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Math Help - spectral thoerem question

  1. #1
    Member Mauritzvdworm's Avatar
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    spectral thoerem question

    Suppose T is a normal operator on a Hilbert space H, show that there exists a self-adjoint operator S on H such that T=f(S), where f is a continuous function form the spectrum of S into C
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  2. #2
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    Opalg's Avatar
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    Quote Originally Posted by Mauritzvdworm View Post
    Suppose T is a normal operator on a Hilbert space H, show that there exists a self-adjoint operator S on H such that T=f(S), where f is a continuous function form the spectrum of S into C
    This is a strange result, and I'm not convinced that it is true.

    Suppose for example that H = L^2(\overline{\mathbb{D}}), the Hilbert space of all square-integrable functions on the closed unit disk \overline{\mathbb{D}} in the complex plane (with respect to 2-dimensional Lebesgue measure). Let M_z be the operator on H defined by M_z\xi(z) = z\xi(z)\ (\xi\in H). In other words, the operator M_z consists of multiplication by the coordinate function. Then M_z is a normal operator on H, and its spectrum is equal to \overline{\mathbb{D}}. If there is a selfadjoint operator S on H such that T=f(S), where f is a continuous function from the spectrum of S into \mathbb{C} then the spectral mapping theorem states that f maps the spectrum of S onto the spectrum of T. So f would have to be something like a space-filling curve, mapping a compact subset of the real line onto the unit disk in the plane.

    I suppose it's possible that every normal operator could come from a selfadjoint operator in this sort of way, but it somehow seems unlikely to me.
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  3. #3
    Member Mauritzvdworm's Avatar
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    I see, so when the operator T is in fact unitary the result is trivial, since then we can find a self-adjoint operator, say S such that T=e^{iS}
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