The Level function in rearrangement invariant spaces

Publ. Mat. 45 (2001), 175–198 THE LEVEL FUNCTION IN REARRANGEMENT INVARIANT SPACES Gord Sinnamon Abstract An exact expression for the down norm is given in terms of the level function on all rearrangement invariant spaces and a useful approximate expression is given for the down norm on all rearrangement invariant spaces whose upper Boyd index is not one. 1. Introduction Let λbe a measure on Rand take Xto be a rearrangement invariant space of λ-measurable functions. We define the down norm of a λ-measurable function fto be fX↓= sup R |f|gdλ:g≥0,gnon-increasing,gX≤1.(1.1) Had we taken the supremum over all gin the unit ball of X, the associate space of the Banach function space X, we would have recovered the norm of fin Xso it is immediate that fX↓≤fX. The significance of the down norm is that the inequality R fgdλ ≤fX↓gX (1.2) holds for all fand all non-negative, non-increasing functions g. Since the down norm of fis smaller than the norm of fin Xthis estimate uses the monotonicity hypothesis on gto improve the usual estimate R fgdλ ≤fXgX. 2000 Mathematics Subject Classification. Primary: 46E30; Secondary: 26D15. Key words. Level function, rearrangement invariant, down norm. Support from the Natural Sciences and Engineering Research Council of Canada is gratefully acknowledged. 176 G. Sinnamon To use inequality (1.2) effectively it is necessary to understand the down norm. In the case that Xis Lebesgue space this has been done in two ways. Halperin [6] and Lorentz [9] gave an exact expression for the down norm when λis a non-negative weight function times Lebesgue measure. Given a function fthey constructed a related function f◦, called the level function of f, and showed that the down norm is precisely the norm of f◦in X.In[13], the level function construction was extended to general (regular) measures on R, the down norm was shown to define a Banach space and the dual space was also constructed. The second approach to understanding the down norm, given in [12] for weighted Lebesgue spaces, was to give an equivalent norm in a more tractable form. The norm in Xof a certain averaging operator applied to fwas shown to be equivalent to the down norm of f. The loss of exactness is more than made up for because the averaging operator is linear. This approach was extended to Orlicz spaces with weights in [8] where the down norm in sequence spaces was also considered. Our object is to look at both of these approaches in the more general setting of rearrangement invariant spaces. As often happens when theorems are examined in their natural generality, the proofs reduce to their essential features and greater understanding is gained. We will see how the averaging operator involved in the results of [12] and [8] arises naturally from the level function construction and how the finiteness of λ(R) affects that operator. We will also see why the level function approach to the down norm remains valid in all rearrangement invariant spaces while a restriction is required for the other approach to be valid. In [12] the restriction was that the Lebesgue index be greater than one and in [8]a∆ 2condition was imposed on the N-functions defining the Orlicz spaces. For definitions and notation involving Banach function spaces and rearrangement invariant spaces we refer to [1]. We adopt the convention that 0·∞ =∞/∞=0. IfAand Bare expressions involving f, we write A≈Bto mean that there exists a positive constant C, not depending on f, such that C−1A≤B≤CA. The range of integration of an integral given with limits is taken to be the closed