In 1968, J. Lindenstrauss and A. Pe{\l}czy'{n}ski posed a problem: Is every complemented subspace $X$ of an $\mathcal{L}_{p}$-space ($1\leq p\leq \infty$) either an $\mathcal{L}_{p}$-space or isomorphic to a Hilbert space? If $p=1$ or $\infty$ and $X$ is infinite-dimensional, $X$ can not be a Hilbert space. In 1969, J. Lindenstrauss and H. P. Rosenthal stated that

Theorem. If $Y$ is an $\mathcal{L}_{p,1+\epsilon}$-space for every $\epsilon>0$ and $X$ is a $1$-complemented subspace of $Y$, then $X$ is an $\mathcal{L}_{p,1+\epsilon}$-space for every $\epsilon>0$.

They mentioned that the proof of the theorem can be seen in J. Lindenstrauss and A. Pe{\l}czy'{n}ski's paper. However, I can not see any of its proof.

Question 1. In what reference is there a proof of Theorem ?

Question 2. If $X$ is a $C$-complemented subspace of $Y$ in Theorem, is $X$ an $\mathcal{L}_{p,\lambda}$-space for some $\lambda$ depending only on $C$ ?

Thank you!


I don't know if there is a good reference for Theorem 1. The Lindenstrauss-Pelczynski proof shows that $Y$ is contractively complemented in $L_p(\mu)$ for some measure $\mu$ (at least for $p$ in the reflexive range), and contractively complemented subspaces of $L_p(\mu)$ are isometrically isomorphic to some $L_p(\nu)$. That is in books, or you can look at Beata Randrianantoanina's 2001 paper. These days we would use ultra products to get the L-P part.

In Question 2, you of course have to assume that $X$ is not isomorphic to a Hilbert space. When $2<p<\infty$, use Kadec-Pelczynski to see that $X$ contains a subspace $1+\epsilon$ isomorphic to $\ell_p$. That condition passes to finite codimensional subspaces, so the argument I outlined in your other thread gives what you want. The case $1<p<2$ follows by duality, and for $p=1$ you use the fact that every non reflexive subspace of $L_1(\mu)$ contains a $1+\epsilon$ copy of $\ell_1$ (this is in Wojtaszczyk's book). In the $1<p<2$ case, you can improve the constant by using the result that every subspace of $L_p(\mu)$ that is isomorphic to $\ell_p$ has a smaller subspace that is $1+\epsilon$ isomorphic to $\ell_p$. The argument for the $p=1$ case almost goes over. Schechtman and I outlined the argument in our 2008 paper in J. Eur. Math. Soc. 10, 1105–1119 because we did not know an adequate reference.


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.