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[SOLVED] continuous extension of homomorphism

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Boromir

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Feb 15, 2014
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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

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Feb 7, 2012
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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 $P(T,T^*)$
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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Boromir

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Feb 15, 2014
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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?
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

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Feb 7, 2012
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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.
Okay, it's the operator norm in B(H). But what norm are you using on the space of polynomials?
 
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Boromir

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Feb 15, 2014
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Okay, it's the operator norm in B(H). But what norm are you using on the space of polynomials?
a polynomial of operators is just an operator so the same norm.
 

Opalg

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Feb 7, 2012
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a polynomial of operators is just an operator so the same norm.
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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Boromir

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Feb 15, 2014
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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).
$x$ is a member of the spectrum of $T$. $p(x,x^*)$ is a polynomial in $x$ and $x$ conjugate, so just a complex number.
 
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Boromir

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Feb 15, 2014
38
forget it