Gelfand map is topologically injective
Let $A$ be a commutative Banach algebra and let
$$G: A rightarrow C_{0}(text{Spec}(A))$$ be a topologically injective map.
Recall that the map $T: X rightarrow Y$ is called topologically injective if there exists a positive $C > 0$ such that $||Tx|| geq C ||x||$ for all $x in X$. One can also show that the latter is equivalent to the definition in which
$$T: X rightarrow text{im}(T)$$ is a homeomorphism (in fact, it is the original definition)
How to prove that the Gelfand map $G$ is topologicall injective if and only if there exists $C > 0$ such that for all $a in A$
$$||a^{2} || geq C ||a||^{2}$$
functional-analysis c-star-algebras banach-algebras gelfand-representation
asked 11 hours ago
Arteom
1,449714
add a comment |
Let $A$ be a commutative Banach algebra and let
$$G: A rightarrow C_{0}(text{Spec}(A))$$ be a topologically injective map.
Recall that the map $T: X rightarrow Y$ is called topologically injective if there exists a positive $C > 0$ such that $||Tx|| geq C ||x||$ for all $x in X$. One can also show that the latter is equivalent to the definition in which
$$T: X rightarrow text{im}(T)$$ is a homeomorphism (in fact, it is the original definition)
How to prove that the Gelfand map $G$ is topologicall injective if and only if there exists $C > 0$ such that for all $a in A$
$$||a^{2} || geq C ||a||^{2}$$
functional-analysis c-star-algebras banach-algebras gelfand-representation
asked 11 hours ago
Arteom
1,449714
1). you forgot a 'y'. 2). what are your thoughts?
– mathworker21
10 hours ago
add a comment |
Let $A$ be a commutative Banach algebra and let
$$G: A rightarrow C_{0}(text{Spec}(A))$$ be a topologically injective map.
Recall that the map $T: X rightarrow Y$ is called topologically injective if there exists a positive $C > 0$ such that $||Tx|| geq C ||x||$ for all $x in X$. One can also show that the latter is equivalent to the definition in which
$$T: X rightarrow text{im}(T)$$ is a homeomorphism (in fact, it is the original definition)
How to prove that the Gelfand map $G$ is topologicall injective if and only if there exists $C > 0$ such that for all $a in A$
$$||a^{2} || geq C ||a||^{2}$$
functional-analysis c-star-algebras banach-algebras gelfand-representation
asked 11 hours ago
Arteom
1,449714
Let $A$ be a commutative Banach algebra and let
$$G: A rightarrow C_{0}(text{Spec}(A))$$ be a topologically injective map.
Recall that the map $T: X rightarrow Y$ is called topologically injective if there exists a positive $C > 0$ such that $||Tx|| geq C ||x||$ for all $x in X$. One can also show that the latter is equivalent to the definition in which
$$T: X rightarrow text{im}(T)$$ is a homeomorphism (in fact, it is the original definition)
How to prove that the Gelfand map $G$ is topologicall injective if and only if there exists $C > 0$ such that for all $a in A$
$$||a^{2} || geq C ||a||^{2}$$
functional-analysis c-star-algebras banach-algebras gelfand-representation
functional-analysis c-star-algebras banach-algebras gelfand-representation
asked 11 hours ago
Arteom
1,449714
asked 11 hours ago
Arteom
1,449714
asked 11 hours ago
Arteom
1,449714
asked 11 hours ago
Arteom
1,449714
asked 11 hours ago
Arteom
1,449714
1,449714
1). you forgot a 'y'. 2). what are your thoughts?
– mathworker21
10 hours ago
add a comment |
1). you forgot a 'y'. 2). what are your thoughts?
– mathworker21
10 hours ago
1). you forgot a 'y'. 2). what are your thoughts?
– mathworker21
10 hours ago
1). you forgot a 'y'. 2). what are your thoughts?
– mathworker21
10 hours ago
add a comment |
1 Answer
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The spectral radius $r(a)=lim_{ntoinfty} |a^{2^n}|^{frac{1}{2^n}}geq lim_{ntoinfty} (C^n|a|^{2^n})^{frac{1}{2^n}}=|a|$, thus $|G(a)|=r(a)=|a|$. So, $G$ is in fact an isometry.
answered 5 hours ago
C.Ding
1,3831321
add a comment |
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1 Answer
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1 Answer
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active
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The spectral radius $r(a)=lim_{ntoinfty} |a^{2^n}|^{frac{1}{2^n}}geq lim_{ntoinfty} (C^n|a|^{2^n})^{frac{1}{2^n}}=|a|$, thus $|G(a)|=r(a)=|a|$. So, $G$ is in fact an isometry.
answered 5 hours ago
C.Ding
1,3831321
add a comment |
The spectral radius $r(a)=lim_{ntoinfty} |a^{2^n}|^{frac{1}{2^n}}geq lim_{ntoinfty} (C^n|a|^{2^n})^{frac{1}{2^n}}=|a|$, thus $|G(a)|=r(a)=|a|$. So, $G$ is in fact an isometry.
answered 5 hours ago
C.Ding
1,3831321
add a comment |
The spectral radius $r(a)=lim_{ntoinfty} |a^{2^n}|^{frac{1}{2^n}}geq lim_{ntoinfty} (C^n|a|^{2^n})^{frac{1}{2^n}}=|a|$, thus $|G(a)|=r(a)=|a|$. So, $G$ is in fact an isometry.
answered 5 hours ago
C.Ding
1,3831321
The spectral radius $r(a)=lim_{ntoinfty} |a^{2^n}|^{frac{1}{2^n}}geq lim_{ntoinfty} (C^n|a|^{2^n})^{frac{1}{2^n}}=|a|$, thus $|G(a)|=r(a)=|a|$. So, $G$ is in fact an isometry.
answered 5 hours ago
C.Ding
1,3831321
answered 5 hours ago
C.Ding
1,3831321
answered 5 hours ago
C.Ding
1,3831321
answered 5 hours ago
C.Ding
1,3831321
1,3831321
add a comment |
add a comment |
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1). you forgot a 'y'. 2). what are your thoughts?
– mathworker21
10 hours ago