## Notes of Gilles Pisier’s lecture nr 3

1. James’ Theorem

1.1. Back to equivalence of superreflexivity and super RNP

Definition 1 Let

$\displaystyle \begin{array}{rcl} \theta_n (B)=\sup\{\theta\,;\,\exists x_i \in B,\,\xi_i \in B^* \textrm{ such that } \xi_j (x_i)=0\textrm{ or }\theta\textrm{ according wether }i

Here, sequences have length ${n}$.

$\displaystyle \begin{array}{rcl} t_n (B)=\sup\{\delta\,;\,\exists \textrm{ a }\delta\textrm{-separated dyadic tree if length }n\textrm{ in }U_B\}. \end{array}$

The equivalence of superreflexivity and super RNP can be formulated as follows.

Theorem 2 For a Banach space ${B}$, the following are equivalent:

1. ${B}$ is superreflexive.
2. ${\theta_n (B)}$ tends to ${0}$ as ${n}$ tends to ${\infty}$.
3. ${t_n (B)}$ tends to ${0}$ as ${n}$ tends to ${\infty}$.

Proof: Note that properties 2 and 3 are super properties. Go to ultraproducts where finite sequences become infinite sequences, and apply Ptak’s or Milman-Milman’s theorem. $\Box$

Linial: is there a quantitative connection between ${\theta_n}$ and ${t_n}$ ? Answer: the proof gives something…

Corollary 3 ${B}$ is superreflexive iff ${B^*}$ is superreflexive.

Indeed, property 2 is symmetric in ${B}$ and ${B^*}$ (up to a change in notation), ${\theta_n (B^*)\geq \theta_n (B)}$. Since superreflexive implies reflexive, ${\theta_n (B)=\theta_n (B^{**})\geq \theta_n (B^{*})}$.

1.2. Variant of James’ Theorem

James’ Theorem

Theorem 4 (James) In ${B}$ is non reflexive, then ${\forall \delta>0}$, there exist ${x_1}$, ${x_2 \in B}$ such that ${|x_1 \pm x_2|>2-\delta}$.

has a variant

Theorem 5 (James) In ${B}$ is non reflexive, then ${\forall n\geq 1}$, ${\forall \delta>0}$, there exist ${x_1 ,\ldots, x_n\in B}$ such that for all ${n}$ choices of signs ${\epsilon_i}$ of the form ${+++\cdots ----}$ (call them admissible), ${|\sum\epsilon_i x_i|>n-\delta}$.

This suggests the following

Definition 6 Say ${B}$ is ${J}$-convex is there exists ${n}$ such that ${\forall \delta>0}$, there exist ${x_1 ,\ldots, x_n\in B}$ such that for all admissible choices of signs ${\epsilon_i}$, ${|\sum\epsilon_i x_i|>n-\delta}$.

This is a super property, so it leads to

Theorem 7 ${B}$ is superreflexive iff ${B}$ is ${J}$-convex.

Proof: If ${B}$ is ${J}$-convex, then ${B}$ is reflexive. But ${J}$-convexity is super, so ${B}$ is superreflexive. Conversely, assume that ${B}$ is not ${J}$-convex. Then in some space ${X}$ finitely representable in ${B}$, there exists a bounded sequence ${x_n}$ such that for all admissible choices of signs,

$\displaystyle \begin{array}{rcl} |\sum\epsilon_i x_i|=n. \end{array}$

By Hahn-Banach, there exist unit functionals ${\xi_j}$ such that

$\displaystyle \begin{array}{rcl} \xi_j (x_1 +\cdots+x_j -x_{j+1}-\cdots-x_n)=n. \end{array}$

This implies that ${\xi_j (x_i)=1}$ or ${-1}$ according wether ${i or ${i\geq j}$. This shows that ${X}$ is not reflexive. $\Box$

Remark 1 There exists ${X}$ finitely representable in ${B}$ which is the completion of ${L^1 [0,1]}$ equipped with a norm that satisfies

$\displaystyle \begin{array}{rcl} \sup_{0

The unit ball of this space contains an infinite ${1}$-separated dyadic tree.

1.3. Proof of James’ Theorems \ref

and 5}

The proof we shall give is due to ergodic theorists Brunel and Sucheston around 1975. They realized that the following two properties clarify the proof a lot.

