Compact operator, functional calculus
$begingroup$
Let $H$ be a Hilbert space, T be a compact operator and $f$ be a bounded function on $sigma(T)$. Now I want to show that the operator $f(T)$ (in the sense of functional calculus) is compact if $dim(H)<infty$. I think the compactness of $f(T)$ follows directly. Under which assumptions do we have the other direction?
Thanks for your help!
functional-analysis operator-theory functional-calculus
asked Jan 2 at 13:51
ShaqAttack1337ShaqAttack1337
110111
$endgroup$
add a comment |
$begingroup$
Let $H$ be a Hilbert space, T be a compact operator and $f$ be a bounded function on $sigma(T)$. Now I want to show that the operator $f(T)$ (in the sense of functional calculus) is compact if $dim(H)<infty$. I think the compactness of $f(T)$ follows directly. Under which assumptions do we have the other direction?
Thanks for your help!
functional-analysis operator-theory functional-calculus
asked Jan 2 at 13:51
ShaqAttack1337ShaqAttack1337
110111
$endgroup$
1
$begingroup$
You fundamentally changed the nature of your question after answers have been added. You should roll your question back to its original state, then open a new question.
$endgroup$
– Aweygan
Jan 3 at 3:14
$begingroup$
How are you forming $f(T)$ for a bounded function only? Are you assuming that $T$ is selfadjoint?
$endgroup$
– DisintegratingByParts
Jan 3 at 5:07
add a comment |
$begingroup$
Let $H$ be a Hilbert space, T be a compact operator and $f$ be a bounded function on $sigma(T)$. Now I want to show that the operator $f(T)$ (in the sense of functional calculus) is compact if $dim(H)<infty$. I think the compactness of $f(T)$ follows directly. Under which assumptions do we have the other direction?
Thanks for your help!
functional-analysis operator-theory functional-calculus
asked Jan 2 at 13:51
ShaqAttack1337ShaqAttack1337
110111
$endgroup$
Let $H$ be a Hilbert space, T be a compact operator and $f$ be a bounded function on $sigma(T)$. Now I want to show that the operator $f(T)$ (in the sense of functional calculus) is compact if $dim(H)<infty$. I think the compactness of $f(T)$ follows directly. Under which assumptions do we have the other direction?
Thanks for your help!
functional-analysis operator-theory functional-calculus
functional-analysis operator-theory functional-calculus
asked Jan 2 at 13:51
ShaqAttack1337ShaqAttack1337
110111
asked Jan 2 at 13:51
ShaqAttack1337ShaqAttack1337
110111
edited Jan 2 at 14:13
ShaqAttack1337
asked Jan 2 at 13:51
ShaqAttack1337ShaqAttack1337
110111
asked Jan 2 at 13:51
ShaqAttack1337ShaqAttack1337
110111
asked Jan 2 at 13:51
ShaqAttack1337ShaqAttack1337
110111
110111
1
$begingroup$
You fundamentally changed the nature of your question after answers have been added. You should roll your question back to its original state, then open a new question.
$endgroup$
– Aweygan
Jan 3 at 3:14
$begingroup$
How are you forming $f(T)$ for a bounded function only? Are you assuming that $T$ is selfadjoint?
$endgroup$
– DisintegratingByParts
Jan 3 at 5:07
add a comment |
1
$begingroup$
You fundamentally changed the nature of your question after answers have been added. You should roll your question back to its original state, then open a new question.
$endgroup$
– Aweygan
Jan 3 at 3:14
$begingroup$
How are you forming $f(T)$ for a bounded function only? Are you assuming that $T$ is selfadjoint?
$endgroup$
– DisintegratingByParts
Jan 3 at 5:07
1
1
$begingroup$
You fundamentally changed the nature of your question after answers have been added. You should roll your question back to its original state, then open a new question.
$endgroup$
– Aweygan
Jan 3 at 3:14
$begingroup$
You fundamentally changed the nature of your question after answers have been added. You should roll your question back to its original state, then open a new question.
$endgroup$
– Aweygan
Jan 3 at 3:14
$begingroup$
How are you forming $f(T)$ for a bounded function only? Are you assuming that $T$ is selfadjoint?
$endgroup$
– DisintegratingByParts
Jan 3 at 5:07
$begingroup$
How are you forming $f(T)$ for a bounded function only? Are you assuming that $T$ is selfadjoint?
$endgroup$
– DisintegratingByParts
Jan 3 at 5:07
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
As stated this is clearly not so. For example let $f(z)=z$; then $f(T)=T$ is compact whether $H$ has finite dimension or not. (Maybe you were supposed to show that if $f(T)$ is compact for all such $f$ then $H$ has finite dimension?)
answered Jan 2 at 14:03
David C. UllrichDavid C. Ullrich
60k43994
$endgroup$
add a comment |
$begingroup$
This is not true. If $T$ is a normal compact operator on $H$, and $f$ is continuous on $sigma(T)$ and $f(0)=0$, then $f(T)$ lies in the $C^*$-algebra generated by $T$, and can be approximated by polynomials in $T,T^*$ with no constant term. The algebra of such polynomials in $T,T^*$ contains only compact operators, hence $f(T)$ is compact (regardless of the dimension of $H$).
