I'm trying to work my way up through various definitions in order to understand the formulation of the spectral theorem for unbounded operators, in which figure projection valued measures. I'm having a hard time, since I've only had an introductory course to functional analysis where the only version of the spectral theorem we touched upon was for compact self adjoint operators.
My first task is then to understand what the "spectrum" of an unbounded operator is. The definition I am working with is as the complement of the resolvent set, defined in the following way
Let $A: D_A\rightarrow H$ be a linear operator defined in a subset $D_A$ of a Hilbert (or at least Banach) space $H$. We construct the map $$A_{z}:=A-z \, \text{id}_H$$ for some $z\in \mathbb{C}$. A complex number is said to be regular if: 1. $A_z$ is injective, 2. Its inverse is bounded, 3. Its inverse is densely defined.
Then the resolvent set is the set of regular complex numbers
My question is: how do these conditions define the spectrum as a suitable generalization of what is usually meant by spectrum of a compact operator? What I mean is this:
Imagine I were a mathematician discovering the fact that bounded, or unbounded, anyway, non compact, operators may not have eigenvalues, but I still want to be able to speak of a notion closely related to that, for these operators. Why do I choose these conditions? The first one seems natural to me. Even if it's not guaranteed, some of these operators may have eigenvalues in the usual sense, which has to do with the fact that the resolvent function is not injective. But what about condition 2? What does the unboundedness of the inverse of that operator tell me about its associated $z$, in relation to $A$, that makes it somehow a generalisation of an eigenvalue?
My intuition tells me that these conditions must capture complex numbers that are "close" to being eigenvalues, in some sense, (which would be what I would try to do if I were trying to come up with the notion of spectrum), and the wikipedia page of the spectral theorem seems to corroborate this general idea. I, however, can't see how, so I was hoping someone could help me!
I realize similar questions have been asked here. I've seen this and this one. But while the questions are in the same spirit of mine, I don't think the answers are in the same spirit of the answers I am looking for.
Best Answer
Firstly the spectrum (at least for bounded operators) is usually introduced before compact operators are analysed and then it is shown that for a compact operator every non zero element of the spectrum is an eigenvalue. So i am not sure what exactly you mean by:
Furthermore i have trouble with the statement:
Compact operators do not need to have eigenvalues either. Consider the Banach space $C[0,1]$ and the operator $T: C[0,1]\to C[0,1]$ defined by $T(f)(s) = \int_0^s f(t) dt$. Then $T$ is compact and does not have any eigenvalues. This operator has spectrum (as i define it below) $\sigma (T) =\{ 0 \}$, but $0$ is not an eigenvalue.
For the remainder i will discuss why the spectrum is a good generalization for the set of eigenvalues of an operator in finite dimensions.
Spectrum For Bounded Operators
Let $X$ be a Banach space and (for simplicity) consider a bounded operator $T: X \to X $.
Define the resolvent as $$\rho(T) = \{ \lambda \in \mathbb{C} : T- \lambda I \ \text{is invertible in} \ L(X) \},$$ where $L(X)$ denotes the bounded and linear operators. Then the spectrum is defined as $\sigma (T) = \mathbb{C} \setminus \rho (T)$.
Because of the open mapping theorem it is true that $T- \lambda I$ is bijective $\Rightarrow$ $T- \lambda I$ is invertible in $L(X)$ (has bounded inverse).
Therefore $\lambda \in \sigma(T)$ if and only if one (or both) of the following two statements are true:
Even in finite dimensions we study the invertibility of $T-\lambda I$ via the determinant when finding eigenvalues for example. Just in finite dimensions not invertible $\Leftrightarrow$ not injective.
So the spectrum really just includes the "new" case in infinite dimensions. Of course we dont know yet that this is the "right" generalization of the eigenvalues (at least for self-adjoint/normal operators on a Hilbert space) yet.
Spectrum For Unbounded Operators
Let $\mathcal{H}$ be a Hilbert space and $T :\mathcal{H} \supset \mathrm{dom} (T) \to \mathcal{H} $ a densely defined operator.
Define the resolvent by $$\rho(T) = \{ \lambda \in \mathbb{C} : T- \lambda I : \mathrm{dom} (T) \to \ \mathcal{H} \ \text{is bijective and} \ (T- \lambda I )^{-1} \in L(\mathcal{H}) \}$$ and the spectrum by $\sigma (T) = \mathbb{C} \setminus \rho(T) $.
The definition of the resolvent is a special case of your definition.
Now if $T$ is closed (for example always true if $T$ is self-adjoint), then by the open mapping theorem: $T- \lambda I $ bijective $\Rightarrow (T- \lambda I )^{-1} \in L(\mathcal{H})$. Therefore the meaning of the spectrum is then exactly as above in the bounded case.
Properties Of Spectrum For Self Adjoint Operators
The first signs that show us that the spectrum is perhaps the right generalization of the eigenvalues for a self-adjoint operator on a Hilbert space $T: \mathcal{H} \supset \mathrm{dom} (T) \to \mathcal{H}$ are the following two propositions (see i.e. Hall Quantum Theory For Mathematicians Theorem 9.17 and 9.18):
The first one tells us that the spectrum is real, just as the eigenvalues are real for a self-adjoint operator in finite dimensions. The second tells us that every member of the spectrum is an approximate eigenvalue, because for every $\varepsilon >0$ there exist a unit vector $\psi \in \mathrm{dom} (T)$ so that $\| T\psi - \lambda \psi \| < \varepsilon$. So $\psi$ is almost an eigenvector. This also affirms your intuition.