# 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. Mar 23, 2019 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. Mar 23, 2019 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. Mar 23, 2019 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. Mar 31, 2019 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! Apr 4, 2019 at 12:22