Doubt in theorem from Conway's book on Functional Analysis
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I am reading John B. Conway. A Course in Functional Analysis. Springer, 1997 (second edition). Theorem X.4.10c states:
If $(X,Omega)$ is a measurable space, $H$ is a Hilbert space and $E$ is a spectral measure on $(X,Omega,H)$, let $Phi(X,Omega)$ be the algebra of all $Omega$-measurable functions $phi:Xtomathbb{C}$ and define $rho:Phi(X,Omega)to C(H)$ by $rho(phi)=intphimbox{d}E$. Then for $phi,psi$ in $Phi(X,Omega)$, if $psi$ is bounded, $rho(phi)rho(psi)=rho(psi)rho(phi)=rho(phipsi)$.
I doubt this claim. If $psi=0$ and $mbox{dom }rho(phi)neq H$, then $mbox{dom }(rho(psi)rho(phi))neq H$, but $mbox{dom }rho(phipsi)=mbox{dom }rho(0)=H$.
Am I missing something?
functional-analysis spectral-theory
asked Dec 11 '18 at 20:38
SmileyCraftSmileyCraft
3,761519
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add a comment |
$begingroup$
I am reading John B. Conway. A Course in Functional Analysis. Springer, 1997 (second edition). Theorem X.4.10c states:
If $(X,Omega)$ is a measurable space, $H$ is a Hilbert space and $E$ is a spectral measure on $(X,Omega,H)$, let $Phi(X,Omega)$ be the algebra of all $Omega$-measurable functions $phi:Xtomathbb{C}$ and define $rho:Phi(X,Omega)to C(H)$ by $rho(phi)=intphimbox{d}E$. Then for $phi,psi$ in $Phi(X,Omega)$, if $psi$ is bounded, $rho(phi)rho(psi)=rho(psi)rho(phi)=rho(phipsi)$.
I doubt this claim. If $psi=0$ and $mbox{dom }rho(phi)neq H$, then $mbox{dom }(rho(psi)rho(phi))neq H$, but $mbox{dom }rho(phipsi)=mbox{dom }rho(0)=H$.
Am I missing something?
functional-analysis spectral-theory
asked Dec 11 '18 at 20:38
SmileyCraftSmileyCraft
3,761519
$endgroup$
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I see your point. Did you go through the proof and see where the mistake was?
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– Stephen Montgomery-Smith
Dec 18 '18 at 18:41
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The proof was left as an exercise to the reader. Therefore, I have no way to check this, unfortunately.
$endgroup$
– SmileyCraft
Dec 18 '18 at 21:18
add a comment |
$begingroup$
I am reading John B. Conway. A Course in Functional Analysis. Springer, 1997 (second edition). Theorem X.4.10c states:
If $(X,Omega)$ is a measurable space, $H$ is a Hilbert space and $E$ is a spectral measure on $(X,Omega,H)$, let $Phi(X,Omega)$ be the algebra of all $Omega$-measurable functions $phi:Xtomathbb{C}$ and define $rho:Phi(X,Omega)to C(H)$ by $rho(phi)=intphimbox{d}E$. Then for $phi,psi$ in $Phi(X,Omega)$, if $psi$ is bounded, $rho(phi)rho(psi)=rho(psi)rho(phi)=rho(phipsi)$.
I doubt this claim. If $psi=0$ and $mbox{dom }rho(phi)neq H$, then $mbox{dom }(rho(psi)rho(phi))neq H$, but $mbox{dom }rho(phipsi)=mbox{dom }rho(0)=H$.
Am I missing something?
functional-analysis spectral-theory
asked Dec 11 '18 at 20:38
SmileyCraftSmileyCraft
3,761519
$endgroup$
I am reading John B. Conway. A Course in Functional Analysis. Springer, 1997 (second edition). Theorem X.4.10c states:
If $(X,Omega)$ is a measurable space, $H$ is a Hilbert space and $E$ is a spectral measure on $(X,Omega,H)$, let $Phi(X,Omega)$ be the algebra of all $Omega$-measurable functions $phi:Xtomathbb{C}$ and define $rho:Phi(X,Omega)to C(H)$ by $rho(phi)=intphimbox{d}E$. Then for $phi,psi$ in $Phi(X,Omega)$, if $psi$ is bounded, $rho(phi)rho(psi)=rho(psi)rho(phi)=rho(phipsi)$.
I doubt this claim. If $psi=0$ and $mbox{dom }rho(phi)neq H$, then $mbox{dom }(rho(psi)rho(phi))neq H$, but $mbox{dom }rho(phipsi)=mbox{dom }rho(0)=H$.
Am I missing something?
functional-analysis spectral-theory
functional-analysis spectral-theory
asked Dec 11 '18 at 20:38
SmileyCraftSmileyCraft
3,761519
asked Dec 11 '18 at 20:38
SmileyCraftSmileyCraft
3,761519
edited Dec 11 '18 at 20:44
SmileyCraft
asked Dec 11 '18 at 20:38
SmileyCraftSmileyCraft
3,761519
asked Dec 11 '18 at 20:38
SmileyCraftSmileyCraft
3,761519
asked Dec 11 '18 at 20:38
SmileyCraftSmileyCraft
3,761519
3,761519
$begingroup$
I see your point. Did you go through the proof and see where the mistake was?
$endgroup$
– Stephen Montgomery-Smith
Dec 18 '18 at 18:41
$begingroup$
The proof was left as an exercise to the reader. Therefore, I have no way to check this, unfortunately.
$endgroup$
– SmileyCraft
Dec 18 '18 at 21:18
add a comment |
$begingroup$
I see your point. Did you go through the proof and see where the mistake was?
$endgroup$
– Stephen Montgomery-Smith
Dec 18 '18 at 18:41
$begingroup$
The proof was left as an exercise to the reader. Therefore, I have no way to check this, unfortunately.
$endgroup$
– SmileyCraft
Dec 18 '18 at 21:18
$begingroup$
I see your point. Did you go through the proof and see where the mistake was?
$endgroup$
– Stephen Montgomery-Smith
Dec 18 '18 at 18:41
$begingroup$
I see your point. Did you go through the proof and see where the mistake was?
$endgroup$
– Stephen Montgomery-Smith
Dec 18 '18 at 18:41
$begingroup$
The proof was left as an exercise to the reader. Therefore, I have no way to check this, unfortunately.
$endgroup$
– SmileyCraft
Dec 18 '18 at 21:18
$begingroup$
The proof was left as an exercise to the reader. Therefore, I have no way to check this, unfortunately.
$endgroup$
– SmileyCraft
Dec 18 '18 at 21:18
add a comment |
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$begingroup$
I see your point. Did you go through the proof and see where the mistake was?
$endgroup$
– Stephen Montgomery-Smith
Dec 18 '18 at 18:41
$begingroup$
The proof was left as an exercise to the reader. Therefore, I have no way to check this, unfortunately.
$endgroup$
– SmileyCraft
Dec 18 '18 at 21:18