Uniqueness of the Solution to the Operator Equation f (A)=A

关于算子方程f(A)=A解的唯一性(英文)

Let H be a complex Hilbert space, and let , where {Bn} is a sequence of normal operators on H comm muting pairwise, such that and (H): A commutes with every Bn and with then T is the unique one in X satisfying the equation, and T must be normal.

设H是复Hilbert空间,又设f(z)=sum form n=0 to ∞(B_nZ^n),z∈Δ={z:|z|<1},其中{B_n}是H上一列两两交换的正常算子,满足条件:级数按范数收敛,‖f(z)‖<1在Δ上处处成立,且1 σ(B_1)又记X={A∈β(H):σ(A)Δ且A与每个B_n交换}。本文证明了,若有T∈X使得f(T)=T,则T是X中满足所论方程的唯一元素。此外,T必须是正常算子。