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What is the usual definition of the spectral measure for a nonnegative self-adjoint operator on a Hilbert space?

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3

$begingroup$

Let $H$ be a $mathbb R$-Hilbert space and $(H_lambda)_{lambdage0}$ be a spectral decomposition of $H$ (see below). Now, let $$mathcal Dleft(A_varphiright):=left{xin H:int_0^inftyvarphi(lambda):{rm d}langlepi_lambda x,xrangle_H<inftyright}$$ and $$langle A_varphi x,yrangle_H:=int_0^inftyvarphi(lambda):{rm d}langlepi_lambda x,yrangle_H;;;text{for all }xinmathcal Dleft(A_varphiright)text{ and }yin Htag1$$ (the integral has to be understood as a Lebesgue-Stieltjes integral) for Borel measurable $varphi:[0,infty)tomathbb R$.

If I got it right, the spectral theorem states that if $(mathcal D(A),A)$ is a nonnegative self-adjoint operator on $H$, $(H_lambda)_{tge0}$ can be chosen such that $A_varphi=A$, where $varphi(lambda):=lambda$ for $lambdain[0,infty)$. However, I've always seen the identity $$A_1=operatorname{id}_H,tag2$$ but this seems to be wrong. By $(1)$ and the definition of the Lebesgue-Stieltjes measure, $$langle A_1x,yrangle_H=lim_{lambdatoinfty}langlepi_lambda x,yrangle_H-langlepi_0x,yrangle_Htag3.$$ Clearly $lim_{lambdatoinfty}langlepi_lambda x,yrangle_H=langle x,yrangle_H$, so it seems like $(1)$ should be replaced by $$langle A_varphi x,yrangle_H:=varphi(0)langlepi_0 x,yrangle_H+int_0^inftyvarphi(lambda):{rm d}langlepi_lambda x,yrangle_H;;;text{for all }xinmathcal Dleft(A_varphiright)text{ and }yin Htag4.$$ On the other hand, we could also extend $H_lambda$ for $lambda<0$ by setting $H_lambda:=left{0right}$ (and hence $pi_lambda=0$) for all $lambda<0$. With this definition we could define $$langle A_varphi x,yrangle_H:=int_{mathbb R}varphi(lambda):{rm d}langlepi_lambda x,yrangle_H;;;text{for all }xinmathcal Dleft(A_varphiright)text{ and }yin Htag5$$ by setting $varphi(lambda)=0$ for all $lambda<0$. With $(4)$ we would again obtain $(2)$.

So, I'm highly confused. Which is the usual definition when one is interested in the spectral measure for a nonnegative self-adjoint operator? (Please note that I'm only interested in Lebesgue-Stieltjes integration; not in Riemann-Stieltjes integration.)

---

Definitions:

$(H_lambda)_{lambdage0}$ is called spectral decomposition of $H$ if

1. 
   $H_lambda$ is a closed subspace of $H$ for all $lambdage0$;


2. 
   $(H_lambda)_{lambdage0}$ is nondecreasing and right-continuous, i.e. $$bigcap_{mu>lambda}H_mu=H_lambda;;;text{for all }lambdage0;$$ and


3. 
   $bigcup_{lambdage0}H_lambda$ is dense.

Let $pi_lambda$ denote the orthogonal projection of $H$ onto $H_lambda$ for $lambdage0$. It can be shown that

4. 
   $[0,infty)nilambdamapstopi_lambda$ is nondecreasing, i.e. $$langlepi_lambda x,xrangle_Hlelanglepi_mu x,xrangle_H;;;text{for all }xin H,$$ and right-continuous (with respect to the strong operator topology)

So,

5. 
   $[0,infty)nilambdamapstolanglepi_lambda x,xrangle_H=left|pi_lambda xright|_H^2$ is bounded (by $left|xright|_H^2)$, nondecreasing and right-continuous for all $xin H$
   


6. 
   $[0,infty)nilambdamapstolanglepi_lambda x,yrangle_H=2^{-1}left(langlepi_lambda(x+y),x+yrangle_H-langlepi_lambda x,xrangle_H-langlepi_lambda y,yrangle_Hright)$ is right-continuous and of bounded variation for all $x,yin H$
   

functional-analysis operator-theory spectral-theory stieltjes-integral self-adjoint-operators

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edited Mar 21 at 9:12

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asked Mar 20 at 13:16

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This question has an open bounty worth +50
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3

$begingroup$

Let $H$ be a $mathbb R$-Hilbert space and $(H_lambda)_{lambdage0}$ be a spectral decomposition of $H$ (see below). Now, let $$mathcal Dleft(A_varphiright):=left{xin H:int_0^inftyvarphi(lambda):{rm d}langlepi_lambda x,xrangle_H<inftyright}$$ and $$langle A_varphi x,yrangle_H:=int_0^inftyvarphi(lambda):{rm d}langlepi_lambda x,yrangle_H;;;text{for all }xinmathcal Dleft(A_varphiright)text{ and }yin Htag1$$ (the integral has to be understood as a Lebesgue-Stieltjes integral) for Borel measurable $varphi:[0,infty)tomathbb R$.