interval so that b a fdλ=[a,b] fdλ but b −∞ fdλ=(−∞,b] f dλ. 2. The down norm Let Xbe a rearrangement invariant space over the measure space (R,λ). For the down norm to be interesting some restrictions on The Level Function in R.I. Spaces 177 λare in order. Since we want non-negative, non-increasing functions to be λ-measurable we assume that sets of the form (−∞,x] and (−∞,x) are λ-measurable which means that all Borel sets are λ-measurable. To ensure that the space Xactually contains non-trivial non-negative, nonincreasing functions we assume that for each x,λ(−∞,x]<∞.For technical reasons the measure λin [13] was assumed to be regular and since we wish to apply those results we make the same assumption here. Finally, in working with rearrangement invariant spaces it is usual to assume that the underlying measure space is resonant [1, Definition II.2.3 and Theorem II.2.7] so that, among other things, the associate space will also be rearrangement invariant. For these reasons we assume henceforth that λis a regular Borel measure on R, λ(−∞,x]<∞for all x∈R,and λis nonatomic or else completely atomic with all atoms having measure 1. (2.1) With these assumptions on the measure λwe can show that X↓, the collection of functions fsatisfying fX↓<∞, is a normed vector space: It is easy to see that X↓is a vector space containing Xand it is clear from (1.1) that · X↓is non-negative, homogeneous, and satisfies the triangle inequality. It remains to show that only the zero function has zero norm in X↓. For each x∈R,λ(−∞,x]<∞so χ(−∞,x]∈Xand hence, if fX↓= 0 we have x −∞ fdλ= 0 for each x. It follows that f=0λ-almost everywhere and we have shown that · X↓is a norm. In fact, X↓is a Banach space as we show in Theorem 5.3. In most applications, the measure λis weighted Lebesgue measure on the half line, dλ(x)=w(x)dx with wa non-negative, locally integrable function on [0,∞). The case that λis counting measure on the positive integers also arises. The rearrangement invariant spaces for the latter measure include lpand Orlicz sequence spaces while those for the former measure are weighted Lebesgue spaces, Orlicz spaces, Lorentz spaces and others. We hasten to point out that while the weighted Lebesgue space Lp w[0,∞) is not rearrangement invariant with respect to Lebesgue measure unless w≡1, it is rearrangement invariant with respect to the measure w(x)dx. We plan to use the level function to relate the norm in the space X↓ to the norm on the original space X. The next proposition introduces the level function as constructed in [13]. For convenience we define B 178 G. Sinnamon to be the collection of λ-measurable functions on Rwhich are bounded and supported in a set of the form (−∞,M] for some M∈R. Proposition 2.1. Suppose λsatisfies (2.1) and f∈B. Then there is a non-negative, non-increasing function f◦∈B, called the level function of fwith respect to λ, and having the following properties. (a) There exists a finite or countable collection of disjoint intervals Ii of finite, non-zero λmeasure such that f=f◦λ-almost everywhere on E=R\∪ iIiand for each i, f◦(x)=(1/λIi)Ii |f|dλ for λ-almost every x∈Ii. (b) If gis non-negative and non-increasing then R] |f|gdλ≤R f◦g dλ. (c) If f1,f 2∈Band |f1|≤|f2|then f◦ 1≤f◦ 2. Proof: The structure of the level function of fis given in [13, Theorem 4.4, Definition 4.6, Corollary 4.8 and Theorem 4.9]. There it is shown that f◦is non-negative and non-increasing and that (a) holds. It is not assumed in [13] that fis supported on (−∞,M] so the possibility of a level interval of infinite λmeasure is considered there. An easy argument shows that if fis