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Are resolvents for self-adjoint operators themselves self-adjoint?

  1. Dec 9, 2013 #1
    Let [itex]T[/itex] be a (possibly unbounded) self-adjoint operator on a Hilbert space [itex]\mathscr H[/itex] with domain [itex]D(T)[/itex], and let [itex]\lambda \in \rho(T)[/itex]. Then we know that [itex](T-\lambda I)^{-1}[/itex] exists as a bounded operator from [itex]\mathscr H[/itex] to [itex]D(T)[/itex]. Question: do we also know that [itex](T-\lambda I)^{-1}[/itex] is self-adjoint? Can someone prove or give a counterexample?
     
  2. jcsd
  3. Dec 9, 2013 #2
    Note: I have reason to believe this is true in the case that [itex]T[/itex] is a positive operator (i.e., [itex](Tx,x) \geq 0[/itex] for all [itex]x\in D(T)[/itex]). Whether it is true in general...well, I'm not sure!
     
  4. Dec 10, 2013 #3
    I believe I have a quick and easy proof of this proposition, provided the resolvent has the form [itex](T-\lambda I)^{-1}[/itex] for [itex]\lambda \in \mathbb R[/itex]. We have
    [tex]
    \langle (T-\lambda I)^{-1}x,y \rangle = \langle x, [(T-\lambda I)^{-1}]^* y \rangle = \langle x, (T^* - \overline \lambda I)^{-1} y \rangle = \langle x, (T-\lambda I)^{-1}y \rangle,
    [/tex]
    since [itex]\lambda = \overline \lambda[/itex] and [itex]T^* = T[/itex] by assumption. This also uses the fact that [itex](T^*)^{-1} = (T^{-1})^*[/itex] for [itex]T[/itex] bounded and invertible. Now, if [itex]\lambda[/itex] is complex, this argument is obviously bogus, and in fact I'm reasonably confident that it's just not true!
     
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