Spectral decomposition of a positive operator
Clash Royale CLAN TAG#URR8PPP
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Let $M:=displaystyleint_0^+infty lambda ,mathrmdE_lambda$ be the spectral decomposition of a positive operator $M$ on a complex Hilbert space $F$.
For $xin F$, we set
$$x_n=displaystyleint_lambda>1/nlambda^-1/2E_lambda(x),mathrmdE_lambda.$$
Why
$$Mx_n=int_lambda> 1/nlambdalambda^-1/2E_lambda(x) ,mathrmdE_lambda?$$
functional-analysis operator-theory
asked Aug 27 at 13:37
Schüler
1,3551321
 |Â
show 10 more comments
up vote
0
down vote
favorite
Let $M:=displaystyleint_0^+infty lambda ,mathrmdE_lambda$ be the spectral decomposition of a positive operator $M$ on a complex Hilbert space $F$.
For $xin F$, we set
$$x_n=displaystyleint_lambda>1/nlambda^-1/2E_lambda(x),mathrmdE_lambda.$$
Why
$$Mx_n=int_lambda> 1/nlambdalambda^-1/2E_lambda(x) ,mathrmdE_lambda?$$
functional-analysis operator-theory
asked Aug 27 at 13:37
Schüler
1,3551321
I don't understand how to interchange between the two integrals
– Schüler
Aug 27 at 13:43
1
In general you have $int f(lambda) dE_lambdaint g(lambda)dE_lambda=int f(lambda)g(lambda) dE_lambda$
– Lorenzo Quarisa
Aug 27 at 13:50
@LorenzoQuarisa It is correct to write $E_lambda(x)$? Is $E_lambda$ an operator which actes on $F$? Thanks a lot.
– Schüler
Aug 27 at 13:53
1
$E_lambdadE_lambdax$ can make sense. However, $E_lambdax dE_lambda$ does not make sense because you have a vector on the left an operator $dE_lambda$.
– DisintegratingByParts
Aug 27 at 18:51
1
How are you making sense of this integral? That seems to be at the heart of the problem.
– DisintegratingByParts
Aug 28 at 16:16
 |Â
show 10 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $M:=displaystyleint_0^+infty lambda ,mathrmdE_lambda$ be the spectral decomposition of a positive operator $M$ on a complex Hilbert space $F$.
For $xin F$, we set
$$x_n=displaystyleint_lambda>1/nlambda^-1/2E_lambda(x),mathrmdE_lambda.$$
Why
$$Mx_n=int_lambda> 1/nlambdalambda^-1/2E_lambda(x) ,mathrmdE_lambda?$$
functional-analysis operator-theory
asked Aug 27 at 13:37
Schüler
1,3551321
Let $M:=displaystyleint_0^+infty lambda ,mathrmdE_lambda$ be the spectral decomposition of a positive operator $M$ on a complex Hilbert space $F$.
For $xin F$, we set
$$x_n=displaystyleint_lambda>1/nlambda^-1/2E_lambda(x),mathrmdE_lambda.$$
Why
$$Mx_n=int_lambda> 1/nlambdalambda^-1/2E_lambda(x) ,mathrmdE_lambda?$$
functional-analysis operator-theory
asked Aug 27 at 13:37
Schüler
1,3551321
asked Aug 27 at 13:37
Schüler
1,3551321
asked Aug 27 at 13:37
Schüler
1,3551321
asked Aug 27 at 13:37
Schüler
1,3551321
1,3551321
I don't understand how to interchange between the two integrals
– Schüler
Aug 27 at 13:43
1
In general you have $int f(lambda) dE_lambdaint g(lambda)dE_lambda=int f(lambda)g(lambda) dE_lambda$
– Lorenzo Quarisa
Aug 27 at 13:50
@LorenzoQuarisa It is correct to write $E_lambda(x)$? Is $E_lambda$ an operator which actes on $F$? Thanks a lot.
– Schüler
Aug 27 at 13:53
1
$E_lambdadE_lambdax$ can make sense. However, $E_lambdax dE_lambda$ does not make sense because you have a vector on the left an operator $dE_lambda$.
– DisintegratingByParts
Aug 27 at 18:51
1
How are you making sense of this integral? That seems to be at the heart of the problem.
– DisintegratingByParts
Aug 28 at 16:16
 |Â
show 10 more comments
I don't understand how to interchange between the two integrals
– Schüler
Aug 27 at 13:43
1
In general you have $int f(lambda) dE_lambdaint g(lambda)dE_lambda=int f(lambda)g(lambda) dE_lambda$
– Lorenzo Quarisa
Aug 27 at 13:50
@LorenzoQuarisa It is correct to write $E_lambda(x)$? Is $E_lambda$ an operator which actes on $F$? Thanks a lot.