supported on (−∞,M] then all the level intervals Iiare contained in (−∞,M] and hence are of finite λmeasure. Clearly we may discard those of zero λ-measure. Part (b) is given in [13, Theorem 4.11] and (c) is proved in [13, Theorem 5.2]. The main result of this section is given in the next theorem for f∈B and in Corollary 2.4 for general f. Theorem 2.2. Suppose λsatisfies (2.1),Xis a rearrangement invariant space over (R,λ),f∈B, and f◦is the level function of fwith respect to λ. Then fX↓=f◦X. Proof: We use the level intervals of fto define the operator Af. Afh=hχE+ i1 λIiIi hdλ χIi. Note that Afis self-adjoint, that is R(Afg)hdλ =Rg(Afh)dλ for appropriate gand h. Also note that by [1, Theorem II.4.8] Afis a contraction on any rearrangement invariant space, in particular AfhX≤ The Level Function in R.I. Spaces 179 hXfor h∈X. It is clear from the definition that Af|f|=f◦and since the sets Iiare intervals, Afhis non-negative and non-increasing whenever his. If gis non-negative and non-increasing and gX≤1 then by Proposition 2.1(b), R |f|gdλ≤R f◦gdλ≤f◦X so we have fX↓≤f◦X. Now we prove the reverse inequality. A simple limiting argument shows that f◦X= sup R f◦hdλ where the supremum is taken over non-negative functions h∈Bsatisfying hX≤1. For such an h, since f◦is non-negative and nonincreasing, we have R f◦hdλ≤R f◦h◦dλ =R (Af|f|)(Ahh)dλ =R |f|(Af(Ahh)) dλ. Set g=Af(Ahh). Since Ahh=h◦is non-negative and non-increasing we see that gis non-negative and non-increasing. Moreover, gX≤ hX≤1 since both Afand Ahare contractions on X. We conclude that f◦X≤sup R |f|gdλ:g≥0,gnon-increasing,gX≤1=fX↓. This completes the proof. To extend the definition of the level function from f∈Bto all λ-measurable functions fwe use Proposition 2.1(c). Definition 2.3. If fis a λ-measurable function let fn= min(|f|,n)χ(−∞,n]and set f◦= limn→∞ f◦ n. Clearly fn∈Bfor each nso f◦ nis defined. By Proposition 2.1(c), {f◦ n}is a non-decreasing sequence so the limit in Definition 2.3 always exists as a function which takes values in [0,∞]. Moreover, if f∈B then fn=ffor sufficiently large nso the new definition of f◦agrees with the original one. It is immediate that, with this definition of the level function, Proposition 2.1(c) remains valid for arbitrary functions. An application of the Monotone Convergence Theorem shows that Part (b) also extends. Part (a) does not hold for arbitrary functions because the rightmost level interval may have infinite λmeasure. To what extent the structure of 180 G. Sinnamon f◦for an arbitrary function can be described in terms of level intervals is not clear. The sequence fn= min(|f|,n)χ(−∞,n]in Definition 2.3 is chosen for convenience, in Section 5 we show that the definition of f◦is independent of the approximating sequence. Corollary 2.4. Suppose λsatisfies (2.1) and Xis a rearrangement invariant space over (R,λ).Iff∈X↓then f◦is finite λ-almost everywhere, belongs to X, and fX↓=f◦X. Proof: Since fn= min(|f|,n)χ(−∞,n]is a non-decreasing sequence, Proposition 2.1(c) shows that f◦ nis also. The Fatou property of the Banach function space X, Theorem 2.2, and the observation that fn≤|f|show that f◦X= lim n→∞ f◦ nX= lim n→∞ fnX↓≤fX↓. Now [1, Lemma I.1.5(i) and Theorem I.1.4] show that f◦∈Xand is therefore finite λ-almost everywhere. For each non-negative, non-increasing function gwith gX≤1we have, by the Monotone Convergence Theorem and Proposition 2.1(b), R |f|gdλ= lim n→∞ R fngdλ≤lim n→∞ R f◦ ngdλ≤lim n→∞ f◦ nX=f◦X. Taking the supremum over all such gyields fX↓≤f◦Xand completes the proof. 