Perpanjangan operator kebisingan


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Dalam masalah yang sedang saya kerjakan, perpanjangan dari operator kebisingan muncul secara alami, dan saya ingin tahu apakah ada pekerjaan sebelumnya. Pertama izinkan saya merevisi operator kebisingan dasar T εTε pada fungsi Boolean yang bernilai nyata. Diberi fungsi f : { 0 , 1 } nRf:{0,1}nR dan εε , pp st 0 ε 10ε1 , ε = 1 - 2 pε=12p , kita mendefinisikan T εRTεR sebagai T ε f ( x) = E y μ p [ f ( x + y ) ]Tεf(x)=Eyμp[f(x+y)]

μ pμp adalah distribusi pada yy diperoleh dengan menetapkan setiap bit dari nn vektor-bit menjadi 11 secara mandiri dengan probabilitas pp dan 00 sebaliknya. Secara setara, kita dapat menganggap proses ini sebagai membalik setiap bit xx dengan probabilitas independen pp . Sekarang operator kebisingan ini memiliki banyak sifat yang berguna, termasuk menjadi multiplikasi T ε 1 T ε 2 = T ε 1 ε 2Tε1Tε2=Tε1ε2 dan memiliki nilai eigen dan vektor eigen yang bagus ( T ε ( χ S )= ε | S | χ S diTε(χS)=ε|S|χS mana χ SχS milik basis paritas).

Biarkan saya sekarang mendefinisikan ekstensi T ε saya , yang saya tunjukkan sebagai R ( p 1 , p 2 ) . R ( p 1 , p 2 )R diberikan oleh R ( p 1 , p 2 ) f ( x ) = E y μ p , x [ f ( x + y ) ] . Tapi di sini distribusi kami μTεR(p1,p2)R(p1,p2)RR(p1,p2)f(x)=Eyμp,x[f(x+y)]p , x sedemikian rupa sehingga kita membalik1bitxke0dengan probabilitas p 1 dan0bitxke1dengan probabilitas p 2 . ( Μ p , x sekarang jelas distribusi tergantung padaxdi mana fungsi tersebut dievaluasi, dan jika p 1 = p 2 maka R ( p 1 , p 2 ) mengurangi ke operator kebisingan 'biasa'.)μp,x1x0p10x1p2μp,xxp1=p2R(p1,p2)

Saya bertanya-tanya, apakah operator ini R ( p 1 , p 2 ) sudah dipelajari dengan baik di suatu tempat dalam literatur? Atau apakah sifat-sifat dasarnya jelas? Saya baru mulai dengan analisis Boolean, jadi ini mungkin langsung bagi seseorang yang lebih akrab dengan teori daripada saya. Khususnya saya tertarik jika vektor eigen dan nilai eigen memiliki karakterisasi yang bagus, atau apakah ada properti multiplikatif.R(p1,p2)

Jawaban:


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I'll answer the second part of the question.

I. Eigenvalues and Eigenfunctions

Let's first consider the one dimensional case n=1n=1. It is easy to check that the operator Rp1,p2Rp1,p2 has two eigenfunctions: 11 and ξ(x)=(p1+p2)xp1={p1, if x=0,p2, if x=1.

ξ(x)=(p1+p2)xp1={p1,p2, if x=0, if x=1.
with eigenvalues 11 and 1p1p21p1p2, respectively.

Now consider the general case. For S{1,,n}S{1,,n}, let ξS(x)=iSξ(xi)ξS(x)=iSξ(xi). Observe that ξSξS is an eigenfunction of Rp1,p2Rp1,p2. Indeed since all variables xixi are independent, we have Rp1,p2(ξ(x))=Rp1,p2(iSξ(xi))=iSRp1,p2(ξ(xi))=iS((1p1p2)ξ(xi))=(1p1p2)|S|ξS(x).

Rp1,p2(ξ(x))=Rp1,p2(iSξ(xi))=iSRp1,p2(ξ(xi))=iS((1p1p2)ξ(xi))=(1p1p2)|S|ξS(x).