– Schüler
Aug 27 at 13:53
1
$E_lambdadE_lambdax$ can make sense. However, $E_lambdax dE_lambda$ does not make sense because you have a vector on the left an operator $dE_lambda$.
– DisintegratingByParts
Aug 27 at 18:51
1
How are you making sense of this integral? That seems to be at the heart of the problem.
– DisintegratingByParts
Aug 28 at 16:16
I don't understand how to interchange between the two integrals
– Schüler
Aug 27 at 13:43
I don't understand how to interchange between the two integrals
– Schüler
Aug 27 at 13:43
1
1
In general you have $int f(lambda) dE_lambdaint g(lambda)dE_lambda=int f(lambda)g(lambda) dE_lambda$
– Lorenzo Quarisa
Aug 27 at 13:50
In general you have $int f(lambda) dE_lambdaint g(lambda)dE_lambda=int f(lambda)g(lambda) dE_lambda$
– Lorenzo Quarisa
Aug 27 at 13:50
@LorenzoQuarisa It is correct to write $E_lambda(x)$? Is $E_lambda$ an operator which actes on $F$? Thanks a lot.
– Schüler
Aug 27 at 13:53
@LorenzoQuarisa It is correct to write $E_lambda(x)$? Is $E_lambda$ an operator which actes on $F$? Thanks a lot.
– Schüler
Aug 27 at 13:53
1
1
$E_lambdadE_lambdax$ can make sense. However, $E_lambdax dE_lambda$ does not make sense because you have a vector on the left an operator $dE_lambda$.
– DisintegratingByParts
Aug 27 at 18:51
$E_lambdadE_lambdax$ can make sense. However, $E_lambdax dE_lambda$ does not make sense because you have a vector on the left an operator $dE_lambda$.
– DisintegratingByParts
Aug 27 at 18:51
1
1
How are you making sense of this integral? That seems to be at the heart of the problem.
– DisintegratingByParts
Aug 28 at 16:16
How are you making sense of this integral? That seems to be at the heart of the problem.
– DisintegratingByParts
Aug 28 at 16:16
 |Â
show 10 more comments
1 Answer
1
active
oldest
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up vote
1
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The Borel functional calculus $fmapsto int_sigma(M)f(lambda) ,d E_lambda$ is a $*$-homomorphism from the set of bounded Borel measurable functions to the set of bounded operators on a Hilbert space. So, you have: $$int_sigma(M)f(lambda) ,d E_lambdaint_sigma(M)g(lambda) ,d E_lambda=int_sigma(M)f(lambda)g(lambda) ,d E_lambda$$ for any bounded Borel measurable functions on $sigma(M)$.
edited Aug 28 at 4:17
Eduardo Longa
1,5052518
answered Aug 28 at 2:47
C.Ding
1,1191321
Is it true that $$left|int_0leqlambdaleq 1/n ,mathrmdE_lambda(x)right|leq int_0^+infty ,mathrmdE_lambda(x)=1?$$ Thank you for your help.
– Schüler
Aug 31 at 11:50
An $*$-homomorphism is automatically contracted，hence $|int_0leq lambdaleq frac1n ,d E_lambda|leq |chi_[0, frac1n]|leq1=int_0^infty ,d E_lambda$.
– C.Ding
Aug 31 at 14:07
Thank you. But please I don't understand why $x_n$ should be $$x_n=displaystyleint_lambda>1/nlambda^-1/2E_lambda,mathrmdE_lambda(x).$$ and not $$x_n=displaystyleint_lambda>1/nlambda^-1/2E_lambda(x),mathrmdE_lambda.$$
– Schüler
Aug 31 at 14:26
In your question, $E_lambda$ is a projection-valued measure or spectral measure. The integrand $f$ in the formula $int f,d E_lambda $ should be a complex-valued function. The correct form of $x_n$ should be $$x_n=(int_lambda>frac1n lambda^-frac12, dE_lambda) circ (E_lambda (x))=int_lambda>frac1ncapx lambda^-frac12, dE_lambda .$$
– C.Ding
Aug 31 at 14:55
Thank you for the comment. By the way $int_lambda> 1/n ,mathrmdE_lambda$ is a complex number, however $E_lambda(x)in F$ so I don't understand how you use $circ$? Also $E_lambda(x)$ depends on $lambda$ so it can not go out of the integral
– Schüler
Aug 31 at 16:27
 |Â
show 3 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
The Borel functional calculus $fmapsto int_sigma(M)f(lambda) ,d E_lambda$ is a $*$-homomorphism from the set of bounded Borel measurable functions to the set of bounded operators on a Hilbert space. So, you have: $$int_sigma(M)f(lambda) ,d E_lambdaint_sigma(M)g(lambda) ,d E_lambda=int_sigma(M)f(lambda)g(lambda) ,d E_lambda$$ for any bounded Borel measurable functions on $sigma(M)$.
