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<j,\,\xi_j (x_i)=0, \quad \forall i\geq j,\,\xi_j (x_i)=\theta. \end{array}
  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<j,\,\xi_j (x_i)=0, \quad \forall i\geq j,\,\xi_j (x_i)=\theta. \end{array}
  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<s<1}|\int_{s}^{1}f(t)\,dt|\leq \|f\|\leq f|_{L_1}. \end{array}

 

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

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

 

 

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About metric2011

metric2011 is a program of Centre Emile Borel, an activity of Institut Henri Poincaré, 11 rue Pierre et Marie Curie, 75005 Paris, France. See http://www.math.ens.fr/metric2011/
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