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A FACTORIZATION CONSTANT FOR ln

arXiv:math/9302213v1 [math.FA] 11 Feb 1993A FACTORIZATION CONSTANT FOR lnp, 0 < p < 1N.T. PeckIn this paper we are concerned with factoring the identity operator on an n-dimensional quasi-normed space X through a space ℓK∞.We seek a good lower bound λ(x) for ∥P ∥∥T ∥over all factorizations IdX =PT, with T : X −→ℓK∞,P : ℓK∞−→X.

When X is ℓnp, 1 ≤p, the constant isknown: see [5, Theorem 32.9] and the references given for that theorem.For p < 1, we will obtain the lower estimate λ(lnp ) ≥Cn1p −12 (log n)−12 . (A Tand P with ∥P ∥∥T ∥≤Cn1p −12 are easily obtained.

)Throughout, C denotes a constant, which may vary from one occurrence to thenext, but which is independent of n.We thank Y. Gordon for valuable conversations.Lemma 1. Let (ri), 1 ≤i ≤n, be the first n Rademacher functions and letα1, .

. .

αn be real. Letting m denote Lebesque measure on (0, 1), we havem(nXi=1αiri > αqlog nXα2i)≤n−Cα2,for any positive α.Proof.

This is well known; for completeness, we sketch a proof, following a sugges-tion of R. Kaufman.Typeset by AMS-TEX1

We can assume Pni=1 α2i = 1. Put f = Pni=1 αiri.

By Khintchine’s inequality,for some constant C and all p ≥1, ∥f∥p ≤Cp12 .¿From this, m{|f| > λ} ≤(C/λ)ppp2 = Kppp2 with K = C/λ.Now minimize Kppp2 in p. At the minimizer, we find p = K−2e, from whichKppp2 = exp(−λ2/eC2). (Note that p > 1 for λ > C.) Finally, put λ = α√log n toget the conclusion.■The space ℓ∞is not of type 2; but the conclusion of the next lemma will sufficefor our purposes.Lemma 2.

Let Y be an n-dimensional subspace of L1(0, 1) and let f1 . .

. fn beelements of Y ∗, of norm at most 1.

Then for some s in (0, 1),sup∥y∥≤1nXi=1ri(s)fi(y) ≤Cpn log n.Proof. For any 0 < ǫ <12, a result of Schechtman [3] implies that there are anN ≤Cǫ−2 log(ǫ−1)n2 and an isomorphism U : Yinto−→ℓN1 such that ∥U∥∥U −1∥≤1 + ǫ.

See also the results of Bourgain–Lindenstrauss–Milman [1] and Talagrand[4]. In particular, taking ǫ = 14, say, we obtain the corresponding U; we can assume∥U∥= 1, so that ∥U −1∥≤54 and N ≤Cn2 (after changing C) ≤n3, if n issufficiently large.Let e1, .

. .

eN be the unit basis vectors in ℓN1 . For each i let Φi be a Hahn–Banachextension to ℓN1 of fiU −1 on U(Y ), with ∥Φi∥≤54; then for each j, Φi(ej)∥≤54.Now fix α with Cα2 > 3, where C is the constant in the conclusion of lemma 1;2

then n3n−Cα2 ≤14 if n is sufficiently large. Sincem(Xφi(ej)ri > 54αpn log n)≤m(Xφi(ej)ri > αqlog nXφ2i (ej))≤n−Cα2,for eachj,there is a set A, m(A) > 34, such that |Pni=1 φi(ej)ri(s)| ≤54α√n log n for each jand each s in A.Now if y ∈Y and ∥y∥≤1, ∥Uy∥≤1.

Write Uy = PNj=1 αjej, PNj=1 |αj| ≤1;applying the above inequality to each j and recalling that fi = fiU −1U on Y , wehavenXi=1fi(y)ri(s) ≤54αpn log n= Cpn log nfor each s in A.■Notation. Let A be an algebra of measurable subsets of (0, 1).

For 0 < p ≤∞,Lp(A) is the space of functions in Lp(0, 1) which are A-measurable. For ease ofargument, we deal with an L∞(A) with “homogeneous” A, rather than ℓK∞.Theorem.

Let A be a finite subalgebra of measurable subsets of (0, 1) containingthe dyadic intervals j−12n , j2n, 1 ≤j ≤2n, and assume the atoms of A all have thesame measure. Let X be an n-dimensional vector space.

Let T : X −→L∞(A) bea linear map; and let (ei)ni=1 be elements of X such that ∥T(ei)∥∞≤1, 1 ≤i ≤n.Let P : L∞(A) −→X be a linear operator such that PT = IdX. Then there is win L∞(A) with ∥w∥∞≤C√n log n such that P(w) = Pni=1 ri(s)ei, for some s in(0, 1).3

Proof. Let (Ij)kj=1 be the atoms of A, and let z1, .

. .

zk−n be a basis for ker P.Define a (k −n) × k matrix by zi,j = constant value of zi on the atom Ij.Now row–reduce the matrix (zi,j). In k −n of the columns there will be one1 with all other entries 0; denote the atoms corresponding to the n remaining“distinguished” columns by Is1, .

. .

