Let $H$ be a Hilbert space with projections $P,Q$, then $lim_{ntoinfty}(PQ)^nx= Rx$ for all $xin H$, where...











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Suppose we have a Hilbert space $mathcal H$, with projections $P,Q$, and let $R$ be the projection onto $Pmathcal Hcap Qmathcal H$, then I would like to show that $(PQ)^nto R$ strongly.



We can clearly reduce to the case where $Pmathcal Hcap Qmathcal H=0$ and $Pmathcal H+Qmathcal H=mathcal H$, in which case we must show that $(PQ)^nto0$ strongly. Now, if the limit $y=lim_{ntoinfty}(PQ)^n x$ exists, then if it is nonzero, then
$$|y| = |PQ y|<|Q y|le| y|$$ which is a contradiction, so once we have convergence, we're done. However, I don't see quite how to show convergence in a general infinite dimensional case.










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  • Do you assume your projections to be orthogonal?
    – Rhys Steele
    Nov 20 at 15:42






  • 1




    (I've just realised there may be some confusion. What I'm asking is if $P$ is the orthogonal projection onto $Pmathcal{H}$ and similarly for $Q$. Not if you are assuming that $P$ is orthogonal to $Q$)
    – Rhys Steele
    Nov 20 at 15:53






  • 1




    Oh I see, then I’m considering non-commuting orthogonal projections, using your terminology.
    – Monstrous Moonshine
    Nov 20 at 15:54















up vote
1
down vote

favorite












Suppose we have a Hilbert space $mathcal H$, with projections $P,Q$, and let $R$ be the projection onto $Pmathcal Hcap Qmathcal H$, then I would like to show that $(PQ)^nto R$ strongly.



We can clearly reduce to the case where $Pmathcal Hcap Qmathcal H=0$ and $Pmathcal H+Qmathcal H=mathcal H$, in which case we must show that $(PQ)^nto0$ strongly. Now, if the limit $y=lim_{ntoinfty}(PQ)^n x$ exists, then if it is nonzero, then
$$|y| = |PQ y|<|Q y|le| y|$$ which is a contradiction, so once we have convergence, we're done. However, I don't see quite how to show convergence in a general infinite dimensional case.










share|cite|improve this question






















  • Do you assume your projections to be orthogonal?
    – Rhys Steele
    Nov 20 at 15:42






  • 1




    (I've just realised there may be some confusion. What I'm asking is if $P$ is the orthogonal projection onto $Pmathcal{H}$ and similarly for $Q$. Not if you are assuming that $P$ is orthogonal to $Q$)
    – Rhys Steele
    Nov 20 at 15:53






  • 1




    Oh I see, then I’m considering non-commuting orthogonal projections, using your terminology.
    – Monstrous Moonshine
    Nov 20 at 15:54













up vote
1
down vote

favorite









up vote
1
down vote

favorite











Suppose we have a Hilbert space $mathcal H$, with projections $P,Q$, and let $R$ be the projection onto $Pmathcal Hcap Qmathcal H$, then I would like to show that $(PQ)^nto R$ strongly.



We can clearly reduce to the case where $Pmathcal Hcap Qmathcal H=0$ and $Pmathcal H+Qmathcal H=mathcal H$, in which case we must show that $(PQ)^nto0$ strongly. Now, if the limit $y=lim_{ntoinfty}(PQ)^n x$ exists, then if it is nonzero, then
$$|y| = |PQ y|<|Q y|le| y|$$ which is a contradiction, so once we have convergence, we're done. However, I don't see quite how to show convergence in a general infinite dimensional case.










share|cite|improve this question













Suppose we have a Hilbert space $mathcal H$, with projections $P,Q$, and let $R$ be the projection onto $Pmathcal Hcap Qmathcal H$, then I would like to show that $(PQ)^nto R$ strongly.



We can clearly reduce to the case where $Pmathcal Hcap Qmathcal H=0$ and $Pmathcal H+Qmathcal H=mathcal H$, in which case we must show that $(PQ)^nto0$ strongly. Now, if the limit $y=lim_{ntoinfty}(PQ)^n x$ exists, then if it is nonzero, then
$$|y| = |PQ y|<|Q y|le| y|$$ which is a contradiction, so once we have convergence, we're done. However, I don't see quite how to show convergence in a general infinite dimensional case.







functional-analysis operator-theory hilbert-spaces






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asked Nov 20 at 15:32









Monstrous Moonshine

2,6941630




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  • Do you assume your projections to be orthogonal?
    – Rhys Steele
    Nov 20 at 15:42






  • 1




    (I've just realised there may be some confusion. What I'm asking is if $P$ is the orthogonal projection onto $Pmathcal{H}$ and similarly for $Q$. Not if you are assuming that $P$ is orthogonal to $Q$)
    – Rhys Steele
    Nov 20 at 15:53






