Subcriticality, positivity, and gaugeability of the Schrödinger operator

Subcriticality, positivity, and gaugeability of the Schrödinger operator
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薛定谔算子的次临界性、正性和可测性

DOI:
10.1090/s0273-0979-1990-15965-8
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发表时间:
1990
影响因子:
1.3
通讯作者:
Z. Zhao
Z. Zhao
中科院分区:
数学1区
文献类型:
--
作者:
Z. Zhao

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我们从以下三个方面研究了R(d>3)中薛定谔算子H:= -(A/2)+ V > 0的性质:(I)次临界性:直觉上,如果H > 0是次临界的,那么应该可以通过小扰动来扰动H,并且仍然保持它的非负性。更确切地说,我们有以下断言:(a)对于任何q e Bc(Bc表示具有紧支集的有界Borel函数类),存在e > 0使得-(A/2)+ V + eq > 0。(b)存在一个函数q e Bc,q < 0且q ^ 0 a.e.使得-(A/2)+ V + q > 0。次临界状态还有另外两种定义:(c)(B)。Simon [7])存在fi > 0使得-(A/2)+(l+fi)V>0。(d)(M. Murata [6])对于f(?),存在一个正的绿色函数G(-,-). (II)强正性:(e)存在Hu = 0的正解u > 0,极限为:lim,^ u(x)> 0。(f)存在Hu = 0的解u,且c > u > c > 0。(g)存在Hu = 0的解u,且u > c > 0。(III)可度量性:设{Xt:t > 0}是R中的布朗运动,设E表示从xeR开始的布朗路径上的期望。设u0(x)= ε *[exp(/0°° V(Xs)ds)]。(h)w0(.x)^ oo在R d . (i)u 0(x)在i中有界?.
We investigate properties of the Schrödinger operator H := -(A/2) + V > 0 in R(d>3) in the following three aspects: (I) Subcriticality: Intuitively, the idea is that if H > 0 is subcritical, then it should be possible to perturb H by small perturbations and still keep its nonnegativity. More precisely, we have the following assertions: (a) For any q e Bc (Bc denotes the class of bounded Borel functions with compact support), there exists an e > 0 such that -(A/2) + V + eq > 0. (b) There exists a function q e Bc, q < 0 and q ^ 0 a.e. such that -(A/2) + V + q > 0. There have been two other definitions of subcriticality: (c) (B. Simon [7]) There exists fi > 0 such that -(A/2) + (l+fi)V>0. (d) (M. Murata [6]) There exists a positive Green function G(-,-) for /ƒ. (II) Strong Positivity: (e) There exists a positive solution u > 0 of Hu = 0 with the limit: lim, , ^ u(x) > 0. (f) There exists a solution u of Hu = 0 with c > u > c > 0. (g) There exists a solution u of Hu = 0 with u > c > 0. (III) Gaugeability: Let {Xt : t > 0} be the Brownian motion in R and let E denote the expectation over the Brownian paths starting from x e R . Put u0(x) := £*[exp(/0°° V(Xs) ds)]. (h) w0(.x) ^ oo in R d . (i) u0(x) is bounded in i? .