Random embedding of pn into rN - 0 < r < p < 2 2pp+2 r 1

0 downloads 0 Views 262KB Size Report
Random embedding of ln p into lN r. 0 < r < p < 2. 2p p+2. ≤ r ≤ 1. Omer FRIEDLAND. Olivier GUÉDON. Université Pierre et Marie CURIE. Université Paris-Est ...
Random embedding of `np into `Nr 0 0 1+ε

`np ,→ `N1 ,

N = C(p, ε)n.

• More precisely, they gave an explicit definition of a random

operator, T : `np → `N1 , and proved that : 1 − ε ≤ |Tα|1 ≤ 1 + ε ,

∀α ∈ Spn−1 .

History

• Figiel – Lindenstrauss – Milman ’77 proved,

following Milman’s approach to Dvoretsky theorem : 1+ε

`n2 ,→ `N1 ,

N = C(ε)n.

• Kashin ’77, with a different approach, proved : C(η)

`n2 ,→ `N1 , where η > 0.

N = (1 + η)n,

History

• Figiel – Lindenstrauss – Milman ’77 proved,

following Milman’s approach to Dvoretsky theorem : 1+ε

`n2 ,→ `N1 ,

N = C(ε)n.

• Kashin ’77, with a different approach, proved : C(η)

`n2 ,→ `N1 , where η > 0.

N = (1 + η)n,

History

almost-isometric ε−embed isomorphic with N = (1 + η)n

1≤p 0.

• Large deviation is useful for almost-isometric results and

for obtaining upper bounds in general.

Ideas behind this result • Fix α ∈ Spn−1 . We have

P{ |Tα|1 − |α|p ≥ t} ≤ 2 exp(−cp Ntq ), note that |α|p = 1. • It means

1 − t ≤ |Tα|1 ≤ 1 + t t=ε>0 ,

N = Cn.

• But

|Tα|1 ≤ 1 + t ,

∀t > 0.

• Large deviation is useful for almost-isometric results and

for obtaining upper bounds in general.

Ideas behind this result • Fix α ∈ Spn−1 . We have

P{ |Tα|1 − |α|p ≥ t} ≤ 2 exp(−cp Ntq ), note that |α|p = 1. • It means

1 − t ≤ |Tα|1 ≤ 1 + t t=ε>0 ,

N = Cn.

• But

|Tα|1 ≤ 1 + t ,

∀t > 0.

• Large deviation is useful for almost-isometric results and

for obtaining upper bounds in general.

Ideas behind this result • Fix α ∈ Spn−1 . We have

P{ |Tα|1 − |α|p ≥ t} ≤ 2 exp(−cp Ntq ), note that |α|p = 1. • It means

1 − t ≤ |Tα|1 ≤ 1 + t t=ε>0 ,

N = Cn.

• But

|Tα|1 ≤ 1 + t ,

∀t > 0.

• Large deviation is useful for almost-isometric results and

for obtaining upper bounds in general.

Ideas behind this result

P{ |Tα|1 − |α|p ≥ t} ≤ 2 exp(−cp Ntq ). • In our situation : N = (1 + η)n and t ∈ (0, 1). • We may assume in addition that α ∈ Spn−1 has a small

support : | supp(α)| ≤ δn. Cn X n (1 + η)n X δn • It means that for such vectors with δ ' C1 , we may use this

large deviation inequality again, and have a lower bound.

Ideas behind this result

P{ |Tα|1 − |α|p ≥ t} ≤ 2 exp(−cp Ntq ). • In our situation : N = (1 + η)n and t ∈ (0, 1). • We may assume in addition that α ∈ Spn−1 has a small

support : | supp(α)| ≤ δn. Cn X n (1 + η)n X δn • It means that for such vectors with δ ' C1 , we may use this

large deviation inequality again, and have a lower bound.

Ideas behind this result

P{ |Tα|1 − |α|p ≥ t} ≤ 2 exp(−cp Ntq ). • In our situation : N = (1 + η)n and t ∈ (0, 1). • We may assume in addition that α ∈ Spn−1 has a small

support : | supp(α)| ≤ δn. Cn X n (1 + η)n X δn • It means that for such vectors with δ ' C1 , we may use this

large deviation inequality again, and have a lower bound.

