operator $T$ from a Banach space $E$ to a Banach space $F$ is called completely continuous (also called Dunford-Pettis) if $\|Tx_{n}\|\rightarrow 0$ for every weakly null sequence $(x_{n})_{n}$ in $E$.

In 1994, W. Wnuk introduced the concept of almost Dunford-Pettis operators. Let $X$ be a Banach lattice and let $X_{+}=\{x\in X:x\geq 0\}$ be the positive cone. An operator $T$ from a Banach lattice $X$ to a Banach space $E$ is called almost Dunford-Pettis if $\|Tx_{n}\|\rightarrow 0$ for every weakly null disjoint sequence $(x_{n})_{n}$ in $X$. He also remarked that $T$ is almost Dunford-Pettis if and only if $\|Tx_{n}\|\rightarrow 0$ for every weakly null disjoint sequence $(x_{n})_{n}$ in $X_{+}$.

It is natural to introduce the concept of positive completely continuous operators. We say that an operator $T$ from a Banach lattice $X$ to a Banach space $E$ is positive completely continuous if $\|Tx_{n}\|\rightarrow 0$ for every weakly null sequence $(x_{n})_{n}$ in $X_{+}$. Clearly, it follows from Wnuk's remark that every positive completely continuous operator is almost Dunford-Pettis. Does the converse hold?

Question. Do positive completely continuous operators and almost Dunford-Pettis operators coincide?

$\endgroup$
  • 2
    $\begingroup$ I'm not sure why this received a downvote. $\endgroup$ – Jochen Glueck Apr 10 at 6:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.