We get that ξS(x)ξS(x) is an eigenfunction of Rp1,p2Rp1,p2 with eigenvalue (1p1p2)|S|(1p1p2)|S| for every S{1,,n}S{1,,n}. Since functions ξS(x)ξS(x) span the whole space, Rp1,p2Rp1,p2 has no other eigenfunctions (that are not linear combinations of ξS(x)ξS(x)).

II. Multiplicative Property

In general, the “multiplicative property” doesn't hold for Rp1,p2Rp1,p2 since the eigenbasis of Rp1,p2Rp1,p2 depends on p1p1 and p2p2. However, we have R2p1,p2=Rp1,p2,

R2p1,p2=Rp1,p2,
where p1=2p1(p1+p2)p1p1=2p1(p1+p2)p1 and p2=2p2(p1+p2)p2p2=2p2(p1+p2)p2. To verify that, first note that Rp1,p2Rp1,p2 and Rp1,p2Rp1,p2 have the same set of eigenfunctions {ξS}{ξS}. We have, R2p1,p2(ξS)=(1p1p2)2|S|ξS=(1p1p2)|S|ξS=Rp1,p2(ξS)
R2p1,p2(ξS)=(1p1p2)2|S|ξS=(1p1p2)|S|ξS=Rp1,p2(ξS)
since 1p1p2=1p1(2(p1+p2))p2(2(p1+p2))=1(p1+p+2)(2(p1+p2))=12(p1+p2)+(p1+p2)2=(1p1p2)2.
1p1p2=1p1(2(p1+p2))p2(2(p1+p2))=1(p1+p+2)(2(p1+p2))=12(p1+p2)+(p1+p2)2=(1p1p2)2.

III. Relation to the Bonami—Beckner operator

Let us think of functions from {0,1}n{0,1}n to RR as polylinear polynomials. Let δ=12p1p2p1+p2δ=12p1p2p1+p2. Consider the operator Aδ(f)=f(x1+δ,,xn+δ).

Aδ(f)=f(x1+δ,,xn+δ).
It maps every multilinear polynomial ff to a multilinear polynomial A[f]A[f]. We have, Rp1,p2(f)=A1δTεAδ(f),
Rp1,p2(f)=A1δTεAδ(f),
where ε=1p1p2ε=1p1p2. Note that parts I and II follow from this formula and properties of the Bonami—Beckner operator.

Yury, thank you for the answer! That's a good starting point for me to work with; I should now be able to work out if there are analogues of the hyper contractive inequality. Will post back here if I get any more interesting analysis.
Amir

This is very long after the fact, but I am curious how you derived the third part and the relation to the Becker Bonami operator?
Amir

(a) It is sufficient to check the identity for f=1f=1 and f=xif=xi. If it holds for 11 and xixi, then it's easy to see that it holds for all characters. By linearity, it holds for all functions. (b) Alternatively, from I, TεTε and Rp1,p2Rp1,p2 have the same set of eigenvalues; eigenvector iSxiiSxi of TT “corresponds” to eigenvector iSξ(xi)iSξ(xi) of RR. Thus R(f)=A1TA(f)R(f)=A1TA(f) where A is a linear map that maps ξ(x)ξ(x) to xx.
Yury

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We were eventually able to analyze hypercontractive properties of Rp1,p2 (http://arxiv.org/abs/1404.1191), building off of the main Fourier analysis of Rp,0 by Ahlberg, Broman, Griffiths and Morris (http://arxiv.org/abs/1108.0310).

To summarize, the effect of a biased operator Rp,0 on a function f can be analyzed as a symmetric noise operator in a biased measure space. This gives a weak form of hypercontractivity, which depends on how the 2 norm of f varies when switching to a choice of biased measure μ dependent on p.


You might want to 'accept' this answer so that the question doesn't keep popping up (disclaimer: I am an author on the linked paper)
Suresh Venkat
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