edited Aug 28 at 4:17
Eduardo Longa
1,5052518
answered Aug 28 at 2:47
C.Ding
1,1191321
Is it true that $$left|int_0leqlambdaleq 1/n ,mathrmdE_lambda(x)right|leq int_0^+infty ,mathrmdE_lambda(x)=1?$$ Thank you for your help.
– Schüler
Aug 31 at 11:50
An $*$-homomorphism is automatically contracted，hence $|int_0leq lambdaleq frac1n ,d E_lambda|leq |chi_[0, frac1n]|leq1=int_0^infty ,d E_lambda$.
– C.Ding
Aug 31 at 14:07
Thank you. But please I don't understand why $x_n$ should be $$x_n=displaystyleint_lambda>1/nlambda^-1/2E_lambda,mathrmdE_lambda(x).$$ and not $$x_n=displaystyleint_lambda>1/nlambda^-1/2E_lambda(x),mathrmdE_lambda.$$
– Schüler
Aug 31 at 14:26
In your question, $E_lambda$ is a projection-valued measure or spectral measure. The integrand $f$ in the formula $int f,d E_lambda $ should be a complex-valued function. The correct form of $x_n$ should be $$x_n=(int_lambda>frac1n lambda^-frac12, dE_lambda) circ (E_lambda (x))=int_lambda>frac1ncapx lambda^-frac12, dE_lambda .$$
– C.Ding
Aug 31 at 14:55
Thank you for the comment. By the way $int_lambda> 1/n ,mathrmdE_lambda$ is a complex number, however $E_lambda(x)in F$ so I don't understand how you use $circ$? Also $E_lambda(x)$ depends on $lambda$ so it can not go out of the integral
– Schüler
Aug 31 at 16:27
 |Â
show 3 more comments
up vote
1
down vote
The Borel functional calculus $fmapsto int_sigma(M)f(lambda) ,d E_lambda$ is a $*$-homomorphism from the set of bounded Borel measurable functions to the set of bounded operators on a Hilbert space. So, you have: $$int_sigma(M)f(lambda) ,d E_lambdaint_sigma(M)g(lambda) ,d E_lambda=int_sigma(M)f(lambda)g(lambda) ,d E_lambda$$ for any bounded Borel measurable functions on $sigma(M)$.
edited Aug 28 at 4:17
Eduardo Longa
1,5052518
answered Aug 28 at 2:47
C.Ding
1,1191321
Is it true that $$left|int_0leqlambdaleq 1/n ,mathrmdE_lambda(x)right|leq int_0^+infty ,mathrmdE_lambda(x)=1?$$ Thank you for your help.
– Schüler
Aug 31 at 11:50
An $*$-homomorphism is automatically contracted，hence $|int_0leq lambdaleq frac1n ,d E_lambda|leq |chi_[0, frac1n]|leq1=int_0^infty ,d E_lambda$.
– C.Ding
Aug 31 at 14:07
Thank you. But please I don't understand why $x_n$ should be $$x_n=displaystyleint_lambda>1/nlambda^-1/2E_lambda,mathrmdE_lambda(x).$$ and not $$x_n=displaystyleint_lambda>1/nlambda^-1/2E_lambda(x),mathrmdE_lambda.$$
– Schüler
Aug 31 at 14:26
In your question, $E_lambda$ is a projection-valued measure or spectral measure. The integrand $f$ in the formula $int f,d E_lambda $ should be a complex-valued function. The correct form of $x_n$ should be $$x_n=(int_lambda>frac1n lambda^-frac12, dE_lambda) circ (E_lambda (x))=int_lambda>frac1ncapx lambda^-frac12, dE_lambda .$$
– C.Ding
Aug 31 at 14:55
Thank you for the comment. By the way $int_lambda> 1/n ,mathrmdE_lambda$ is a complex number, however $E_lambda(x)in F$ so I don't understand how you use $circ$? Also $E_lambda(x)$ depends on $lambda$ so it can not go out of the integral
– Schüler
Aug 31 at 16:27
 |Â
show 3 more comments
up vote
1
down vote
up vote
1
down vote
The Borel functional calculus $fmapsto int_sigma(M)f(lambda) ,d E_lambda$ is a $*$-homomorphism from the set of bounded Borel measurable functions to the set of bounded operators on a Hilbert space. So, you have: $$int_sigma(M)f(lambda) ,d E_lambdaint_sigma(M)g(lambda) ,d E_lambda=int_sigma(M)f(lambda)g(lambda) ,d E_lambda$$ for any bounded Borel measurable functions on $sigma(M)$.
