# Complemented subspaces of Lorentz sequence spaces?

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!

• The strategy you described using constant coefficient block sequences won't work in general. If the weight function is submultiplicative, then every block sequence has a subsequence either equivalent to the original basis or to $\ell_p$. Of course, this doesn't rule out existence of other complemented subspaces without a symmetric basis. – Bunyamin Sari Mar 23 '19 at 17:56
• @BunyaminSari Thanks for the info. Yes, I realize it won't work in general for classifying all complemented subspaces. However it is a start. Also, somebody named Randrianantoanina (whom I've never met) claims to have proved that certain 1-complemented subspaces (in particular, the "disjointly supported" ones?) are always spanned by constant coefficient block bases. So it is, I guess, a fairly nice thing to take care of. – Ben W Mar 23 '19 at 18:20
• @Ben W. You can meet her at the AMS meeting in Hartford in April. Beata is one of the organizers of the Special Session in Banach space theory. – Bill Johnson Mar 23 '19 at 20:37

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 $$L_p(0,1)$$.
• I hate to admit it but I don't really understand very much of this answer. I see how $E$ is a subspace of the $L_p$ space, but is there an easy way to see how the corresponding copies of $E_n$ are uniformly complemented? And even then I don't know very much about abstract $L_p$ spaces, so I don't see how to get $E$ itself complemented. And then how does it follow that $E$ is isomorphic to a complemented subspace of $L_p(0,1)$ just because it's complemented in an abstract $L_p$ space? Sorry to be so needy. Maybe you could just refer me to a book to read. – Ben W Mar 31 '19 at 0:13
• Thanks for putting in the time to explain the details. That's a huge help. Actually, I think there's a sticking point in the above argument where we try to say that the identity on $X$ factors through $\ell_p^\mathcal{U}$,but since I need only consider the reflexive case slight modifications together with the results from the Heinrich paper will take care of it. I should have known about passing to a separable sublattice. Not sure why I didn't think of that one. Thanks again! – Ben W Apr 4 '19 at 12:22