In the second section, the non-linear case (a general Hardy-type inequality) is handled with a direct and analytic proof. In the last section, it is illustrated that the basic estimates presented in the first two sections can still be improved considerably....
POINTWISE HARDY INEQUALITIES 421 Proof of Theorem 1. Fix u ∈C∞ (Ω) and 0 |u(x)|≤C1,(x)Mq g(x)holds for all x∈Ωt. Since q (Mqg)Lp(Rn)≤C(n,p/q)(g)Lp(Rn) ,and hence the inequality(8) follows with a=0....
(Ω) and all x ∈ Ω. Using this lemma we can easily adopt the proof of [5, Theorem 1] to prove Hardy’s inequality. 3.3. Theorem. Let Ω be an open and bounded subset of R n . Let p: Ω → [1, ∞) be log-H¨older continuous in Ω with 1 < p ...
Using inequality: \left\lvert F(z)\right\rvert\leq\frac{1}{(1-\left\lvert z\right\rvert)^{\frac{1}{p}}}\lVert F\rVert_{H^p}\\ for F\in H^p with 0<p<\infty , we can prove H^p is a complete space. Thus H^p is closed subspace of L^p in isometry sense. Remark...
Second, for the caseα=n/pit is enough to replace inequality (13) with the inequality〈x;∇g(s(x))〉≤−1and the restof the proof is similar to the caseα6=n/p.In the proof ofε-sharpness of inequality (3) we will choose the functionuεsuch that the ratio of the left-...
he Hardy inequality A probabilistic proof of the Hardy inequalityA probabilistic proof of the Hardy inequalityJensen inequalityScale distributionThis short note provides a fully probabilistic proof to the Hardy inequality.doi:10.1016/j.spl.2015.03.007Walker...
Lemma(Minkowski's integral inequality) 设(X,μ),(T,ν) 为两个 σ -有限的测度空间, 1≤p<∞ . 那么对乘积空间 (X,μ)×(T,ν) 上任意非负的可测函数 F=F(x,t) , 有 ‖∫XFx(⋅)dμ(x)‖Lp≤∫X‖Fx‖Lpdμ(x).这里Fx(t)=F(x,t). Proof. 记 g(t)=∫XF(x,t)dμ(x)....
An improved Hardy–Sobolev inequality and its applications, Proc. Amer. Math. Soc., 130 (2002), pp. 489-505 View in ScopusGoogle Scholar [3] Adimurthi, M.J. Esteban An improved Hardy–Sobolev inequality in W1,p and its applications to Schrödinger operators NoDEA Nonlin. Differential Equa...
4 Korn’s Inequality in Other Spaces In this section we note that our proof of Korn’s inequality in a Hardy-Sobolev space yields a similar result in other spaces upon which the Riesz transforms are bounded linear operators. The proof is the same as the one we have given, modulo a chang...
Then the inequality(2.1)‖u‖Lp,q(Rn)≤A‖(−△)s/2u‖Lp1,q1(Rn)αq‖u‖Lp2,q2(Rn)q−αqholds forαq1+q−αq2=1,α(1p1−sn)+(q−α)1p2=qpandA≤(3np1n−p1s(ωn−1n)n−sn)αq. Proof Applying the Hölder inequality and simple computation yields ‖u‖Lp,q...