Optimal domain for the Hardy operator [Póster]

VIII JORNADAS DE MATEM ´ ATICA APLICADA Optimal domain for the Hardy operator∗ Olvido Delgado Instituto Universitario de Matem´atica Pura y Aplicada Universidad Polit´ecnica de Valencia Introduction In the 1920’s, Hardy proved what today is one of the classical inequalities on integration: Z∞ 01 xZx 0 f(s)ds pdx1/p ≤p p−1Z∞ 0 f(x)pdx1/p for 1 < p < ∞and fany positive measurable real function. This inequality has been studied by a lot of mathematicians producing a great variety of Hardy type inequalities which apply in different areas of mathematics. Considering the average of functions as an operator S, i.e. Sf(x) := 1 xZx 0 f(s)ds, x > 0, for f∈L1 loc[0,∞), the Hardy’s inequality establishes that S:Lp[0,∞)→Lp[0,∞) is well defined and continuous. This Sis called the Hardy operator. •Is there any space Ylarger than Lp[0,∞)such that S:Y→Lp[0,∞)? •Which is the largest space Ysuch that S:Y→Lp[0,∞)? The general case Let Xbe a Banach function space (B.f.s.), i.e. a Banach space of measurable real functions on [0,∞), satisfying |f| ≤ |g|a.e. and g∈X⇒f∈Xand kfkX≤ kgkX. •Is there any B.f.s. Ysuch that S:Y→X? •Which is the largest B.f.s. Ysuch that S:Y→X? Note that if S:Y→Xis well defined is automatically continuous, since it is a positive operator between Banach lattices. Optimal domain for S The space [S, X] := {f: [0,∞)→Rmeasurable: S|f| ∈ X} is a B.f.s. with the norm kfk[S,X]=kS|f|kX. Moreover, •S: [S, X]→X, •[S, X] is the largest B.f.s. Ysatisfying S:Y→X. So, [S, X] is the optimal domain for Sconsidered with values in X. Is it possible to give a precise description for [S, X]? Remark. For a positive kernel operator Tsatisfying some suitable conditions, it is possible to describe [T, X] in terms of interpolation spaces (see [1] and [2]). Unfortunately, the Hardy operator Sdoes not satisfies these conditions. However, we can describe [S, X] for some particulars 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,∞)=nf: [0,∞)→Rmeasurable : sup t≥0tf∗(t)<∞o, where f∗is the decreasing rearrangement of f, i.e. f∗(t) = inf{c > 0 : m{x∈[0,∞] : |f(x)|> c}≤t} for mbeing the Lebesgue measure. •Let ϕ: [0,∞)→[0,∞) be an increasing concave function with ϕ(0) = 0 and consider the Lorentz space Λϕ=nf: [0,∞)→Rmeasurable : Z∞ 0 f∗(t)dϕ(t)<∞o. Theorem. Suppose that ϕsatisfies Xϕ(0+) = 0, Xθϕ(t) := R∞ t ϕ′(s) sds < ∞for all t > 0, X∃C > 0 such that ϕ(t) t≤C θϕ(t) for all t > 0. Then, [S, Λϕ] = L1 θϕ, where L1 θϕdenotes the space of integrable functions with respect to the Lebesgue measure with density θϕ. In particular S, Lp,1[0,∞)=L1 t1/p−1,for 1 < p < ∞, since Lp,1[0,∞)= Λϕfor ϕ(t) = t1/p and θϕ(t) = 1 (p−1) 1 t1−1/p . [S, X]may not be r.i. Recall that a B.f.s. Xis rearrangement invariant (r.i.), if whenever fand ghave the same distribution function and f∈Xit follows that g∈X. Theorem. If Xis 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. Xsatisfies L1∩L∞⊂X⊂L1+L∞. Then, the claim of the above theorem follows, since [S, X]*L1+L∞. A function g∈L1+L∞\[S, X] is given by g(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−tfor 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 for p=∞. R.i. optimal domain for S Since [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,∞)→Rmeasurable : Sf∗∈X} is an r.i. B.f.s. with the norm kfk[S,X]r.i. =kSf∗kX. 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 Sconsidered 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= Λϕand θϕ(s) = R∞ s ϕ′(y) ydy. Then, •[S, Λϕ]r.i. = ΛRt 0θϕ(s)ds , whenever ϕ(0+) = 0, •[S, Λϕ]r.i. =L∞∩ΛRt 0θϕ(s)ds , whenever ϕ(0+)>0. When does Xcoincide 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 Xalso satisfies S: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 Xis obviously r.i. and, since |Sf| ≤ S|f| ≤ Sf∗, it follows that S:X→X. Therefore, the following proposition holds. Proposition. Xr.i. and S:X→X⇔[S, X]r.i. =X In particular, S, Lp[0,∞)r.i. =Lp[0,∞) for all 1 < p ≤ ∞. ∗A Joint work with Javier Soria (Dpto. Matem´atica Aplicada y An´alisis, Universidad de Barcelona), published at Journal of Functional Analysis, 244 (2007) 119–133.