# Tsirelson Space Lecture #1

This is the inaugural blog post containing notes from my lectures given at UVa in the Fall of 2016. This first lecture was on Friday September 9th. Please comment or email if you have any questions or corrections.

Let ${c_{00}}$ be the space of all finitely supported sequences of real numbers. After defining a norm ${\|\cdot\|}$ on ${c_{00}}$ the resulting Banach space will be the completion of ${c_{00}}$ with respect to this norm. The standard unit vectors are denoted ${(e_i)}$ and will be a Schauder basis for this Banach space. We begin with the definition of the Schreier familes of finite order.

Let ${\mathcal{S}_{0}=\{\{n\}:n \in \mathbb{N}\}\cup \{\emptyset\}}$ (the singleton subsets of ${\mathbb{N}}$) and for each ${k \in \mathbb{N}}$ let

$\displaystyle \mathcal{S}_{k+1} =\{ \cup_{i=1}^n E_i : n\in \mathbb{N}, n \leqslant E_1 < E_2 < \cdots

Here ${E < F}$ means the maximum of ${E}$ is less than the minimum of ${F}$ (by convention ${\emptyset < E < \emptyset}$). For each ${k \in \mathbb{N}}$ define the following norm on ${c_{00}}$

$\displaystyle \|x\|_{\mathcal{S}_k} = \max_{ F \in\mathcal{S}_k} \sum_{i \in F} |x(i)|.$

Let ${X_{k}}$ be the completion of ${c_{00}}$ with the norm ${\|\cdot\|_{\mathcal{S}_k}}$. It is known that every (always infinite dimensional) subspace of ${X_k}$ contains an isomorphic copy of ${c_0}$ (this is called ${c_0}$-saturated). This proof is somewhat complicated but we can note that for ${X_1}$ the following block sequence is isomorphic to ${c_0}$: Let ${x_1= e_1}$, ${x_2 = \frac{1}{2}(e_2+e_3)}$, ${x_3 = \frac{1}{4}(e_4 + e_5 + e_6 +e_7)}$, etc. It is (relatively) easy to show that for each ${n \in \mathbb{N}}$ we have ${\| \sum_{i=1}^n x_i\|=1}$.

As a warm up exercise we will prove that ${X_{1}}$ does not contain ${\ell_1}$. We need the following theorem due to R.C. James. Our general Banach space ${X}$ is the completion of ${c_{00}}$ with respect to a norm and having as a Schauder basis ${(e_i)}$.

Theorem 1 The Banach space ${X}$ contains an isomorphic copy of ${\ell_1}$ if and only if for each ${\varepsilon>0}$ there is a normalized block sequence ${(x_i)}$ in ${X}$ so that for all ${(a_i) \in c_{00}}$ we have

$\displaystyle (1-\varepsilon) \sum_{i } |a_i| \leqslant \|\sum_i a_i x_i\|$

Moreover the analogous statement holds for ${c_0}$.

We have included the proof of the above theorem at the end of this post. We first prove the following.

Theorem 2 ${X_{1}}$ does not contain ${\ell_1}$.

This follows from the statement above that ${X_k}$ is ${c_0}$ saturated, however, it is not hard to prove so we will is supply the proof.

Proof: Let ${(x_i)_{i=0}^\infty}$ be the normalized block sequence equivalent to ${\ell_1}$ with ${\varepsilon=1/10}$. Let ${n \geqslant 2 \max \mbox{supp}~ x_0}$ and ${x=x_0 + \frac{1}{n} \sum_{i=1}^n x_i}$. Note that by our assumption ${18/10 \leqslant \|x\| \leqslant 2}$. If ${\mathcal{S}_1 \ni F \geqslant \max\mbox{supp} ~x_0}$ then ${\sum_{i \in F} |x(i)| \leqslant \|\frac{1}{n} \sum_{i=1}^n x_i\|_{\mathcal{S}_1} \leqslant 1}$. Thus ${|F|\leqslant \min F < \max\mbox{supp} ~x_0 \leqslant n/2}$. In this case

$\displaystyle \|x\| \leqslant \|x_0\| + \frac{1}{n} \sum_{k \in F} \sum_{i=1}^n |x_i(k)| \leqslant 1+ \frac{|F|}{n} \leqslant \frac{3}{2}.$

This contradicts the lower bound of ${18/10}$. $\Box$

We will now define Tsirelson’s norm and prove Tsirelson’s space ${T}$ is ${2-\varepsilon}$ distortable for each ${\varepsilon>0}$. Recall the definition of distortable.

