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Isometric stability property of certain Banach spaces

arXiv:math/9306208v1 [math.FA] 9 Jun 1993Isometric stability property of certain Banach spacesAlexander KoldobskyAbstract. Let E be one of the spaces C(K) and L1, F be an arbitrary Banach space,p > 1, and (X, σ) be a space with a finite measure.

We prove that E is isometric to asubspace of the Lebesgue-Bochner space Lp(X; F) only if E is isometric to a subspace of F.Moreover, every isometry T from E into Lp(X; F) has the form Te(x) = h(x)U(x)e, e ∈E,where h : X →R is a measurable function and, for every x ∈X, U(x) is an isometry fromE to F.Mailing address:.Till August 1,1993:Department of MathematicsUniversity of Missouri-ColumbiaColumbia, MO 65211, USAAfter August 1,1993:Division of Mathematics, Computer Science and StatisticsUniversity of Texas at San AntonioSan Antonio, TX 78249, USAAMS classification:. 46B04, 47B80Key words:.

isometries, Lebesgue-Bochner spaces, random operatorsRunning title:. Isometric embeddingsi

Isometric stability property of certain Banach spacesAlexander Koldobsky1Department of MathematicsUniversity of Missouri-ColumbiaColumbia, MO 652111.Introduction.Let E and F be Banach spaces, p ≥1, (X, σ) be a finite measure space, and Lp(X, F)be the Lebesque-Bochner space of (equivalence classes of) strongly measurable functionsf : X 7→F with∥f∥p =ZX∥f(x)∥pdσ(x) < ∞.We show that if E = C(K) (with K being a compact metric space) or E = L1 thenthe space E can be isometric to a subspace of Lp(X, F) with p > 1 only if E is isometricto a subspace of F.The isomorphic version of this result has been proved for the spaces E = c0 (S.Kwapien[5] and J.Bourgain [1]), E = l1 (G.Pisier [7]) and E = l∞(J.Mendoza [6], see also [2] forgeneralizations to Kothe spaces of vector valued functions). E.Saab and P.Saab [9] haveproved the isomorphic version for the space E = L1 under the additional assumption thatF is a dual space.For any 1 ≤p ≤q ≤r, the space Lq is isometric to a subspace of Lp(X, Lr) (see [8]).On the other hand, if r > 2, r ̸= q, q ̸= 2 the space Lq is not isomorphic to a subspaceof Lr.

Thus, both isometric and isomorphic versions fail to be true in the case whereE = Lq, q > 1. (The space L2 is isometric to a subspace of Lp for any p > 0, so it isisometric to a subspace of Lp(X, F) for any space F)Besides proving the above mentioned result for the spaces C(K) and L1, we completelycharacterize isometric embeddings of these spaces into Lp-spaces of vector valued functions.Denote by I(E, F) the set of isometries from E to F. A mapping U : X 7→I(E, F) iscalled strongly measurable if, for each e ∈E, the function ∥U(x)e∥is measurable on X.1After August 1, 1993 the author’s address will be:Division of Mathematics, ComputerScience and Statistics, University of Texas at San Antonio, San Antonio, TX 78249.1

If the set I(E, F) is non-empty then, obviously, for every strongly measurable mappingU : X 7→I(E, F) and every function h : X 7→R with ∥h∥Lp(X) = 1, the operatorT : E 7→Lp(X, F) defined by(1)Te(x) = h(x)U(x)e, e ∈Eis an isometry.We prove below that, for E = C(K) or E = L1 and for an arbitrary space F, everyisometry from E to Lp(X, F) has the form (1). The question if all isometries from E toLp(X, F) have the form (1) makes sense and has some applications even if E = F. Forexample, if E = F = Lq, q > 1, p > 1, then every isometry from Lq to Lp(X, Lq) has theform (1) if and only if p ̸= q, q ̸= 2, q ̸∈(p, 2).

This result was proved and applied to thedescription of isometries of Lebesgue-Bochner spaces in the paper [3].As a consequence of the characterization of isometries from E = C(K) or E = L1to Lp(X, F) we obtain the following result on random operators. Suppose that (X, σ) isa probability space, let p > 1, L(E, F) be the space of linear operators from E to F andconsider a random operator S : X 7→L(E, F) which is an isometry in average, i.e.

