# Equivalent definitions of Tsirelson type norms

I want to show that given $\mathcal M$ a compact family of $\wp(\mathbb{N}) = \{0,1\}^\mathbb N$ containing all the singletons and $0 < \theta < 1$, the norms (on $c_{00}$) defined by $$\lVert x\rVert_{(\mathcal M,\theta)} = \max \left\{\lVert x\rVert_\infty, \theta\sup\sum_{i=1}^n \lVert E_ix\rVert_{(\mathcal M,\theta)}\right\},$$ where the supremum is taken over all of the $\mathcal M$-admissible families of $\mathbb N$, and $$\lVert x\rVert_{W(\mathcal M,\theta)}=\sup\{f(x) : f\in W(\mathcal M,\theta)\}$$ are equal for every $x$.

Attempts:

In $(\geq)$, I managed to show a slightly weaker inequality:

Let $x = (x(n))_n\in c_{00}$. Given $E_1<\cdots<E_n$ an $\mathcal M$-admissible family, for each $1\leq i\leq n$, define $$f_i = \theta\sum_{k\in E_i}\varepsilon_ke_k^\ast\in W(\mathcal M,\theta),$$ where $\varepsilon_k = \text{sgn}(e_k^\ast(x))$. Hence, $$f_i(x) = \theta\sum_{k\in E_i}\lvert x(k)\rvert \geq\theta\lVert{E_ix}\rVert_{(\mathcal M,\theta)},$$ and since $f = \theta\sum_{i=1}^n f_i\in W(\mathcal M,\theta)$, we have $\lVert x\rVert_{W(\mathcal M,\theta)}\geq \theta\lVert x\rVert_{(\mathcal M,\theta)}$.

In the reverse inequality I tried an induction argument:

Again, let $x\in c_{00}$. It is easy to show that for each $\varepsilon e_k^\ast$, eith $\varepsilon = \pm1$, the inequality holds. Then, given a functional $f = \theta(f_1+\cdots f_n)\in W(\mathcal M,\theta)$, with $\{f_i\}_{i=1}^n$ an $\mathcal M$-admissible family of functionals, suppose that for each $f_i$ there is an $\mathcal M$-admissible family $\mathcal E_i = \left\{E_j^i\right\}_{j=1}^{n_i}$ such that $$f_i(x)\leq\theta\sum_{j=1}^{n_i}\lVert E_j^ix\rVert_{(\mathcal M,\theta)}.$$ Notice we can suppose, without loss of generality that, $\mathcal E_i\subseteq\text{supp} f$. For each $1\leq i\leq n$, let $\{m_j^i\}_{j=1}^{n_i}\in\mathcal M$ satisfying $$m_1^i\leq E_1^i < \cdots < m_{n_i}^i \leq E_{n_i}^i.$$ Since we have $\text{supp} f_2 <\cdots \text{supp} f_n$, we have $$m_1^1\leq E_1^1 < m_2^1\leq E_2^1 < \cdots < m_{n_1}^1\leq E_{n_1}^1 < m_1^2\leq E_1^2 <\cdots < m_j^i\leq E_j^i <\cdots < m_{n_n}^n \leq E_{n_n}^n.$$ If, somehow, I could be sure that $\left\{m_j^i\right\}_{i,j}\in\mathcal M$, i.e., if the collection $\left\{E_j^i\right\}_{i,j}$ were $\mathcal M$-admissible then the induction would be complete.

Basic Definitions:

1. A family $E_1 < \cdots < E_n$ is said to be $\mathcal M$-admissible if there exists $\{m_i\}_{i=1}^n\in\mathcal M$ such that $m_1\leq E_1 <\cdots < m_n \leq E_n$.
2. If we see $c_{00} = \text{span } \{e_n^\ast\}_{n\in\mathbb N}$, then $W(\mathcal M,\theta)$ is the minimal subset of $c_{00}$ containing $\pm e_n^\ast$ and if $f_1 < \cdots < f_n$ is $\mathcal M$-admissible (i.e., $\text{supp }f_1 < \cdots <\text{supp }f_n$ is $\mathcal M$-admissible) then $f = \theta(f_1+\cdots+f_n)\in W(\mathcal M,\theta)$.

Please check every step to make sure that I didn't miss anything. If you inductively define a sequence of sets $(W_n)_{n\in\mathbb{N}}$ as \begin{eqnarray}W_0&=&\{\pm e_n^*: n\in\mathbb{N}\} \\ W_{n+1}&=&W_n\cup\Bigg\{\theta \sum_{i=1}^m f_i: f_1<\ldots< f_m \ \text{admissible family with each } f_i\in W_n \Bigg\},\end{eqnarray} then $W=\bigcup_{n=0}^\infty W_n$. Lets denote the norm induced by $W$ by $\|\cdot\|_W$ and the other one by simply $\|\cdot\|$. Usually in these types of arguments you need to write down the inductive hypothesis carefully.

