Gelfand map is topologically injective
$begingroup$
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 Dec 25 '18 at 1:50
hyperkahlerhyperkahler
1,487714
$endgroup$
add a comment |
$begingroup$
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 Dec 25 '18 at 1:50
hyperkahlerhyperkahler
1,487714
$endgroup$
$begingroup$
1). you forgot a 'y'. 2). what are your thoughts?
$endgroup$
– mathworker21
Dec 25 '18 at 2:17
add a comment |
$begingroup$
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 Dec 25 '18 at 1:50
hyperkahlerhyperkahler
1,487714
$endgroup$
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 Dec 25 '18 at 1:50
hyperkahlerhyperkahler
1,487714
asked Dec 25 '18 at 1:50
hyperkahlerhyperkahler
1,487714
asked Dec 25 '18 at 1:50
hyperkahlerhyperkahler
1,487714
asked Dec 25 '18 at 1:50
hyperkahlerhyperkahler
1,487714
asked Dec 25 '18 at 1:50
hyperkahlerhyperkahler
1,487714
1,487714
$begingroup$
1). you forgot a 'y'. 2). what are your thoughts?
$endgroup$
– mathworker21
Dec 25 '18 at 2:17
add a comment |
$begingroup$
1). you forgot a 'y'. 2). what are your thoughts?
$endgroup$
– mathworker21
Dec 25 '18 at 2:17
$begingroup$
1). you forgot a 'y'. 2). what are your thoughts?
$endgroup$
– mathworker21
Dec 25 '18 at 2:17
$begingroup$
1). you forgot a 'y'. 2). what are your thoughts?
$endgroup$
– mathworker21
Dec 25 '18 at 2:17
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 Dec 25 '18 at 7:34
C.DingC.Ding
1,3911421
$endgroup$
1
$begingroup$
Indeed, it turned out to be much easier than i had expected. The "only if" part can be proved, i guess, in the similar fashion: since $G$ is topologically injective, $||Gx|| geq C ||x||$ which implies $r(x) geq C ||x||$, whilst it's well-known that $r(x) leq ||x^{n}||^{frac{1}{n}}$ for any $n$, plugging the $n = 2$ into the last line yields $||x^{2}|| geq C ||x||^{2}$
$endgroup$
– hyperkahler
Dec 26 '18 at 1:01
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 Dec 25 '18 at 7:34
C.DingC.Ding
1,3911421
$endgroup$
1
$begingroup$
Indeed, it turned out to be much easier than i had expected. The "only if" part can be proved, i guess, in the similar fashion: since $G$ is topologically injective, $||Gx|| geq C ||x||$ which implies $r(x) geq C ||x||$, whilst it's well-known that $r(x) leq ||x^{n}||^{frac{1}{n}}$ for any $n$, plugging the $n = 2$ into the last line yields $||x^{2}|| geq C ||x||^{2}$
$endgroup$
– hyperkahler
Dec 26 '18 at 1:01
add a comment |
$begingroup$
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 Dec 25 '18 at 7:34
C.DingC.Ding
1,3911421
$endgroup$
1
$begingroup$
Indeed, it turned out to be much easier than i had expected. The "only if" part can be proved, i guess, in the similar fashion: since $G$ is topologically injective, $||Gx|| geq C ||x||$ which implies $r(x) geq C ||x||$, whilst it's well-known that $r(x) leq ||x^{n}||^{frac{1}{n}}$ for any $n$, plugging the $n = 2$ into the last line yields $||x^{2}|| geq C ||x||^{2}$
$endgroup$
– hyperkahler
Dec 26 '18 at 1:01
add a comment |
$begingroup$
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 Dec 25 '18 at 7:34
C.DingC.Ding
1,3911421
$endgroup$
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 Dec 25 '18 at 7:34
C.DingC.Ding
1,3911421
answered Dec 25 '18 at 7:34
C.DingC.Ding
1,3911421
answered Dec 25 '18 at 7:34
C.DingC.Ding
1,3911421
answered Dec 25 '18 at 7:34
C.DingC.Ding
1,3911421
1,3911421
1
$begingroup$
Indeed, it turned out to be much easier than i had expected. The "only if" part can be proved, i guess, in the similar fashion: since $G$ is topologically injective, $||Gx|| geq C ||x||$ which implies $r(x) geq C ||x||$, whilst it's well-known that $r(x) leq ||x^{n}||^{frac{1}{n}}$ for any $n$, plugging the $n = 2$ into the last line yields $||x^{2}|| geq C ||x||^{2}$
$endgroup$
– hyperkahler
Dec 26 '18 at 1:01
add a comment |
1
$begingroup$
Indeed, it turned out to be much easier than i had expected. The "only if" part can be proved, i guess, in the similar fashion: since $G$ is topologically injective, $||Gx|| geq C ||x||$ which implies $r(x) geq C ||x||$, whilst it's well-known that $r(x) leq ||x^{n}||^{frac{1}{n}}$ for any $n$, plugging the $n = 2$ into the last line yields $||x^{2}|| geq C ||x||^{2}$
$endgroup$
– hyperkahler
Dec 26 '18 at 1:01
1
1
$begingroup$
Indeed, it turned out to be much easier than i had expected. The "only if" part can be proved, i guess, in the similar fashion: since $G$ is topologically injective, $||Gx|| geq C ||x||$ which implies $r(x) geq C ||x||$, whilst it's well-known that $r(x) leq ||x^{n}||^{frac{1}{n}}$ for any $n$, plugging the $n = 2$ into the last line yields $||x^{2}|| geq C ||x||^{2}$
$endgroup$
– hyperkahler
Dec 26 '18 at 1:01
$begingroup$
Indeed, it turned out to be much easier than i had expected. The "only if" part can be proved, i guess, in the similar fashion: since $G$ is topologically injective, $||Gx|| geq C ||x||$ which implies $r(x) geq C ||x||$, whilst it's well-known that $r(x) leq ||x^{n}||^{frac{1}{n}}$ for any $n$, plugging the $n = 2$ into the last line yields $||x^{2}|| geq C ||x||^{2}$
$endgroup$
– hyperkahler
Dec 26 '18 at 1:01
add a comment |
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$begingroup$
1). you forgot a 'y'. 2). what are your thoughts?
$endgroup$
– mathworker21
Dec 25 '18 at 2:17