Products of free spaces and applications

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Products of free spaces and applications

Pedro L. Kaufmann

I BWB - Maresias 2014

Pedro L. Kaufmann Products of free spaces and applications

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Spaces of Lipschitz functions

Let (M,d) be a metric space, 0∈M.

Notation

Lip₀(M) :={f :M →R|f is Lipschitz, f(0) = 0}

is a Banach space when equipped with the norm

kfk_Lip := sup

x6=y

|f(x)−f(y)|

d(x,y) .

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A predual for Lip

₀

(M )

For eachx ∈M, consider the evaluation functional δ_x ∈Lip₀(M)^∗ porδxf :=f(x).

Definition/Proposition

F(M) :=span{δ_x|x ∈M}

is thefree space over M, and it is an isometric predual toLip0(M).

• Geometric interpretation: µ, ν finitely supported probabilities

⇒ kµ−νk_F is the earthmover distance betweenµ andν.

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A predual for Lip

₀

(M )

For eachx ∈M, consider the evaluation functional δ_x ∈Lip₀(M)^∗ porδxf :=f(x).

Definition/Proposition

F(M) :=span{δ_x|x ∈M}

is thefree space over M, and it is an isometric predual toLip0(M).

• Geometric interpretation: µ, ν finitely supported probabilities

⇒ kµ−νk_F is the earthmover distance betweenµ andν.

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Relationship between M and F (M )

Linear interpretation property

∀L:M →N Lipschitz with L(0_M) = 0_N ∃! ˆL:F(M)→ F(N) linearsuch that te following diagram commutes:

M −−−−→^L N



 y^δ

M



 y^δ

N

F(M) −−−−→ F^ˆ^L (N)

•In particular, M ∼^L N⇒ F(M)' F(N). The converse does not hold in general.

•(Godefroy, Kalton 2003) If X is Banach andλ≥1,X isλ-BAP

⇔ F(X) is λ-BAP.

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Relationship between M and F (M )

M −−−−→^L N



 y^δ

M



 y^δ

N

F(M) −−−−→ F^ˆ^L (N)

⇔ F(X) is λ-BAP.

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Relationship between M and F (M )

M −−−−→^L N



 y^δ

M



 y^δ

N

F(M) −−−−→ F^ˆ^L (N)

⇔ F(X) is λ-BAP.

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The structure of the free spaces is still a big mystery

•(Godard 2010) F(M) is isometric to a subspace of L1⇔M is a subset of anR-tree;

•(Naor, Schechtman 2007)F(R²) is not isomorphic to any subspace ofL₁;

•Let (X,k · k) be a finite dimensional Banach space. Then (Godefroy, Kalton 2003)F(X) has MAP and (H´ajek, Perneck´a 2013)F(X) admits a Schauder basis;

•Problem: (posed by H´ajek, Perneck´a)F ⊂Rⁿ⇒ F(F) has Schauder basis?

•Problem: F(R²)' F(R³)???

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The structure of the free spaces is still a big mystery

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The structure of the free spaces is still a big mystery

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The structure of the free spaces is still a big mystery

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The structure of the free spaces is still a big mystery

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The structure of the free spaces is still a big mystery

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Products of free spaces

Main Result

LetX be a Banach space. ThenF(X)'(P∞

n=1F(X))_`

1. Recall: LetM be a metric space,N ⊂M. N is a Lipschitz retract ofM if there is a Lipschitz functionL:M →N (called Lipschitz retraction) such that L|_N =Id. M is anabsolute Lipschitz retract if it is a Lipschitz retract of any metric space containing it.

Consequence 1: nonlinear Pe lczy´nski’s method for free spaces LetX be a Banach space andM be a metric space, and suppose thatX andM admit Lipschitz retractsN1 and N2, respectively, such thatX is Lipschitz equivalent toN₂ andM is Lipschitz equivalent toN1. Then F(X)' F(M).

Proof: Linear interpretation property + Main Result + classic Pe lczy´nski’s method applied to the free spaces.

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Products of free spaces

Main Result

n=1F(X))_`

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Products of free spaces

Main Result

n=1F(X))_`

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Products of free spaces

Consequence 2: free space of balls

LetX be a Banach space. ThenF(B_X)' F(X).

Proof: an adaptation of the proof of the main result.

Consequence 3: aboutF(c₀)

LetM be a separable metric space which is an absolute Lipschitz retract andF ⊂M a Lipschitz retract ofM such thatBc0

∼L F. ThenF(M)' F(c₀). In particular, if K is an infinite compact metric space, thenF(C(K))' F(c₀).