Definition 3 Let ${X}$ be a Banach space with a norm ${\|\cdot\|}$. Then ${X}$ is ${C}$-distortable (for ${C > 1}$) if there is an equivalent norm ${|\cdot |}$ on ${X}$ so that in every infinite dimensional subspace ${Y}$ of ${X}$ there are vectors ${x,y \in Y}$ with ${\|x\|=\|y\|=1}$ and ${|x|/|y|\geqslant C}$.

The theorem of R.C. James from above can be rephrased as stating the ${\ell_1}$ and ${c_0}$ are not ${1+\varepsilon}$ distortable for any ${\varepsilon>0}$ (or simply not distortable). Also note that if ${X}$ is ${C}$-distortable for some ${C>1}$ then it every subspace is ${C}$-distortable. Therefore in order to show that a space does not contain ${\ell_1}$ is suffices to show that it is distortable (although it is a bit overkill!).

We will now construct Tsirelson’s space. We construct a norm on ${c_{00}}$ by constructing a set of vectors in ${c_{00}}$ are that will constitute the so-called norming set. A set ${W \subset c_{00}}$ is called a norming set if it contains ${e^*_i}$ (the standard unit vectors) and satisfies ${f \in W}$ implies ${-f \in W}$. If ${W}$ is a norming set then ${\|x\|_W= \sup\{ \langle f, x\rangle : f \in W \}}$ is a norm on ${c_{00}}$. Here ${\langle f, x\rangle}$ is the dot product and will also be denoted ${f(x)}$. Norming sets offer an generic way for constructing a space whose basis is the unit vector basis ${(e_i)}$.

We now construct the norming set of Tsirelson’s space ${T}$. Let ${W_0=\{e_i^*: i\in \mathbb{N}\}}$ and

$\displaystyle W_{n+1} = W_n \cup \{ \frac{1}{2} \sum_{i=1}^d f_i : (f_i)_{i=1}^d \mbox{ is a block sequence in } W_n \mbox{ and }(\min \mbox{supp} f_i)_{i=1}^d\in S_1\}.$

Then ${W_T= \cup_{n=1}^\infty W_n}$. We list is few important facts regarding ${W_T}$. The Tsirelson norm is ${\|x\|_T =\|x\|_{W_T}}$.

• (i) If ${f \in W_T}$ then either ${f \in W_0}$ or there is a block seqquence ${(f_i)_{i=1}^n \in W_T}$ so that ${f= \frac{1}{2} \sum_{i=1}^n f_i}$ and ${(\min \mbox{supp} f_i)_{i=1}^n \in S_1}$. Such block sequences ${(f_i)_{i=1}^n}$ are called admissible.
• (ii) If ${(f_i)_{i=1}^n}$ is an admissible block sequence in ${W_T}$ then ${f= \frac{1}{2}\sum_{i=1}^n f_i \in W_T}$.

The Tsirelson norm satifies the following implicit equation.

$\displaystyle \|x \|_T = \max\{\|x\|_\infty, \sup \frac{1}{2} \sum_{i=1}^n \|E_i\|_T: n\leqslant E_1

We note that we can replace the constant ${\frac{1}{2}}$ by any ${\theta\in (0,1)}$ and the resulting spaces will have all of the same structural properties, but will be totally incomparable for different ${\theta}$‘s.

It is useful for us to record the following somewhat technical notation. If ${f=\frac{1}{2}\sum_{i=1}^n f_i \in W_T}$ such that ${(f_i)_{i=1}^n}$ is an admissible block we say ${length(f)=n}$ (note that this is not uniquely defined). Suppose further that ${(x_j)_{j=1}^m}$ is a normalized block sequence and define the following two sets:

$\displaystyle A_1 = \{j : |\{i : \mbox{supp} f_i \cap \mbox{supp} x_j \not=\emptyset \}|\leqslant 1\}\mbox{ and }A_2 = \{1,\ldots ,m\} \setminus A_1.$

For ${j\in A_2}$ let ${I_j = \{i : \mbox{supp} x_j \cap \mbox{supp} f_i\not=\emptyset\}}$. As a convention, we will try to keep ${j,m}$ as indices for the vectors and ${i,n}$ as indices for functionals. Then

$\displaystyle f(\sum_{j=1}^m x_i) = \frac{1}{2} \sum_{i=1}^n f_i ( \sum_{j \in A_1} x_j) + \sum_{j\in A_2} \frac{1}{2} \sum_{i\in I_j}f_i ( x_j).$

