Delgado Garrido, Olvido

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idUS. Depósito de Investigación Universidad de Sevilla

VIII JORNADAS DE MATEMA´TICA APLICADAOptimal domain for the Hardy operator∗Olvido DelgadoInstituto Universitario de Matema´tica Pura y AplicadaUniversidad Polite´cnica de ValenciaIntroductionIn the 1920’s, Hardy proved what today is one of the classical inequalities on integration:(∫ ∞0( 1x∫ x0f (s) ds)pdx)1/p≤ pp− 1(∫ ∞0f (x)pdx)1/pfor 1 < p <∞ and f any positive measurable real function. This inequality has been studiedby a lot of mathematicians producing a great variety of Hardy type inequalities which applyin different areas of mathematics.Considering the average of functions as an operator S, i.e.Sf (x) :=1x∫ x0f (s) ds, x > 0,for f ∈ L1loc([0,∞)), the Hardy’s inequality establishes that S:Lp([0,∞)) → Lp([0,∞))is well defined and continuous. This S is called the Hardy operator .• Is there any space Y larger than Lp([0,∞)) such that S:Y → Lp([0,∞))?• Which is the largest space Y such that S:Y → Lp([0,∞))?The general caseLet X be a Banach function space (B.f.s.), i.e. a Banach space of measurable real functionson [0,∞), satisfying|f | ≤ |g| a.e. and g ∈ X ⇒ f ∈ X and ‖f‖X ≤ ‖g‖X .• Is there any B.f.s. Y such that S:Y → X?• Which is the largest B.f.s. Y such that S:Y → X?Note that if S:Y → X is well defined is automatically continuous, since it is a positiveoperator between Banach lattices.Optimal domain for SThe space[S,X ] := { f : [0,∞)→ R measurable: S|f | ∈ X }is a B.f.s. with the norm ‖f‖[S,X ] = ‖S|f |‖X . Moreover,• S: [S,X ]→ X ,• [S,X ] is the largest B.f.s. Y satisfying S:Y → X .So, [S,X ] is the optimal domain for S considered with values in X .Is it possible to give a precise description for [S,X ]?Remark. For a positive kernel operator T satisfying some suitable conditions, it is possibleto describe [T,X ] in terms of interpolation spaces (see [1] and [2]). Unfortunately, the Hardyoperator S does not satisfies these conditions. However, we can describe [S,X ] for someparticulars X .[ 1 ] G. P. Curbera and W. J. Ricker, Optimal domains for kernel operators via interpolation, Math. Nachr. 244 (2002), 47–63.[ 2 ] O. Delgado, Optimal domains for kernel operators on [0,∞)× [0,∞), Studia Math. 174 (2006), 131–145.Particular cases of [S,X ]• [S, L1([0,∞))] = {0}.• [S, L1,∞([0,∞))] = L1([0,∞)), for the quasi-B.f.s.L1,∞([0,∞)) = {f : [0,∞)→ R measurable : supt≥0{tf∗(t)}<∞},where f∗ is the decreasing rearrangement of f , i.e.f∗(t) = inf{c > 0 : m({x ∈ [0,∞] : |f (x)| > c}) ≤ t}for m being the Lebesgue measure.• Let ϕ: [0,∞)→ [0,∞) be an increasing concave function with ϕ(0) = 0 and consider theLorentz spaceΛϕ ={f : [0,∞)→ R measurable :∫ ∞0f∗(t) dϕ(t) <∞}.Theorem. Suppose that ϕ satisfiesX ϕ(0+) = 0,X θϕ(t) :=∫∞tϕ′(s)s ds <∞ for all t > 0,X ∃ C > 0 such that ϕ(t)t ≤ C θϕ(t) for all t > 0.Then,[S,Λϕ] = L1θϕ,where L1θϕdenotes the space of integrable functions with respect to the Lebesguemeasure with density θϕ.In particular [S, L p,1([0,∞))] = L1t 1/p−1, for 1 < p <∞,since L p,1([0,∞)) = Λϕ for ϕ(t) = t1/p and θϕ(t) = 1(p−1) 1t1−1/p .[S,X ] may not be r.i.Recall that a B.f.s. X is rearrangement invariant (r.i.), if whenever f and g have the samedistribution function and f ∈ X it follows that g ∈ X .Theorem. If X is an r.i. B.f.s. and S:X → X , then[S,X ] is not r.i.In particular X ( [S,X ].Note that every r.i. B.f.s. X satisfies L1 ∩ L∞ ⊂ X ⊂ L1 + L∞. Then, the claim of theabove theorem follows, since [S,X ] * L1 + L∞. A function g ∈ L1 + L∞\[S,X ] is givenbyg(t) =h′(t− (tk+1 − 1))if t ∈ (tk+1 − 1, tk+1)0 in other case,where tk+1 = ek, h(t) = 1−√1− t for t ∈ [0, 1] and h′ is the derivative of h.From the Hardy inequality it follows that S:Lp([0,∞)) → Lp([0,∞)) for 1 < p < ∞.Then [S, Lp([0,∞))] is not r.i. and Lp([0,∞))  [S, Lp([0,∞))]. This holds also forp =∞.R.i. optimal domain for SSince [S,X ] may not be r.i., now a natural question is:Which is the largest r.i. B.f.s. contained in [S,X ]?The space[S,X ]r.i. := {f : [0,∞)→ R measurable : Sf∗ ∈ X}is an r.i. B.f.s. with the norm ‖f‖[S,X ]r.i. = ‖Sf∗‖X . Moreover,• [S,X ]r.i. ⊂ [S,X ], since S|f | ≤ Sf∗,• [S,X ]r.i. is the largest r.i. B.f.s. contained in [S,X ].So, [S,X ] is the r.i. optimal domain for S considered with values in X .Is it possible to give a precise description for [S,X ]r.i.?Particular cases of [S,X ]r.i.• [S, L1([0,∞))]r.i. = {0}.• [S, L1,∞([0,∞))]r.i. = L1([0,∞)).Consider the Lorentz space X = Λϕ and θϕ(s) =∫∞sϕ′(y)y dy. Then,• [S,Λϕ]r.i. = Λ∫ t0θϕ(s) ds, whenever ϕ(0+) = 0,• [S,Λϕ]r.i. = L∞ ∩ Λ∫ t0θϕ(s) ds, whenever ϕ(0+) > 0.When does X coincide with [S,X ]r.i.?For every r.i. B.f.s. X , since f∗ ≤ Sf∗, it follows that [S,X ]r.i. ⊂ X . If X also satisfiesS:X → X , then[S,X ]r.i. ⊂ X ⊂ [S,X ].So, since [S,X ]r.i. is the largest r.i. B.f.s. contained in [S,X ], we have that [S,X ]r.i. = X .Conversely, suppose [S,X ]r.i. = X . Then X is obviously r.i. and, since |Sf | ≤ S|f | ≤ Sf∗,it follows that S:X → X . Therefore, the following proposition holds.Proposition. X r.i. and S:X → X ⇔ [S,X ]r.i. = XIn particular, [S, Lp([0,∞))]r.i. = Lp([0,∞))for all 1 < p ≤ ∞.∗ A Joint work with Javier Soria (Dpto. Matema´tica Aplicada y Ana´lisis, Universidad deBarcelona), published at Journal of Functional Analysis, 244 (2007) 119–133.

Optimal domain for the Hardy operator

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Optimal domain for the Hardy operator

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