If I got it right, the spectral theorem states that if $(mathcal D(A),A)$ is a nonnegative self-adjoint operator on $H$, $(H_lambda)_{tge0}$ can be chosen such that $A_varphi=A$, where $varphi(lambda):=lambda$ for $lambdain[0,infty)$. However, I've always seen the identity $$A_1=operatorname{id}_H,tag2$$ but this seems to be wrong. By $(1)$ and the definition of the Lebesgue-Stieltjes measure, $$langle A_1x,yrangle_H=lim_{lambdatoinfty}langlepi_lambda x,yrangle_H-langlepi_0x,yrangle_Htag3.$$ Clearly $lim_{lambdatoinfty}langlepi_lambda x,yrangle_H=langle x,yrangle_H$, so it seems like $(1)$ should be replaced by $$langle A_varphi x,yrangle_H:=varphi(0)langlepi_0 x,yrangle_H+int_0^inftyvarphi(lambda):{rm d}langlepi_lambda x,yrangle_H;;;text{for all }xinmathcal Dleft(A_varphiright)text{ and }yin Htag4.$$ On the other hand, we could also extend $H_lambda$ for $lambda<0$ by setting $H_lambda:=left{0right}$ (and hence $pi_lambda=0$) for all $lambda<0$. With this definition we could define $$langle A_varphi x,yrangle_H:=int_{mathbb R}varphi(lambda):{rm d}langlepi_lambda x,yrangle_H;;;text{for all }xinmathcal Dleft(A_varphiright)text{ and }yin Htag5$$ by setting $varphi(lambda)=0$ for all $lambda<0$. With $(4)$ we would again obtain $(2)$.

So, I'm highly confused. Which is the usual definition when one is interested in the spectral measure for a nonnegative self-adjoint operator? (Please note that I'm only interested in Lebesgue-Stieltjes integration; not in Riemann-Stieltjes integration.)

---

Definitions:

$(H_lambda)_{lambdage0}$ is called spectral decomposition of $H$ if

1. 
   $H_lambda$ is a closed subspace of $H$ for all $lambdage0$;


2. 
   $(H_lambda)_{lambdage0}$ is nondecreasing and right-continuous, i.e. $$bigcap_{mu>lambda}H_mu=H_lambda;;;text{for all }lambdage0;$$ and


3. 
   $bigcup_{lambdage0}H_lambda$ is dense.

Let $pi_lambda$ denote the orthogonal projection of $H$ onto $H_lambda$ for $lambdage0$. It can be shown that

4. 
   $[0,infty)nilambdamapstopi_lambda$ is nondecreasing, i.e. $$langlepi_lambda x,xrangle_Hlelanglepi_mu x,xrangle_H;;;text{for all }xin H,$$ and right-continuous (with respect to the strong operator topology)

So,

5. 
   $[0,infty)nilambdamapstolanglepi_lambda x,xrangle_H=left|pi_lambda xright|_H^2$ is bounded (by $left|xright|_H^2)$, nondecreasing and right-continuous for all $xin H$
   


6. 
   $[0,infty)nilambdamapstolanglepi_lambda x,yrangle_H=2^{-1}left(langlepi_lambda(x+y),x+yrangle_H-langlepi_lambda x,xrangle_H-langlepi_lambda y,yrangle_Hright)$ is right-continuous and of bounded variation for all $x,yin H$
   

functional-analysis operator-theory spectral-theory stieltjes-integral self-adjoint-operators

share|cite|improve this question

edited Mar 21 at 9:12

0xbadf00d

asked Mar 20 at 13:16

0xbadf00d0xbadf00d

1,74141534

$endgroup$

This question has an open bounty worth +50
reputation from 0xbadf00d ending in 3 hours.