Ideas behind this result

P{ |Tα|1 − |α|p ≥ t} ≤ 2 exp(−cp Ntq ). • In our situation : N = (1 + η)n and t ∈ (0, 1). • We may assume in addition that α ∈ Spn−1 has a small

support : | supp(α)| ≤ δn. Cn X n (1 + η)n X δn • It means that for such vectors with δ ' C1 , we may use this

large deviation inequality again, and have a lower bound.

Ideas behind this result

P{ |Tα|1 − |α|p ≥ t} ≤ 2 exp(−cp Ntq ). • In our situation : N = (1 + η)n and t ∈ (0, 1). • We may assume in addition that α ∈ Spn−1 has a small

support : | supp(α)| ≤ δn. Cn X n (1 + η)n X δn • It means that for such vectors with δ ' C1 , we may use this

large deviation inequality again, and have a lower bound.

Division of Spn−1 (following Rudelson-Vershynin) • Let δ, ρ ∈ (0, 1). • We define

Sparse(δ) = {α ∈ `np : | supp(α)| ≤ δn}. • We partition Spn−1 into two sets with respect to Sparse(δ)

and ρ. • We define the following sets :

AS(δ, ρ) = {α ∈ Spn−1 : distp (α, Sparse(δ)) ≤ ρ}, NAS(δ, ρ) = Spn−1 \ AS(δ, ρ), where AS(δ, ρ) is the ρ-enlargement (for the `np metric) of the set of sparse vectors intersected with Spn−1 .

Division of Spn−1 (following Rudelson-Vershynin) • Let δ, ρ ∈ (0, 1). • We define

Sparse(δ) = {α ∈ `np : | supp(α)| ≤ δn}. • We partition Spn−1 into two sets with respect to Sparse(δ)

and ρ. • We define the following sets :

AS(δ, ρ) = {α ∈ Spn−1 : distp (α, Sparse(δ)) ≤ ρ}, NAS(δ, ρ) = Spn−1 \ AS(δ, ρ), where AS(δ, ρ) is the ρ-enlargement (for the `np metric) of the set of sparse vectors intersected with Spn−1 .

Division of Spn−1 (following Rudelson-Vershynin) • Let δ, ρ ∈ (0, 1). • We define

Sparse(δ) = {α ∈ `np : | supp(α)| ≤ δn}. • We partition Spn−1 into two sets with respect to Sparse(δ)

and ρ. • We define the following sets :

AS(δ, ρ) = {α ∈ Spn−1 : distp (α, Sparse(δ)) ≤ ρ}, NAS(δ, ρ) = Spn−1 \ AS(δ, ρ), where AS(δ, ρ) is the ρ-enlargement (for the `np metric) of the set of sparse vectors intersected with Spn−1 .

Small ball estimate • For α ∈ NAS(δ, ρ)

P{|Tα|1 ≤ t} ≤ (cp t)N , • It means

t ≤ |Tα|1

t > 0.

w.h.p

• Basic properties of NAS vector :

∃I ⊆ {1, . . . , n} such that |I| ≥ 12 δnρp and ∀k ∈ I we have ρ 1 ≤ |αk | ≤ . 1/p (2n) (δn)1/p

Small ball estimate • For α ∈ NAS(δ, ρ)

P{|Tα|1 ≤ t} ≤ (cp t)N , • It means

t ≤ |Tα|1

t > 0.

w.h.p

• Basic properties of NAS vector :

∃I ⊆ {1, . . . , n} such that |I| ≥ 12 δnρp and ∀k ∈ I we have ρ 1 ≤ |αk | ≤ . 1/p (2n) (δn)1/p

Small ball estimate • For α ∈ NAS(δ, ρ)

P{|Tα|1 ≤ t} ≤ (cp t)N , • It means

t ≤ |Tα|1

t > 0.

w.h.p

• Basic properties of NAS vector :

∃I ⊆ {1, . . . , n} such that |I| ≥ 12 δnρp and ∀k ∈ I we have ρ 1 ≤ |αk | ≤ . 1/p (2n) (δn)1/p

Small ball estimate • For α ∈ NAS(δ, ρ)

P{|Tα|1 ≤ t} ≤ (cp t)N , • It means

t ≤ |Tα|1

t > 0.

w.h.p

• Basic properties of NAS vector :

∃I ⊆ {1, . . . , n} such that |I| ≥ 12 δnρp and ∀k ∈ I we have ρ 1 ≤ |αk | ≤ . 1/p (2n) (δn)1/p

Small ball estimate • For α ∈ NAS(δ, ρ)

P{|Tα|1 ≤ t} ≤ (cp t)N , t ≤ |Tα|1

• It means

t > 0.

w.h.p

• Basic properties of NAS vector :

∃I ⊆ {1, . . . , n} such that |I| ≥ 12 δnρp and ∀k ∈ I we have

ρ,δ

|αk | ∼

1 n1/p

.

