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<j\textrm{ or }i\geq j\}. \end{array}

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<j} 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<s<1}|\int_{0}^{s}f(t)\,dt|+|\int_{s}^{1}f(t)\,dt|\leq \|f\|\leq f|_{L_1}. \end{array}

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.


\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<p\leq 2}.

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

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}}.


\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


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