3. An equivalent norm Expressing the down norm of a function fin terms of the level function of f, although exact, has a major drawback. The map f→f◦ is not linear, in fact, it is not even sublinear. In the Lebesgue space case it was shown in [12] that the space X↓has an equivalent norm which can be expressed in terms of a linear averaging operator applied to f. The same averaging operator was shown to work in Orlicz spaces in [8]. The linearity of this averaging operator leads to a duality principle which reduces weighted inequalities for a general operator considered over monotone functions to weighted inequalities for a modified operator considered over all functions. In this section we show that the equivalent norm and the duality principle remain valid for a wide range of rearrangement invariant spaces. Since the techniques involved are quite different this also provides new proofs of some results of [12] and [8]. The Level Function in R.I. Spaces 181 Since λsatisfies (2.1), its cumulative distribution function is finite on R. Let Λ(x)=x −∞ dλ for x∈[−∞,∞] and define the averaging operator Pby Pf(x)=Λ(x)−1x −∞ fdλ+Λ(∞)−1∞ −∞ f dλ. By our convention the second term is absent, regardless of f, when Λ(∞)=∞. Theorem 3.1. Suppose λsatisfies (2.1) and Xis a rearrangement invariant space over (R,λ). Then fX↓≈PfXfor all f≥0if and only if P:X→Xis bounded. Proof: Suppose first that fX↓≈PfXfor all f≥0. Then there exists a constant Csuch that for any f∈X, PfX≤CfX↓≤CfX so P:X→Xis bounded. Conversely, suppose that P:X→Xis bounded and hence continuous. Then there exists a constant Csuch that PfX≤CfXfor all f∈X.Iff≥0 then set fn= min(f,n)χ(−∞,n]so that f◦is the pointwise limit of the increasing sequence {f◦ n}. By Proposition 2.1(b) with g=χ(−∞,x]we have x −∞ fndλ ≤x −∞ f◦ ndλ for each nand each x∈R.Thus PfnX≤P(f◦ n)X≤Cf◦ nX and so, using the Monotone Convergence Theorem and the Fatou property of X, we have PfX≤Cf◦X=CfX↓. On the other hand, since f◦ nis non-negative and non-increasing, f◦ n≤ P(f◦ n) and hence fX↓=f◦X= lim n→∞ f◦ nX≤lim n→∞ P(f◦ n)X. To complete the proof it will suffice to prove the following lemma since then we will have fX↓≤3 lim n→∞ PfnX=3PfX. Lemma 3.2. Suppose λsatisfies (2.1) and Xis a rearrangement invariant space over (R,λ). Then for any non-negative f∈Bwe have P(f◦)X≤3PfX. 182 G. Sinnamon Proof: We show that P(f◦)−PfX≤2PfX, from which the result is immediate. For this argument we need a few details from [13, Definition 4.6] in addition to those presented in Proposition 2.1. With Iiand Eas in Proposition 2.1, if we define aiand biby (ai,b i)⊂Ii⊂[ai,b i] then we have the following: The point x∈Eif and only if (−∞,x) f◦dλ =(−∞,x) fdλ and (−∞,x] f◦dλ =(−∞,x] f dλ. The point xis interior to one of the intervals Iiif and only if (−∞,x) f◦dλ > (−∞,x) fdλ and (−∞,x] f◦dλ > (−∞,x] f dλ. The left endpoint, ai∈Iiif and only if (−∞,ai] f◦dλ > (−∞,ai] f dλ. The right endpoint, bi∈Iiif and only if (−∞,bi) f◦dλ > (−∞,bi) f dλ. It follows that P(f◦)(x)−Pf(x) = i Λ(x)−1x −∞ f◦dλ −x −∞ fdλ χIi(x) = i Λ(x)−1Ii∩(−∞,x] f◦dλ −Ii∩(−∞,x] fdλ χIi(x) ≤ i Λ(x)−1Ii∩(−∞,x] f◦dλχIi(x). The second equality above is easy to prove in two cases depending on whether ai∈Iior not. We use Proposition 2.1(a) to continue the calculation. P(f◦)(x)−Pf(x) ≤ i Λ(x)−1Ii∩(−∞,x] dλ Ii dλ−1Ii f dλχIi(x) = i λ(−∞,x]−1λ(Ii∩(−∞,x])λ(Ii)−1Ii f dλχIi(x). The Level Function in R.I. Spaces 183 Now we use the obvious inequality λ(Ii∩(−∞,x])λ(Ii∪(−∞,x]) ≤λ(−∞,x]λ(Ii) to get P(f◦)(x)−Pf(x)≤ i λ(Ii∪(−∞,x])−1Ii f dλχIi(x).