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Let $X$ be a Banach space with a $1$-subsymmetric basis $(e_i)$. I'm trying to understand why it is the case that for any $x = \sum_{i=1}^\infty x_i e_i \in X$, any strictly increasing sequence $(n_i) \subset \mathbb{N}$, and any $(\epsilon_i) \subset \{-1,1\}$ it follows that

$$\|\sum_{i=1}^\infty x_i e_i\| = \|\sum_{i=1}^\infty \epsilon_i x_i e_{n_i}\|.$$

By definition of the subsymmetric constant we have that

$$\|\sum_{i=1}^\infty x_i e_i\| \geq \|\sum_{i=1}^\infty \epsilon_i x_i e_{n_i}\|.$$

I'm having trouble getting the reverse inequality. I guess that's sort of trivial, but I don't see it. I know that the fact that $(e_i)$ is $1$-subsymmetric implies that it is $1$-unconditional, and this should be enough to get the reverse inequality... Any hint would be appreciated.

ragrigg
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  • If $(e_n)$ is $1$-unconditional, if for any $N$ and any $|a_1|\le|b_1|$, $\ldots$, $|a_N|\le|b_N|$, you have, $\Vert\sum_{i=1}^N a_i e_i\Vert\le \Vert\sum_{i=1}^N b_i e_i\Vert$. The equality follows easily from this. (Just use the preceding twice. Once with $a_i=\epsilon_i x_i$, $b_i=x_i$. Then with the $a_i$ and $b_i$ interchanged.) – David Mitra Dec 25 '13 at 00:25
  • You might find Proposition 3.4.4 here useful. – David Mitra Dec 25 '13 at 00:51
  • @DavidMitra: I see what you pointed out. However, I had a typo in the equality. On the right hand side I should've written $e_{n_i}$, not $e_i$. I think in that case your argument does not work anymore. Any further ideas? – ragrigg Dec 26 '13 at 04:34

1 Answers1

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Here's an outline (perhaps more than needed) taken from Lindenstrauss and Tzafriri, Classical Banach Spaces I:

Define a new norm $\Vert\ \Vert_0$ on $X$ via, for $x=\sum\limits_{n=1}^\infty a_i e_i$ $$ \Vert x\Vert_0=\sup_{(\epsilon_i), \{n_i\}}\Bigl\Vert\sum_{i=1}^\infty a_i\epsilon_i e_{n_i}\Bigr\Vert $$ where $(\epsilon_i)$ is a sequence of signs and $\{n_i\}$ is a strictly increasing sequence of positive integers.

Trivially, $\Vert x\Vert\le \Vert x\Vert_0$. Using this and the $1$-subsymmetry of $(e_i)$ (that is your definition), one concludes $$ \Vert x\Vert =\Vert x\Vert_0,\ \ x\in X.$$ (more generally, if a symmetric basis $(x_n)$ has sub-symmetric constant $C$, then $\Vert x\Vert\le\Vert x\Vert_0\le C\Vert x\Vert$).

Now for a fixed sequence of signs $(\epsilon_i)$ and a fixed strictly increasing sequence of positive integers $(n_i)$ verify that for $x=\sum\limits_{n=1}^\infty a_n e_n$ and $y=\sum\limits_{i=1}^\infty a_i \epsilon_i e_{n_i}$ one has $\Vert x\Vert_0 = \Vert y\Vert_0$.

David Mitra
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  • I see, David. The way I verified that $| x |0 = | y |_0$ is basically by using the fact (since our basis is unconditional) that, for any permutation $\pi: \mathbb{N} \to \mathbb{N}$, $| \sum{i=1}^\infty a_i \epsilon_i e_{\pi(i)} | = | \sum_{i=1}^\infty a_i \epsilon_i e_i |$. Do you see it in a different way? – ragrigg Dec 27 '13 at 17:03
  • @ragrigg Yes. You can write $y=\sum_{i=1}^\infty b_i e_i$ where $b_j= a_j \epsilon_j$ for $j=n_i$ and $b_j=0$ otherwise. Then it's easy to see $\Vert x \Vert_0 =\sup_{(\sigma_i),{m_i}}\Vert \sum_{i=1}^\infty \sigma_i a_i e_{m_i}\Vert =\Vert y\Vert_0$ (of course, $(\sigma_i)$ is a sequence of signs and ${m_i}$ is a strictly increasing sequence of positive integers). – David Mitra Dec 27 '13 at 17:45