2
$\begingroup$

Sobczyk's theorem states that if a separable Banach space $X$ contains a subspace isometric to $c_{0}$, then $X$ contains a subspace $Z$ which is isometric to $c_{0}$ and is $2$-complemented in $X$. Since every projection from $c$ onto its subspace $c_{0}$ has norm at least two, the projection constant in Sobczyk's theorem cannot in general be improved. In 1999, P. N. Dowling, N. Randrianantoanina and B. Turett (Remarks on James's distortion theorems II, Bull. Austral. Math. Soc. 59(1999), 515-522) proved the following theorem:

Theorem: Let $X$ be a Banach space so that the unit ball $B_{X^{*}}$ is weak*-sequentially compact. If $X$ contains a subspace isomorphic to $c_{0}$, then, for every $\epsilon>0$, there exist a subspace $Z$ of $X$ and a projection $P$ from $X$ onto $Z$ such that $Z$ is $(1+\epsilon)$-isomorphic to $c_{0}$ and $\|P\|\leq 1+\epsilon$.

A result similar to this theorem was proved by S. Diaz and A. Fernandez (S. Diaz and A. Fernandez, Reflexivity in Banach lattices, Arch Math. 63(1994), 549-552) in the setting of Banach spaces not containing isomorphic copy of $l_{1}$. P. N. Dowling, N. Randrianantoanina and B. Turett posed the following natural question:

Question: If a Banach space $X$ contains a complemented copy of $c_{0}$, then, for every $\epsilon>0$, do there exist a subspace $Z$ of $X$ and a projection $P$ from $X$ onto $Z$ such that $Z$ is $(1+\epsilon)$-isomorphic to $c_{0}$ and $\|P\|\leq 1+\epsilon$?

I do not know whether this question is solved or there are some results about this question.

Thank you!

$\endgroup$
1
$\begingroup$

Narcisse and I gave a counterexample in

Johnson, William B.(1-TXAM); Randrianantoanina, Narcisse(1-MMOH) On complemented versions of James's distortion theorems. (English summary) Proc. Amer. Math. Soc. 135 (2007), no. 9, 2751–2757.

You could have discovered this just by checking on MathSciNet reviews of papers that referenced the [DRT] paper.

$\endgroup$
  • $\begingroup$ Thanks, Bill. You are right. $\endgroup$ – Dongyang Chen Oct 24 '20 at 0:56

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.