Observe that ${|A_2| \leqslant \min\{n,m\}}$ (${m}$ is clear and ${n}$ because there is at least one unique ${i}$ for each ${j}$ in ${A_2}$). In addition,

$\displaystyle \sum_{j\in A_2} \frac{1}{2} \sum_{i\in I_j}f_i ( x_j) \leqslant \sum_{j \in A_2} \|x_j\| \leqslant \min\{n,m\}.$

In the other case apply the triangle inequality to see

$\displaystyle \frac{1}{2} \sum_{i=1}^n f_i ( \sum_{j \in A_1} x_j) \leqslant \frac{1}{2} \sum_{j \in A_1} \|x_j\| \leqslant \frac{|A_1|}{2}.$

Therefore

$\displaystyle f(\sum_{j=1}^m x_i) \leqslant \frac{|A_1|}{2} + |A_2| \leqslant \frac{m}{2} + \min\{m,n\}. \ \ \ \ \ (3)$

Theorem 4 Tsirelson’s space does not contain ${\ell_p}$ for any ${p}$ or ${c_0}$.

Proof: If a normalized block ${(x_j)_{j=0}^\infty}$ was ${C}$ isomorphic to ${\ell_p}$ for ${p>1}$ then, in particular, ${\| \sum_{j=n}^{2n-1}x_j\|\leqslant Cn^{1/p}.}$ However we can find ${f_i \in W_T}$ for each ${i}$ so that ${\mbox{supp} f_i = \mbox{supp} x_i}$ and ${f_i(x_i)=1}$. Then ${f= \frac{1}{2}\sum_{i=n}^{2n-1} f_i\in W_T}$ and

$\displaystyle f( \sum_{j=n}^{2n-1}x_j)\geqslant \frac{n}{2}.$

This is a contradiction of sufficiently large ${n}$.

In the case of ${\ell_1}$ we suppose that ${(x_i)_{i=0}^m}$ is a normalized block sequence with ${m/6 \geqslant \max \mbox{supp} x_0}$ and isomorphic to ${\ell_1}$ with ${\varepsilon = 1/10}$. Then applying the above equation for an arbitrary ${f}$ with length ${n}$

$\displaystyle \frac{18}{10} \leqslant f(x_0 + \frac{1}{m} \sum_{j=1}^m x_j) \leqslant f(x_0)+ \frac{1}{2}+ \frac{\min\{m,n\}}{m}.$

If ${f(x_0) =0}$ then the estimate is ${3/2}$. If ${f(x_0)\not=0}$ then ${n \leqslant \max \mbox{supp} x_0 \leqslant m/6}$ and so we have the estimate ${3/2 + 1/6}$ which is a contradiction. $\Box$

The final theorem for this post is to show that ${T}$ is distortable.

Theorem 5 Tsirelson’s space is ${2-\varepsilon}$ distortable for each ${\varepsilon>0}$.

Consider the following equivalent norm on ${T}$

$\displaystyle \|x\|_n = \sup\{ f(x) : f\in W_T,~ \mbox{length}(f) =n\}$

It is clear that ${\|x\|\geqslant \|x\|_n}$. By letting ${f_1}$ be the norming functional of ${x}$ and ${f=1/2(f_1)}$ we can see that ${\|x\|\leqslant 2 \|x\|_n}$. Therefore each norm ${\|\cdot\|_n}$ is ${2}$ equivalent.

Fix ${\varepsilon>0}$ and consider a block basic sequence ${(z_i)}$ in ${T}$. Krivine’s Theorem states that for any Banach space with a basis there is a ${p\in [1,\infty]}$ so that ${\ell_p}$ is block finitely representable in the space. For Tsirelson’s space it follows from our arguments from before that no ${\ell_p}$ other than ${p=1}$ can be a Krivine $p$. Thus for each ${m}$ and ${\eta>0}$ there is a block ${(x_j)_{j=1}^m}$ that is ${1+\eta}$ equivalent to the unit vector basis of ${\ell_1^m}$. Let ${x= \frac{1}{m} \sum_{j=1}^m x_j}$. Then ${\|x\| \geqslant 1/(1+\eta)}$. We need to estimate ${\|x\|_n}$ from above. Of course we have

$\displaystyle \|x\|_n \leqslant \frac{|A_1|}{2m}+ \frac{\min\{m,n\}}{m} \leqslant \frac{1}{2} + \frac{n}{m}.$

Choosing ${n/m<\eta}$ and ${y= x/\|x\|}$, we have the ${\|x/\|x\|\|_n \leqslant (1/2+\eta)(1+\eta)}$.