This question has not received enough attention.

add a comment | 

$begingroup$

Let $H$ be a $mathbb R$-Hilbert space and $(H_lambda)_{lambdage0}$ be a spectral decomposition of $H$ (see below). Now, let $$mathcal Dleft(A_varphiright):=left{xin H:int_0^inftyvarphi(lambda):{rm d}langlepi_lambda x,xrangle_H<inftyright}$$ and $$langle A_varphi x,yrangle_H:=int_0^inftyvarphi(lambda):{rm d}langlepi_lambda x,yrangle_H;;;text{for all }xinmathcal Dleft(A_varphiright)text{ and }yin Htag1$$ (the integral has to be understood as a Lebesgue-Stieltjes integral) for Borel measurable $varphi:[0,infty)tomathbb R$.

If I got it right, the spectral theorem states that if $(mathcal D(A),A)$ is a nonnegative self-adjoint operator on $H$, $(H_lambda)_{tge0}$ can be chosen such that $A_varphi=A$, where $varphi(lambda):=lambda$ for $lambdain[0,infty)$. However, I've always seen the identity $$A_1=operatorname{id}_H,tag2$$ but this seems to be wrong. By $(1)$ and the definition of the Lebesgue-Stieltjes measure, $$langle A_1x,yrangle_H=lim_{lambdatoinfty}langlepi_lambda x,yrangle_H-langlepi_0x,yrangle_Htag3.$$ Clearly $lim_{lambdatoinfty}langlepi_lambda x,yrangle_H=langle x,yrangle_H$, so it seems like $(1)$ should be replaced by $$langle A_varphi x,yrangle_H:=varphi(0)langlepi_0 x,yrangle_H+int_0^inftyvarphi(lambda):{rm d}langlepi_lambda x,yrangle_H;;;text{for all }xinmathcal Dleft(A_varphiright)text{ and }yin Htag4.$$ On the other hand, we could also extend $H_lambda$ for $lambda<0$ by setting $H_lambda:=left{0right}$ (and hence $pi_lambda=0$) for all $lambda<0$. With this definition we could define $$langle A_varphi x,yrangle_H:=int_{mathbb R}varphi(lambda):{rm d}langlepi_lambda x,yrangle_H;;;text{for all }xinmathcal Dleft(A_varphiright)text{ and }yin Htag5$$ by setting $varphi(lambda)=0$ for all $lambda<0$. With $(4)$ we would again obtain $(2)$.

So, I'm highly confused. Which is the usual definition when one is interested in the spectral measure for a nonnegative self-adjoint operator? (Please note that I'm only interested in Lebesgue-Stieltjes integration; not in Riemann-Stieltjes integration.)

---

Definitions:

$(H_lambda)_{lambdage0}$ is called spectral decomposition of $H$ if

1. 
   $H_lambda$ is a closed subspace of $H$ for all $lambdage0$;


2. 
   $(H_lambda)_{lambdage0}$ is nondecreasing and right-continuous, i.e. $$bigcap_{mu>lambda}H_mu=H_lambda;;;text{for all }lambdage0;$$ and


3. 
   $bigcup_{lambdage0}H_lambda$ is dense.

Let $pi_lambda$ denote the orthogonal projection of $H$ onto $H_lambda$ for $lambdage0$. It can be shown that

4. 
   $[0,infty)nilambdamapstopi_lambda$ is nondecreasing, i.e. $$langlepi_lambda x,xrangle_Hlelanglepi_mu x,xrangle_H;;;text{for all }xin H,$$ and right-continuous (with respect to the strong operator topology)

So,

5. 
   $[0,infty)nilambdamapstolanglepi_lambda x,xrangle_H=left|pi_lambda xright|_H^2$ is bounded (by $left|xright|_H^2)$, nondecreasing and right-continuous for all $xin H$
   


6. 
   $[0,infty)nilambdamapstolanglepi_lambda x,yrangle_H=2^{-1}left(langlepi_lambda(x+y),x+yrangle_H-langlepi_lambda x,xrangle_H-langlepi_lambda y,yrangle_Hright)$ is right-continuous and of bounded variation for all $x,yin H$
   

functional-analysis operator-theory spectral-theory stieltjes-integral self-adjoint-operators

share|cite|improve this question

edited Mar 21 at 9:12

0xbadf00d

asked Mar 20 at 13:16

0xbadf00d0xbadf00d

1,74141534

$endgroup$

share|cite|improve this question

edited Mar 21 at 9:12

0xbadf00d

asked Mar 20 at 13:16

0xbadf00d0xbadf00d

1,74141534

share|cite|improve this question

edited Mar 21 at 9:12

0xbadf00d

asked Mar 20 at 13:16

0xbadf00d0xbadf00d

1,74141534

asked Mar 20 at 13:16

0xbadf00d0xbadf00d

1,74141534

asked Mar 20 at 13:16

0xbadf00d0xbadf00d

1,74141534