Can this self-adjoint operator have an infinite-dimensional compression with compact inverse?

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The following might be quite straightforward, but I very rarely work in detail with unbounded operators, so I thought it would be worth seeing quickly if I have overlooked an example that is obvious from the right point of view.



Let $H=L^2(-infty,infty)$ and let $S:H supset rm dom(S) to H$ be the densely-defined operator $(Sf)(x)=e^xf(x)$, with domain
$$
rm dom(S)=^2,dx <infty
$$
Then $S$ is self-adjoint.



Question. Does there exists a projection $P:Hto H$, with range $V$, satisfying the following properties?



  1. $V$ is infinite-dimensional and $V':=rm dom(S)cap V$ is dense in $V$.


  2. there is a compact operator $R:Vto V$ such that $RPSPvert_V'= I_V'$?


Some remarks:



(a) If we weakened 2. to merely require $R$ being bounded, then this should be easy by taking $V=L^2[0,infty)$ and $R:Vto V$ to be multiplication by $e^-x$. But of course $R$ has continuous spectrum, so it can't be compact.



(b) If we intertwine with the Fourier transform or something similar, perhaps we can we obtain such a $V$ as the solution space to a suitable differential equation? (I have not looked at this vague idea in any detail yet.)



(c) If the answer to the question is positive, can we even get $R$ being trace-class?










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  • 1




    "Projection" here means "orthogonal projection"?
    – Nate Eldredge
    Aug 17 at 3:09










  • @NateEldredge oops, yes. (Had my op alg hat on, where that is the usual shorthand)
    – Yemon Choi
    Aug 17 at 3:25














up vote
4
down vote

favorite
2












The following might be quite straightforward, but I very rarely work in detail with unbounded operators, so I thought it would be worth seeing quickly if I have overlooked an example that is obvious from the right point of view.



Let $H=L^2(-infty,infty)$ and let $S:H supset rm dom(S) to H$ be the densely-defined operator $(Sf)(x)=e^xf(x)$, with domain
$$
rm dom(S)=^2,dx <infty
$$
Then $S$ is self-adjoint.



Question. Does there exists a projection $P:Hto H$, with range $V$, satisfying the following properties?



  1. $V$ is infinite-dimensional and $V':=rm dom(S)cap V$ is dense in $V$.


  2. there is a compact operator $R:Vto V$ such that $RPSPvert_V'= I_V'$?


Some remarks:



(a) If we weakened 2. to merely require $R$ being bounded, then this should be easy by taking $V=L^2[0,infty)$ and $R:Vto V$ to be multiplication by $e^-x$. But of course $R$ has continuous spectrum, so it can't be compact.



(b) If we intertwine with the Fourier transform or something similar, perhaps we can we obtain such a $V$ as the solution space to a suitable differential equation? (I have not looked at this vague idea in any detail yet.)



(c) If the answer to the question is positive, can we even get $R$ being trace-class?










share|cite|improve this question

















  • 1




    "Projection" here means "orthogonal projection"?
    – Nate Eldredge
    Aug 17 at 3:09










  • @NateEldredge oops, yes. (Had my op alg hat on, where that is the usual shorthand)
    – Yemon Choi
    Aug 17 at 3:25












up vote
4
down vote

favorite
2









up vote
4
down vote

favorite
2






2





The following might be quite straightforward, but I very rarely work in detail with unbounded operators, so I thought it would be worth seeing quickly if I have overlooked an example that is obvious from the right point of view.



Let $H=L^2(-infty,infty)$ and let $S:H supset rm dom(S) to H$ be the densely-defined operator $(Sf)(x)=e^xf(x)$, with domain
$$
rm dom(S)=^2,dx <infty
$$
Then $S$ is self-adjoint.



Question. Does there exists a projection $P:Hto H$, with range $V$, satisfying the following properties?



  1. $V$ is infinite-dimensional and $V':=rm dom(S)cap V$ is dense in $V$.


  2. there is a compact operator $R:Vto V$ such that $RPSPvert_V'= I_V'$?


Some remarks:



(a) If we weakened 2. to merely require $R$ being bounded, then this should be easy by taking $V=L^2[0,infty)$ and $R:Vto V$ to be multiplication by $e^-x$. But of course $R$ has continuous spectrum, so it can't be compact.



