## Notes of Gilles Pisier’s lecture nr 2

Today, I want to prove equivalence of superreflexibility and super RNP. Tomorrow, I will continue with the equivalence of these properties with the following statement.

There exist ${C>0}$, ${p>1}$, ${q<\infty}$ such that for all square integrable martingales ${f_n}$ with increments ${d_n :=f_{n}-f_{n-1}}$.

$\displaystyle \begin{array}{rcl} \frac{1}{C}\,(\sum\|d_n\|_{L_2 (B)}^{p})^{1/p}\leq\|\sum_{n=0}^{\infty}d_n\L_{L_2 (B)}\leq C\,(\sum\|d_n\|_{L_2 (B)}^{p})^{1/p}. \end{array}$

1. Super properties

Definition 1 Let ${X}$, ${B}$ be Banach spaces. Say ${X}$ is finitely representable in ${B}$ if ${\forall \epsilon>0}$, every finite dimensional subspace ${E\subset X}$ can be embedded in ${B}$ with distorsion ${<1+\epsilon}$.

Example 1 ${L_p}$ is finitely representable in ${\ell_p}$, whereas these spaces

Definition 2 Let ${P}$ be a property of Banach spaces. We say that ${B}$ has super-${P}$ if every space finitely representable in ${B}$ has ${P}$.

Say ${P}$ is a super property if ${P\Leftrightarrow}$ super-${P}$.

Example 2 Type ${p}$ (and cotype ${q}$) are super properties.

1.1. Ultraproducts

Definition 3 Let ${(B_i)_{i\in I}}$ be a family of Banach spaces. Let ${\mathcal{U}}$ be an ultrafilter on ${I}$. The ultraproduct

$\displaystyle \begin{array}{rcl} B_{\mathcal{U}}=(\prod_{i\in I}B_i)^{\mathrm{restr}}/\sim \end{array}$

where in the restricted product one considers bounded sequences ${(x_i)}$, and one mods out sequences such that ${\lim_{\mathcal{U}}|x_i|_{B_i}=0}$. ${\lim_{\mathcal{U}}|x_i|_{B_i}}$ defines a norm on it.

When ${B_i}$ are all the same, one speaks of the ultrapower ${B^{I}/\mathcal{U}}$.

Proposition 4 ${X}$ is finitely representable in ${B}$ iff there exist an index set ${I}$ and an ultra${X}$ embeds isometrically in the ultraproduct ${B^{I}/\mathcal{U}}$.

Ultraproducts are nice for clearing out ${\epsilon}$‘s from definitions and statements. They hide estimates, which may be a dommage since estimates are at the heart of functional analysis and its applications.

Proof requires two lemmas. The first one is easy.

Lemma 5 Let ${E}$ be a finite dimensional Banach space. Let ${\epsilon>0}$, ${\delta>0}$. Let ${u:E\rightarrow Y}$ be a linear map. Assume ${u}$ is ${1+\delta}$-isometric on an ${\epsilon}$-net ${S}$ of the unit sphere of ${E}$. Then ${u}$ is ${\frac{1+\delta}{1-\epsilon}}$-isometric on the whole of ${E}$.

Lemma 6 ${B^I /\mathcal{U}}$ is finitely representable in ${B}$.

Proof: Let ${E\subset B^I /\mathcal{U}}$ be finite dimensional. Pick a finite ${\epsilon}$-net ${S}$ in its unit sphere. Let ${e_1 ,\ldots,e_n}$ be a basis of ${E}$. View ${e_j}$ as the limit of a bounded sequence ${e_j (i)\in B}$. By definition, for all ${x=\sum_{j}\alpha_{j}e_j \in S}$,

$\displaystyle \begin{array}{rcl} |x|_{B_{\mathcal{U}}}\lim_{\mathcal{U}}|\sum_{j}\alpha_{j}e_j (i)|_{B_i}, \end{array}$

so for large enough ${i}$, ${u:x\mapsto \sum_{j}\alpha_{j}e_j (i)}$ is an ${1+\delta}$-isometric embedding on ${S}$, so ${u}$ is nearly isometric on ${E}$. $\Box$

Proof: of Proposition 4. In one direction, by transitivity of finite representability. Conversely, if ${X}$ is finitely representable in ${B}$, take ${I=}$ the set of pairs ${(E,\epsilon)}$ where ${E}$ runs through finite dimensional subspaces of ${X}$. For each ${i=(E,\epsilon)}$, choose a ${1+\epsilon}$-isometric embedding ${u_i}$ of ${E}$ in ${B}$. Then for every ${x\in X}$, the sequence ${(u_i (x))}$ represents a point ${u(x)\in B^I /\mathcal{U}}$, and ${u}$ is isometric. $\Box$

Proposition 7 If a property ${P}$ is stable under isomorphisms and subspaces, so does super-${P}$.