Theorem [Multi-dimensional Esseen type inequality] Let X be a random vector in RN , such that the function ξ 7→ E exp(ihξ, Xi) belongs to L1 (RN ). Then for any compact star-shape K ⊂ RN , for any t > 0  t N Z P {kXkK ≤ t} ≤ |K| |E exp (ihξ, Xi)| dξ. 2π RN

Remarks • We generalize the classical Esseen inequality to the

multi-dimensional case, and to an arbitrary norm. • The proof is an application of Fourier analysis.

Theorem [Multi-dimensional Esseen type inequality] Let X be a random vector in RN , such that the function ξ 7→ E exp(ihξ, Xi) belongs to L1 (RN ). Then for any compact star-shape K ⊂ RN , for any t > 0  t N Z P {kXkK ≤ t} ≤ |K| |E exp (ihξ, Xi)| dξ. 2π RN

Remarks • We generalize the classical Esseen inequality to the

multi-dimensional case, and to an arbitrary norm. • The proof is an application of Fourier analysis.

Theorem [Multi-dimensional Esseen type inequality] Let X be a random vector in RN , such that the function ξ 7→ E exp(ihξ, Xi) belongs to L1 (RN ). Then for any compact star-shape K ⊂ RN , for any t > 0  t N Z P {kXkK ≤ t} ≤ |K| |E exp (ihξ, Xi)| dξ. 2π RN

Remarks • We generalize the classical Esseen inequality to the

multi-dimensional case, and to an arbitrary norm. • The proof is an application of Fourier analysis.

Application P{ |Tα|1 ≤ t} ≤ (cp t)N .  R t N P {kXkK ≤ t} ≤ |K| 2π RN |E exp (ihξ, Xi)| dξ.

• For α ∈ NAS(δ, ρ) • Recall :

Set K = N · BN1 and X = N · Tα. Then |Tα|1 = kXkK .

• Lemma : For any vector α ∈ NAS(δ, ρ), the function

ξ 7→ E exp(i N hξ, Tαi), belongs to L1 (RN ). Moreover, Z |E exp(i N hξ, Tαi)|dξ ≤ C(p, δ, ρ)N . RN

Application P{ |Tα|1 ≤ t} ≤ (cp t)N .  R t N P {kXkK ≤ t} ≤ |K| 2π RN |E exp (ihξ, Xi)| dξ.

• For α ∈ NAS(δ, ρ) • Recall :

Set K = N · BN1 and X = N · Tα. Then |Tα|1 = kXkK .

• Lemma : For any vector α ∈ NAS(δ, ρ), the function

ξ 7→ E exp(i N hξ, Tαi), belongs to L1 (RN ). Moreover, Z |E exp(i N hξ, Tαi)|dξ ≤ C(p, δ, ρ)N . RN

Application P{ |Tα|1 ≤ t} ≤ (cp t)N .  R t N P {kXkK ≤ t} ≤ |K| 2π RN |E exp (ihξ, Xi)| dξ.

• For α ∈ NAS(δ, ρ) • Recall :

Set K = N · BN1 and X = N · Tα. Then |Tα|1 = kXkK .

• Lemma : For any vector α ∈ NAS(δ, ρ), the function

ξ 7→ E exp(i N hξ, Tαi), belongs to L1 (RN ). Moreover, Z |E exp(i N hξ, Tαi)|dξ ≤ C(p, δ, ρ)N . RN

Application P{ |Tα|1 ≤ t} ≤ (cp t)N .  R t N P {kXkK ≤ t} ≤ |K| 2π RN |E exp (ihξ, Xi)| dξ.

• For α ∈ NAS(δ, ρ) • Recall :

Set K = N · BN1 and X = N · Tα. Then |Tα|1 = kXkK .

• Lemma : For any vector α ∈ NAS(δ, ρ), the function

ξ 7→ E exp(i N hξ, Tαi), belongs to L1 (RN ). Moreover, Z |E exp(i N hξ, Tαi)|dξ ≤ C(p, δ, ρ)N . RN