Abstract
If psi :[0,ell ]rightarrow [0,infty [ is absolutely continuous, nondecreasing, and such that psi (ell )>psi (0), psi (t)>0 for t>0, then for fin L^1(0,ell ), we have ‖f‖1,ψ,(0,ℓ):=∫0ℓψ′(t)ψ(t)2∫0tf∗(s)ψ(s)dsdt≈∫0ℓ|f(x)|dx=:‖f‖L1(0,ℓ),(∗)\\documentclass[12pt]{minimal}\t\t\t\t\\usepackage{amsmath}\t\t\t\t\\usepackage{wasysym}\t\t\t\t\\usepackage{amsfonts}\t\t\t\t\\usepackage{amssymb}\t\t\t\t\\usepackage{amsbsy}\t\t\t\t\\usepackage{mathrsfs}\t\t\t\t\\usepackage{upgreek}\t\t\t\t\\setlength{\\oddsidemargin}{-69pt}\t\t\t\t\\begin{document}$$\\begin{aligned} \\Vert f\\Vert _{1,\\psi ,(0,\\ell )}:=\\int \\limits _0^\\ell \\frac{\\psi '(t)}{\\psi (t)^2}\\int \\limits _0^tf^*(s)\\psi (s)dsdt\\approx \\int \\limits _0^\\ell |f(x)|dx=:\\Vert f\\Vert _{L^1(0,\\ell )},\\quad (*) \\end{aligned}$$\\end{document}where the constant in > rsim depends on psi and ell . Here by f^* we denote the decreasing rearrangement of f. When applied with f replaced by |f|^p, 1<p<infty , there exist functions psi so that the inequality Vert |f|^pVert _{1,psi ,(0,ell )}le Vert |f|^pVert _{L^1(0,ell )} is not rougher than the classical one-dimensional integral Hardy inequality over bounded intervals (0,ell ). We make an analysis on the validity of (*) under much weaker assumptions on the regularity of psi , and we get a version of Hardy’s inequality which generalizes and/or improves existing results.
Highlights
As a by-product of a characterization of an interpolation space between grand and small Lebesgue spaces, in [7, Theorems 6.2 and 6.4](see [1] for recent developments), it has been shown, in particular, that for 1 < p < ∞, ⎡ 1⎛ t ⎞ ⎤1 p ⎛1 ⎞1 p⎣ ⎝ (1 − log s)−1f ∗(s)pds⎠ dt ⎦ ≈ ⎝ |f (x)|pdx⎠, t (1.1)where f ∗ denotes the decreasing rearrangement of f
The goal of this note is to give a direct proof of a generalized version of this equivalence for functions
It may be of interest to observe that the factor 1/t in the integrand in the left hand side is exactly ψ (t)/ψ2(t) if we set ψ(t) = log Strictly speaking, this choice is not allowed in Theorem 2.1 because in t = 1 the function ψ diverges and continuity in the closed interval [0, 1] is lost
Summary
Note that the constant in the right hand side of the classical Hardy inequality (1.2) is optimal The optimality of the constant in the right hand side of the classical Hardy inequality (1.2) can be expressed writing sup f ≡0
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