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Spectrum of a linear operator on a Banach space

  1. May 6, 2013 #1
    I'm trying to understand the spectrum and resolvent of a linear operator on a Banach space in as much generality as I possibly can.

    It seems that the furthest the concept can be "pulled back" is to a linear operator [itex]T: D(T) \to X[/itex], where [itex]X[/itex] is a Banach space and [itex]D(T)\subseteq X[/itex]. But here are a few questions:

    (1) Doesn't [itex]D(T)[/itex] necessarily have to be a subspace of [itex]X[/itex] for this concept to make any sense? For instance, if it's not a subspace, there can be elements [itex]x,y[/itex] for which [itex]Tx[/itex] and [itex]Ty[/itex] make sense, but for which [itex]T(x+y)[/itex] makes no sense, since [itex]x+y[/itex] might not be an element of [itex]D(T)[/itex].

    (2) The Wikipedia article here says that, in order for [itex]\lambda \in \rho(T)[/itex], we need both (1) [itex](T-\lambda)^{-1}[/itex] exists and (2) [itex](T-\lambda)^{-1}[/itex] is defined on a dense subset of [itex]X[/itex]. Is this equivalent to saying that [itex]\text{Ran} (T-\lambda)[/itex] must be a dense subset of [itex]X[/itex] and that [itex](T-\lambda): D(T) \to \text{Ran} (T - \lambda)[/itex] must be a bijection?
     
  2. jcsd
  3. May 6, 2013 #2

    micromass

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    Yes. And we usually require ##D(T)## to be dense in ##T## as well. It doesn't really matter, if it's not dense, then we can always restrict ##X## to ##\overline{D(T)}## and work with that.

    Yes, this is correct. We also want ##(T- \lambda I)^{-1}## to be bounded though.

    If you want to see a very general definition of spectrum, then you should study Banach algebras. But this is a spectrum that coincides with the spectrum of bounded operators.
     
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