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arXiv 2007.14349math.FAmath.OA
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具有自伴幂或正规幂的无界算子及其相关结果

Unbounded operators having self-adjoint or normal powers and some related results

Souheyb Dehimi, Mohammed Hichem Mortad

摘要:

我们证明了一个稠密定义的可闭算子$A$使得$A^2$的预解集不为空必然是闭的。然后将该结果推广到多项式$p(A)$的情况。我们还推广了Sebestyén-Tarcsay最近关于J. von Neumann结果的逆的结果。其他有趣的结果也给出了,其中之一是一个证明,如果$T$是一个拟正规(无界)运营商,使$T^n$是正常的一些$n\geq2$,然后$T$是正常的。利用Pietrzycki-Stochel最近的一个结果,我们推论了使$T^n$正规的闭次正规算子必正规.另一个值得注意的结果是,一个亚正规算子$A$,有界或无界,使得$A^p$和$A^q$是自伴的一些互素数$p$和$q$,是自伴的。证明了可逆算子(有界或无界)$A$是正规的,对于某些互质数$p$和$q$,$A^p$和$A^q$是正规的.这两个结果是使用算术中的Bézout定理来证明的。

英文摘要:

We show that a densely defined closable operator $A$ such that the resolvent set of $A^2$ is not empty is necessarily closed. This result is then extended to the case of a polynomial $p(A)$. We also generalize a recent result by Sebestyén-Tarcsay concerning the converse of a result by J. von Neumann. Other interesting consequences are also given, one of them being a proof that if $T$ is a quasinormal (unbounded) operator such that $T^n$ is normal for some $n\geq2$, then $T$ is normal. By a recent result by Pietrzycki-Stochel, we infer that a closed subnormal operator such that $T^n$ is normal, must be normal. Another remarkable result is the fact that a hyponormal operator $A$, bounded or not, such that $A^p$ and $A^q$ are self-adjoint for some co-prime numbers $p$ and $q$, is self-adjoint. It is also shown that an invertible operator (bounded or not) $A$ for which $A^p$ and $A^q$ are normal for some co-prime numbers $p$ and $q$, is normal. These two results are shown using Bézout's theorem in arithmetic.

机构、总结、中文摘要均由AI分析而来,可能存在错误,请以论文正文为准。

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