(b) If we intertwine with the Fourier transform or something similar, perhaps we can we obtain such a $V$ as the solution space to a suitable differential equation? (I have not looked at this vague idea in any detail yet.)



(c) If the answer to the question is positive, can we even get $R$ being trace-class?










share|cite|improve this question













The following might be quite straightforward, but I very rarely work in detail with unbounded operators, so I thought it would be worth seeing quickly if I have overlooked an example that is obvious from the right point of view.



Let $H=L^2(-infty,infty)$ and let $S:H supset rm dom(S) to H$ be the densely-defined operator $(Sf)(x)=e^xf(x)$, with domain
$$
rm dom(S)=^2,dx <infty
$$
Then $S$ is self-adjoint.



Question. Does there exists a projection $P:Hto H$, with range $V$, satisfying the following properties?



  1. $V$ is infinite-dimensional and $V':=rm dom(S)cap V$ is dense in $V$.


  2. there is a compact operator $R:Vto V$ such that $RPSPvert_V'= I_V'$?


Some remarks:



(a) If we weakened 2. to merely require $R$ being bounded, then this should be easy by taking $V=L^2[0,infty)$ and $R:Vto V$ to be multiplication by $e^-x$. But of course $R$ has continuous spectrum, so it can't be compact.



(b) If we intertwine with the Fourier transform or something similar, perhaps we can we obtain such a $V$ as the solution space to a suitable differential equation? (I have not looked at this vague idea in any detail yet.)



(c) If the answer to the question is positive, can we even get $R$ being trace-class?







fa.functional-analysis operator-theory sp.spectral-theory






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asked Aug 17 at 2:28









Yemon Choi

16.2k54699




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  • 1




    "Projection" here means "orthogonal projection"?
    – Nate Eldredge
    Aug 17 at 3:09










  • @NateEldredge oops, yes. (Had my op alg hat on, where that is the usual shorthand)
    – Yemon Choi
    Aug 17 at 3:25












  • 1




    "Projection" here means "orthogonal projection"?
    – Nate Eldredge
    Aug 17 at 3:09










  • @NateEldredge oops, yes. (Had my op alg hat on, where that is the usual shorthand)
    – Yemon Choi
    Aug 17 at 3:25







1




1




"Projection" here means "orthogonal projection"?
– Nate Eldredge
Aug 17 at 3:09




"Projection" here means "orthogonal projection"?
– Nate Eldredge
Aug 17 at 3:09












@NateEldredge oops, yes. (Had my op alg hat on, where that is the usual shorthand)
– Yemon Choi
Aug 17 at 3:25




@NateEldredge oops, yes. (Had my op alg hat on, where that is the usual shorthand)
– Yemon Choi
Aug 17 at 3:25










1 Answer
1






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up vote
7
down vote



accepted










Sure, for instance let $P$ be the orthogonal projection onto the closed span of the characteristic functions $chi_[n,n+1)$ for $n in mathbbN$. You get property 1 because each of these functions is in the domain of $S$, and you get property 2 because, identifying $V$ with $l^2$ in the obvious way, the operator $PSP$ becomes multiplication by, I think, $e^n(e - 1)$. So this compression has compact, even trace-class inverse, namely multiplication by $e^-n(e-1)^-1$.






share|cite|improve this answer
















  • 1




    Kicking myself now for not thinking of trying to discretize the spectrum by brute force. Looks good - I'll come back tomorrow and double-check the details
    – Yemon Choi
    Aug 17 at 3:42






  • 1




    BTW, by comprssing to an increasing sequence $(n_k)$ you can make the eigenvalues of $PSP$ go to infinity as fast as you want.
    – Nik Weaver
    Aug 17 at 13:52










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
7
down vote



accepted










Sure, for instance let $P$ be the orthogonal projection onto the closed span of the characteristic functions $chi_[n,n+1)$ for $n in mathbbN$. You get property 1 because each of these functions is in the domain of $S$, and you get property 2 because, identifying $V$ with $l^2$ in the obvious way, the operator $PSP$ becomes multiplication by, I think, $e^n(e - 1)$. So this compression has compact, even trace-class inverse, namely multiplication by $e^-n(e-1)^-1$.