edited Aug 28 at 4:17
Eduardo Longa
1,5052518
answered Aug 28 at 2:47
C.Ding
1,1191321
The Borel functional calculus $fmapsto int_sigma(M)f(lambda) ,d E_lambda$ is a $*$-homomorphism from the set of bounded Borel measurable functions to the set of bounded operators on a Hilbert space. So, you have: $$int_sigma(M)f(lambda) ,d E_lambdaint_sigma(M)g(lambda) ,d E_lambda=int_sigma(M)f(lambda)g(lambda) ,d E_lambda$$ for any bounded Borel measurable functions on $sigma(M)$.
edited Aug 28 at 4:17
Eduardo Longa
1,5052518
answered Aug 28 at 2:47
C.Ding
1,1191321
edited Aug 28 at 4:17
Eduardo Longa
1,5052518
edited Aug 28 at 4:17
Eduardo Longa
1,5052518
edited Aug 28 at 4:17
Eduardo Longa
1,5052518
1,5052518
answered Aug 28 at 2:47
C.Ding
1,1191321
answered Aug 28 at 2:47
C.Ding
1,1191321
answered Aug 28 at 2:47
C.Ding
1,1191321
1,1191321
Is it true that $$left|int_0leqlambdaleq 1/n ,mathrmdE_lambda(x)right|leq int_0^+infty ,mathrmdE_lambda(x)=1?$$ Thank you for your help.
– Schüler
Aug 31 at 11:50
An $*$-homomorphism is automatically contracted，hence $|int_0leq lambdaleq frac1n ,d E_lambda|leq |chi_[0, frac1n]|leq1=int_0^infty ,d E_lambda$.
– C.Ding
Aug 31 at 14:07
Thank you. But please I don't understand why $x_n$ should be $$x_n=displaystyleint_lambda>1/nlambda^-1/2E_lambda,mathrmdE_lambda(x).$$ and not $$x_n=displaystyleint_lambda>1/nlambda^-1/2E_lambda(x),mathrmdE_lambda.$$
– Schüler
Aug 31 at 14:26
In your question, $E_lambda$ is a projection-valued measure or spectral measure. The integrand $f$ in the formula $int f,d E_lambda $ should be a complex-valued function. The correct form of $x_n$ should be $$x_n=(int_lambda>frac1n lambda^-frac12, dE_lambda) circ (E_lambda (x))=int_lambda>frac1ncapx lambda^-frac12, dE_lambda .$$
– C.Ding
Aug 31 at 14:55
Thank you for the comment. By the way $int_lambda> 1/n ,mathrmdE_lambda$ is a complex number, however $E_lambda(x)in F$ so I don't understand how you use $circ$? Also $E_lambda(x)$ depends on $lambda$ so it can not go out of the integral
– Schüler
Aug 31 at 16:27
 |Â
show 3 more comments
Is it true that $$left|int_0leqlambdaleq 1/n ,mathrmdE_lambda(x)right|leq int_0^+infty ,mathrmdE_lambda(x)=1?$$ Thank you for your help.
– Schüler
Aug 31 at 11:50
An $*$-homomorphism is automatically contracted，hence $|int_0leq lambdaleq frac1n ,d E_lambda|leq |chi_[0, frac1n]|leq1=int_0^infty ,d E_lambda$.