• We first show inductively that $|f(x)|\leq \|x\|$ for every $x\in c_{00}$ and every $f\in W$. If $f\in W_0$, then this clearly holds. Inductive Hypothesis: Suppose that $|f(x)|\leq \|x\|$ for every $x\in c_{00}$ and every $f\in W_{n}$. We will show that the same holds for every $f\in W_{n+1}$: Let $f=\theta \sum_{i=1}^m f_i$ with $f_1<\ldots<f_m \in W_n$ admissible. Lets also set $E_i=\mathrm{spt} f_i$ for $i=1,\ldots, m.$ By the inductive hypothesis, $|f_i(E_ix)|\leq \|E_ix\|$ for every $i$, so \begin{eqnarray} |f(x)|\leq \theta \sum_{i=1}^m |f_i(x)| = \sum_{i=1}^m |f_i(E_ix)| \leq \theta\sum_{i=1}^m \|E_ix\|\leq \|x\|. \end{eqnarray} Therefore, $\|x\|_W\leq \|x\|$ for every $x\in c_{00}$.
• We will show that $\|x\|\leq \|x\|_W$ for every $x\in c_{00}$, using induction on the cardinality of the support of $x$, $|\mathrm{spt}x|$. If $|\mathrm{spt}x|=1$, then it clearly holds. Suppose that it holds for every $x$ with $|\mathrm{spt}x|\leq n$ and pick an $x$ with $|\mathrm{spt}x|=n+1$. If $\|x\|=\|x\|_0$ it clearly holds. If not, then for every $\varepsilon>0$, there exists an admissible family $E_1<\ldots<E_m$, such that $$\theta \sum_{i=1}^m\|E_ix\|>\|x\|-\varepsilon.$$ Without loss of generality we may also assume that $m\geq 2$ and that all the $E_ix$'s are non empty, so we can apply the inductive hypothesis on them to obtain a family of functionals $f_1, \ldots, f_m \in W$ such that $$|f_i(E_ix)|>\|E_ix\|-\frac{\varepsilon}{\theta m}, \ \forall i=1, \ldots, m.$$ We may also assume that $\textrm{spt} f_i=E_i$ for every $i$, thus the family $f_1<\ldots< f_m$ is admissible and the function $f=\theta \sum_{i=1}^m f_i$ belongs in $W$. You can also make all the $f_i(E_ix)$ positive. So, \begin{eqnarray} \|x\|_W\geq |f(x)| =\big|\theta \sum_{i=1}^m f_i(x)\big| = \theta \sum_{i=1}^m |f_i(E_ix)| >\theta \sum_{i=1}^m\|E_ix\|-\varepsilon>\|x\|-2\varepsilon. \end{eqnarray} Since this holds for every $\varepsilon>0$, you obtain the inverse inequality.

Edit: (Addressing why we may assume that $m\geq 2$.)

If your family $\mathcal{M}$ is hereditary (which means that for every $M\in \mathcal{M}$ and $\emptyset \neq N\subseteq M$, then $N\in \mathcal{M}$ as well) we can show that for every $x\in c_{00}$ and $E\subseteq \mathbb{N}$, $\|Ex\|\leq \|x\|$: While computing the norm of $Ex$, pick an admissible family $E_1<\ldots<E_n$ and for every $i$ set $A_i=E_i\cap E$. Consider only the $A_i$'s which are nonempty (if any). Then the family $A_{k_1}<\ldots<A_{k_m}$, $m\leq n$ is also admissible (because $\mathcal{M}$ is hereditary), so $$\sum_{i=1}^n\|E_i (Ex)\|=\sum_{i=1}^m\|(E_{k_i}\cap E)x\|=\sum_{i=1}^m\|A_ix\|$$ and the last expression appears in the computation of the norm of $x$. Also $\|Ex\|_\infty \leq \|x\|_\infty$, so if you combine these two observations you get that $\|Ex\|\leq \|x\|$.

Now, for the argument, if $m=1$, then it would have to be that $\theta \|E_1x\|>\|x\|-\varepsilon$. But since $\theta\|E_1x\|\leq \theta \|x\|$, this would imply that $\varepsilon >(1-\theta)\|x\|$. So if you pick a sufficiently small $\varepsilon$ in the beginning of the argument, you can make sure that $m\geq 2$.

• It seems ok, I just need a bit more of time to check the second inequality before accepting the answer Sep 20, 2017 at 1:17
• Absolutely, these arguments can be very tricky, especially when you start building more complex Tsirelson type norms. Sep 20, 2017 at 9:14
• I'm sorry for taking so long, but I don't think we can suppose that $m\geq 2$ on the second inequality, since the family $\mathcal{M} = \{ \{n\} : n\in\mathbb{N}\}$ is compact and there is no $\mathcal M$-admissible family with more than 2 elements. However, I think this can be fixed by simply assuming that $|\text{supp } E_ix| < |\text{supp } x|$, for all $i$, since by induction on the support of $x$, we can show that $\|x\| = \|x\|'$ ... Sep 22, 2017 at 13:46
• where $$\|x\|' = \|x\|_0,\text{ if } |\text{supp x}| = 1$$ $$\|x\|' = \max\{\|x\|_0,\theta\sup\sum_{i=1}^n\|E_ix\|'\}$$ where the sup is taken over all the admissible families $E_1 < \cdots < E_n$ such that $|\text{supp }E_ix| < |\text{supp } x|$, for all $i$. Sep 22, 2017 at 13:46
• @user8485 In the case you wrote ($\mathcal{M}=\mathcal{A}_1$), we have that $\|x\|=\|x\|_0$, so we don't need to examine the second case at all. The whole argument with the $m$ was for the case where $\|x\|>\|x\|_0$. I I have made an edit to address it in more detail. From what I remember these families are assumed to be compact, hereditary and consisting of finite sets. Since you don't mention it, maybe you don't need the hereditary property here, but I am not sure. Try to see if it works with it and then you revisit it to see if you can omit it. Sep 22, 2017 at 19:49