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A question on Hilbert space theory

  1. Sep 25, 2006 #1

    dextercioby

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    Let's say we have a self-adjoint, densly defined closed linear operator acting on a separable Hilbert space [itex] H [/itex]

    [tex] A:D_{A}\rightarrow H [/tex]

    Let [itex] \lambda [/itex] be an eigenvalue of A and let

    [tex] \Delta_{A}\left(\lambda\right) = \{\left(A-\lambda \hat{1}_{H}\right)f, \ f\in D_{A}\} [/tex]

    How do i prove that

    [tex] D_{A}\perp \Delta_{A}(\lambda) \Leftrightarrow \bar{\Delta_{A}(\lambda)} \neq H [/tex].

    Daniel.
     
  2. jcsd
  3. Sep 25, 2006 #2

    matt grime

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    Unless you define D_A I don't think we can do the question.
     
  4. Sep 25, 2006 #3

    Hurkyl

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    I presume D_A is the domain of A.
     
  5. Sep 26, 2006 #4

    matt grime

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    The domain of A is H.
     
  6. Sep 26, 2006 #5

    George Jones

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    No, Hurkyl is right. By the Hellinger-Toeplitz theorem, any self-adjoint operator whose domain is all of H is bounded. So, the best that one can expect for an unbounded self-adjoint operator A is that its domain D_A is a dense subset of H, as Daniel says.
     
  7. Sep 26, 2006 #6

    matt grime

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    Then why did the OP say we had an operator acting on H, if H is not the domain? Must just be a convention thing, but seems odd to me.
     
  8. Sep 26, 2006 #7

    Hurkyl

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    It's not that much different from talking about meromorphic functions on C, or rational functions on an an algebraic variety, is it?
     
  9. Sep 26, 2006 #8

    matt grime

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    As in they have poles? No, I think that is different, slightly, since we implicitly extend the codomain to have a point at infinity thus the functions are defined at all points of the domain. I get the impression here that there is no possible extension to an operator defined on the whole of H, though I may be mistaken in this assumption.
     
  10. Sep 27, 2006 #9

    dextercioby

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    George Jones nailed it on the head. It's assumed unbounded, so by Hellinger-Toeplitz theorem

    [tex] D_{A}\neq H [/tex]

    but in order to be self-adjoint one must have that the closure of [itex] D_{A} [/itex] is H.

    I guess the question still remains open. :uhh:

    Daniel.
     
  11. Sep 27, 2006 #10

    Hurkyl

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    That can't be right.

    Let A be any projection operator. So [itex]D_A = H[/itex]. Then, [itex]\Delta_A(0)[/itex] is the image of A. For most such operators, [itex]D_A[/itex] is clearly not orthogonal to [itex]\Delta_A(0)[/itex]... and yet [itex]\Delta_A(0)[/itex] is not dense in H.

    Or, did you mean to only consider unbounded operators?
     
  12. Sep 29, 2006 #11

    dextercioby

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    Yes, that operator is meant to be unbounded. Look at it this way, if it were bounded, why would one consider [itex]D_{A}[/itex] (the domain of A) dense in H but yet different from H ? It would have been simply H, as a bounded operator could be extended by continuity on all H...

    Daniel.
     
    Last edited: Sep 29, 2006
  13. Sep 29, 2006 #12

    matt grime

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    But the case Hurkyl gave is a special case of what you wrote and it breaks down in the special case so why should it be true if we exclude the special case above? I think you should taken that post as a hint as to what you need to do. (Not that I know what the proof is, before you ask.) There are a lot of hypotheses on A up there, and H, so how are you attempting to use them?
     
  14. Sep 29, 2006 #13

    dextercioby

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    Perhaps the problem that i stated is not quite correct. It would be fair to post the whole context. Here's a scanned version of page #89 of Ahiezer & Glazman's book. It's part of the 43-rd section which i'm trying to fully understand.

    The thing i didn't understand is the argument he uses in the proof of theorem #2 namely that under the assumptions i stated in my first post (which appear on page 88).

    He asserts that if [itex] Af=\lambda f [/itex] for an "f" both in the domain and in the range of A (closed, densly-defined self-adjoint linear operator) then

    [tex] f\perp \Delta_{A}\left(\lambda\right) \Leftrightarrow \Delta_{A} \ \mbox{ is not everywhere dense in H} [/tex].

    Huh? :confused:

    Daniel.
     
    Last edited: Nov 22, 2006
  15. Sep 29, 2006 #14

    dextercioby

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    So i was basically wrong in the first post, since "f" is not arbitrary in D_{A} so i can't assert that generally D_{A} \perp \Delta_{A}\left(\lambda\right)

    Daniel.
     
  16. Sep 29, 2006 #15

    Hurkyl

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    Let me write [itex]D = D_A[/itex] and [itex]\Delta = \Delta_A(\lambda)[/itex] for simplicity.

    Incidentally, it's easy to see that you can't have both [itex]D \perp \Delta[/itex] and [itex]\bar{\Delta} = H[/itex]: the inner product is continuous, and this implies that it's identically zero.
     
  17. Sep 29, 2006 #16

    dextercioby

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    So lemme see why you think it's trivial. If i assume

    [tex] \bar{\Delta} =H [/tex]

    , then

    [tex] \bar{\Delta}\supset H \supset D[/itex]

    But [itex] D\perp \Delta [/itex] , so considering a sequence [itex] g_{n} [/itex] from [itex] \Delta [/itex] convergent to "g" in [itex] \bar{\Delta} [/itex], then from

    [tex] \langle f,g_{n}\rangle =0 [/tex]

    it follows that

    [tex] \langle f,g\rangle =0 [/tex]

    ,using continuity of the scalar product.

    So i proved that [itex]D \perp \bar{\Delta} [/itex]. Since above i assumed that [itex] D\subset \bar{\Delta} [/itex], it follows that [itex] D={0}_{H} [/itex] which is false. Ergo, the initial assumption is false.

    Daniel.
     
  18. Sep 29, 2006 #17

    George Jones

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    Let my try and elaborate a bit on Hurky's last post.

    Let [itex]U[/itex] be any any dense subset of [itex]H[/itex], and let [itex]v \in H[/itex] be such that [itex]\left< v , u \right> = 0[/itex] for every [itex]u \in U[/itex]. Then, by continuity and density (insert small analysis argument), [itex]v = 0[/itex].

    Now, in the first half of the proof, one has [itex]\left< f , u \right> =0[/itex] for every [itex]u \in \Delta[/itex]. Therefore, if [itex]\Delta[/itex] were dense in [itex]H[/itex], then [itex]f[/itex] would have to be zero. But this can't be, since [itex]f[/itex] is an eigenvector.

    Oops, I didn't see Daniel's last post.
     
    Last edited: Sep 29, 2006
  19. Sep 29, 2006 #18

    dextercioby

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    I see, and there's no reverse of that statement. So i'm supposed to get that

    [tex] \bar{\Delta}\neq H \nRightarrow D\perp \Delta [/tex]

    , but instead one can find a nonzero vector from D which is orthogonal to [itex] \Delta [/itex].

    Daniel.
     
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