Isn. Enlarge the matrix (zi,j) to a k × k matrix(yi,j) by adding n rows of zeros in each of rows s1 through sn.We can obviously regard yi,j as an A × A-measurable function y(s, t) on (0, 1) ×(0, 1), which satisfies these properties:(1) if s /∈Uj Isj and if s and t are in the same atom, y(s, t) = 1;(2) if s /∈Uj Isj and if s and t are in different atoms, y(s, t) = 0;(3) if t /∈Uj Isj, y(s, t) = 0 for all s;(4) if we define yt on (0, 1) by yt(s) = y(s, t), then yt ∈ker P;(5) a function f in L∞(A) is in ker P if and only if there is a function β(t) inL∞(A) so that f(s) =Ryt(s)β(t)dt for all s.Properties (1) −(5) are evident from the description of ker P and the propertiesof a row-reduced matrix.Let s∗i be a point of the atom Isi, 1 ≤i ≤n, and let gi = T(ei), 1 ≤i ≤n; then∥gi∥∞≤1.Now defineR(s, t) =nXi=1ri(s)gi(t),and for a function y(t), defineψs(y) =ZR(s, t)y(t)dt=nXi=1(Zy(t)gi(t)dt)ri(s).4

Let Y be the span of the functions yt(s∗i ), 1 ≤i ≤n, regarded as functions of t.Since y −→Ry(t)gi(t)dt is of norm at most 1, Lemma 2 implies that there is a setA of measure > 34 such that for any s in A,|ψs(y)| ≤Cpn log n ∥y∥1for all y in Y .Take any norm-preserving extension of ψs to all of L1( , dt); then there is afunction hs(t) in L∞(., dt) with ∥hs∥∞≤C√n log n and such thatψs(y) =Zy(t)hs(t)dtfor all y in Y . Restating this,(6)Z(Rs(t) −hs(t))y(t)dt = 0for all y in Y .

Now putαs(s) = 1mZ(Rs(t) −hs(t))yt(s)dt,where m is the measure of an atom of A.Then αs ∈ker P by property (5),αs(s) = Rs(s) −hs(s), for s not in Sni=1 Isi, by (1) and (2), and αs(si) = 0,1 ≤i ≤n, by (6).The set A in the proof of Lemma 2 has measure ≥34. vvApplying Lemma 1again, we can require that|Rs(s∗i )| ≤Cpn log nfor each i, 1 ≤i ≤n, for all s in a set B with m(B) ≥34.

Thus A ∩B has positivemeasure, so choose s in A ∩B.5

Now, to finish the proof, set w(s) = Rs(s)−αs(s). If s /∈Sni=1 Isi, w(s) = Rs(s)−αs(s) = hs(s), so |w(s)| ≤C√n log n. Also, |w(s∗i )| = |Rs(s∗i )| ≤C√n log n; thus∥w∥∞≤C√n log n. Finally, Pw = Pni=1 ri(s)ei since αs ∈ker P. This completesthe proof.■Corollary 1.

Let X be n-dimensional, and let XT−→ℓK∞P−→X be a factorizationof IdX through ℓK∞. Then ∥P∥∥T∥≥Csn(n log n)12 , wheresn =sup∥T ei∥≤11≤i≤ninf±1 ∥nXi=1±ei∥.In particular, if ∥T∥≤1, ∥P∥≥Cbn(n log n)−12 , where bn = sup∥ei∥≤11≤i≤ninf±1 ∥nXi=1±ei∥.

(See [2].)Proof. For K ≥n , this is an immediate consequence of the Theorem.

Assumenow that K < n . Let j = n −K , and define ˜T: X −→ℓn∞= ℓK∞⊗ℓj∞by˜T(x) = (T(x), 0) .

Define ˜P : ℓn∞−→X by ˜P(w, z) = P(w). Note that ˜P ˜T = IdXand that ∥˜P∥= ∥P∥, ∥˜T ∥= ∥T∥.

The result now follows from the Theorem.■Corollary 2. Suppose ℓnpT−→ℓK∞P−→ℓnp is any factorization of the identity on ℓnpthrough ℓK∞, 0 < p ≤1, with ∥T∥= 1.

Then ∥P∥≥Cn1p −12 (log n)−12 .Proof. Take X = ℓnp, ei the usual ith basis vector in ℓnp, 1 ≤i ≤n, for 0 < p ≤1.Now apply Corollary 1.■Remark.

For 0 < p < 1, define T : ℓnp −→L∞(A) by defining T(ei) = ri,1 ≤i ≤n and extending linearly. Then ∥T∥= 1.

Define P : L∞(A) −→ℓnp byT(x) = Pni=1(Rxri)ei. It is easily checked that ∥T : L∞(A) −→ℓn1∥≤√n; since∥Id : ℓn1 −→ℓnp∥= n1p −1, it follows that ∥P∥≤n1p −12 .

Obviously PT = Id, so upto a logarithmic factor, the order of λ(ℓnp) is correct.6

References1. J. Bourgain, J. Lindenstrauss, and V. Milman, Approximation of zonoids by zonotopes, ActaMath.

162 (1989), 73–141.2. N.J. Kalton, The three–space problem for locally bounded F–spaces, Compositio Math.

37(1978), 243–276.3. G. Schechtman, More on embedding subspaces of ℓnr , Compositio Math.

61 (1987), 159-170.4. M. Talagrand, Embedding subspaces of L1 into ℓN1 , Proc.

Amer. Math.

Soc. 108 (1990),363–369.5.

N. Tomczak–Jaegermann, Banach-Mazur distances and finite-dimensional operator ideals,Pitman monograph 38, Longman, 1989.University of IllinoisUrbana, Illinois 618017

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