Definition 8 1. Say a sequence ${(x_i)\in B}$ is subsymmetric if for every sequence ${n(1)< n(2),\ldots}$ and ${\alpha_i \in{\mathbb R}}$,

$\displaystyle \begin{array}{rcl} |\sum\alpha_i x_i|=|\sum\alpha_i x_{n(i)}|. \end{array}$

2. Say a sequence ${(x_i)\in B}$ is additive if for every ${m\geq 1}$,

$\displaystyle \begin{array}{rcl} |\sum _i \alpha_i \frac{1}{m}(x_{mi+1}+\cdots+x_{mi+m})|\leq|\sum\alpha_i x_{n(i)}|. \end{array}$

We are given a sequence ${x_j}$ such that

$\displaystyle \begin{array}{rcl} \theta\sup_j |\sum_{i\geq j}\alpha_i|\leq|\sum\alpha_i x_i|\leq \sum|\alpha_i|. \end{array}$

Call this a James sequence. We start with the end of the proof.

Step 3. Assuming that ${(x_j)}$ is subsymmetric and additive, we proceed to extracting squares.

Let

$\displaystyle \begin{array}{rcl} x=x_1 -x_3 +x_5 -x_7 +\cdots, \end{array}$

${2m}$ terms, which we symbolically denote by ${(+0-0)}$. Similarly, ${y=(0+0-)^m}$, i.e.

$\displaystyle \begin{array}{rcl} y=x_2 -x_4 +x_6 -x_8 +\cdots. \end{array}$

Let ${r(m)=|x|}$. Then ${|x+y|=2r(m)}$, ${|x-y|\geq 2r(m)-2|x_1 -x_2|}$, so ${\hat{x}=x/r(m)}$ and ${\hat{y}=y/r(m)}$ do the job provided ${r(m)}$ tends to infinity.

Using sequences ${(+00-00)^m}$, ${(0+00-0)^m}$ and ${(00+00-)^m}$, one gets 3 vectors such that all ${x+y+z}$, ${x+y-z}$, ${x-y-z}$ have norm close to ${|x|}$ (but not other combinations of signs). And it works the same for ${n}$-tuples.

If ${r(m)}$ stays bounded, set ${d_m =x_{m-1}-x_m}$.

Claim: ${|\sum_{k\notin A }d_k|\leq |\sum_{k}d_k|}$ for all ${A\subset[m]}$.

Indeed, when summing, signs create cancellations. It follows that ${d_k}$ behaves unconditionally, i.e. for arbitrary signs ${\epsilon_k}$,

$\displaystyle \begin{array}{rcl} |\sum\epsilon_k \alpha_k d_k|\leq 2|\sum\alpha_k d_k|. \end{array}$

So the closure of span${(d_k)}$ is isomorphic to ${c_0}$. So ${c_0}$, and in particular, ${\ell_1}$, is coarsely finitely representable in ${B}$. This implies that ${\ell_1}$ is finitely representable in ${B}$, i.e. ${B}$ is not superreflexive.

Step 1. We can upgrade the given sequence to be subsymmetric.

Consider all ${|\sum\alpha_i x_{n(i)}|}$. By the James assumption, they stay bounded. In some ultraproduct, limits exist, leading to a norm on ${{\mathbb R}^{{\mathbb N}}}$. The completion is a Banach space finitely representable in ${B}$ where the canonical basis is James and subsymmetric.

Step 2. We can upgrade the subsymmetric sequence to be subsymmetric and additive.

In the same manner, consider all ${|\sum _i \alpha_i \frac{1}{m}(x_{mi+1}+\cdots+x_{mi+m})|}$. By subsymmetry, they form a nondecreasing family of norms on ${{\mathbb R}^{{\mathbb N}}}$. The limiting Banach space is finitely representable in ${B}$, and the canonical basis is James, subsymmetric and additive.