Proof: Let ${B}$ have super-${P}$, let ${X}$ be isomorphic to ${B}$. Then ${X^I /\mathcal{U}}$ is isomorphic to ${B^I /\mathcal{U}}$. ${B^I /\mathcal{U}}$ is finitely representable in ${B}$, so it has ${P}$. So does ${X^I /\mathcal{U}}$. Let ${Y}$ be finitely representable in ${X}$. Then ${Y}$ embeds isometrically in ${X^I /\mathcal{U}}$, so it has ${P}$. This proves that ${X}$ has super-${P}$. $\Box$

1.2. Equivalence of super reflexivity and super RNP

The following theorem describes non reflexivity as a gap between bounded variation and boundedness.

Theorem 8 (Ptak, Milman-Milman) For ${B}$ a Banach space, the following are equivalent:

1. ${B}$ is not reflexive.
2. For all ${0<\theta<1}$, there exists sequences ${x_n \in U_B}$ and ${\xi_n \in U_{B^*}}$ such that $\displaystyle \begin{array}{rcl} \forall i
3. There exists ${0<\theta<1}$, there exists sequences ${x_n \in U_B}$ and ${\xi_n \in U_{B^*}}$ such that $\displaystyle \begin{array}{rcl} \forall i
4. For all ${0<\theta<1}$, there exists a sequence ${x_n \in B}$ such that for all finitely supported sequences ${\alpha_n}$, $\displaystyle \begin{array}{rcl} \theta \sup_j |\sum_{i\geq j}\alpha_i|\leq|\sum\alpha_i x_i|\leq \sum|\alpha_i|. \end{array}$
5. There exists ${0<\theta<1}$, there exists a sequence ${x_n \in B}$ such that for all finitely supported sequences ${\alpha_n}$, $\displaystyle \begin{array}{rcl} \theta \sup_j |\sum_{i\geq j}\alpha_i|\leq|\sum\alpha_i x_i|\leq \sum|\alpha_i|. \end{array}$
6. For all ${0<\theta<1}$, there exists a sequence ${y_n \in B}$ such that for all finitely supported sequences ${\beta_n}$, $\displaystyle \begin{array}{rcl} \theta \sup_n |\beta_n|\leq|\sum\beta_n y_n|\leq \sum|\beta_n -\beta_{n+1}|. \end{array}$
7. There exists ${0<\theta<1}$, there exists a sequence ${y_n \in B}$ such that for all finitely supported sequences ${\beta_n}$, $\displaystyle \begin{array}{rcl} \theta \sup_n |\beta_n|\leq|\sum\beta_n y_n|\leq \sum|\beta_n -\beta_{n+1}|. \end{array}$
8. The inclusion ${\mathcal{V}_1 \rightarrow \ell_{\infty}}$ factors through ${B}$.

Proof: 6${\Leftrightarrow}$8: Extend each functional

$\displaystyle \begin{array}{rcl} \sum\beta_n y_n \rightarrow (\beta_n)_n \end{array}$

to ${B}$ by Hahn-Banach. This gives a map ${B\rightarrow\ell_{\infty}}$.

4${\Leftrightarrow}$6: Abel summation.

2${\Leftrightarrow}$4: Because ${\xi_j (\sum\alpha_i x_i)=\theta(\sum_{i\geq j}\alpha_i)}$.

1${\Leftarrow}$2: If ${B}$ is reflexive, one can assume that ${x_n}$ tends to ${x}$ in ${B}$ and ${\xi_n}$ tends to ${\xi}$ in ${B^*}$, then ${\xi(x)}$ is the limit of ${0}$ and ${\theta}$, contradiction.