(3.1) Note that for x∈Ii,Ii∪(−∞,x] does not depend on x. It is either (−∞,b i) or (−∞,b i] depending on whether or not biis in Ii. Set Bi= λ(Ii∪(−∞,x]). Define I0and I1by I0={i:2Bi<Λ(∞)}and I1={i:2Bi≥Λ(∞)}. For each i∈I 0choose ci∈Rsuch that Λ(ci)=2Bi. This is possible if λis non-atomic because Λ is continuous in that case. It is also possible if λconsists of equal atoms because the condition λ(−∞,x]<∞ensures that the atoms do not cluster. The reason for choosing such a ciis so that the set I i=(−∞,c i]\(Ii∪(−∞,x]) has λ-measure Bifor each x∈Ii. For each i∈I 1set I i=Ii. We claim that      i B−1 iIi f dλχIi    X ≤     i B−1 iIi f dλχI i    X .(3.2) This is a familiar calculation in rearrangement invariant spaces which follows from Lemma 3.3 below. Now, if i∈I 0and x∈I ithen x≤ciso 2Bi=Λ(ci)≥Λ(x). It follows that  i∈I0 B−1 iIi f dλχI i(x)≤2Λ(x)−1 i∈I0Ii fdλχ I i(x)≤2Λ(x)−1x −∞ fdλ since the intervals Iiare disjoint. If i∈I 1then I i=Iiand 2Bi≥Λ(∞) so, once again using disjointness, we have  i∈I1 B−1 iIi fdλχ I i(x)≤2Λ(∞)−1∞ −∞ f dλ. Combining these last two estimates with (3.1) and (3.2) yields the desired inequality P(f◦)−PfX≤2PfX and completes the proof. 190 G. Sinnamon Proof: Since ¯αX<1, Theorem 4.4 shows that there exists a positive constant Csuch that C−1PϕX≤ϕX↓≤CPϕX for all ϕ∈X↓. If a C1exists satisfying (4.5) then for any f∈Yand any non-negative, non-increasing g∈Xwe have M f(T∗g)dµ =R (Tf)gdλ≤TfX↓gX ≤CPTfXgX≤CC1fYgX. Taking the supremum over all fwith fY≤1 yields (4.6) with C2= CC1. Conversely, if there exists a C2satisfying (4.6) then for any f∈Y and any non-negative, non-increasing g∈Xwe have R (Tf)gdλ=M f(T∗g)dµ ≤fYT∗gY≤C2fYgX. Taking the supremum over all non-negative, non-increasing g∈Xwith gX≤1 we have TfX↓≤C2fYand hence PTfX≤CTfX↓≤CC2fY and so (4.5) holds with C1=CC2. The Boyd indices are known for many classes of rearrangement invariant spaces. The simplest is the class of Lebesgue spaces. For 1 ≤p≤ ∞let Lp λdenote the collection of λ-measurable functions fsuch that fLp λ<∞where fLp λ≡R |f|pdλ1/p for p<∞and fL∞ λ≡ess supλ x∈R |f(x)|. It is well known that the upper Boyd index of Lp λis 1/p. Theorem 4.4 reduces to the following. Proposition 4.6. Suppose λsatisfies (2.1) and 1≤p≤∞. Then P:Lp λ→Lp λif and only if 1<p≤∞if and only if sup R fgdλ :g≥0,gnon-increasing,gLp λ ≤1 ≡fLp λ↓≈PfLp λ. The Level Function in R.I. Spaces 191 Note that since λmay be counting measure on the set of positive integers this includes the case Lp λ=lp. If 1 <p<∞and vis a non-negative weight defined on (0,∞) then we may define λby dλ(x)=χ(0,∞)(x)v(x)dx and replace fby f/v to obtain sup ∞ 0 fg :g≥0,gnon-increasing,gLp v≤1 ≈∞ 0x 0 fpx 0 v−p v(x)dx1/p +∞ 0 f∞ 0 v−1/p which was proved in [12, Theorem 1]. Considerable progress has been made on determining the Boyd indices of Orlicz spaces in [3], [4], [5], [10] and others but only a small portion of this theory is required for our purposes. We refer to [11] for the definitions of a Young’s function Φ, its complementary Young’s function Ψ, and the Orlicz space LΦ λ. We say a Young’s function satisfies the ∆2condition and write Φ ∈∆2provided there exists a constant C>1 such that Φ(2x)≤CΦ(x) for all x>0. We say that Φ satisfies the ∆∞ 2 condition and write Φ ∈∆∞ 2provided there exist constants N>0 and C>1 such that Φ(2x)≤CΦ(x) for all x>N. Proposition 4.7. Suppose λsatisfies (2.1),Φis a Young’s function, and Ψis its complementary Young’s function. Then P:LΦ λ→LΦ λif and only if Ψ∈∆∞ 2if and only if (4.7) sup R fgdλ :g≥0,gnon-increasing,gLΨ λ≤1 ≡fLΦ λ↓≈PfLΦ λ. Proof: The associate space of LΦ λis LΨ λwith equivalent norms so all that is needed to deduce this result from Theorem 4.4 is to verify that the upper Boyd index of LΦ λis less than one if and only if Ψ ∈∆∞ 2. Since the upper Boyd index of LΦ λis one minus