Why a spectral resolution should be left-continuous?

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The usual definition I can find (almost) everywhere on the literature of a spectral resolution on a Hilbert space $mathscrH$ is this: a family of orthogonal projections $(E_t)_t in mathbbR$ such that
- $t leq s implies E_t leq E_s$
- $lim_t rightarrow -infty E_t = 0 , quad lim_t rightarrow +infty E_t$
- $lim_t searrow s E_t = E_s$,
where the limit is always taken in the strong topology sense. I wonder why one would ask the property 3 of a spectral resolution and not left-continuity. I've heard that right-continuity should actually be a better choice for a measure-theoretical reason: we can always take the limit of a measure of a sequence of increasing measurable sets and that doesn't work always if the sequence is decreasing, but I'm not convinced of that.
functional-analysis spectral-theory
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up vote
3
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The usual definition I can find (almost) everywhere on the literature of a spectral resolution on a Hilbert space $mathscrH$ is this: a family of orthogonal projections $(E_t)_t in mathbbR$ such that
- $t leq s implies E_t leq E_s$
- $lim_t rightarrow -infty E_t = 0 , quad lim_t rightarrow +infty E_t$
- $lim_t searrow s E_t = E_s$,
where the limit is always taken in the strong topology sense. I wonder why one would ask the property 3 of a spectral resolution and not left-continuity. I've heard that right-continuity should actually be a better choice for a measure-theoretical reason: we can always take the limit of a measure of a sequence of increasing measurable sets and that doesn't work always if the sequence is decreasing, but I'm not convinced of that.
functional-analysis spectral-theory
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
The usual definition I can find (almost) everywhere on the literature of a spectral resolution on a Hilbert space $mathscrH$ is this: a family of orthogonal projections $(E_t)_t in mathbbR$ such that
- $t leq s implies E_t leq E_s$
- $lim_t rightarrow -infty E_t = 0 , quad lim_t rightarrow +infty E_t$
- $lim_t searrow s E_t = E_s$,
where the limit is always taken in the strong topology sense. I wonder why one would ask the property 3 of a spectral resolution and not left-continuity. I've heard that right-continuity should actually be a better choice for a measure-theoretical reason: we can always take the limit of a measure of a sequence of increasing measurable sets and that doesn't work always if the sequence is decreasing, but I'm not convinced of that.
functional-analysis spectral-theory
The usual definition I can find (almost) everywhere on the literature of a spectral resolution on a Hilbert space $mathscrH$ is this: a family of orthogonal projections $(E_t)_t in mathbbR$ such that
- $t leq s implies E_t leq E_s$
- $lim_t rightarrow -infty E_t = 0 , quad lim_t rightarrow +infty E_t$
- $lim_t searrow s E_t = E_s$,
where the limit is always taken in the strong topology sense. I wonder why one would ask the property 3 of a spectral resolution and not left-continuity. I've heard that right-continuity should actually be a better choice for a measure-theoretical reason: we can always take the limit of a measure of a sequence of increasing measurable sets and that doesn't work always if the sequence is decreasing, but I'm not convinced of that.
functional-analysis spectral-theory
functional-analysis spectral-theory
edited Aug 30 at 1:59
asked Aug 30 at 1:48
B. Chinaski
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462211
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It doesn't make a difference whether you require left-continuity or right-continuity. The two definitions are equivalent, up to a reparameterization. Namely, if $(E_t)$ satisfies the definition with right-continuity and you define $F_t=limlimits_tnearrow s E_s$ then $(F_t)$ will satisfy the definition with left-continuity. Conversely, if $(F_t)$ satisfies the definition with right-continuity, then $E_t=limlimits_tsearrow sF_s$ satisfies the definition with left-continuity, and these two operations are inverse to each other.
The idea here is that $E_t$ represents the "measure" of $(-infty,t]$, while $F_t$ represents the "measure" of $(-infty,t)$. This is equivalent data, since a measure on all Borel sets can be recovered from either one. (Note that the issue you allude to with descending sequences of measurable sets is not relevant, since that's only an issue for measures that could be infinite due to not being able to subtract $infty-infty$, whereas our measures are projection-valued and we can always subtract projections.)
I thought you might end up on a issue like the one I mentioned when dealing with unbounded self-adjoint operators, for example... but what you said makes sense.
â B. Chinaski
Aug 30 at 3:06
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
It doesn't make a difference whether you require left-continuity or right-continuity. The two definitions are equivalent, up to a reparameterization. Namely, if $(E_t)$ satisfies the definition with right-continuity and you define $F_t=limlimits_tnearrow s E_s$ then $(F_t)$ will satisfy the definition with left-continuity. Conversely, if $(F_t)$ satisfies the definition with right-continuity, then $E_t=limlimits_tsearrow sF_s$ satisfies the definition with left-continuity, and these two operations are inverse to each other.
