超常可测算子,隶属于半有限冯-诺依曼代数

IF 0.8 Q2 MATHEMATICS
Airat Bikchentaev
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引用次数: 0

摘要

让 \(\mathcal {M}\) 是希尔伯特空间 \(\mathcal {H}\) 上的冯-诺依曼算子代数,并且 \(\tau \) 是 \(\mathcal {M}\) 上的忠实正态半有限迹、\S(\mathcal {M}, \tau )\) 是所有 \(\tau\) 可测算子的 \( ^*\)- 代数。假设S(\mathcal {M}, \tau )\ 中的算子\(T)是超常的或\( ^*\)-paranormal 的。如果\(T^n\)对于某个\(nin \mathbb {N}\)是\(\tau \)-紧凑的,那么T就是\(\tau \)-紧凑的;如果\(T^n=0\)对于某个\(nin \mathbb {N}\),那么\(T=0\);if (T^3=T) then\(T=T^*\); if (T^2\in L_1(\mathcal {M}, \tau )\) then\(T\in L_2(\mathcal {M}, \tau )\) and\(\Vert T\Vert _2^2=\Vert T^2\Vert _1\)。如果一个算子 \(T\in S(\mathcal {M}, \tau )\) 是下正则的,并且 \(T^{*p}T^q\) 对于某个 \(p, q \in \mathbb {N}\cup \{0\}), \(p+q \ge 1\) 是紧凑的,那么 T 就是正则的。如果 \(T\in S(\mathcal {M}, \tau )\) 对于某个 \(0<ple 1\) 是 p-hyponormal 的,那么算子 \((T^*T)^p-(TT^*)^p\) 在 \( \mathcal {M}\) 中不可能有逆。如果算子\(T\in S(\mathcal {M}, \tau )\) 是下正则(或共正则),并且算子\(T^2\)是赫米特的,那么T就是正则的。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
Hyponormal measurable operators, affiliated to a semifinite von Neumann algebra

Let \(\mathcal {M}\) be a von Neumann algebra of operators on a Hilbert space \(\mathcal {H}\) and \(\tau \) be a faithful normal semifinite trace on \(\mathcal {M}\), \(S(\mathcal {M}, \tau )\) be the \( ^*\)-algebra of all \(\tau \)-measurable operators. Assume that an operator \(T\in S(\mathcal {M}, \tau )\) is paranormal or \( ^*\)-paranormal. If \(T^n\) is \(\tau \)-compact for some \(n\in \mathbb {N}\) then T is \(\tau \)-compact; if \(T^n=0\) for some \(n\in \mathbb {N}\) then \(T=0\); if \(T^3=T\) then \(T=T^*\); if \(T^2\in L_1(\mathcal {M}, \tau )\) then \(T\in L_2(\mathcal {M}, \tau )\) and \(\Vert T\Vert _2^2=\Vert T^2\Vert _1\). If an operator \(T\in S(\mathcal {M}, \tau )\) is hyponormal and \(T^{*p}T^q\) is \(\tau \)-compact for some \(p, q \in \mathbb {N}\cup \{0\}\), \(p+q \ge 1\) then T is normal. If \(T\in S(\mathcal {M}, \tau )\) is p-hyponormal for some \(0<p\le 1\) then the operator \((T^*T)^p-(TT^*)^p\) cannot have the inverse in \( \mathcal {M}\). If an operator \(T\in S(\mathcal {M}, \tau )\) is hyponormal (or cohyponormal) and the operator \(T^2\) is Hermitian then T is normal.

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CiteScore
1.60
自引率
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发文量
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