---

Common subspaces of Lp-spaces

arXiv:math/9211209v1 [math.FA] 5 Nov 1992Common subspaces of Lp-spacesAlexander KoldobskyDepartment of MathematicsUniversity of Missouri-ColumbiaColumbia, MO 65211Abstract. For n ≥2, p < 2 and q > 2, does there exist an n-dimensional Banach spacedifferent from Hilbert spaces which is isometric to subspaces of both Lp and Lq?

Generaliz-ing the construction from the paper ”Zonoids whose polars are zonoids” by R.Schneider wegive examples of such spaces. Moreover, for any compact subset Q of (0, ∞)\{2k, k ∈N},we can construct a space isometric to subspaces of Lq for all q ∈Q simultaneously.AMS classification: Primary 46B04, Secondary 46E30Key words: isometries, positive definite functions, spherical harmonics.e-mail: mathak@mizzou1.missouri.edui

Common subspaces of Lp-spacesAlexander KoldobskyDepartment of MathematicsUniversity of Missouri-ColumbiaColumbia, MO 652111. Introduction.This work started with the following question: For given n ≥2, p ∈(0, 2) and q > 2,does there exist an n-dimensional Banach space which is different from Hilbert spaces andwhich is isometric to subspaces of both Lp and Lq?It is a well-known fact first noticed by P.Levy that Hilbert spaces are isometric tosubspaces of Lq for all q > 0.

On the other hand, it was proved in [4] that, for n ≥3, q >2, p > 0, the function exp(−∥x∥pq) is not positive definite where ∥x∥q = (|x1|q+...+|xn|q)1/q. (This result gave an answer to a question posed by I.J.Schoenberg [10] in 1938) In 1966,J.Bretagnolle, D.Dacunha-Castelle and J.L.Krivine [1] proved that,for 0 < p < 2, a space(E, ∥· ∥) is isometric to a subspace of Lp if and only if the function exp(−∥x∥p) is positivedefinite.

Thus, in the language of isometries, the above mentioned result from [4] meansthat, for every n ≥3, q > 2, p ∈(0, 2), the space lqn is not isometric to a subspace ofLp. (For p ≥1, this fact was first proved in [2]) The initial purpose of this work was to finda non-Hilbertian subspace (E, ∥· ∥) of Lq with q > 2 of the dimension at least 3 such thatthe function exp(−∥x∥p) is positive definite.

The latter problem is equivalent to that atthe beginning of the paper.We prove, however, a more general fact: For every n ≥2 and every compact subset Qof (0, ∞) \ {2k, k ∈N}, there exists an n-dimensional Banach space different from Hilbertspaces which is isometric to subspaces of Lq for all q ∈Q simultaneously.In 1975, R.Schneider [9] proved that there exist non-trivial zonoids whose polars arezonoids or, in other words, there exist non-Hilbertian Banach spaces X such that X andX∗are isometric to subspaces of L1. It turns out that Schneider’s construction of specialsubspaces of L1 can be extended to all numbers q > 0 which are not even integers and inthis way we obtain our main result.1

2. Some properties of spherical harmonics.We start with some properties of spherical harmonics (see [6] for details).Let Pm denote the space of sperical harmonics of degree m on the unit sphere Ωn inRn.

Remind that sperical harmonics of degree m are restrictions to the sphere of harmonichomogeneous polynomials of degree m. We consider spherical harmonics as functions fromthe space L2(Ωn). Any two spherical harmonics of different degrees are orthogonal inL2(Ωn) [6, p.2]The dimension N(n, m) of the space Pm can easily be calculated [6, p.4]:(1)N(n, m) = (2m + n −2)Γ(n + m −2)Γ(m + 1)Γ(n −1)Let {Ymj : j = 1, ..., N(n, m)} be an orthonormal basis of the space Pm.

By theAddition Theorem [ 6, p.9], for every x ∈Ωn,(2)N(n,m)Xj=1Y 2mj(x) = N(n, m)ωnwhere ωn = 2πn/2/Γ(n/2) is the surface of the sphere Ωn.Linear combinations of functions Ymj are dense in the space L2(Ωn) [ 6, p.43]. There-fore, if F is a continuous function on Ωn and (F, Ymj) = 0 for every m = 0, 1, 2, ... andevery j = 1, ....N(n, m) then F ≡0 on Ωn.

