functional calculus
$begingroup$
Suppose $A$ is a non-unital $C^*$ algebra,$B$ is another $C^*$ algebra.Suppose $phi:Arightarrow B$ is a non-zero $*$ homomorphism and $x_0$ is a normal elememt in $A$,by continuous functional calculus,we have $phi(f(x_0))=f(phi(x_0))$ for any $fin C_0（sigma_{A}(x_0))$ .My question is:can we choose a function $fin C_0（sigma_{A}(x_0))$ such that $|phi(f(x_0))|>1$?
operator-theory operator-algebras c-star-algebras von-neumann-algebras
asked Dec 3 '18 at 17:46
mathrookiemathrookie
918512
$endgroup$
add a comment |
$begingroup$
Suppose $A$ is a non-unital $C^*$ algebra,$B$ is another $C^*$ algebra.Suppose $phi:Arightarrow B$ is a non-zero $*$ homomorphism and $x_0$ is a normal elememt in $A$,by continuous functional calculus,we have $phi(f(x_0))=f(phi(x_0))$ for any $fin C_0（sigma_{A}(x_0))$ .My question is:can we choose a function $fin C_0（sigma_{A}(x_0))$ such that $|phi(f(x_0))|>1$?
operator-theory operator-algebras c-star-algebras von-neumann-algebras
asked Dec 3 '18 at 17:46
mathrookiemathrookie
918512
$endgroup$
$begingroup$
If you can find a function $f$ such that $phi(f)neq0$, can you see why this is true?
$endgroup$
– Aweygan
Dec 3 '18 at 18:30
$begingroup$
If $phi(x_0)neq0$,you mean that $f(z)=z,zin C_0（sigma_{A}(x_0))$ is suitable?But how to ensure that $|phi(f(x_0))|geq1$?
$endgroup$
– mathrookie
Dec 4 '18 at 2:16
add a comment |
$begingroup$
Suppose $A$ is a non-unital $C^*$ algebra,$B$ is another $C^*$ algebra.Suppose $phi:Arightarrow B$ is a non-zero $*$ homomorphism and $x_0$ is a normal elememt in $A$,by continuous functional calculus,we have $phi(f(x_0))=f(phi(x_0))$ for any $fin C_0（sigma_{A}(x_0))$ .My question is:can we choose a function $fin C_0（sigma_{A}(x_0))$ such that $|phi(f(x_0))|>1$?
operator-theory operator-algebras c-star-algebras von-neumann-algebras
asked Dec 3 '18 at 17:46
mathrookiemathrookie
918512
$endgroup$
Suppose $A$ is a non-unital $C^*$ algebra,$B$ is another $C^*$ algebra.Suppose $phi:Arightarrow B$ is a non-zero $*$ homomorphism and $x_0$ is a normal elememt in $A$,by continuous functional calculus,we have $phi(f(x_0))=f(phi(x_0))$ for any $fin C_0（sigma_{A}(x_0))$ .My question is:can we choose a function $fin C_0（sigma_{A}(x_0))$ such that $|phi(f(x_0))|>1$?
operator-theory operator-algebras c-star-algebras von-neumann-algebras
operator-theory operator-algebras c-star-algebras von-neumann-algebras
asked Dec 3 '18 at 17:46
mathrookiemathrookie
918512
asked Dec 3 '18 at 17:46
mathrookiemathrookie
918512
edited Dec 4 '18 at 2:34
mathrookie
asked Dec 3 '18 at 17:46
mathrookiemathrookie
918512
asked Dec 3 '18 at 17:46
mathrookiemathrookie
918512
asked Dec 3 '18 at 17:46
mathrookiemathrookie
918512
918512
$begingroup$
If you can find a function $f$ such that $phi(f)neq0$, can you see why this is true?
$endgroup$
– Aweygan
Dec 3 '18 at 18:30
$begingroup$
If $phi(x_0)neq0$,you mean that $f(z)=z,zin C_0（sigma_{A}(x_0))$ is suitable?But how to ensure that $|phi(f(x_0))|geq1$?
$endgroup$
– mathrookie
Dec 4 '18 at 2:16
add a comment |
$begingroup$
If you can find a function $f$ such that $phi(f)neq0$, can you see why this is true?
$endgroup$
– Aweygan
Dec 3 '18 at 18:30
$begingroup$
If $phi(x_0)neq0$,you mean that $f(z)=z,zin C_0（sigma_{A}(x_0))$ is suitable?But how to ensure that $|phi(f(x_0))|geq1$?
$endgroup$
– mathrookie
Dec 4 '18 at 2:16
$begingroup$
If you can find a function $f$ such that $phi(f)neq0$, can you see why this is true?
$endgroup$
– Aweygan
Dec 3 '18 at 18:30
$begingroup$
If you can find a function $f$ such that $phi(f)neq0$, can you see why this is true?
$endgroup$
– Aweygan
Dec 3 '18 at 18:30
$begingroup$
If $phi(x_0)neq0$,you mean that $f(z)=z,zin C_0（sigma_{A}(x_0))$ is suitable?But how to ensure that $|phi(f(x_0))|geq1$?
