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I want to define the Tsirelson norm $\lVert\,\cdot\,\rVert_\mathcal{T}$on $c_{00}$ $$\lVert x\rVert_\mathcal{T} = \max\left\{\lVert x\rVert_0, \frac12\sup_{n\leq E_1 <\cdots <E_n}\sum_{i=1}^n\lVert E_ix\rVert_\mathcal{T}\right\},\quad\forall x\in c_{00},$$ as the limit of the sequence of norms $(\lVert\,\cdot\,\rVert_n)_{n\geq 0}$ where $$\lVert x\rVert_0 = \lVert x\rVert_\infty$$ $$\lVert x\rVert_{n+1} = \max\left\{\lVert x\rVert_0,\frac12\sup_{n\leq E_1 <\cdots <E_n}\sum_{i=1}^n\lVert E_ix\rVert_n\right\},$$ but I'm having troubles showing that this sequence is increasing. For context, this is an outline of what I'm trying:

I'm trying to prove the inequality inductively, and the base case $\lVert\,\cdot\,\rVert_1 \geq \lVert\,\cdot\,\rVert_0$ is immediate. Suppose $n+1$ is such that $\lVert\,\cdot\,\rVert_{n+1}\geq\lVert\,\cdot\,\rVert_n$. Then, given a family $n\leq E_1<\cdots<E_n$ of natural numbers the case where $n < E_1$ is easy, because it is sufficient to choose an arbitrary $E_{n+1} > E_n$ and we have $$\sum_{i=1}^{n+1}\lVert E_ix\rVert_{n+1}\geq\sum_{i=1}^n\lVert E_ix\rVert_{n+1}\geq\sum_{i=1}^{n}\lVert E_ix\rVert_n,$$ where the last equality follows from the induction hypothesis, and hence $\lVert x\rVert_{n+2}\geq\lVert x\rVert_{n+1}$. The problem is the case where $n\in E_1$. I even tried to separate into further cases, where $E_1 = \{n\}$ and $\{n\}\varsubsetneq E_1$, but it was useless. I'd appreciate some help with this.


Basic definitions:

  1. Given two sets $E,\,F$ of $\mathbb{N}$, we say that $E\leq F$ if $\sup E\leq \min F$.
  2. Given $E\subseteq\mathbb{N}$ and $x = (x(n))_n\in c_{00}$, we denote by $Ex$ the sequence $$ Ex = \left(x(n)\chi_E(n)\right)_n = \sum_{j\in E} x(j)e_j \in c_{00},$$ where $e_j$ denotes the $j$-th unit vector.
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  • $\begingroup$ Is the supremum taken over all $E_1, \ldots, E_n \subseteq \mathbb{N}$? Like this: $$\lVert x\rVert_{n+1} = \max\left\{\lVert x\rVert_0,\frac12\sup\left\{\sum_{i=1}^n\lVert E_ix\rVert_n : E_1, \ldots, E_n \subseteq \mathbb{N}\text{ such that }\{n\}\leq E_1 <\cdots <E_n\right\}\right\}$$ $\endgroup$ Sep 19, 2017 at 14:43
  • $\begingroup$ @mechanodroid yes $\endgroup$ Sep 19, 2017 at 14:48
  • $\begingroup$ Also, I don't get how exactly the recursion for $\|x\|_\mathcal{T}$ works. My guess: the support of $E_ix$ is a subset of the support of $x$, so in finitely many iterations you get a sequence with only one nonzero element, which you can then calculate with $\|x\|_\infty$. Did you mean something like this: $$\lVert x\rVert_\mathcal{T} = \max\left\{\lVert x\rVert_\infty, \frac12\sup_{n\in\mathbb{N}}\left(\sup_{n\leq E_1 <\cdots <E_n}\sum_{i=1}^n\lVert E_ix\rVert_\mathcal{T}\right)\right\}$$ $\endgroup$ Sep 19, 2017 at 14:49

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In the Figiel and Johnson construction the recursive formula is $$\lVert x\rVert_{n+1} = \max\left\{\lVert x\rVert_n,\frac12\sup_{k\leq E_1 <\cdots <E_k}\sum_{i=1}^n\lVert E_ix\rVert_n\right\},$$ where both your admissible family $E_1, \ldots, E_k$ and $k\in \mathbb{N}$ vary. Also, if I am not mistaken, this is equal to $$\lVert x\rVert_{n+1} = \max\left\{\lVert x\rVert_0,\frac12\sup_{k\leq E_1 <\cdots <E_k}\sum_{i=1}^n\lVert E_ix\rVert_n\right\},$$ so we can replace $\|x\|_n$ by $\|x\|_0$ in the inductive formula.

What looks awkward in the formula you wrote, is that $n$ also restricts your admissible family: For every $x\in c_{00}$ if you pick an $n>\text{maxsupp} x$, then none of the sets $E_1<\ldots <E_n$ meets the support of $x$, so it has to be that $\|x\|_m=\|x\|_0$, for every $m\geq n$, right? If so, then by taking the limit, your norm becomes just the supremum norm.

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    $\begingroup$ You're absolutely right. I got my definition from Argyros and Todorcevic's Ramsey Methods in Analysis, and I suppose it was just a typo since the $n$-index is free to vary on the Tsirelson norm but fixed on the $n$-norm. With the right definition it is easy to show that the norms are increasing. $\endgroup$ Sep 19, 2017 at 18:07

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