Proof: all separable metric spaces are Lipschitz equivalent to subsets ofc₀ (Aharoni 1974) + linear interpretation property + main result + classic Pe lczy´nski’s method.

•The statement in redwas already proved by Dutrieux and Ferenczi via a different method in 2006.

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Products of free spaces

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Products of free spaces

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Ingredient 1 to prove that F (X ) ' ( P

∞

n=1

F (X ))

_`

1

: linear extensions of Lipschitz functions

Definition

Given a pointed metric space (M,d,0) and a subset F containing 0, let us denote byExt₀(F,M)the set of all extensions

E :Lip0(F)→Lip0(M) which are linearand continuous. Let Ext₀^pt(F,M) be the subset of Ext0(F,M) consisting of all pointwise-to-pointwise continuous elements.

•(Brudnyi, Brudnyi 2007) There exists a 2-dimensional

Riemannian manifoldM and a subset F such thatExt₀(F,M) =∅.

•(Banach space example) Let X ⊂c0 be a subspace failing AP.

ThenF(X) is not complemented inF(c₀), thus Ext₀^pt(X,c₀) =∅.

•(Lancien, Perneck´a 2013/Lee Naor 2005) 0∈F ⊂Rⁿ⇒ Ext₀^pt(F,Rⁿ)6=∅.

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Ingredient 1 to prove that F (X ) ' ( P

∞

n=1

F (X ))

_`

1

: linear extensions of Lipschitz functions

Definition

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Ingredient 1 to prove that F (X ) ' ( P

∞

n=1

F (X ))

_`

1

: linear extensions of Lipschitz functions

Definition

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Ingredient 1 to prove that F (X ) ' ( P

∞

n=1

F (X ))

_`

1

: linear extensions of Lipschitz functions

Definition

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Ingredient 2: Metric quotients and a decomposition result

Definition: metric quotient

Let (M,d) be a metric space, F ⊂M be closed and nonempty, and let∼_F the equivalence relation on M which identifies all elements ofF. Then

d˜(˜x,y˜) := min{d(x,y),d(x,F) +d(y,F)},x˜,˜y ∈M/∼_F is a distance onM/∼_F, and (M/∼_F,d˜) is called thequotient metric space ofM by∼_F, which we denote byM/F.

•Lip₀(M/F)∼={f ∈Lip₀(M) :f|_F =constant}.

Lemma (quotient decomposition)

Let (M,d,0) be a pointed metric space andF be a subset

containing 0, and suppose that there existsE ∈Ext₀^pt(F,M). Then F(M)' F(F)⊕₁F(M/F).

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Ingredient 2: Metric quotients and a decomposition result

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Ingredient 2: Metric quotients and a decomposition result

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Ingredient 3: Kalton’s approximation results

Let (M,d,0) be a pointed metric space, and denoteB_r :=B_r(0).

K1

Givenr1, . . . ,rn,s1, . . . ,sn ∈Z,r1<s1 <r2 <· · ·<sn and γ_k ∈ F(B₂sk \B₂rk) and writingθ:= min_k=1,...,n−1{r_k+1−s_k}, then

kγ₁+· · ·+γ_nk_F ≥ 2^θ−1 2^θ+ 1

n

X

k=1

kγ_kk_F.

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Ingredient 3: Kalton’s approximation results

K2

Consider, for eachk ∈Z, the linear operator Tk :F(M)→ F(B₂k+1\B₂^k−1) defined by

Tkδx :=











0, if x∈B₂^k−1;

(log₂d(x,0)−k+ 1)δ_x, if x∈B₂k \B₂k−1; (k+ 1−log₂d(x,0))δx, if x∈B₂^k+1 \B₂^k;

0, if x6∈B₂k+1.

Then, for eachγ ∈ F(M), we have thatγ =P

k∈ZT_kγ unconditionally and

X

k∈Z

kT_kγk_F ≤72kγk_F.

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Proof that X Banach ⇒ F (X ) ' ( P

∞

n=1

F (X ))

_`

1

•First, note that, for each k ∈Z,

F(B₂k+1\B₂k)∼=F(B₂\B₁)' F(B₄\B₁)∼=F(B₂k+1\B₂k−1).

Strategy: Show that F(X),→^c (P∞

n=1F(B₂\B1))_`

1 and that F(X)←^c-(P∞

n=1F(B₂\B1))_`

1, then apply Pe lczy´nski’s method.

•(,→) Define^c T andS as follows:

F(X) ,→^T (P∞

n=1F(B₂k+1)\ F(B₂k−1))_`

1

S F(X)

(γ_k) 7→ P

k∈Zγ_k

γ 7→ (Tkγ)

ThenT ◦S is a projection onto T(F(X))' F(X).