$endgroup$
– mathrookie
Dec 4 '18 at 2:16
$begingroup$
If $phi(x_0)neq0$,you mean that $f(z)=z,zin C_0（sigma_{A}(x_0))$ is suitable?But how to ensure that $|phi(f(x_0))|geq1$?
$endgroup$
– mathrookie
Dec 4 '18 at 2:16
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
By Urysohn's lemma, we can always choose a continuous function that vanishes at infinity with $f(z)=10000$ at a fixed point $z$. So, $|phi(f(x_0))|=|f(phi(x_0))|=|f|>9999$.
answered Dec 4 '18 at 2:56
C.DingC.Ding
1,3911321
$endgroup$
$begingroup$
Probably, the OP wants that $f(phi(x_0)) in B$ and so one should also require that $f(0) = 0$. This however is no problem, since the spectrum of $phi(x_0)$ contains points other than zero. E.g. $f(z) = lambda z$ for some large $lambda > 0$.
$endgroup$
– user42761
Dec 4 '18 at 13:11
add a comment |
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1 Answer
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oldest
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1 Answer
1
active
oldest
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active
oldest
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active
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$begingroup$
By Urysohn's lemma, we can always choose a continuous function that vanishes at infinity with $f(z)=10000$ at a fixed point $z$. So, $|phi(f(x_0))|=|f(phi(x_0))|=|f|>9999$.
answered Dec 4 '18 at 2:56
C.DingC.Ding
1,3911321
$endgroup$
$begingroup$
Probably, the OP wants that $f(phi(x_0)) in B$ and so one should also require that $f(0) = 0$. This however is no problem, since the spectrum of $phi(x_0)$ contains points other than zero. E.g. $f(z) = lambda z$ for some large $lambda > 0$.
$endgroup$
– user42761
Dec 4 '18 at 13:11
add a comment |
$begingroup$
By Urysohn's lemma, we can always choose a continuous function that vanishes at infinity with $f(z)=10000$ at a fixed point $z$. So, $|phi(f(x_0))|=|f(phi(x_0))|=|f|>9999$.
answered Dec 4 '18 at 2:56
C.DingC.Ding
1,3911321
$endgroup$
$begingroup$
Probably, the OP wants that $f(phi(x_0)) in B$ and so one should also require that $f(0) = 0$. This however is no problem, since the spectrum of $phi(x_0)$ contains points other than zero. E.g. $f(z) = lambda z$ for some large $lambda > 0$.
$endgroup$
– user42761
Dec 4 '18 at 13:11
add a comment |
$begingroup$
By Urysohn's lemma, we can always choose a continuous function that vanishes at infinity with $f(z)=10000$ at a fixed point $z$. So, $|phi(f(x_0))|=|f(phi(x_0))|=|f|>9999$.
answered Dec 4 '18 at 2:56
C.DingC.Ding
1,3911321
$endgroup$
By Urysohn's lemma, we can always choose a continuous function that vanishes at infinity with $f(z)=10000$ at a fixed point $z$. So, $|phi(f(x_0))|=|f(phi(x_0))|=|f|>9999$.
answered Dec 4 '18 at 2:56
C.DingC.Ding
1,3911321
answered Dec 4 '18 at 2:56
C.DingC.Ding
1,3911321
answered Dec 4 '18 at 2:56
C.DingC.Ding
1,3911321
answered Dec 4 '18 at 2:56
C.DingC.Ding
1,3911321
1,3911321
$begingroup$
Probably, the OP wants that $f(phi(x_0)) in B$ and so one should also require that $f(0) = 0$. This however is no problem, since the spectrum of $phi(x_0)$ contains points other than zero. E.g. $f(z) = lambda z$ for some large $lambda > 0$.
$endgroup$
– user42761
Dec 4 '18 at 13:11
add a comment |
$begingroup$
Probably, the OP wants that $f(phi(x_0)) in B$ and so one should also require that $f(0) = 0$. This however is no problem, since the spectrum of $phi(x_0)$ contains points other than zero. E.g. $f(z) = lambda z$ for some large $lambda > 0$.
$endgroup$
– user42761
Dec 4 '18 at 13:11
$begingroup$
Probably, the OP wants that $f(phi(x_0)) in B$ and so one should also require that $f(0) = 0$. This however is no problem, since the spectrum of $phi(x_0)$ contains points other than zero. E.g. $f(z) = lambda z$ for some large $lambda > 0$.
$endgroup$
– user42761
Dec 4 '18 at 13:11
$begingroup$
Probably, the OP wants that $f(phi(x_0)) in B$ and so one should also require that $f(0) = 0$. This however is no problem, since the spectrum of $phi(x_0)$ contains points other than zero. E.g. $f(z) = lambda z$ for some large $lambda > 0$.
$endgroup$
– user42761
Dec 4 '18 at 13:11
add a comment |
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$begingroup$
If you can find a function $f$ such that $phi(f)neq0$, can you see why this is true?
$endgroup$
– Aweygan
Dec 3 '18 at 18:30
$begingroup$
If $phi(x_0)neq0$,you mean that $f(z)=z,zin C_0（sigma_{A}(x_0))$ is suitable?But how to ensure that $|phi(f(x_0))|geq1$?
$endgroup$
– mathrookie
Dec 4 '18 at 2:16