Here (F, Y ) stands for the scalar product inL2(Ωn).Let ∆be the Laplace-Beltrami operator on the sphere Ωn. Then for every Ym ∈Pmwe have [ 6, p.39](3)∆Ym + m(m + n −2)Ym ≡0An immediate consequence of (3) ( and a well-known fact) is that ∆is a symmetricoperator and we can aplly Green’s formula: for every function H from the class C2r, r ∈Nof functions on Ωn having continuous partial derivatives of order 2r and for every Ym ∈Pm, m ≥1,(4)(−m(m + n −2))r(H, Ym) = (H, ∆rYm) = (∆rH, Ym)We also need the Funk-Henke formula [ 6, p.20]:for every Ym ∈Pm, every continuousfunction f on [−1, 1] and every x ∈Ωn,(5)ZΩnf(⟨x, ξ⟩)Ym(ξ)dξ = λmYm(x)where ⟨·, ·⟩stands for the scalar product in Rn and(6)λm =(−1)mπ(n−1)/22m−1Γ(m + (n −1)/2)Z 1−1f(t) dmdtm (1 −t2)m+(n−3)/2dtLet us calculate λm in the case where f(t) = |t|q, q > 0.2

Lemma 1. If q > 0, q ̸= 2k, k ∈N and f(t) = |t|q then(7)λm = πn/2−1Γ(q + 1) sin(π(m −q)/2)Γ((m −q)/2)2q−1Γ((m + n + q)/2)Proof: Assume first that q > m and calculate the integral from (6) by parts m times.Then use the formulaR 1−1 t2α−1(1 −t2)β−1dt = Γ(α)Γ(β)/Γ(α + β) and formulas for Γ-function: Γ(2x) = 22x−1Γ(x)Γ(x + 1/2)/π1/2 and Γ(1 −x)Γ(x) = π/ sin(πx).

We get(7) for q > m. Note that both sides of (7) are analytic functions of q in the domainℜq > 0, q ̸= 2k, k ∈N. Because of the uniqueness of analytic extension, (7) holds for everyq from this domain.

We are done.3. Main result.Let X be an n-dimensional subspace of Lq = Lq([0, 1]) with q > 0.

Let f1, ..., fn bea basis in X and µ be the joint distribution of the functions f1, ..., fn with respect toLebesgue measure (µ is a finite measure on Rn). Then, for every x ∈Rn,(11) ∥x∥q = ∥nXk=1xkfk∥q =Z 10|nXk=1xkfk(t)|qdt =ZRn |⟨x, ξ⟩|qdµ(ξ) =ZΩn|⟨x, ξ⟩|qdν(ξ)where ν is the projection of µ to the sphere.

(For every Borel subset A of Ωn, ν(A) =R{tA,t∈R} ∥x∥q2dµ(x)) The representation (11) of the norm is usially called the Levy repre-sentation. It is clear now that a norm in an n-dimensional Banach space admits the Levyrepresentation with a probability measure on the sphere if and only if this space is isomet-ric to a subspace of Lq.

(Given the Levy representation we can choose functions f1, ..., fnon [0, 1] with the joint distribution ν and define an isometry by x →Pnk=1 xkfk, x ∈Rn)If we replace the measure ν by an arbitrary continuous (not necessarily non-negative)function on the sphere then a representation similar to the Levy representation is possiblefor a large class of Banach spaces (see [5] for the Levy representation with distributionsinstead of measures; such representation is possible for any Banach space and any q whichis not an even integer) This is an idea going back to W.Blaschke that any smooth enoughfunction on the sphere can be represented in the form (11) with a continuous functioninstead of a measure on the sphere. However, W.Blaschke and then R.Schneider [8] re-stricted themselves to the case q = 1 which is particularly important in the theory ofconvex bodies.

The following theorem is an extension of R.Schneider’s results from [ 8,p.77] and [ 9, p.367] to all positive numbers q which are not even integersTheorem 1. Let q > 0, q ̸= 2k, k ∈N and let H be an even function of the class C2r onΩn where r ∈N and 2r > n + q.