The idea here is that $E_t$ represents the "measure" of $(-infty,t]$, while $F_t$ represents the "measure" of $(-infty,t)$. This is equivalent data, since a measure on all Borel sets can be recovered from either one. (Note that the issue you allude to with descending sequences of measurable sets is not relevant, since that's only an issue for measures that could be infinite due to not being able to subtract $infty-infty$, whereas our measures are projection-valued and we can always subtract projections.)
I thought you might end up on a issue like the one I mentioned when dealing with unbounded self-adjoint operators, for example... but what you said makes sense.
â B. Chinaski
Aug 30 at 3:06
add a comment |Â
up vote
2
down vote
accepted
It doesn't make a difference whether you require left-continuity or right-continuity. The two definitions are equivalent, up to a reparameterization. Namely, if $(E_t)$ satisfies the definition with right-continuity and you define $F_t=limlimits_tnearrow s E_s$ then $(F_t)$ will satisfy the definition with left-continuity. Conversely, if $(F_t)$ satisfies the definition with right-continuity, then $E_t=limlimits_tsearrow sF_s$ satisfies the definition with left-continuity, and these two operations are inverse to each other.
The idea here is that $E_t$ represents the "measure" of $(-infty,t]$, while $F_t$ represents the "measure" of $(-infty,t)$. This is equivalent data, since a measure on all Borel sets can be recovered from either one. (Note that the issue you allude to with descending sequences of measurable sets is not relevant, since that's only an issue for measures that could be infinite due to not being able to subtract $infty-infty$, whereas our measures are projection-valued and we can always subtract projections.)
I thought you might end up on a issue like the one I mentioned when dealing with unbounded self-adjoint operators, for example... but what you said makes sense.
â B. Chinaski
Aug 30 at 3:06
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
It doesn't make a difference whether you require left-continuity or right-continuity. The two definitions are equivalent, up to a reparameterization. Namely, if $(E_t)$ satisfies the definition with right-continuity and you define $F_t=limlimits_tnearrow s E_s$ then $(F_t)$ will satisfy the definition with left-continuity. Conversely, if $(F_t)$ satisfies the definition with right-continuity, then $E_t=limlimits_tsearrow sF_s$ satisfies the definition with left-continuity, and these two operations are inverse to each other.
The idea here is that $E_t$ represents the "measure" of $(-infty,t]$, while $F_t$ represents the "measure" of $(-infty,t)$. This is equivalent data, since a measure on all Borel sets can be recovered from either one. (Note that the issue you allude to with descending sequences of measurable sets is not relevant, since that's only an issue for measures that could be infinite due to not being able to subtract $infty-infty$, whereas our measures are projection-valued and we can always subtract projections.)
It doesn't make a difference whether you require left-continuity or right-continuity. The two definitions are equivalent, up to a reparameterization. Namely, if $(E_t)$ satisfies the definition with right-continuity and you define $F_t=limlimits_tnearrow s E_s$ then $(F_t)$ will satisfy the definition with left-continuity. Conversely, if $(F_t)$ satisfies the definition with right-continuity, then $E_t=limlimits_tsearrow sF_s$ satisfies the definition with left-continuity, and these two operations are inverse to each other.
The idea here is that $E_t$ represents the "measure" of $(-infty,t]$, while $F_t$ represents the "measure" of $(-infty,t)$. This is equivalent data, since a measure on all Borel sets can be recovered from either one. (Note that the issue you allude to with descending sequences of measurable sets is not relevant, since that's only an issue for measures that could be infinite due to not being able to subtract $infty-infty$, whereas our measures are projection-valued and we can always subtract projections.)
answered Aug 30 at 2:48
Eric Wofsey
166k12195308
166k12195308
I thought you might end up on a issue like the one I mentioned when dealing with unbounded self-adjoint operators, for example... but what you said makes sense.
â B. Chinaski
Aug 30 at 3:06
add a comment |Â
I thought you might end up on a issue like the one I mentioned when dealing with unbounded self-adjoint operators, for example... but what you said makes sense.
â B. Chinaski
Aug 30 at 3:06
I thought you might end up on a issue like the one I mentioned when dealing with unbounded self-adjoint operators, for example... but what you said makes sense.
â B. Chinaski
Aug 30 at 3:06
I thought you might end up on a issue like the one I mentioned when dealing with unbounded self-adjoint operators, for example... but what you said makes sense.
â B. Chinaski
Aug 30 at 3:06
add a comment |Â
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