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I've seen a variaty of slightly different definitions for the spectrum and its division into pure point spectrum residual spectrum and so on.
Thus I'm wondering what could be an appropriate definition.

Moreover, when does the spectral theorem apply in principle?
I have heard for normal operators and read once (I don't remember where) positive operators I guess...

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For the spectrum of a densely defined operator $T$ in a Hilbert space $H$, one usually defines the resolvent set $\rho(T)$ as the set of $\lambda \in \mathbb{C}$ such that the densely defined operator $T- \lambda$ is bijective and the inverse operator $(T- \lambda)^{-1}$ is a bounded operator. This can be slightly simplified for closed operators.

The spectrum $\sigma(T)$ of $T$ is defined as $\mathbb{C}\setminus \rho(T)$, and here are some possible partitions of the spectrum for selfadjoint operators:

  1. point spectrum/continous spectrum/residual spectrum. This partition can be translated as "Why isn't the inverse a bounded operator?":

    • The point spectrum is the set of eigenvalues $\lambda$, i.e. those points where $T-\lambda$ is not injective. In this case, the inverse is not an operator, but a linear relation, i.e. a multivalued operator.
    • The continuous spectrum is the set of $\lambda$s where $T-\lambda$ is injective, (i.e. $(T-\lambda)^{-1}$ is a linear operator), but not surjective, but the range of $T- \lambda$ is dense in $H$. This gives rise to an inverse operator $(T-\lambda)^{-1}$ which is densely defined, but it's not a bounded operator. For an explanation, see below.
    • The residual spectrum is literally the rest: It's the set of $\lambda$s where $T-\lambda$ is injective, but not surjective and the range of $T-\lambda$ is not dense, i.e. the inverse is a linear operator which is not densely defined.
  2. For selfadjoint operators, the residual spectrum is empty and there is a decompositionof the spectrum which corresponds to the spectral measure: If $\mu$ is the spectral measure of a selfadjoint operator $T$, then this measure admits a decomposition into three measures with $\mu = \mu_{\rm ac} + \mu_{\rm sc} + \mu_{\rm pp}$, where $\mu_{\rm ac}$ is absolutely continuous and $\mu_{\rm ac}$ is singular continuous with respect to the lebesgue measure, $\mu_{\rm pp}$ is a pure point measure and singular w.r.t the Lebesgue measure. It's worth noting that there may be eigenvalues in the support of the continuous measures if there are eigenvalues embedded in the continuous spectrum.

  3. From a perturbation point of view, there is another (familiy of) spectral decomposition(s), which reduces for self-adjoint operators in Hilbert spaces to the following:

    • The essential spectrum is defined as the points $\lambda$ such that $T- \lambda$ is not a Fredholm operator, i.e. it either doesn't have a closed range, or the kernel and/or cokernel are not finite-dimensional.
    • The discrete spectrum is defined as all spectral points which are not in the essential spectrum, which means that these are the isolated eigenvalues with finite multiplicity.

For the spectral theorem, there is a number of spectral theorems for different families of operators in Hilbert spaces: Compact operators (no normality/selfadjointness needed); bounded, self-adjoint operators; bounded normal operators, unbounded self-adjoint operators, unbounded normal operators. The more information you have on your operator, the better and more descriptive is your spectral theorem.

Edit: For a closed operator $T$ in a Banach space $H$, why does $\lambda \in \sigma_c(T)$ imply that $(T-\lambda)^{-1}$ is not bounded?*

Let $\lambda \in \sigma_c(T)$, i.e. $T$ is injective and the range of $T-\lambda$ is dense, but not the whole space. Suppose $(T-\lambda)^{-1}$ is bounded, i.e. there exists $M>0$ s.th. $\Vert (T-\lambda)^{-1} y \Vert\leq M\Vert y\Vert$ for all $y$ in the range of $T-\lambda$, thus we have $$\Vert x \Vert \leq M \Vert(A-\lambda)x\Vert$$ for all $x$ in the domain of $T$.

Since the range of $T-\lambda$ is not the whole space, but dense, there is a $y \in H$ which isn't in range of $T-\lambda$, but there is a sequence of $(x_n)$ in the domain of $T$ such that $(T-\lambda)x_n$ converges to $y$. Since this sequence converges, it's a Cauchy sequence and the displayed formula we got from the boundedness yields that $x_n$ is a Cauchy sequence as well. Since $H$ is complete, there exists some $x\in H$ such that $x_n$ converges to $x$. Now closedness of $T$ yields that $x$ is in the domain of $T$, and furthermore $(T-\lambda)x=y$, i.e. $y$ is in the range of $T-\lambda$, a contradiction.

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    $\begingroup$ To point two in 1. how do you know that the resolvent cannot be bounded? $\endgroup$ Feb 4 '14 at 17:06
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    $\begingroup$ In short, because this would contradict the closedness of $T$. In long, see my recent edit. :) Btw: Personally, I wouldn't call $(T-\lambda)^{-1}$ the resolvent unless it's bounded and everywhere defined. $\endgroup$
    – Roland
    Feb 4 '14 at 18:22

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