the lower Boyd index of LΨ λwe wish to show that the lower Boyd index of Ψ is greater than zero if and only if Ψ ∈∆∞ 2. This follows from [10, Theorem 3.2b and Theorem 4.2]. When λis weighted Lebesgue measure on the half line, or λis counting measure on the positive integers (4.7) was established in [8, Theorem 2.2 and Theorem 3.2] under the assumption that both Φ and Ψ satisfy the ∆2condition. For sequence spaces Heinig and Kufner give somewhat 192 G. Sinnamon more. Their Theorem 3.2 includes a weighted version of the down norm which suggests the following problem. Problem 4.8. Suppose λsatisfies (2.1), Xis a rearrangement invariant space over (R,λ), and vis a non-negative, λ-measurable function. Characterize the norm fX↓ v= sup R |f|gdλ:g≥0,gnon-increasing,gvX≤1. 5. Completeness and duality We have seen that X↓is a normed vector space. In this section we show that X↓is a Banach space of functions which is not, in general, a Banach function space. We also characterize the dual space of X↓.To begin we show that the map f→f◦preserves increasing limits. Proposition 5.1. Suppose that λsatisfies (2.1) and f∈B.If0≤fn↑ |f|then f◦ n↑f◦. Proof: Since f∈B,fn∈Bfor all nand hence f,fn∈L2 λ⊂L2 λ↓for all n.By[13, Theorem 5.4] f◦is the unique 2-level function of fand f◦ n is the unique 2-level function of fn.Now[13, Lemma 5.3] with hn=f◦ n shows that limn→∞ f◦ nis also a 2-level function of f. We conclude that limn→∞ f◦ n=f◦as required. Theorem 5.2. Suppose that λsatisfies (2.1) and Xis a rearrangement invariant space over (R,λ).If0≤fn↑|f|then f◦ n↑f◦and fnX↓↑ fX↓. Proof: First note that Proposition 2.1(c) easily extends to arbitrary functions and therefore fn≤|f|implies f◦ n≤f◦and we have limn→∞ f◦ n≤ f◦. To prove the other inequality let h=|f|, set hn= min(h, n)χ(−∞,n] and define mn,k = min(fn,h k). Since fn↑h≥hkfor all k, we have limn→∞ mn,k =hkfor all k. Since hk∈B, Proposition 5.1 shows that limn→∞ m◦ n,k =h◦ kfor all k.Now by Definition 2.3 f◦= lim k→∞ h◦ k= lim k→∞ lim n→∞ m◦ n,k ≤lim k→∞ lim n→∞ f◦ n= lim n→∞ f◦ n. The Level Function in R.I. Spaces 193 Thus we have f◦ n↑f◦. Now we apply Corollary 2.4 and the Fatou property in Xto get lim n→∞ fnX↓= lim n→∞ f◦ nX=f◦X=fX↓. This completes the proof. Theorem 5.3. If λsatisfies (2.1) and Xis a rearrangement invariant space over (R,λ)then X↓is a Banach space. Proof: We have already shown that X↓is a normed linear space, it remains to prove completeness. To do this we show that every absolutely summable sequence in X↓is summable in X↓. Suppose that fn∈X↓for all nand ∞ n=1 fnX↓<∞. Then |fn|∈X↓and so SN≡N n=1 |fn|∈X↓for each N. Let Sbe the pointwise limit of the non-decreasing sequence SN, that is, S=∞ n=1 |fn|. Since SN↑Sand lim N→∞ SNX↓≤ N  n=1 fnX↓≤ ∞  n=1 fnX↓<∞ we have SX↓<∞by Theorem 5.2 and hence S∈X↓. In particular this implies that Sis finite λ-almost everywhere because for any M∈R, χ(−∞,M]is non-increasing so M −∞ Sdλ≤SX↓χ(−∞,M]X<∞. (Since λ(−∞,M] is finite, χ(−∞,M]∈X.) Thus, Sis finite λ-almost everywhere on (−∞,M] but since Mwas arbitrary, Sis finite λ-almost everywhere on R. We have shown that ∞ n=1 |fn|converges pointwise λ-almost everywhere and it follows that ∞ n=1 fnconverges pointwise λ-almost everywhere. Let FN=N n=1 fnand F=∞ n=1 fn. Fix K, set IN= infn≥N|Fn−FK|≤|FN−FK|for N>Kand note that IN∈X↓ with INX↓≤N n=K+1 fnX↓≤∞ n=K+1 fnX↓. The sequence IN is non-decreasing and converges pointwise to |F−FK|. Thus, applying Theorem 5.2 again, F−FKX↓= lim N→∞ INX↓≤ ∞  n=K+1 fnX↓ and so F−FKX↓tends to zero as K→∞. That is, FK→Fin X↓ as K→∞. This completes the proof. 