– C.Ding
Aug 31 at 14:07
Thank you. But please I don't understand why $x_n$ should be $$x_n=displaystyleint_lambda>1/nlambda^-1/2E_lambda,mathrmdE_lambda(x).$$ and not $$x_n=displaystyleint_lambda>1/nlambda^-1/2E_lambda(x),mathrmdE_lambda.$$
– Schüler
Aug 31 at 14:26
In your question, $E_lambda$ is a projection-valued measure or spectral measure. The integrand $f$ in the formula $int f,d E_lambda $ should be a complex-valued function. The correct form of $x_n$ should be $$x_n=(int_lambda>frac1n lambda^-frac12, dE_lambda) circ (E_lambda (x))=int_lambda>frac1ncapx lambda^-frac12, dE_lambda .$$
– C.Ding
Aug 31 at 14:55
Thank you for the comment. By the way $int_lambda> 1/n ,mathrmdE_lambda$ is a complex number, however $E_lambda(x)in F$ so I don't understand how you use $circ$? Also $E_lambda(x)$ depends on $lambda$ so it can not go out of the integral
– Schüler
Aug 31 at 16:27
Is it true that $$left|int_0leqlambdaleq 1/n ,mathrmdE_lambda(x)right|leq int_0^+infty ,mathrmdE_lambda(x)=1?$$ Thank you for your help.
– Schüler
Aug 31 at 11:50
Is it true that $$left|int_0leqlambdaleq 1/n ,mathrmdE_lambda(x)right|leq int_0^+infty ,mathrmdE_lambda(x)=1?$$ Thank you for your help.
– Schüler
Aug 31 at 11:50
An $*$-homomorphism is automatically contracted，hence $|int_0leq lambdaleq frac1n ,d E_lambda|leq |chi_[0, frac1n]|leq1=int_0^infty ,d E_lambda$.
– C.Ding
Aug 31 at 14:07
An $*$-homomorphism is automatically contracted，hence $|int_0leq lambdaleq frac1n ,d E_lambda|leq |chi_[0, frac1n]|leq1=int_0^infty ,d E_lambda$.
– C.Ding
Aug 31 at 14:07
Thank you. But please I don't understand why $x_n$ should be $$x_n=displaystyleint_lambda>1/nlambda^-1/2E_lambda,mathrmdE_lambda(x).$$ and not $$x_n=displaystyleint_lambda>1/nlambda^-1/2E_lambda(x),mathrmdE_lambda.$$
– Schüler
Aug 31 at 14:26
Thank you. But please I don't understand why $x_n$ should be $$x_n=displaystyleint_lambda>1/nlambda^-1/2E_lambda,mathrmdE_lambda(x).$$ and not $$x_n=displaystyleint_lambda>1/nlambda^-1/2E_lambda(x),mathrmdE_lambda.$$
– Schüler
Aug 31 at 14:26
In your question, $E_lambda$ is a projection-valued measure or spectral measure. The integrand $f$ in the formula $int f,d E_lambda $ should be a complex-valued function. The correct form of $x_n$ should be $$x_n=(int_lambda>frac1n lambda^-frac12, dE_lambda) circ (E_lambda (x))=int_lambda>frac1ncapx lambda^-frac12, dE_lambda .$$
– C.Ding
Aug 31 at 14:55
In your question, $E_lambda$ is a projection-valued measure or spectral measure. The integrand $f$ in the formula $int f,d E_lambda $ should be a complex-valued function. The correct form of $x_n$ should be $$x_n=(int_lambda>frac1n lambda^-frac12, dE_lambda) circ (E_lambda (x))=int_lambda>frac1ncapx lambda^-frac12, dE_lambda .$$
– C.Ding
Aug 31 at 14:55
Thank you for the comment. By the way $int_lambda> 1/n ,mathrmdE_lambda$ is a complex number, however $E_lambda(x)in F$ so I don't understand how you use $circ$? Also $E_lambda(x)$ depends on $lambda$ so it can not go out of the integral
– Schüler
Aug 31 at 16:27
Thank you for the comment. By the way $int_lambda> 1/n ,mathrmdE_lambda$ is a complex number, however $E_lambda(x)in F$ so I don't understand how you use $circ$? Also $E_lambda(x)$ depends on $lambda$ so it can not go out of the integral
– Schüler
Aug 31 at 16:27
 |Â
show 3 more comments
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I don't understand how to interchange between the two integrals
– Schüler
Aug 27 at 13:43
1
In general you have $int f(lambda) dE_lambdaint g(lambda)dE_lambda=int f(lambda)g(lambda) dE_lambda$
– Lorenzo Quarisa
Aug 27 at 13:50
@LorenzoQuarisa It is correct to write $E_lambda(x)$? Is $E_lambda$ an operator which actes on $F$? Thanks a lot.
– Schüler
Aug 27 at 13:53
1
$E_lambdadE_lambdax$ can make sense. However, $E_lambdax dE_lambda$ does not make sense because you have a vector on the left an operator $dE_lambda$.
– DisintegratingByParts
Aug 27 at 18:51
1
How are you making sense of this integral? That seems to be at the heart of the problem.
– DisintegratingByParts
Aug 28 at 16:16