Then there exists a continuous function bH on the sphere3

Ωn such that, for every x ∈Ωn,(8)H(x) =ZΩn|⟨x, ξ⟩|qbH(ξ)dξBesides that, there exist constants K(q) and L(q) depending on n and q only such that,for every x ∈Ωn,(9)|bH(x)| ≤K(q)∥H∥L2(Ωn) + L(q)∥∆2rH∥L2(Ωn)Proof: Define a function bH on Ωn by(10)bH(x) =∞Xm=0λ−1mN(n,m)Xj=1(N, Ymj)Ymj(x)Since H is an even function and Ymj are odd functions if m is odd the sum is, actually,taken over even integers m only.Let us prove that the series in the right-hand side of (10) converges uniformly on Ωn.By the Cauchy-Schwartz inequality, (2) and the fact that Ymj form an orthonormal basisin Pm, we get|N(n,m)Xj=1(∆rN, Ymj)Ymj(x)| ≤(N(n,m)Xj=1(∆rN, Ymj)2)1/2(N(n,m)Xj=1Y 2mj(x))1/2≤∥∆rH∥L2(Ωn)N(n, m)ωn1/2It follows from (4) and the latter inequality that|bH(x)| ≤|λ−10 (H, Y0)Y0(x)| +∞Xm=2;2|mλ−1m (−1m(m + n −2))r|N(n,m)Xj=1(∆rH, Ymj)Ymj(x)| ≤|λ0|−1ω−1/2n∥H∥L2(Ωn) +∞Xm=2;2|mλ−1m m−2rN(n, m)ωn1/2∥∆rH∥L2(Ωn)Let us show that the series P∞m=2;2|m λ−1m m−2r(N(n, m)/ωn)1/2 converges.In fact, itfollows from (1) that N(n, m) = O(mn−2) and it is an easy consequence of (7) and theStirling formula that λ−1m = O(m(n+2q)/2). Since 2r > n + q = (n + 2q)/2 + (n −1)/2 + 1we get λ−1m m−2r(N(n, m)/ωn)1/2 = o(m−1−ǫ) for some ǫ > 0, and the series is convergent.We denote the sum of this series by L(q) and put K(q) = |λ0|−1ω−1/2n, so we get (9).4

We have proved that the series in (10) converges uniformly and defines a continuousfunction on Ωn. It follows from (5) and the fact that all functions Ymj are orthogonal that(H, Ymj) = (RΩn |⟨x, ξ⟩|qbH(ξ)dξ, Ymj(x)) for every m = 0, 1, 2, ... and j = 1, ..., N(n, m).Hence, the function bH satisfies (8).Let X be an n-dimensional Banach space, q > 0, q ̸= 2k, k ∈N.

Let c(q) = Γ((n +q)/2)/(2Γ((q + 1)/2)π(n−1)/2) be a constant such that 1 = c(q)RΩn |⟨x, ξ⟩|qdξ for everyx ∈Ωn. (The latter integral does not depend on the choice of x ∈Ωn; it means that thenorm of the space ln2 admits the Levy representation with the uniform measure on thesphere and the space ln2 is isometric to a subspace of Lq for every q)Denote by H(x), x ∈Ωn the restriction of the function ∥x∥q to the sphere Ωn.

Assumethat the function H belongs to the class C2r on Ωn where 2r > n + q, r ∈N. Let bH bethe function corresponding to H by Theorem 1.Lemma 2.

If the number K(q)∥H −1∥L2(Ωn) + L(q)∥∆rH∥L2(Ωn) is less than c(q) thenthe space X is isometric to a subspace of Lq.Proof: By (8) and definition of the number c(q),H(x) −1 =ZΩn|⟨x, ξ⟩|q(bH(ξ) −c(q))dξfor every x ∈Ωn. By (9), |bH(x) −c(q)| < c(q) for every x ∈Ωn.

It means that thefunction bH is positive on the sphere. The equality (8) means that the space X admits theLevy representation with a non-negative measure and, by the reasoning at the beginningof Section 3, X is isometric to a subspace of Lq.Now we are able to prove the main result of this paper.

Let us only note that, forevery function f of the class C2 on the sphere Ωn and for a small enough number λ, thefunction N(x) = 1 + λf(x), x ∈Ωn is the restriction to the sphere of some norm in Rn.This is an easy consequence of the following one-dimensional fact: If a, b ∈R, g is a convexfunction on [a, b] with g′′ > δ > 0 on [a, b] for some δ and h ∈C2[a, b] then functions g+λhhave positive second derivatives on [a, b] for sufficiently small λ’s, and, hence, are convexon [a, b].Theorem 2. Let Q be a compact subset of (0, ∞) \ {2k, k ∈N}.