•(←^c-) The speaker will explain.

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Proof that X Banach ⇒ F (X ) ' ( P

∞

n=1

F (X ))

_`

1

n=1F(B₂\B1))_`

F(X) ,→^T (P∞

n=1F(B₂k+1)\ F(B₂k−1))_`

1

S F(X)

(γ_k) 7→ P

k∈Zγ_k

γ 7→ (Tkγ)

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Proof that X Banach ⇒ F (X ) ' ( P

∞

n=1

F (X ))

_`

1

n=1F(B₂\B1))_`

F(X) ,→^T (P∞

n=1F(B₂k+1)\ F(B₂k−1))_`

1

S F(X)

(γ_k) 7→ P

k∈Zγ_k

γ 7→ (Tkγ)

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Proof that X Banach ⇒ F (X ) ' ( P

∞

n=1

F (X ))

_`

1

n=1F(B₂\B1))_`

F(X) ,→^T (P∞

n=1F(B₂k+1)\ F(B₂k−1))_`

1

S F(X)

(γ_k) 7→ P

k∈Zγ_k

γ 7→ (Tkγ)

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An application

Theorem (free spaces over compact riemannian manifolds)

LetM be a compact metric space such that eachx ∈M admits a neighborhood which is bi-Lipschitz embeddable inRⁿ. Then there is a complemented copy ofF(M) in F(Rⁿ).

If moreover the unit ball ofRⁿ is bi-Lipschitz equivalent to a Lipschitz retract ofM, then F(M)' F(Rⁿ). In particular, the Lipschitz-free space over anyn-dimensional compact Riemannian manifold equipped with its geodesic metric is isomorphic toF(Rⁿ).

For the proof we make use of the following:

Lang, Plaut 2001 (bi-Lipchitz embeddability intoRⁿ)

LetM be a compact metric space such that each point of M admits a neighborhood which is bi-Lipschitz embeddable inRⁿ. ThenM is bi-Lipschitz embeddable in Rⁿ.

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An application

Theorem (free spaces over compact riemannian manifolds)

LetM be a compact metric space such that eachx ∈M admits a neighborhood which is bi-Lipschitz embeddable inRⁿ. Then there is a complemented copy ofF(M) in F(Rⁿ).

If moreover the unit ball ofRⁿ is bi-Lipschitz equivalent to a Lipschitz retract ofM, then F(M)' F(Rⁿ). In particular, the Lipschitz-free space over anyn-dimensional compact Riemannian manifold equipped with its geodesic metric is isomorphic toF(Rⁿ).

For the proof we make use of the following:

Lang, Plaut 2001 (bi-Lipchitz embeddability intoRⁿ)

LetM be a compact metric space such that each point of M admits a neighborhood which is bi-Lipschitz embeddable inRⁿ. ThenM is bi-Lipschitz embeddable in Rⁿ.

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Products of free spaces and applications

Products of free spaces and applications

Spaces of Lipschitz functions

A predual for Lip

(M )

A predual for Lip

(M )

Relationship between M and F (M )

Relationship between M and F (M )

Relationship between M and F (M )

The structure of the free spaces is still a big mystery

The structure of the free spaces is still a big mystery

The structure of the free spaces is still a big mystery

The structure of the free spaces is still a big mystery

The structure of the free spaces is still a big mystery

The structure of the free spaces is still a big mystery

Products of free spaces

Products of free spaces

Products of free spaces

Products of free spaces

Products of free spaces

Products of free spaces

Ingredient 1 to prove that F (X ) ' ( P

F (X ))

: linear extensions of Lipschitz functions

Ingredient 1 to prove that F (X ) ' ( P

F (X ))

: linear extensions of Lipschitz functions

Ingredient 1 to prove that F (X ) ' ( P

F (X ))

: linear extensions of Lipschitz functions

Ingredient 1 to prove that F (X ) ' ( P

F (X ))

: linear extensions of Lipschitz functions

Ingredient 2: Metric quotients and a decomposition result

Ingredient 2: Metric quotients and a decomposition result

Ingredient 2: Metric quotients and a decomposition result

Ingredient 3: Kalton’s approximation results

Ingredient 3: Kalton’s approximation results

Proof that X Banach ⇒ F (X ) ' ( P

F (X ))

Proof that X Banach ⇒ F (X ) ' ( P

F (X ))

Proof that X Banach ⇒ F (X ) ' ( P

F (X ))

Proof that X Banach ⇒ F (X ) ' ( P

F (X ))

An application

An application

Muito obrigado!