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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!

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  • $\begingroup$ 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. $\endgroup$ – Bunyamin Sari Mar 23 '19 at 17:56
  • $\begingroup$ @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. $\endgroup$ – Ben W Mar 23 '19 at 18:20
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    $\begingroup$ @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. $\endgroup$ – Bill Johnson Mar 23 '19 at 20:37
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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.

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  • $\begingroup$ Cool, I was hoping it had a negative answer. Thank you! $\endgroup$ – Ben W Mar 23 '19 at 16:55
  • $\begingroup$ 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. $\endgroup$ – Ben W Mar 31 '19 at 0:13
  • $\begingroup$ 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! $\endgroup$ – Ben W Apr 4 '19 at 12:22

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