We build what is called a rapidly increading sequence (RIS) of ${\ell_1}$ averages ${(x_j)_{j=1}^n}$. That is, ${x_j = \frac{1}{m_j} \sum_{k=1}^{m_j} x_{j,k}}$ is an ${\ell_1^{m_j}}$ average with constant ${(1+\eta)}$ and

$\displaystyle m_j> \max \mbox{supp} x_{j-1} (n/\varepsilon).$

Let ${x= \frac{2}{n} \sum_{j=1}^n x_j}$. Then it is easily seen that ${\|x\|_n\geqslant 1/(1+\eta)}$. We need an upper estimate on ${\|x\|}$. Consider an arbitrary functional ${g \in W_T}$ of length ${d}$. Suppose $j_0$ is the smallest index so that ${g}$ intersects ${x_{j_0}}$. Then

$\displaystyle g(x) \leqslant \frac{2}{n} g(x_{j_0}) + \frac{2}{n}\sum_{j={j_0}+1}^n g(\frac{1}{m_j}\sum_{k=1}^{m_j} x_{j,k})$

Then ${d \leqslant \max \mbox{supp} x_{1} \leqslant m_j \varepsilon/n}$ for each ${j >{j_0}}$. Then using the inequality (3)

$\displaystyle g(\frac{1}{m_j}\sum_{k=1}^{m_j} x_{j,k}) \leqslant \frac{m_j}{2m_j} + \frac{d}{m_j} \leqslant \frac{1}{2} + \frac{\varepsilon}{n}.$

Simplifying and summing up yields

$\displaystyle g(x) \leqslant \frac{2}{n} + \frac{2}{n} ( \frac{n}{2} + \varepsilon) = 1+\frac{2}{n} + \eta.$

Now for sufficiently large ${n}$ we have ${\|x\|\leqslant 1+2\eta}$. Let ${z=x/\|x\|}$. Then ${\|z\|_n \geqslant (1+2\eta)/(1+\eta)}$. To finish the proof note that ${\|y\|=\|z\|=1}$ and

$\displaystyle \frac{\|z\|_n}{\|y\|_n} \geqslant \frac{(1+2\eta)}{(1/2+\eta)(1+\eta)^2} >2-\varepsilon$

for sufficiently small ${\eta}$.

In the above we use two facts: (1) For fixed ${n}$ we can find an ${m}$ large so that and ${\ell_1^m}$ average is not normed by a functional of length ${n}$. It misses it by the most it can which is ${2}$. (2) For this fixed ${n}$ we can build a vector that is normed by and functional of length ${n}$. This vector is an average of averages each of which are not normed by a functional of length ${n}$.

Proof of James’ Non-Distortion of ${\ell_1}$ Theorem: Let ${\varepsilon >0}$. Suppose that there are positive constants ${C,c}$ and ${(x_i)}$ is in a block sequence in a Banach space ${X}$ that satisfies

$\displaystyle c \sum_i |a_i| \leqslant \| \sum_i a_i x_i\| \leqslant C \sum_i |a_i|$

for all ${(a_i) \in c_{00}}$. Let ${K_n=\inf\{\|\sum_{i=n}^\infty a_i x_i\|: \sum_{i=n}^\infty |a_i|=1\}}$. Then ${K_n}$ is increasing to some ${K}$. Find a block sequence ${y_k=\sum_{j\in F_k} a_j x_j}$ that witnesses this stabilization. Let ${1>\delta>0}$ with ${1-\varepsilon < \delta^2}$ satisfying ${K_{\min F_1} \geqslant \delta K}$ and ${\|y_k\| < K/\delta}$ for all ${k\in \mathbb{N}}$. Let ${u_k=y_k/\|y_k\|}$. Then,

$\displaystyle \|\sum b_k u_k\|\geqslant \frac{\delta}{K} \|\sum b_k y_k\|= \frac{\delta}{K} \|\sum b_k \sum_{j\in F_k} a_j x_j\| \geqslant \frac{\delta K_{\min F_1}}{K} \sum |b_k|\sum_{j \in F_k} |a_j| = (1-\varepsilon ) \sum |b_k|.$

For the James’ ${c_0}$ the distortion theorem we we consider the values

$\displaystyle L_n = \sup \{\|\sum_{i=n} a_i x_i\| : \max|a_i|=1\}.$

and proceed somewhat analogously.