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Characterization of spectral measures: Error in exercise?
Clash Royale CLAN TAG#URR8PPP
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An exercise in a book on unbounded selfadjoint operators reads as follows:
Let $mathfrak A$ be a $sigma$-algebra on $Omega$ and $E$ be a mapping of $mathfrak A$ into the projections of a Hilbert space $H$ such that $E(Omega)=textId$. Show that $E$ is a spectral measure if and only if the following is satisfied:
$E(bigcup_n=1^infty M_n) = s$-$lim_ntoinftyE(M_n)$ for all sequences $(M_n)subsetmathfrak A$ with $M_nsubseteq M_n+1$ for all $n$. (where $s$-$lim$ denotes the strong limit in $H$).
My question: This statement is wrong, isn't it? My counterexample: Take $E(M):=textId$ for all $Minmathfrak A$. Then $E(M)$ is a projection for each $M$ and $E(Omega)=textId$ and the limit condition above is clearly satisfied. But $E$ is not a spectral measure, since it is not additive.
Is my reasoning correct? Does anybody know what this exercise should actually be?
operator-theory hilbert-spaces spectral-theory
asked Jul 17 at 10:35
Frank
15512
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1
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An exercise in a book on unbounded selfadjoint operators reads as follows:
Let $mathfrak A$ be a $sigma$-algebra on $Omega$ and $E$ be a mapping of $mathfrak A$ into the projections of a Hilbert space $H$ such that $E(Omega)=textId$. Show that $E$ is a spectral measure if and only if the following is satisfied:
$E(bigcup_n=1^infty M_n) = s$-$lim_ntoinftyE(M_n)$ for all sequences $(M_n)subsetmathfrak A$ with $M_nsubseteq M_n+1$ for all $n$. (where $s$-$lim$ denotes the strong limit in $H$).
My question: This statement is wrong, isn't it? My counterexample: Take $E(M):=textId$ for all $Minmathfrak A$. Then $E(M)$ is a projection for each $M$ and $E(Omega)=textId$ and the limit condition above is clearly satisfied. But $E$ is not a spectral measure, since it is not additive.
Is my reasoning correct? Does anybody know what this exercise should actually be?
operator-theory hilbert-spaces spectral-theory
asked Jul 17 at 10:35
Frank
15512
What is $E( emptyset)$ in your example ?????
– Fred
Jul 17 at 11:12
Also identity. This still fulfills the requirements.
– Frank
Jul 17 at 13:30
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
An exercise in a book on unbounded selfadjoint operators reads as follows:
Let $mathfrak A$ be a $sigma$-algebra on $Omega$ and $E$ be a mapping of $mathfrak A$ into the projections of a Hilbert space $H$ such that $E(Omega)=textId$. Show that $E$ is a spectral measure if and only if the following is satisfied:
$E(bigcup_n=1^infty M_n) = s$-$lim_ntoinftyE(M_n)$ for all sequences $(M_n)subsetmathfrak A$ with $M_nsubseteq M_n+1$ for all $n$. (where $s$-$lim$ denotes the strong limit in $H$).
My question: This statement is wrong, isn't it? My counterexample: Take $E(M):=textId$ for all $Minmathfrak A$. Then $E(M)$ is a projection for each $M$ and $E(Omega)=textId$ and the limit condition above is clearly satisfied. But $E$ is not a spectral measure, since it is not additive.
Is my reasoning correct? Does anybody know what this exercise should actually be?
operator-theory hilbert-spaces spectral-theory
asked Jul 17 at 10:35
Frank
15512
An exercise in a book on unbounded selfadjoint operators reads as follows:
Let $mathfrak A$ be a $sigma$-algebra on $Omega$ and $E$ be a mapping of $mathfrak A$ into the projections of a Hilbert space $H$ such that $E(Omega)=textId$. Show that $E$ is a spectral measure if and only if the following is satisfied:
$E(bigcup_n=1^infty M_n) = s$-$lim_ntoinftyE(M_n)$ for all sequences $(M_n)subsetmathfrak A$ with $M_nsubseteq M_n+1$ for all $n$. (where $s$-$lim$ denotes the strong limit in $H$).
My question: This statement is wrong, isn't it? My counterexample: Take $E(M):=textId$ for all $Minmathfrak A$. Then $E(M)$ is a projection for each $M$ and $E(Omega)=textId$ and the limit condition above is clearly satisfied. But $E$ is not a spectral measure, since it is not additive.
Is my reasoning correct? Does anybody know what this exercise should actually be?
operator-theory hilbert-spaces spectral-theory
asked Jul 17 at 10:35
Frank
15512
asked Jul 17 at 10:35
Frank
15512
asked Jul 17 at 10:35
Frank
15512
asked Jul 17 at 10:35
Frank
15512
15512
What is $E( emptyset)$ in your example ?????
– Fred
Jul 17 at 11:12
Also identity. This still fulfills the requirements.
– Frank
Jul 17 at 13:30
add a comment |Â
What is $E( emptyset)$ in your example ?????
– Fred
Jul 17 at 11:12
Also identity. This still fulfills the requirements.
– Frank
Jul 17 at 13:30
What is $E( emptyset)$ in your example ?????
– Fred
Jul 17 at 11:12
What is $E( emptyset)$ in your example ?????
– Fred
Jul 17 at 11:12
Also identity. This still fulfills the requirements.
– Frank
Jul 17 at 13:30
Also identity. This still fulfills the requirements.
– Frank
Jul 17 at 13:30
add a comment |Â
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What is $E( emptyset)$ in your example ?????
– Fred
Jul 17 at 11:12
Also identity. This still fulfills the requirements.
– Frank
Jul 17 at 13:30