  • 1




    Oh I see, then I’m considering non-commuting orthogonal projections, using your terminology.
    – Monstrous Moonshine
    Nov 20 at 15:54


















  • Do you assume your projections to be orthogonal?
    – Rhys Steele
    Nov 20 at 15:42






  • 1




    (I've just realised there may be some confusion. What I'm asking is if $P$ is the orthogonal projection onto $Pmathcal{H}$ and similarly for $Q$. Not if you are assuming that $P$ is orthogonal to $Q$)
    – Rhys Steele
    Nov 20 at 15:53






  • 1




    Oh I see, then I’m considering non-commuting orthogonal projections, using your terminology.
    – Monstrous Moonshine
    Nov 20 at 15:54
















Do you assume your projections to be orthogonal?
– Rhys Steele
Nov 20 at 15:42




Do you assume your projections to be orthogonal?
– Rhys Steele
Nov 20 at 15:42




1




1




(I've just realised there may be some confusion. What I'm asking is if $P$ is the orthogonal projection onto $Pmathcal{H}$ and similarly for $Q$. Not if you are assuming that $P$ is orthogonal to $Q$)
– Rhys Steele
Nov 20 at 15:53




(I've just realised there may be some confusion. What I'm asking is if $P$ is the orthogonal projection onto $Pmathcal{H}$ and similarly for $Q$. Not if you are assuming that $P$ is orthogonal to $Q$)
– Rhys Steele
Nov 20 at 15:53




1




1




Oh I see, then I’m considering non-commuting orthogonal projections, using your terminology.
– Monstrous Moonshine
Nov 20 at 15:54




Oh I see, then I’m considering non-commuting orthogonal projections, using your terminology.
– Monstrous Moonshine
Nov 20 at 15:54










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I work in your reduced case and consider instead the operator $A = QPQ$ and note that
$$|(PQ)^nx| = |PA^{n-1}x| leq |A^{n-1}x|$$
so it will suffice to show that $A^n to 0$ strongly as $n to infty$.



The reason it is more convenient to consider $A$ is that $A$ is a self-adjoint operator and so we can apply the spectral theorem in the following form.




Theorem: Let $A$ be a bounded, self-adjoint operator on a Hilbert space $mathcal{H}$. Then there is a finite measure space $(Omega,Sigma,mu)$ and a real valued, essentially bounded function $f$ on $Omega$ and a unitary operator $U:mathcal{H} to L_{mu}^2(Omega)$ such that $$U^*MU = A$$ where $M$ is the multiplication operator defined by $$[Mphi](x)=f(x)phi (x)$$ We also have $|M| = |f|_{infty}$.




It is not so hard to check that if $A$ is additionally a positive operator, as in our case, then we must have $f geq 0$ a.e. and so since for $A = QPQ$ we have that $$1 geq |A| = |M| = |f|_infty,$$ we have here that $f(x) in [0,1]$ a.e.



Now it is clear that $P mathcal{H} cap Q mathcal{H} subseteq ker(A-I)$. In fact, this is an equality since $QPQx = x$ implies that
$$|x| = |QPQx| leq |PQ x| leq |Qx| leq |x|$$
and so $|Qx| = |x|$ and hence by, by pythagoras, we must have $Qx = x$. Then by a similar argument with
$$|x| = |QPQx| = |QPx| leq |Px| leq |x|$$
we see that $Px = x$ also so that $x in ker(A-I)$ implies that $x in P mathcal{H} cap Q mathcal{H}$.



In particular, $P mathcal{H} cap Q mathcal{H} = emptyset$ implies that $ker(M-I) = emptyset$ where $M$ is the multiplication operator associated to $A$. This means that $mu({s: f(s) = 1}) = 0$ so now $f(x) in [0,1)$ a.e.



We are finally ready to complete the proof. By conjugating with our unitary operators it is enough to show that $|M^k phi| to 0$ for every $phi in L_mu^2(Omega)$. This is easy.



$$|M^k phi|^2 = int_Omega |f^k phi|^2 to 0$$
by the D.C.T. since $|f^k phi| to 0$ and $|f^k phi| leq |phi| in L_mu^2(Omega)$.






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    I work in your reduced case and consider instead the operator $A = QPQ$ and note that
    $$|(PQ)^nx| = |PA^{n-1}x| leq |A^{n-1}x|$$
    so it will suffice to show that $A^n to 0$ strongly as $n to infty$.



    The reason it is more convenient to consider $A$ is that $A$ is a self-adjoint operator and so we can apply the spectral theorem in the following form.