share|cite|improve this answer
















  • 1




    Kicking myself now for not thinking of trying to discretize the spectrum by brute force. Looks good - I'll come back tomorrow and double-check the details
    – Yemon Choi
    Aug 17 at 3:42






  • 1




    BTW, by comprssing to an increasing sequence $(n_k)$ you can make the eigenvalues of $PSP$ go to infinity as fast as you want.
    – Nik Weaver
    Aug 17 at 13:52














up vote
7
down vote



accepted










Sure, for instance let $P$ be the orthogonal projection onto the closed span of the characteristic functions $chi_[n,n+1)$ for $n in mathbbN$. You get property 1 because each of these functions is in the domain of $S$, and you get property 2 because, identifying $V$ with $l^2$ in the obvious way, the operator $PSP$ becomes multiplication by, I think, $e^n(e - 1)$. So this compression has compact, even trace-class inverse, namely multiplication by $e^-n(e-1)^-1$.






share|cite|improve this answer
















  • 1




    Kicking myself now for not thinking of trying to discretize the spectrum by brute force. Looks good - I'll come back tomorrow and double-check the details
    – Yemon Choi
    Aug 17 at 3:42






  • 1




    BTW, by comprssing to an increasing sequence $(n_k)$ you can make the eigenvalues of $PSP$ go to infinity as fast as you want.
    – Nik Weaver
    Aug 17 at 13:52












up vote
7
down vote



accepted







up vote
7
down vote



accepted






Sure, for instance let $P$ be the orthogonal projection onto the closed span of the characteristic functions $chi_[n,n+1)$ for $n in mathbbN$. You get property 1 because each of these functions is in the domain of $S$, and you get property 2 because, identifying $V$ with $l^2$ in the obvious way, the operator $PSP$ becomes multiplication by, I think, $e^n(e - 1)$. So this compression has compact, even trace-class inverse, namely multiplication by $e^-n(e-1)^-1$.






share|cite|improve this answer












Sure, for instance let $P$ be the orthogonal projection onto the closed span of the characteristic functions $chi_[n,n+1)$ for $n in mathbbN$. You get property 1 because each of these functions is in the domain of $S$, and you get property 2 because, identifying $V$ with $l^2$ in the obvious way, the operator $PSP$ becomes multiplication by, I think, $e^n(e - 1)$. So this compression has compact, even trace-class inverse, namely multiplication by $e^-n(e-1)^-1$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Aug 17 at 3:26









Nik Weaver

17.9k143114




17.9k143114







  • 1




    Kicking myself now for not thinking of trying to discretize the spectrum by brute force. Looks good - I'll come back tomorrow and double-check the details
    – Yemon Choi
    Aug 17 at 3:42






  • 1




    BTW, by comprssing to an increasing sequence $(n_k)$ you can make the eigenvalues of $PSP$ go to infinity as fast as you want.
    – Nik Weaver
    Aug 17 at 13:52












  • 1




    Kicking myself now for not thinking of trying to discretize the spectrum by brute force. Looks good - I'll come back tomorrow and double-check the details
    – Yemon Choi
    Aug 17 at 3:42






  • 1




    BTW, by comprssing to an increasing sequence $(n_k)$ you can make the eigenvalues of $PSP$ go to infinity as fast as you want.
    – Nik Weaver
    Aug 17 at 13:52







1




1




Kicking myself now for not thinking of trying to discretize the spectrum by brute force. Looks good - I'll come back tomorrow and double-check the details
– Yemon Choi
Aug 17 at 3:42




Kicking myself now for not thinking of trying to discretize the spectrum by brute force. Looks good - I'll come back tomorrow and double-check the details
– Yemon Choi
Aug 17 at 3:42




1




1




BTW, by comprssing to an increasing sequence $(n_k)$ you can make the eigenvalues of $PSP$ go to infinity as fast as you want.
– Nik Weaver
Aug 17 at 13:52




BTW, by comprssing to an increasing sequence $(n_k)$ you can make the eigenvalues of $PSP$ go to infinity as fast as you want.
– Nik Weaver
Aug 17 at 13:52

















 

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