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Integral of self-adjoint operator
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Let $H$ be a Hilbert space over $mathbbC$ and $T$ be a self-adjoint operator. Suppose there exists $c>0$ s.t., for every $xin H$, $(Tx,x)geq c(x,x)$
Then $S:=int_-infty^infty (y^2+T^2)^-1 , dy$ is well-defined, and represent $S$ specifically.
I know that spectrum theorem $T=int lambda , dE_lambda$. But I can't prove well-defined of $S$ and represent.
functional-analysis operator-theory hilbert-spaces
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up vote
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down vote
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Let $H$ be a Hilbert space over $mathbbC$ and $T$ be a self-adjoint operator. Suppose there exists $c>0$ s.t., for every $xin H$, $(Tx,x)geq c(x,x)$
Then $S:=int_-infty^infty (y^2+T^2)^-1 , dy$ is well-defined, and represent $S$ specifically.
I know that spectrum theorem $T=int lambda , dE_lambda$. But I can't prove well-defined of $S$ and represent.
functional-analysis operator-theory hilbert-spaces
If $H=mathbb C$ and $T=c=1$ then the integral does not exist.
– Kavi Rama Murthy
Aug 6 at 5:43
1
You are using $x$ with two different meanings.
– Kavi Rama Murthy
Aug 6 at 5:44
1
In your example, $T$ is not self-adjoint.
– B.T.O
Aug 6 at 12:10
1
Are you sure that you aren't asked to look at $int_-infty^infty(y^2+T^2)^-1dy$?
– DisintegratingByParts
Aug 10 at 15:36
1
@DisintegratingByParts Thanks. I have mistake. I edit.
– user577983
Aug 11 at 6:42
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $H$ be a Hilbert space over $mathbbC$ and $T$ be a self-adjoint operator. Suppose there exists $c>0$ s.t., for every $xin H$, $(Tx,x)geq c(x,x)$
Then $S:=int_-infty^infty (y^2+T^2)^-1 , dy$ is well-defined, and represent $S$ specifically.
I know that spectrum theorem $T=int lambda , dE_lambda$. But I can't prove well-defined of $S$ and represent.
functional-analysis operator-theory hilbert-spaces
Let $H$ be a Hilbert space over $mathbbC$ and $T$ be a self-adjoint operator. Suppose there exists $c>0$ s.t., for every $xin H$, $(Tx,x)geq c(x,x)$
Then $S:=int_-infty^infty (y^2+T^2)^-1 , dy$ is well-defined, and represent $S$ specifically.
I know that spectrum theorem $T=int lambda , dE_lambda$. But I can't prove well-defined of $S$ and represent.
functional-analysis operator-theory hilbert-spaces
edited Aug 11 at 6:39
asked Aug 6 at 4:08
user577983
If $H=mathbb C$ and $T=c=1$ then the integral does not exist.
– Kavi Rama Murthy
Aug 6 at 5:43
1
You are using $x$ with two different meanings.
– Kavi Rama Murthy
Aug 6 at 5:44
1
In your example, $T$ is not self-adjoint.
– B.T.O
Aug 6 at 12:10
1
Are you sure that you aren't asked to look at $int_-infty^infty(y^2+T^2)^-1dy$?
– DisintegratingByParts
Aug 10 at 15:36
1
@DisintegratingByParts Thanks. I have mistake. I edit.
– user577983
Aug 11 at 6:42
add a comment |Â
If $H=mathbb C$ and $T=c=1$ then the integral does not exist.
– Kavi Rama Murthy
Aug 6 at 5:43
1
You are using $x$ with two different meanings.
– Kavi Rama Murthy
Aug 6 at 5:44
1
In your example, $T$ is not self-adjoint.
– B.T.O
Aug 6 at 12:10
1
Are you sure that you aren't asked to look at $int_-infty^infty(y^2+T^2)^-1dy$?
– DisintegratingByParts
Aug 10 at 15:36
1
@DisintegratingByParts Thanks. I have mistake. I edit.
– user577983
Aug 11 at 6:42
If $H=mathbb C$ and $T=c=1$ then the integral does not exist.
– Kavi Rama Murthy
Aug 6 at 5:43
If $H=mathbb C$ and $T=c=1$ then the integral does not exist.
– Kavi Rama Murthy
Aug 6 at 5:43
1
1
You are using $x$ with two different meanings.
– Kavi Rama Murthy
Aug 6 at 5:44
You are using $x$ with two different meanings.
– Kavi Rama Murthy
Aug 6 at 5:44
1
1
In your example, $T$ is not self-adjoint.
– B.T.O
Aug 6 at 12:10
In your example, $T$ is not self-adjoint.
– B.T.O
Aug 6 at 12:10
1
1
Are you sure that you aren't asked to look at $int_-infty^infty(y^2+T^2)^-1dy$?
– DisintegratingByParts
Aug 10 at 15:36
Are you sure that you aren't asked to look at $int_-infty^infty(y^2+T^2)^-1dy$?
– DisintegratingByParts
Aug 10 at 15:36
1
1
@DisintegratingByParts Thanks. I have mistake. I edit.
– user577983
Aug 11 at 6:42
@DisintegratingByParts Thanks. I have mistake. I edit.
– user577983
Aug 11 at 6:42
add a comment |Â
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If $H=mathbb C$ and $T=c=1$ then the integral does not exist.
– Kavi Rama Murthy
Aug 6 at 5:43
1
You are using $x$ with two different meanings.
– Kavi Rama Murthy
Aug 6 at 5:44
1
In your example, $T$ is not self-adjoint.
– B.T.O
Aug 6 at 12:10
1
Are you sure that you aren't asked to look at $int_-infty^infty(y^2+T^2)^-1dy$?
– DisintegratingByParts
Aug 10 at 15:36
1
@DisintegratingByParts Thanks. I have mistake. I edit.
– user577983
Aug 11 at 6:42