194 G. Sinnamon Although X↓is a Banach space, it is not a Banach function space in general as the following example shows: Take λto be Lebesgue measure on the half line. We show that condition [1, Definition I.1.1(P5)] fails for the space L2 λ↓. To do this we exhibit a set Eof finite measure and a sequence of functions {fn}in L2 λ↓such that Efn/fnL2 λ↓is unbounded. Set E= ∞  n=1 [n−2−n,n] and note that λ(E)=∞ n=1 2−n<∞.Iffn=2 nχ[n−2−n,n]we compute Efn= 1 and f◦ n=(1/n)χ[0,n].ThusfnL2 λ↓=f◦ nL2 λ=n−1/2and so Efn/fnL2 λ↓is unbounded for large n. For the remainder of this section we investigate the dual space of X↓. Definition 5.4. Suppose that gis a λ-measurable function. Define ¯gby ¯g(x) = ess supt≥x|g(t)|, set gX↓=¯gX, and let X↓be the collection of functions gfor which gX↓<∞. Note that ¯gis non-negative and non-increasing and that, by a standard measure theory argument, ¯g≥|g|λ-almost everywhere. The space X↓is a subspace of Xsince we have gX≤gX↓.Itis easy to see that · X↓is a norm. Although the notation X↓suggests the associate space of X↓this is not asserted here. In fact, since X↓is not necessarily a Banach function space, it is not clear that it has a well-defined associate space. The space X↓does behave like an associate space, however, as we see in Theorems 5.6 and 5.7 below. Theorem 5.8 shows that the dual space of X↓often coincides with X↓. To prepare for these three theorems we need another result from [13]. Proposition 5.5. Suppose λsatisfies (2.1),α∈(0,1), and f,g are λ-measurable functions such that f◦and ¯gare finite λ-almost everywhere. Then there exists a non-negative λ-measurable function hsuch that R h|g|dλ ≥α2R |f|¯gdλ and R hϕ dλ ≤R |f|ϕdλ for all non-negative, non-increasing, λ-measurable functions ϕ. Proof: This is proved in [13, Lemma 6.5] under the assumption that f◦∈Lp λand ¯g∈Lp λfor some p∈(1,∞]. Only the weaker assumption that f◦and ¯gare finite λ-almost everywhere is used in the proof. It remains valid in this more general situation without alteration. The Level Function in R.I. Spaces 195 Theorem 5.6. Suppose λsatisfies (2.1),Xis a rearrangement invariant space over (R,λ), and f∈X↓. Then fX↓= sup R fgdλ :gX↓≤1.(5.1) Proof: Since ¯gis non-increasing and ¯g≥|g|λ-almost everywhere we have (5.2) R fgdλ ≤R |f||g|dλ ≤R |f|¯gdλ≤R f◦¯gdλ ≤f◦X¯gX=fX↓gX↓. This proves that the left side of (5.1) is no less than the right side. To prove the other inequality note that if gis non-negative and nonincreasing with gX≤1 then sgn(f)gX↓≤1 and R|f|gdλ = Rfsgn(f)gdλ so by (1.1) fX↓= sup R |f|gdλ:g≥0,gnon-increasing,gX≤1 ≤sup R fgdλ :gX↓≤1. This completes the proof. Theorem 5.7. Suppose λsatisfies (2.1),Xis a rearrangement invariant space over (R,λ), and g∈X↓. Then gX↓= sup R fgdλ :fX↓≤1.(5.3) Proof: The calculation in (5.2) shows that the left hand side of (5.3) is no less than the right hand side. To prove the other inequality we require Proposition 5.5. Fix g∈X↓. Then ¯g∈Xso ¯gis finite λ-almost everywhere. Fix α∈(0,1) and choose a non-negative function fwith fX≤1 such that gX↓=¯gX≤1 αR f¯g dλ. Since fX≤1, f∈X⊂X↓so f◦is finite λ-almost everywhere by Corollary 2.4. The function hof Proposition 5.5 satisfies hX↓= sup R hϕ dλ ≤sup R |f|ϕdλ≤fX≤1 196 G. Sinnamon where the suprema are taken over all non-negative, non-increasing functions ϕwith ϕX≤1. Therefore α3gX↓≤α2R f¯gdλ≤R h|g|dλ ≤sup R fgdλ :fX↓≤1. Since this holds for all α∈(0,1) we may let α→1 to obtain the remaining inequality in (5.3). Definition 5.8. Suppose that Ais a Banach space of functions. We say the space Ahas absolutely continuous norm provided every nonincreasing sequence of functions in Awhich converges to zero pointwise, converges to zero in