The discussion focuses on the continuous extension of a homomorphism defined on polynomials in a bounded normal operator \( T \) and its adjoint \( T^* \). The homomorphism \( h \) is expressed as \( h(p(T,T^*))=p(x,x^*) \), where \( x \) is a member of the spectrum of \( T \). Participants emphasize the need to demonstrate that the limit \( \lim p_n(x,x^*) \) exists as the corresponding sequence in \( B(H) \) converges, and that the mapping is well-defined under the operator norm in \( B(H) \).
PREREQUISITESMathematicians, particularly those specializing in functional analysis, operator theory, and anyone studying the properties of bounded normal operators and their spectra.
What have you tried so far? When you use words like "continuously" and "closure", you are implying the existence of a topology in the space B(H) and also in the space of polynomials. How do you define these topologies?Boromir said:Let $T$ be a bounded normal operator and let $x$ be a member of the spectrum. Consider the homomorphism defined on the set of polynomials in $T$ and $T^{*}$ given by $h(p(T,T^*))=p(x,x^*)$ Prove that this map can be continuosly extended to the closure of $P(T,T^*)$
Opalg said:What have you tried so far? When you use words like "continuously" and "closure", you are implying the existence of a topology in the space B(H) and also in the space of polynomials. How do you define these topologies?
Okay, it's the operator norm in B(H). But what norm are you using on the space of polynomials?Boromir said:just the usual one given by the operator norm. I can see that I need to show 2 things, namely that lim$p_{n}(x,x^*)$ exist given that the corresponding sequence in B(H) converges, and that the map is well defined.
Opalg said:Okay, it's the operator norm in B(H). But what norm are you using on the space of polynomials?
The range of the mapping $h$ consists of expressions of the form $p(x,x^*)$. What is $x$ supposed to mean there, and what is $p(x,x^*)$? (It's not an operator).Boromir said:a polynomial of operators is just an operator so the same norm.
Opalg said:The range of the mapping $h$ consists of expressions of the form $p(x,x^*)$. What is $x$ supposed to mean there, and what is $p(x,x^*)$? (It's not an operator).