Results 1 to 7 of 7

Math Help - Unbounded Operators

  1. #1
    Junior Member
    Joined
    Mar 2009
    Posts
    41

    Exclamation Unbounded Operators

    L_+ = ker(A*-i)=[ran(A+i)^perp
    L_- = ker(A*+1) = [ran(A-i)]^perp

    K_+ = {f directsum if: f in L_+}
    K_- = {g directsum (-ig):g in L_-}

    Show that gra A, K_+ and K_- are pairwise orthogonal.

    How do I do this? Please help
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by Nusc View Post
    L_+ = ker(A*-i)=[ran(A+i)^perp
    L_- = ker(A*+1) = [ran(A-i)]^perp

    K_+ = {f directsum if: f in L_+}
    K_- = {g directsum (-ig):g in L_-}

    Show that gra A, K_+ and K_- are pairwise orthogonal.

    How do I do this?
    This is a problem about extensions of a symmetric unbounded operator A on some Hilbert space H, and it's not hard provided that you understand all the definitions.

    For a start, A has a domain D(A) (which is a dense subspace of H). It has an adjoint A* with domain D(A*), such that \langle Af,g\rangle = \langle f,A^*g\rangle for all f in D(A) and g in D(A*). To say that A is symmetric means that D(A)\subseteq D(A^*) and A*(f)=A(f) for all f in D(A).

    The notation gra(A) means the graph of A, which is the set \{f\oplus Af \in H\oplus H : f\in D(A)\}.

    The spaces \mathcal{L}_\pm are defined by \mathcal{L}_\pm = \{f\in D(A^*) : A^*f = \pm if\}. The kernel of an adjoint operator is always the orthogonal complement of the range of the operator, so \mathcal{L}_\pm = \text{ker}(A^*\mp i) = (\text{ran}(A\pm i))^\perp (because -iI is the adjoint of iI).

    The spaces \mathcal{K}_\pm = \{f\oplus(\pm if) : f\in\mathcal{L}_\pm\} are by definition contained in the graph of A*. To see that they are orthogonal to the graph of A, let f\oplus Af \in \text{gra}(A) and g\oplus(ig)\in\mathcal{K}_+. Then

    \langle f\oplus Af, g\oplus ig\rangle = \langle f,g\rangle + \langle Af,ig\rangle (definition of inner product in direct sum)
    {\color{white}\langle f\oplus Af, g\oplus ig\rangle} = \langle f,g\rangle - i\langle Af,g\rangle (conjugate-linear property of inner product)
    {\color{white}\langle f\oplus Af, g\oplus ig\rangle} = -i\langle (A+i)f,g\rangle
    {\color{white}\langle f\oplus Af, g\oplus ig\rangle} = 0 (because g\in(\text{ran}(A+ i))^\perp.

    A similar argument shows that gra(A) and \mathcal{K}_- are orthogonal.

    The fact that \mathcal{K}_+ and \mathcal{K}_- are orthogonal is more or less immediate, because if f\oplus(if)\in\mathcal{K}_+ and g\oplus(ig)\in\mathcal{K}_- then \langle f\oplus if,g\oplus ig \rangle = \langle f,g\rangle + \langle if,-ig\rangle = \langle f,g\rangle - \langle f,g\rangle = 0.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Mar 2009
    Posts
    41

    Question

    I see thank you.

    I'm not sure why it's clear that:

    <br /> <br />
gra A \oplus \mathcal{K}_- \oplus \mathcal{K}_- \subseteq gra A*.<br />


    Then how would I show that this direct sum is dense in gra A*.

    Thanks for your concern,
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by Nusc View Post
    I'm not sure why it's clear that:

    <br /> <br />
gra A \oplus \mathcal{K}_- \oplus \mathcal{K}_- \subseteq gra A*.<br />
    gra(A) is contained in gra(A*) because A is symmetric. (If you're unsure about that, go back and check the definition of a symmetric operator.)

    \mathcal{K}+ and \mathcal{K}_- are contained in gra(A*) by the way in which they are defined. (If you're unsure about that, go back and check the definitions of \mathcal{K}_{\pm}.)

    The sum is a direct one because we have seen that these three spaces are all orthogonal to each other.

    Quote Originally Posted by Nusc View Post
    Then how would I show that this direct sum is dense in gra A*.
    This is proved in Lemma 2.13 of Conway's Course in functional analysis, Chapter X. The idea is that if \text{gra}(A) \oplus \mathcal{K}_- \oplus \mathcal{K}_- is not the whole of gra(A*) then there will be a nonzero vector in its orthogonal complement. An element of gra(A*) is of the form h\oplus A^*(h) for some h in the domain of A*. So you take a vector h in D(A*) such that h\oplus A^*(h) is in that orthogonal complement, and you need to show that h=0. Go slowly and carefully through the proof of Lemma 2.13, and you'll see how it works.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Mar 2009
    Posts
    41

    Question

    Example 2.21:

    Let A and   \mathcal{D} be as in Example 1.11 (   \mathcal{D} = all functions  f:[0,1]\rightarrow C that are absolutely continuous; so A is symmetric (how does this imply that?). The operator B of Example 1.12 (  B \in   \mathcal{C}(L^{2}(0,1))  - it's not even what he meant by   \mathcal{C} ) is a self-adjoint extension of A. Let us determine all self-adjoint extensions of A. To do this it is necessary to determine  \mathcal{L}_\pm . Now f \in \mathcal{L}_\pm if and only if f \in  dom A^* and  \pm if=A^*f=if', so  \mathcal{L}_\pm = \{\alpha e^{\pm x} : \alpha \in C \} . Hence  n_\pm =1 . Also, the isomorphisms of  \mathcal{L}_+ onto  \mathcal{L}_- are all of the form W_\lambda e^{x} = \lambda e^{-x} where   |\lambda| = e.