    Theorem: Let $A$ be a bounded, self-adjoint operator on a Hilbert space $mathcal{H}$. Then there is a finite measure space $(Omega,Sigma,mu)$ and a real valued, essentially bounded function $f$ on $Omega$ and a unitary operator $U:mathcal{H} to L_{mu}^2(Omega)$ such that $$U^*MU = A$$ where $M$ is the multiplication operator defined by $$[Mphi](x)=f(x)phi (x)$$ We also have $|M| = |f|_{infty}$.




    It is not so hard to check that if $A$ is additionally a positive operator, as in our case, then we must have $f geq 0$ a.e. and so since for $A = QPQ$ we have that $$1 geq |A| = |M| = |f|_infty,$$ we have here that $f(x) in [0,1]$ a.e.



    Now it is clear that $P mathcal{H} cap Q mathcal{H} subseteq ker(A-I)$. In fact, this is an equality since $QPQx = x$ implies that
    $$|x| = |QPQx| leq |PQ x| leq |Qx| leq |x|$$
    and so $|Qx| = |x|$ and hence by, by pythagoras, we must have $Qx = x$. Then by a similar argument with
    $$|x| = |QPQx| = |QPx| leq |Px| leq |x|$$
    we see that $Px = x$ also so that $x in ker(A-I)$ implies that $x in P mathcal{H} cap Q mathcal{H}$.



    In particular, $P mathcal{H} cap Q mathcal{H} = emptyset$ implies that $ker(M-I) = emptyset$ where $M$ is the multiplication operator associated to $A$. This means that $mu({s: f(s) = 1}) = 0$ so now $f(x) in [0,1)$ a.e.



    We are finally ready to complete the proof. By conjugating with our unitary operators it is enough to show that $|M^k phi| to 0$ for every $phi in L_mu^2(Omega)$. This is easy.



    $$|M^k phi|^2 = int_Omega |f^k phi|^2 to 0$$
    by the D.C.T. since $|f^k phi| to 0$ and $|f^k phi| leq |phi| in L_mu^2(Omega)$.






    share|cite|improve this answer



























      up vote
      3
      down vote



      accepted










      I work in your reduced case and consider instead the operator $A = QPQ$ and note that
      $$|(PQ)^nx| = |PA^{n-1}x| leq |A^{n-1}x|$$
      so it will suffice to show that $A^n to 0$ strongly as $n to infty$.



      The reason it is more convenient to consider $A$ is that $A$ is a self-adjoint operator and so we can apply the spectral theorem in the following form.




      Theorem: Let $A$ be a bounded, self-adjoint operator on a Hilbert space $mathcal{H}$. Then there is a finite measure space $(Omega,Sigma,mu)$ and a real valued, essentially bounded function $f$ on $Omega$ and a unitary operator $U:mathcal{H} to L_{mu}^2(Omega)$ such that $$U^*MU = A$$ where $M$ is the multiplication operator defined by $$[Mphi](x)=f(x)phi (x)$$ We also have $|M| = |f|_{infty}$.




      It is not so hard to check that if $A$ is additionally a positive operator, as in our case, then we must have $f geq 0$ a.e. and so since for $A = QPQ$ we have that $$1 geq |A| = |M| = |f|_infty,$$ we have here that $f(x) in [0,1]$ a.e.



      Now it is clear that $P mathcal{H} cap Q mathcal{H} subseteq ker(A-I)$. In fact, this is an equality since $QPQx = x$ implies that
      $$|x| = |QPQx| leq |PQ x| leq |Qx| leq |x|$$
      and so $|Qx| = |x|$ and hence by, by pythagoras, we must have $Qx = x$. Then by a similar argument with
      $$|x| = |QPQx| = |QPx| leq |Px| leq |x|$$
      we see that $Px = x$ also so that $x in ker(A-I)$ implies that $x in P mathcal{H} cap Q mathcal{H}$.



      In particular, $P mathcal{H} cap Q mathcal{H} = emptyset$ implies that $ker(M-I) = emptyset$ where $M$ is the multiplication operator associated to $A$. This means that $mu({s: f(s) = 1}) = 0$ so now $f(x) in [0,1)$ a.e.



      We are finally ready to complete the proof. By conjugating with our unitary operators it is enough to show that $|M^k phi| to 0$ for every $phi in L_mu^2(Omega)$. This is easy.



      $$|M^k phi|^2 = int_Omega |f^k phi|^2 to 0$$
      by the D.C.T. since $|f^k phi| to 0$ and $|f^k phi| leq |phi| in L_mu^2(Omega)$.






      share|cite|improve this answer

























        up vote
        3
        down vote



        accepted







        up vote
        3
        down vote



        accepted






        I work in your reduced case and consider instead the operator $A = QPQ$ and note that
        $$|(PQ)^nx| = |PA^{n-1}x| leq |A^{n-1}x|$$
        so it will suffice to show that $A^n to 0$ strongly as $n to infty$.