EDIT
As your question currently stands, it is trivial. If $dim H<infty$, every operator on $H$ is compact, hence if $Tin B(H)$, then $f(T)in B(H)$ is compact a prioru.
answered Jan 2 at 14:06
AweyganAweygan
14k21441
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
As stated this is clearly not so. For example let $f(z)=z$; then $f(T)=T$ is compact whether $H$ has finite dimension or not. (Maybe you were supposed to show that if $f(T)$ is compact for all such $f$ then $H$ has finite dimension?)
answered Jan 2 at 14:03
David C. UllrichDavid C. Ullrich
60k43994
$endgroup$
add a comment |
$begingroup$
As stated this is clearly not so. For example let $f(z)=z$; then $f(T)=T$ is compact whether $H$ has finite dimension or not. (Maybe you were supposed to show that if $f(T)$ is compact for all such $f$ then $H$ has finite dimension?)
answered Jan 2 at 14:03
David C. UllrichDavid C. Ullrich
60k43994
$endgroup$
add a comment |
$begingroup$
As stated this is clearly not so. For example let $f(z)=z$; then $f(T)=T$ is compact whether $H$ has finite dimension or not. (Maybe you were supposed to show that if $f(T)$ is compact for all such $f$ then $H$ has finite dimension?)
answered Jan 2 at 14:03
David C. UllrichDavid C. Ullrich
60k43994
$endgroup$
As stated this is clearly not so. For example let $f(z)=z$; then $f(T)=T$ is compact whether $H$ has finite dimension or not. (Maybe you were supposed to show that if $f(T)$ is compact for all such $f$ then $H$ has finite dimension?)
answered Jan 2 at 14:03
David C. UllrichDavid C. Ullrich
60k43994
answered Jan 2 at 14:03
David C. UllrichDavid C. Ullrich
60k43994
answered Jan 2 at 14:03
David C. UllrichDavid C. Ullrich
60k43994
answered Jan 2 at 14:03
David C. UllrichDavid C. Ullrich
60k43994
60k43994
add a comment |
add a comment |
$begingroup$
This is not true. If $T$ is a normal compact operator on $H$, and $f$ is continuous on $sigma(T)$ and $f(0)=0$, then $f(T)$ lies in the $C^*$-algebra generated by $T$, and can be approximated by polynomials in $T,T^*$ with no constant term. The algebra of such polynomials in $T,T^*$ contains only compact operators, hence $f(T)$ is compact (regardless of the dimension of $H$).
EDIT
As your question currently stands, it is trivial. If $dim H<infty$, every operator on $H$ is compact, hence if $Tin B(H)$, then $f(T)in B(H)$ is compact a prioru.
answered Jan 2 at 14:06
AweyganAweygan
14k21441
$endgroup$
add a comment |
$begingroup$
This is not true. If $T$ is a normal compact operator on $H$, and $f$ is continuous on $sigma(T)$ and $f(0)=0$, then $f(T)$ lies in the $C^*$-algebra generated by $T$, and can be approximated by polynomials in $T,T^*$ with no constant term. The algebra of such polynomials in $T,T^*$ contains only compact operators, hence $f(T)$ is compact (regardless of the dimension of $H$).
EDIT
As your question currently stands, it is trivial. If $dim H<infty$, every operator on $H$ is compact, hence if $Tin B(H)$, then $f(T)in B(H)$ is compact a prioru.
answered Jan 2 at 14:06
AweyganAweygan
14k21441
$endgroup$
add a comment |
$begingroup$
This is not true. If $T$ is a normal compact operator on $H$, and $f$ is continuous on $sigma(T)$ and $f(0)=0$, then $f(T)$ lies in the $C^*$-algebra generated by $T$, and can be approximated by polynomials in $T,T^*$ with no constant term. The algebra of such polynomials in $T,T^*$ contains only compact operators, hence $f(T)$ is compact (regardless of the dimension of $H$).
EDIT
As your question currently stands, it is trivial. If $dim H<infty$, every operator on $H$ is compact, hence if $Tin B(H)$, then $f(T)in B(H)$ is compact a prioru.
answered Jan 2 at 14:06
AweyganAweygan
14k21441
$endgroup$
This is not true. If $T$ is a normal compact operator on $H$, and $f$ is continuous on $sigma(T)$ and $f(0)=0$, then $f(T)$ lies in the $C^*$-algebra generated by $T$, and can be approximated by polynomials in $T,T^*$ with no constant term. The algebra of such polynomials in $T,T^*$ contains only compact operators, hence $f(T)$ is compact (regardless of the dimension of $H$).
EDIT
As your question currently stands, it is trivial. If $dim H<infty$, every operator on $H$ is compact, hence if $Tin B(H)$, then $f(T)in B(H)$ is compact a prioru.
answered Jan 2 at 14:06
AweyganAweygan
14k21441
edited Jan 3 at 3:13
answered Jan 2 at 14:06
AweyganAweygan
14k21441
answered Jan 2 at 14:06
AweyganAweygan
14k21441
answered Jan 2 at 14:06
AweyganAweygan
14k21441
14k21441
add a comment |
add a comment |
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1
$begingroup$
You fundamentally changed the nature of your question after answers have been added. You should roll your question back to its original state, then open a new question.
$endgroup$
– Aweygan
Jan 3 at 3:14
$begingroup$
How are you forming $f(T)$ for a bounded function only? Are you assuming that $T$ is selfadjoint?
$endgroup$
– DisintegratingByParts
Jan 3 at 5:07