S isstrongly measurable and, for every e ∈E,∥e∥p =ZX∥S(x)e∥pdσ(x).Then the random operator S is an isometry (up to a constant) with probability 1 (namely,operators S(x) are isometries multiplied by constants for almost all x ∈X). In fact, if wedefine an operator T : E 7→Lp(X, F) by Te(x) = S(x)e then T is an isometry and (1)implies the desired result.

In the case E = C(K), the result about random operators hasbeen proved before [4] under the assumption that the operators S(x) are bounded.2.Main results.The proofs are based on the following simple fact.Lemma 1. Let E, F be Banach spaces, p > 1, (X, σ) be a finite measure space and T be anisometry from E to Lp(X, F).

If e, f are elements from E with ∥e∥= ∥f∥= 1 and havinga common tangent functional (i.e. there exists x∗∈E∗with ∥x∗∥= 1, x∗(e) = x∗(f) = 1)then, for almost all (with respect to σ) x ∈X, we have(i) ∥Te(x)∥= ∥Tf(x)∥,(ii) for every α > 0, ∥Te(x) + αTf(x)∥= ∥Te(x)∥+ α∥Tf(x)∥.Proof: Obviously, ∥e + αf∥= 1 + α for every α > 0, and we have(1 + α)p = ∥e + αf∥p =ZX∥Te(x) + αTf(x)∥pdσ(x) ≤2

(2)ZX(∥Te(x)∥+ α∥Tf(x)∥)pdσ(x)For α = 0, (2) turns into an equality. Therefore, we get a correct inequality if wetake in both sides of (2) the right-hand derivatives at the point α = 0 and apply Holder’sinequality:1 ≤ZX∥Te(x)∥p−1∥Tf(x)∥dσ(x) ≤ZX∥Te(x)∥pdσ(x) p−1p ZX∥Tf(x)∥pdσ(x)1/p= 1¿From the conditions for equality in Holder’s inequality, we conclude that, for almostall (with respect to σ) x ∈X, ∥Te(x)∥= c∥Tf(x)∥where c is a constant.

Further,1 =ZX∥Te(x)∥pdσ(x) =ZXcp∥Tf(x)∥pdσ(x) = cpand, hence, c = 1.It is clear now that (2) is, in fact, an equality. Hence, for almost all x ∈X, we have(3)∥Te(x) + αTf(x)∥= ∥Te(x)∥+ α∥Tf(x)∥for every α > 0 and the proof is complete.Now we are able to prove the main result.Theorem 1.

Let p > 1, K be a compact metric space, (X, σ), (Y, ν) be spaces with finitemeasures, and F be an arbitrary Banach space. Let E be either the space L1 = L1(Y, ν)or any subspace of C(K) containing the function 1(k) ≡1.

Then(i) If E is isometric to a subspace of Lp(X; F) then E is isometric to a subspace of Fand the set I(E, F) is non-empty. (ii) If T is an isometry from E into Lp(X; F) then there exist a measurable functionh : X →R and a strongly measurable mapping U : X →I(E, F) such that Te(x) =h(x)U(x)e for every e ∈E.Proof: We start with the case E = L1.Any two functions e and f from L1 with disjoint supports in Y have a commontangent functional so we can apply Lemma 1 to any pair of normalized functions withdisjoint supports.Decompose the set Y into two parts Y = Y1 ∪Y2, Y1 ∩Y2 = ∅, ν(Yi) > 0, i = 1, 2.

Fixa function e0 ∈L1(Y1), ∥e0∥= 1 and put h(x) = ∥Te0(x)∥, x ∈X.3

Let fk, k ∈N be a sequence of linearly independent functions with supports in Y2such that their linear span is dense in L1(Y2). Denote by D the set of linear combinationsof functions fk with rational coefficients.

Given fixed representatives Tfk from the cor-responding equivalence classes of functions from the space Lp(X; F), define an operatorT(x) : D 7→F for every x ∈X by T(x)(P λifi) = P λiTfi(x), λi ∈Q.It follows from the statement (i) of Lemma 1 and the fact that D is countable thatthere exists a set X0 ⊂X with σ(X \ X0) = 0 such that, for every x ∈X0 and everyf ∈D,∥T(x)f∥= ∥Tf(x)∥= ∥Te0(x)∥∥f∥= h(x)∥f∥Hence, for every x ∈X0, either h(x) = 0 or the operator U2(x) = T(x)/h(x) is anisometry from D to F.The operators U2(x) can be uniquely extended to isometries on the whole spaceL1(Y2). In fact, given a ∈L1(Y2) and a sequence ak →a, ak ∈D, put U2(x)(a) =limk→∞U2(x)(ak).Further, for any a ∈L1(Y2),∥ak −a∥p =ZX∥Tak(x) −Ta(x)∥pdσ(x) =(4)Zh(x)=0∥Ta(x)∥pdσ(x) +Zh(x)̸=0∥h(x)U2(x)ak −Ta(x)∥pdσ(x) →0as k →∞.