    If  |\lambda|=e , let   \mathcal{D} \equiv \{f+ \alpha e^{x} + \lambda \alpha e^{-x} : \alpha \in C, f \in \mathcal{D}\} .

    A_\lambda(f+\alpha e^{x} + \lambda \alpha e^{-x})= if' + \alpha i e^{x} - i\lambda \alpha e^{-x},

    if f \in \mathcal{D}, \alpha \in C.

    What is the significance of this example? What was Conway trying to illustrate?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by Nusc View Post
    Example 2.21:

    Let A and   \mathcal{D} be as in Example 1.11 (   \mathcal{D} = all functions  f:[0,1]\rightarrow C that are absolutely continuous; so A is symmetric (how does this imply that?).
    To say that A is symmetric means that \langle Af,g\rangle = \langle f,Ag\rangle for all functions f,g in the domain of A (which should consist of all differentiable functions on [0,1] whose derivatives are in L^2[0,1] and which vanish at 0 and 1). The operator A is defined by Af=if'.

    The inner product is given by \langle Af,g\rangle = \int_0^1\!\!if'(t)\overline{g(t)}\,dt. To see that this is equal to \langle f,Ag\rangle, you need to integrate by parts. I'll leave you to do that. Study the calculation very carefully, and you will see that only f (not g) needs to satisfy that condition f(0)=f(1)=0. In fact, the adjoint of A is the operator A* given by the same formula A*g=ig' but on a larger domain, namely the set of all differentiable functions on [0,1] whose derivatives are in L^2[0,1] (but which need not vanish at the endpoints).

    The fundamental idea of von Neumann's theorem is that if A is symmetric (so that A* is an extension of A) then as you increase the domain of A, the domain of its adjoint decreases. If you're lucky, there is an intermediate domain which is the same for the operator and its adjoint. That operator is then selfadjoint.

    In the case of the operator Af=if', that point occurs when the domain is the set of all differentiable functions f on [0,1] whose derivatives are in L^2[0,1] and which satisfy f(0)=f(1). You can check that if f and g both satisfy that condition then the integration by parts calculations works. This salfadjoint extension of A is what Canway calls B in his Example 1.12.

    Quote Originally Posted by Nusc View Post
    The operator B of Example 1.12 (  B \in   \mathcal{C}(L^{2}(0,1))  - it's not even what he meant by   \mathcal{C} ) is a self-adjoint extension of A. Let us determine all self-adjoint extensions of A. To do this it is necessary to determine  \mathcal{L}_\pm . Now f \in \mathcal{L}_\pm if and only if f \in  dom A^* and  \pm if=A^*f=if', so  \mathcal{L}_\pm = \{\alpha e^{\pm x} : \alpha \in C \} . Hence  n_\pm =1 . Also, the isomorphisms of  \mathcal{L}_+ onto  \mathcal{L}_- are all of the form W_\lambda e^{x} = \lambda e^{-x} where   |\lambda| = e.

    If  |\lambda|=e , let   \mathcal{D} \equiv \{f+ \alpha e^{x} + \lambda \alpha e^{-x} : \alpha \in C, f \in \mathcal{D}\} .

    A_\lambda(f+\alpha e^{x} + \lambda \alpha e^{-x})= if' + \alpha i e^{x} - i\lambda \alpha e^{-x},

    if f \in \mathcal{D}, \alpha \in C.

    What is the significance of this example? What was Conway trying to illustrate?
    He is showing how von Neumann's theorem applies to this example. You get a selfadjoint extension of A by increasing its domain, or equivalently by increasing its graph through incorporating part of the graph of A*. It turns out that the way to do this is to add to the graph of A a subspace of \mathcal{K}_+\oplus\mathcal{K}_- which in turn is defined in terms of an isometric map from a subspace of \mathcal{L}_+ to a subspace of \mathcal{L}_-. If this map takes the whole of \mathcal{L}_+ to the whole of \mathcal{L}_- then the extension will be selfadjoint. The condition for that to be possible is that \mathcal{L}_+ and \mathcal{L}_- have the same dimension.
    Last edited by Opalg; March 23rd 2009 at 12:30 AM. Reason: corrected mistake
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Mar 2009
    Posts
    41

    Question

    I see, von Neumann's theorem is not stated in Conway or in Wikipedia.

    It says:

    Let A be a symmetric operator and suppose that there exists a conjugation C with C: D(A)->D(A) and AC = CA. Then A has equal deficiency indices and therefore has self-adjoint extensions.

    Do you know of a more precise way of writing that theorem?
    Last edited by Nusc; March 22nd 2009 at 08:24 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. unbounded component
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: August 29th 2011, 11:20 AM
  2. Unbounded?
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: February 7th 2010, 12:56 PM
  3. unbounded sequence
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: December 9th 2009, 07:56 AM
  4. Unbounded Integral
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 11th 2009, 02:09 PM
  5. The UnBounded Function
    Posted in the Calculus Forum
    Replies: 9
    Last Post: April 18th 2007, 07:21 PM

Search Tags


/mathhelpforum @mathhelpforum