单一等式 $A^{*n}A^n = (A^*A)^n$ 不蕴含无根有向树上加权移位的拟正规性
The single equality $A^{*n}A^n = (A^*A)^n$ does not imply the quasinormality of weighted shifts on rootless directed trees
AI总结：
本文证明了该等式蕴含双边加权移位的拟正规性，但在无根有向树上构造了满足该等式的非拟规范加权移位反例，并指出该反例亦存在于 $L^2$-空间的复合算子类中。
摘要：
证明了每个满足等式 $W^{*n}W^{n}=(W^{*}W)^{n}$（对于某个整数 $n\geq 2$）的有界单射双边加权移位 $W$ 是拟正规的。对于任意整数 $n\geq 2$，构造了一个无根有向树上具有一个分支顶点的有界非拟正规加权移位 $A$ 的例子，它满足等式 $A^{*n}A^{n}=(A^{*}A)^{n}$。还证明了这样的例子可以在 $\sigma$-有限测度空间上的 $L^2$-空间中的复合算子类中构造出来。
英文摘要：
It is proved that each bounded injective bilateral weighted shift $W$ satisfying the equality $W^{*n}W^{n}=(W^{*}W)^{n}$ for some integer $n\geq 2$ is quasinormal. For any integer $n\geq 2$, an example of a bounded non-quasinormal weighted shift $A$ on a rootless directed tree with one branching vertex which satisfies the equality $A^{*n}A^{n}=(A^{*}A)^{n}$ is constructed. It is also shown that such an example can be constructed in the class of composition operators in $L^2$-spaces over $σ$-finite measure spaces.
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。