Therefore, Ta(x) = 0 for almost all x ∈X with h(x) = 0, and Ta(x) =h(x)U2(x)a for almost all x ∈X (if h(x) = 0 we put U2(x) = 0. )Similarly, for almost all x ∈X, we find isometries U1(x) from L1(Y1) to F such thatTb(x) = h(x)U1(x)b for every b ∈L1(Y2).Consider an arbitrary function f ∈L1(Y ).

This function can be uniquely representedas a sum f = f1 + f2 of functions f1 ∈L1(Y1) and f2 ∈L1(Y2). For all x ∈X withh(x) ̸= 0, define operators U(x) from L1(Y ) to F by U(x)f = U1(x)f1 + U2(x)f2.

By thestatement (ii) of Lemma 1, for almost every x with h(x) ̸= 0,∥U(x)f∥= (1/h(x))∥Tf1(x)+Tf2(x)∥= ∥U1(x)f1∥+∥U2(x)f2∥= ∥f1∥+∥f2∥= ∥f∥Thus, operators U(x) are isometries for almost all x ∈X with h(x) ̸= 0. In particular,the set I(L1, F) is non-empty.

Fix any U ∈I(L1, F) and put U(x) = U for every x withh(x) = 0.To prove the second statement of Theorem 1 note that, for every f ∈L1(Y ), we haveTf(x) = Tf1(x) + Tf2(x) = h(x)(U1(x)f1 + U2(x)f2) = h(x)U(x)f for almost all x ∈X,so Tf and h(x)U(x)f are equal elements of the space Lp(X, F).Now let E be a subspace of C(K) containing the function 1(k) ≡1.4

Any function e ∈C(K) has a common tangent functional either with the function1 or with the function −1. Setting, correspondingly, f = 1 or f = −1 in Lemma 1 weobtain that, for an arbitrary e ∈E, ∥Te(x)∥= ∥T1(x)∥∥e∥for almost all x ∈X.

Leth(x) = ∥T1(x)∥.Let ek, k ∈N be a sequence of linearly independent functions from E such that theirlinear span is dense in E and denote by D the set of linear combinations of functions ek withrational coefficients. Given fixed representatives Tek from the corresponding equivalenceclasses of functions from the space Lp(X; F), define an operator T(x) on D for every x ∈Xby T(x)(P λiei) = P λiTei(x), λi ∈R.Since the set D is countable there exists a set X0 ⊂X with σ(X \ X0) = 0 suchthat, for every x ∈X0 and every e ∈D, ∥T(x)e∥= ∥Te(x)∥= h(x)∥e∥.

Hence, for everyx ∈X0, either h(x) = 0 or the operator U(x) = T(x)/h(x) is an isometry from D to F.The operators U(x) can be uniquely extended to isometries on the whole space E,therefore, we have proved the first statement of Theorem 1.Now an argument similar to (4) proves the second statement.Remark. If p = 1 the statement of Theorem 1 is not true.For instance, the two-dimensional space l2∞is isometric to a subspace of L1([0, 1]).

Thus, for any Banach spaceF, l2∞is isometric to a subspace of L1([0, 1]; F), and Theorem 1 would have implied thatl2∞is isometric to a subspace of any Banach space.5

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Belgique, Vol.B30(1978), 83-872. Emmanuele, G.: Copies of l∞in Kothe spaces of vector valued functions, Illinois J.Math.

36(1992), 293-2963. Koldobsky, A.: Isometries of Lp(X; Lq) and equimeasurability, Indiana Univ.

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Koldobsky, A.: Measures on spaces of operators and isometries, J. Soviet Math.42(1988), 1628-16365. Kwapien, S.: On Banach spaces containing c0, Studia Math.

52(1974), 187-1886. Mendoza, J.: Copies of l∞in Lp(µ; X), Proc.

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Pisier, G.: Une propriete de stabilite de la classe des espaces ne contenant pas l1, C.R.Acad. Sci.

Paris Ser A 86(1978), 747-7498. Raynaud, Y.: Sous espaces lr et geometrie des espaces Lp(Lq) et Lφ, C.R.

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