        The reason it is more convenient to consider $A$ is that $A$ is a self-adjoint operator and so we can apply the spectral theorem in the following form.




        Theorem: Let $A$ be a bounded, self-adjoint operator on a Hilbert space $mathcal{H}$. Then there is a finite measure space $(Omega,Sigma,mu)$ and a real valued, essentially bounded function $f$ on $Omega$ and a unitary operator $U:mathcal{H} to L_{mu}^2(Omega)$ such that $$U^*MU = A$$ where $M$ is the multiplication operator defined by $$[Mphi](x)=f(x)phi (x)$$ We also have $|M| = |f|_{infty}$.




        It is not so hard to check that if $A$ is additionally a positive operator, as in our case, then we must have $f geq 0$ a.e. and so since for $A = QPQ$ we have that $$1 geq |A| = |M| = |f|_infty,$$ we have here that $f(x) in [0,1]$ a.e.



        Now it is clear that $P mathcal{H} cap Q mathcal{H} subseteq ker(A-I)$. In fact, this is an equality since $QPQx = x$ implies that
        $$|x| = |QPQx| leq |PQ x| leq |Qx| leq |x|$$
        and so $|Qx| = |x|$ and hence by, by pythagoras, we must have $Qx = x$. Then by a similar argument with
        $$|x| = |QPQx| = |QPx| leq |Px| leq |x|$$
        we see that $Px = x$ also so that $x in ker(A-I)$ implies that $x in P mathcal{H} cap Q mathcal{H}$.



        In particular, $P mathcal{H} cap Q mathcal{H} = emptyset$ implies that $ker(M-I) = emptyset$ where $M$ is the multiplication operator associated to $A$. This means that $mu({s: f(s) = 1}) = 0$ so now $f(x) in [0,1)$ a.e.



        We are finally ready to complete the proof. By conjugating with our unitary operators it is enough to show that $|M^k phi| to 0$ for every $phi in L_mu^2(Omega)$. This is easy.



        $$|M^k phi|^2 = int_Omega |f^k phi|^2 to 0$$
        by the D.C.T. since $|f^k phi| to 0$ and $|f^k phi| leq |phi| in L_mu^2(Omega)$.






        share|cite|improve this answer














        I work in your reduced case and consider instead the operator $A = QPQ$ and note that
        $$|(PQ)^nx| = |PA^{n-1}x| leq |A^{n-1}x|$$
        so it will suffice to show that $A^n to 0$ strongly as $n to infty$.



        The reason it is more convenient to consider $A$ is that $A$ is a self-adjoint operator and so we can apply the spectral theorem in the following form.




        Theorem: Let $A$ be a bounded, self-adjoint operator on a Hilbert space $mathcal{H}$. Then there is a finite measure space $(Omega,Sigma,mu)$ and a real valued, essentially bounded function $f$ on $Omega$ and a unitary operator $U:mathcal{H} to L_{mu}^2(Omega)$ such that $$U^*MU = A$$ where $M$ is the multiplication operator defined by $$[Mphi](x)=f(x)phi (x)$$ We also have $|M| = |f|_{infty}$.




        It is not so hard to check that if $A$ is additionally a positive operator, as in our case, then we must have $f geq 0$ a.e. and so since for $A = QPQ$ we have that $$1 geq |A| = |M| = |f|_infty,$$ we have here that $f(x) in [0,1]$ a.e.



        Now it is clear that $P mathcal{H} cap Q mathcal{H} subseteq ker(A-I)$. In fact, this is an equality since $QPQx = x$ implies that
        $$|x| = |QPQx| leq |PQ x| leq |Qx| leq |x|$$
        and so $|Qx| = |x|$ and hence by, by pythagoras, we must have $Qx = x$. Then by a similar argument with
        $$|x| = |QPQx| = |QPx| leq |Px| leq |x|$$
        we see that $Px = x$ also so that $x in ker(A-I)$ implies that $x in P mathcal{H} cap Q mathcal{H}$.



        In particular, $P mathcal{H} cap Q mathcal{H} = emptyset$ implies that $ker(M-I) = emptyset$ where $M$ is the multiplication operator associated to $A$. This means that $mu({s: f(s) = 1}) = 0$ so now $f(x) in [0,1)$ a.e.



        We are finally ready to complete the proof. By conjugating with our unitary operators it is enough to show that $|M^k phi| to 0$ for every $phi in L_mu^2(Omega)$. This is easy.



        $$|M^k phi|^2 = int_Omega |f^k phi|^2 to 0$$
        by the D.C.T. since $|f^k phi| to 0$ and $|f^k phi| leq |phi| in L_mu^2(Omega)$.







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        edited Nov 20 at 16:50

























        answered Nov 20 at 16:43









        Rhys Steele

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        6,0431829






























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