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- Thread starter Boromir
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- Feb 7, 2012

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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?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 $P(T,T^*)$

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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.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?

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- Feb 7, 2012

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Okay, it's the operator norm in B(H). But what norm are you using on the space of polynomials?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.

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a polynomial of operators is just an operator so the same norm.Okay, it's the operator norm in B(H). But what norm are you using on the space of polynomials?

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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).a polynomial of operators is just an operator so the same norm.

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$x$ is a member of the spectrum of $T$. $p(x,x^*)$ is a polynomial in $x$ and $x$ conjugate, so just a complex number.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).

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