2. Uniform convexity and smoothness

Definition 9 The modulus of convexity of a Banach space ${B}$ is

$\displaystyle \begin{array}{rcl} \delta_{B}(\epsilon)=\inf\{1-|\frac{x+y}{2}|\,;\,x,\,y\in U_B,\,|x-y|\geq\epsilon\}. \end{array}$

The modulus of smoothness of ${B}$ is

$\displaystyle \begin{array}{rcl} \rho_{B}(t)=\sup\{|\frac{|x+ty|+|x-ty|}{2}-1|\,;\,x,\,y\in S_B\}. \end{array}$

Say ${B}$ is uniformly convex if ${\delta_{B}(\epsilon)>0}$ for all ${\epsilon>0}$. Say ${B}$ is uniformly smooth if ${\lim_{t\rightarrow 0}\rho_{B}(t)/t =0}$.

Remark 2 ${B}$ is UC iff ${B^*}$ is US.

2.1. UC versus superreflexivity

Theorem 10 (David Milman) UC implies superreflexive.

Proof: Let ${x^{**}\in U_{B^{**}}}$. Let ${x_i \in B}$ be a sequence (net) converging to ${x^{**}}$. Then ${|x_i|}$ tends to ${1}$. For the same reason, ${|\frac{x_{i}+x_{i+1}}{2}|}$ tends to ${1}$. UC implies So UC implies reflexive. Since UC is super, $\Box$

Note that UC is not isomorphism invariant. Answering a question of James, Enflo proved that

Theorem 11 (Enflo) ${B}$ is superreflexive iff ${B}$ is isomorphic to a UC space.

2.2. ${p}$-uniform convexity

We are heading towards a sharpening of Enflo’s theorem.

Definition 12 Say ${B}$ is ${q}$-UC if ${\exists C}$ such that for all ${\epsilon>0}$, ${\delta_B (\epsilon)\geq C\epsilon^q}$.

Say ${B}$ is ${p}$-US if ${\exists C}$ such that for all ${t}$, ${\rho_B (\epsilon)\leq C\epsilon^p}$.

Example 1 ${L_p}$ is ${p}$-UC for ${2\leq p<\infty}$, ${2}$-UC for ${1.

${L_p}$ is ${2}$-US for ${p\geq 2}$, ${p}$-US for ${1.

This is known as Clarkson inequality.

Theorem 13 (Pisier) The following are equivalent:

1. ${B}$ is superreflexive.
2. ${\exists q<\infty}$ such that ${B}$ is isomorphic to a ${q}$-UC space.
3. ${\exists p>1}$ such that ${B}$ is isomorphic to a ${p}$-US space.
4. ${\exists p>1}$, ${\exists q<\infty}$ such that ${B}$ is isomorphic to a space which is both ${q}$-UC and ${p}$-US.

Open problem: can one have ${p=q}$ in 2 and 3 ?

Lemma 14 Given ${\alpha\leq1}$ and ${C>0}$, the following are equivalent:

1. There exists a norm ${\|\cdot\|}$ on ${B}$ such that ${\alpha|x|\leq |x|\leq \|x\|}$ and $\displaystyle \begin{array}{rcl} \|\frac{x+y}{2}\|+\frac{1}{C^q}|\frac{x-y}{2}|\leq\frac{\|x\|^q +\|y\|^q}{2}. \end{array}$
2. For every ${B}$-valued dyadic martingale ${(f_n)}$ such that $\displaystyle \begin{array}{rcl} \alpha\mathop{\mathbb E}(|f_0|^q) +\frac{1}{C^q}\sum\mathop{\mathbb E}|f_n -f_{n-1}|^q \leq \sup_n \mathop{\mathbb E}(|f_n|^q). \end{array}$

2. implies the same statement for all martingales (but with a different ${C}$).

Example 2 Take ${\alpha=1}$. Then 1. is an equivalent definition of ${q}$-UC.

Proof: Dyadic martingale means that ${\Omega=\{\pm 1\}^{{\mathbb N}}}$ and ${f_{n-1}(\omega)=\frac{f_n (\omega,1)+f_n (\omega,-1)}{2}}$.

Define

$\displaystyle \begin{array}{rcl} \|x\|^q =\inf\{\sup \mathop{\mathbb E}|f_n|^q -\frac{1}{C}\sum \mathop{\mathbb E}|d_n|^q ,\,f_0 =x\}. \end{array}$

infimum over all dyadic martingales starting at ${x}$. To prove triangle inequality, given martingales starting at ${x}$ and ${y}$, one uses the martingale starting at ${y}$ which jumps to ${\frac{x+y}{2}}$ and continues with the ${x}$ martingale. $\Box$