Then there exists aBanach space different from Hilbert spaces which is isometric to a subspace of Lq for everyq ∈Q.Proof: Let f be any infinitely differentiable function on Ωn and fix a number r ∈N sothat 2r > n+q for every q ∈Q. Choose a sufficiently small number λ such that the functionN(x) = 1 + λf(x), x ∈Ωn is the restriction to the sphere of some norm in Rn ( see the5

remark before Theorem 2) and such that, for every q ∈Q, the function H(x) = (N(x))qsatisfies the condition of Lemma 2. The possibility of such choice of λ follows from the factsthat K(q), L(q) and c(q) are continuous functions of q on the set Q and that ∥H −1∥L2(Ωn)and ∥∆2rH∥L2(Ωn) tend to zero uniformly with resrect to q ∈Q as λ tends to zero.

Nowwe can apply Lemma 2 to complete the proof.Finally, let us consider the case where q is an even integer. It is easy to see that, forany fixed number 2k, k ∈N, k > 1, we can make the space X constructed in Theorem2 isometric to a subspace of L2k.

In fact, let N(x) = (1 + λ(x2k1+ ... + x2kn ))1/4. Forsufficiently small numbers λ, N is the restriction to the sphere of some norm in Rn and thecorresponding space X is isometric to a subspace of Lq for every q ∈Q.

On the other hand,X is isometric to a subspace of L2k because the norm admits the Levy representation witha measure on the sphere:1 + λ(x2k1 + ... + x2kn ) =ZΩn|⟨x, ξ⟩|2k(c(2k)dξ + λdδ1(ξ) + ... + λdδn(ξ))where δi is a unit mass at the point ξ ∈Rn with ξi = 1, ξj = 0, j ̸= i.Let us show that one can not make the space X isometric to subspaces of L2p andL2q if p, q ∈N and do not have common factors. In fact, if (X, ∥· ∥) is such a space then,for every x ∈Rn,∥x∥4pq = (ZΩn|⟨x, ξ⟩|2pdµ(ξ))2q = (ZΩn|⟨x, ξ⟩|2qdν(ξ))2pfor some measures µ, ν on Ωn.

The functions in the latter equality are polynomials and,since the polynomial ring has the unique factorization property , we conclude that ∥x∥2 isa homogeneous polynomial of the second order and X is a Hilbert space.The situation is not clear if p and q have common factors. One can find some inter-esting results on Banach spaces with polynomial norms and on the structure of subspacesof L2k, k ∈N in the paper [7].Acknowledgements.

I wish to thank Prof. Nigel Kalton for valuable remarks and helpfuldiscussions during the work on this problem. I am grateful to Prof. Hermann Konig forbringing the paper [9] to my attention.6

References.1.Bretagnolle, J., Dacunha-Castelle, D. and Krivine, J.L. : Lois stables et espaces Lp,Ann.

Inst. H.Poincare, Ser.B 2(1966), 231-259.2.Dor, L.: Potentials and isometric embeddings in L1, Israel J.

Math. 24 (1976), 260-268.3.Grzaslewicz, R.: Plane sections of the unit ball of Lp, Acta Math.

Hung. 52(1988),219-225.4.Koldobsky, A.: Schoenberg’s problem on positive definite functions, Algebra and Anal-ysis 3 (1991), No.

3, 78-85 (Russian); English translation in St.Petersburg Math. J.3 (1992), 563-5705.Koldobsky, A.: Generalized Levy representation of norms and isometric embeddingsinto Lp-spaces, Ann.

Inst. H.Poincare (Prob.

and Stat.) 28 (1992), 335-3536.Muller, C.: Spherical Harmonics, Lect.

Notes in Math. 17 , Springer-Verlag, Berlin,1966.7.Reznick, B.: Banach spaces with polynomial norms, Pacific J.

Math. 82 (1979), 223-235.8.Schneider, R.: Zu einem problem von Shephard uber die projektionen konvexer korper,Math.

Zeitschrift 101 (1967), 71-82.9.Schneider, R.: Zonoids whose polars are zonoids, Proc. Amer.

Math. Soc.

50 (1975),365-368.10.Schoenberg, I.J. : Metric spaces and positive definite functions, Trans.

Amer. Math.Soc.

44 (1938), 522-536.7

---

출처: arXiv:9211.209 • 원문 보기