1${\Rightarrow}$2 is the serious part. We use the following fact (local reflexivity, related to the fact that ${B^{**}}$ is finitely representable in ${B}$ for all ${B}$): For all ${x^{**}\in B^{**}}$, for all ${\epsilon>0}$, for all ${\xi_1 ,\ldots,\xi_n \in B^{*}}$, there exists ${x\in B}$ such that ${|x|<(1+\epsilon)|x^{**}}$ and for all ${i}$, ${x(\xi_i)=x^{**}(\xi_i)}$.

Assume ${B}$ is not reflexive. There exists ${x^{**}\in B^{**}}$ with norm ${1}$ and ${d(x^{**},B)>\theta}$. Since ${|x^{**}=1|}$, there exists ${\xi_1 \in U_{B^*}}$ such that ${x^{**}(\xi_1)>\theta}$. Then above fact yields ${x_1}$ such that ${|x_1|<(1+\epsilon)}$ and ${\xi_1 (x_1)>\theta}$. Since ${d(x^{**},\mathrm{span}(x_1))=\theta}$, by Hahn-Banach, there exists ${\xi_2 \in U_{B^*}}$ such that ${\xi_2 (x_1)=0}$ and ${x^{**}(\xi_2)=\theta}$. Again, there is ${x_2 \in B}$ such that ${|x_2|<(1+\epsilon)}$ and ${\xi_2 (x_2)=\theta}$, and so on… Observe that in addition, we can require that ${\xi_n (x)=0}$ for ${x}$ in any prescribed finite set ${A_n \subset B}$ (replace ${\mathrm{span}(x_1)}$ by ${\mathrm{span}(x_1 ,A_n)}$, and so on). It follows that, given ${\lambda>1}$, we can require that for all sequences ${\alpha_n}$,

$\displaystyle \begin{array}{rcl} \sup_{k}|\sum_{n=1}^{k}\alpha_n x_n|\leq\lambda|\sum\alpha_n x_n|. \end{array}$

Indeed, bad choices would live in a finite dimensional space, the unit ball of which can be discretized. So the argument gives 4. as well. $\Box$

Proposition 9 Super reflexivity is equivalent to super RNP.

Proof: Since reflexivity implies RNP, super reflexivity implies RNP. Conversely, need a lemma.

Lemma 10 Assume ${B}$ is not reflexive. Then there exists a space ${X}$, finitely representable in ${B}$, such that ${X}$ is the completion of ${L_1 [0,1]}$ with respect to a norm ${\|\cdot\|}$ such that

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

Proof: of Lemma. The finite dimensional analogue of this would be

$\displaystyle \begin{array}{rcl} \frac{1}{N}\sup_j |\sum_{i\geq j}\alpha_i|\leq |\sum\alpha_j 1_{I_j}|\leq \frac{1}{N} \sum|\alpha_i|. \end{array}$

which is one of the above equivalent properties. Now take an ultralimit ${X=\prod X_{N,\theta}/\mathcal{U}}$ in which finite sums converge to integrals. By construction, ${X}$ is finitely representable in ${B}$. $\Box$

Now any ${X}$ as in the Lemma fails to have RNP. Indeed, the following martingale diverges. Given an interval ${I}$, let ${I^-}$ and ${I^+}$ be its left and right halves. Given a sequence ${\omega=(\epsilon_n) \in \{\pm 1\}^{{\mathbb N}}}$, define recursively ${I_{\epsilon_1 ,\ldots,\epsilon_n}=I_{\epsilon_1 ,\ldots,\epsilon_{n-1}}^{\epsilon_n}}$. Start with ${f_0 =1}$, set ${f_n (\omega)= 2^n 1_{I_{\epsilon_1 ,\ldots,\epsilon_n}}}$. Then ${f_n \in X}$. ${f_n (\omega)-f_{n-1} (\omega)}$ is supported on ${I_{\epsilon_1 ,\ldots,\epsilon_{n-1}}}$, so ${\|f_n -f_{n-1}\|\geq 1/2}$. $\Box$

In Lemma 10, one can even require (with a different ${X}$)

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

We shall need this tomorrow when proving the following

Theorem 11 (James) ${B}$ non reflexive implies that ${B}$ contains squares, i.e. for all ${\delta>0}$, there exist unit vectors ${x}$, ${y}$ such that ${|x\pm y|\geq 2-\delta}$.

It follows that span${(x,y)}$ is ${1+\epsilon(\delta)}$-isometric to ${\ell_1^2}$. One cannot go beyond dimension ${2}$. Instead, existence of higher-dimensional cubes has to do with type, see Maurey’s course.

References