Let $d(\textbf{w},p)$, $1\leq p<\infty$, denote the Lorentz sequence space, where $\textbf{w}=(w\_n)\_{n=1}^\infty\in c\_0\setminus\ell\_1$ is a normalized decreasing weight.


Is there very much known about the complemented subspaces of $d(\textbf{w},p)$? In general (i.e., without any restrictions on $\textbf{w}$ or $p$), I can only find two: $\ell\_p$ and $d(\textbf{w},p)$ itself.


**Question 1.** What complemented subspaces of $d(\textbf{w},p)$, besides $\ell\_p$ and itself, are already known?


There has been found a third distinct complemented subspace in case $\textbf{w}$ satisfies a certain (NUC) condition, namely that $\inf\_k(\sum\_{i=1}^{2k}w\_n)/(\sum\_{i=1}^kw\_n)=1$. In this case $d(\textbf{w},p)$ contains a 1-complemented subspace isomorphic to $\oplus\_p(\ell\_\infty^n)$.


But can we find other complemented subspaces in the general case?


The obvious thing to try first is to look for constant-coefficient block basic sequences. If the length of the blocks is bounded then they span another copy of $d(\textbf{w},p)$. However, if their lengths tend to infinity, then their span will be distinct from $d(\textbf{w},p)$. In this case, the problem is to show that the resulting space is not isomorphic to $\ell\_p$.


This is pretty easy when $p=1$ or $p=2$. In these cases, since $\ell\_1$ and $\ell\_2$ each admit a unique unconditional basis, it's sufficient to make sure that the constant-coefficient blocks in $d(\textbf{w},p)$ aren't equivalent to those bases. This can be done by taking constant-coefficient block sequences $(d\_i^{(k)})\_{i=1}^{N\_k}$ of fixed length $k$, choosing $N\_k$ sufficiently large that they fail increasingly badly to dominate $\ell\_p^{N\_k}$. Then glue those sequences together for all $k\in\mathbb{N}$, pushing them out far enough so that they're disjoint.


I suspect that this will work for all $1\leq p<\infty$, but proving it is not as straightforward as the above. However, if it could be shown that $\ell\_p$ does not contain uniformly complemented copies of $\text{span}(d\_n)\_{n=1}^N$, where $(d\_n)\_{n=1}^\infty$ is the basis for a Lorentz sequence space, then that would do the trick. Thus:


**Question 2.** Does $\ell\_p$ contain uniformly complemented copies of $\text{span}(d\_n)\_{n=1}^N$?


Thanks guys!



Complemented subspaces of Lorentz sequence spaces?
Question 2 has a negative answer. Suppose $\ell\_p$ contains uniformly complemented copies $E\_N$ of $\text{span}(d\_n)\_{n=1}^N$. Take an ultra power to get a copy $E$ of the completion of $\text{span}(d\_n)\_{n=1}^\infty$ in an $L\_p$ space. The corresponding copies of $E\_n$ in the $L\_p$ space are uniformly complemented in the $L\_p$ space. When $p>1$, this makes $E$ itself complemented. In fact, even when $E$ is not reflexive, it is complemented because $E$ has non trivial cotype and hence is complemented in $E^{\*\*}$. Consequently, you get that $E$ is isomorphic to a complemented subspace of $L\_p(0,1)$, which is impossible (a complemented subspace of $L\_p(0,1)$ that has a symmetric basis must be isomorphic to $\ell\_p$ or $\ell\_2$).


EDIT March 31, 2019. Ben W asks for more details. First, if $X$ is a Banach space and $X^{\cal{U}}$ is an ultra power of $X$ (where ${\cal{U}}$ is a free ultrafilter over the natural numbers, say), then the canonical injection from $X$ into $X^{\*\*}$ factors through $X^{\cal{U}}$. This is easy from the definition of Banach space ultra products and probably is in most books that treat them. 


Let $X$ be the Lorentz space, let $F\_N = \text{span}(d\_n)\_{n=1}^N$, and let $P\_N$ be the basis projection from $X$ onto $F\_N$. Your hypothesis is that there are norm one operators $A\_N: F\_N \to \ell\_p$ and uniformly bounded operators $B\_N:\ell\_p \to F\_N$ s.t. $B\_NA\_N$ is the identity on $F\_N$. The family $A\_NP\_N$ induces a contractive mapping $A$ from $X$ into $\ell\_p^{\cal{U}}$; $Ax:= (A\_NP\_Nx)$, and the $B\_N$ induce an operator $B:\ell\_p^{\cal{U}} \to X^{\cal{U}}$. $X$ is naturally embedded into $X^{\cal{U}}$ as the diagonal, and it is easy to check that $BA$ is the identity on each $F\_N$ and hence on $X$. By the comment in the previous paragraph, we deduce that the injection from $X$ into $X^{\*\*}$ factors through $\ell\_p^{\cal{U}}$. But $X$ is complemented in $X^{\*\*}$, so the identity on $X$ factors through $\ell\_p^{\cal{U}}$; that is, $X$ is isomorphic to a complemented subspace (which we also call $X$) of the abstract $L\_p$ space $\ell\_p^{\cal{U}}$. Let $Y$ be the closed sublattice of $\ell\_p^{\cal{U}}$ generated by $X$. $Y$ is again an abstract $L\_p$ space and hence isometrically isomorphic to $L\_p(\mu)$ for some measure $\mu$. But the isometric characterization of separable $L\_p(\mu)$ is in many books and yield that $X$ is isomorphic to a complemented subspace of $L\_p(0,1)$. 


Most of the above is essentially in the book of Lindenstrauss and Tzafriri; mostly in volume 2.