A. In view of [1, Proposition I.3.5] this definition agrees with [1, Definition I.3.1] when Ais a Banach Function Space. Theorem 5.9. Suppose λsatisfies (2.1),Xis a rearrangement invariant space over (R,λ)and both Xand X↓have absolutely continuous norm. Then the dual space of X↓is X↓. More precisely, each function g∈X↓gives rise to a continuous linear functional Lgon X↓given by Lg(f)=Rfgdλ. The norm of Lgis gX↓and every continuous linear functional on X↓is Lgfor some g∈X↓. Proof: By [1, Corollary I.4.3] X=X∗.Ifg∈X↓then Lgis a clearly linear and Theorem 5.7 shows that Lgis continuous on X↓, having norm gX↓. Suppose now that Lis a continuous, linear functional on X↓. We wish to show that L=Lgfor some g∈X↓. Since Xis a subspace of X↓(with · X↓≤· X) we may consider Las a continuous linear functional on X. The hypothesis that X∗=X shows that there is a function g∈Xsuch that Lf =Rfgdλ for all f∈X. To complete the proof we show that Lf =Rfgdλ for all f∈X↓and that g∈X↓. To do the first we fix f∈X↓, set fn= min(n, max(−n, f))χ(−∞,n], and consider the sequence {|fng|}. This increases pointwise to |fg|. The Monotone Convergence Theorem yields R |fg|dλ = lim n→∞ R |fng|dλ = lim n→∞ L(|fn|sgn(g)) ≤LX→Rlim n→∞ fnX↓≤LX→RfX↓<∞. The Level Function in R.I. Spaces 197 Thus fg ∈L1 λ. Now consider {fn}as a sequence in X↓. Since {|f−fn|} decreases to zero pointwise and X↓has absolutely continuous norm we see that {fn}converges to fin X↓. Since Lis continuous, Lf = lim n→∞ L(fn) = lim n→∞ R fngdλ=R fgdλ where the last inequality follows from the Dominated Convergence Theorem using our observation that fg ∈L1 λ. The second task is to show that g∈X↓. Set gn(x) = min(n, |g(x)|)χ(∞,n]and note that gn∈X↓and {gn}increases pointwise to |g|.Thus{¯gn}increases pointwise to ¯g. The Fatou property of the Banach function space Ximplies that lim n→∞ gnX↓= lim n→∞ ¯gnX=¯gX=gX↓. But gnX↓= sup R f¯gndλ ≤sup R |f||g|dλ = sup L(|f|sgn(g)) ≤LX↓→R. Here the suprema are taken over all functions fwith fX↓≤1. The conclusion is that gX↓≤LX↓→Rso that g∈X↓as required. Corollary 5.10. If λsatisfies (2.1),Xis a rearrangement invariant space over (R,λ)and both Xand X↓have absolutely continuous norm then X↓is complete. Proof: The dual space of any normed linear space is complete. See [13, Example 6.9] for an example to show that X↓need not be reflexive even when both Xand X↓have absolutely continuous norm. It may be that if Xhas absolutely continuous norm then so does X↓ but we have no proof or counterexample. In very many cases, however, it is true. We leave the following as a (non-trivial) exercise: Suppose λ satisfies (2.1), Λ(x)=x −∞ dλ, and hM(x) = min(M,1/Λ(x)) for M>0. If Xhas absolutely continuous norm and hM∈Xfor all M>0 then X↓has absolutely continuous norm. 198 G. Sinnamon References [1] C. Bennett and R. Sharpley,“Interpolation of operators”, Academic Press Inc., Boston, MA, 1988. [2] D. W. Boyd, Indices of function spaces and their relationship to interpolation, Canad. J. Math. 21 (1969), 1245–1254. [3] D. W. Boyd, Indices for the Orlicz spaces, Pacific J. Math. 38 (1971), 315–323. [4] A. Fiorenza and M. Krbec, Indices of Orlicz spaces and some applications, Comment. Math. Univ. Carolin. 38(3) (1997), 433–451. [5] A. Fiorenza and M. Krbec, A formula for the Boyd indices in Orlicz spaces, Funct. Approx. Comment. Math. 26 (1998), 173–179. [6] I. 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Department of Mathematics University of Western Ontario London, Ontario, N6A 5B7 Canada E-mail address:[email protected] Primera versi´o rebuda el 17 de maig de 2000, darrera versi´o rebuda el 10 de juliol de 2000.