Are there some criteria to tell when the point spectrum of a linear operator is discrete? In general it is not the same (take the spectrum of the "annihilation" operator). More specifically, what are the conditions that should be satisfied by a symmetric (or even self-adjoint) operator to have "point spectrum" = "discrete spectrum"?
[Math] When the point spectrum is discrete
functional-analysisspectral-theory
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Yes. Let $f(z) = z + z^{-1} - 2$. Since $\Delta = f(S)$, $\sigma(\Delta) = f(\sigma(S))$. And yes, $\sigma(S) = \{z: |z| = 1\}$. Now note that if $z = e^{i\theta}$, $f(z) = 2 \cos(\theta) - 2$, so $\sigma(\Delta) = [-4, 0]$.
The spectrum of $\Delta$ is all continuous: it is easy to see that $\Delta$ has no eigenvalues in $\ell^2({\mathbb Z})$. In fact, it is absolutely continuous, and this follows from the fact that the inverse image under $f$ of any set of measure 0 in $\mathbb R$ has measure 0 in the unit circle.
Question 1: If $\lambda$ is in the continuous spectrum, then $R_{\lambda}(T)$ is unbounded. That means there exists a sequence $\{ x_n \}$ such that $\|x_n\|=1$ and $\|R_{\lambda}(T)x_n\|\rightarrow\infty$. So, $y_n=\frac{1}{\|R_{\lambda}(T)x_n\|}R_{\lambda}(T)x_n$ is a sequence of unit vectors in $X$ such that $(T-\lambda I)y_n\rightarrow 0$ as $n\rightarrow\infty$. It is typical to refer to this sequence $\{ y_n \}$ of unit vectors as an "approximate eigenvector."
The reason one refers to this as "continuous spectrum" Historically had nothing to do with continuity; such spectrum was found to fill a continuum, rather than being discrete. For example, the multiplication operator $(Mf)(x)=xf(x)$ defined on $L^2[0,1]$ has no eigenvalues, but $[0,1]$ is in the continuous spectrum of $M$. You can see this because the step function $f_{\lambda,\delta}=\frac{1}{\sqrt{2\delta}}\chi_{[\lambda-\delta,\lambda+\delta]}$ is a unit vector for which $$ Mf_{\lambda,\delta} \approx \lambda f_{\lambda,\delta}, $$ in the sense that $$ \|(M-\lambda I)f_{\lambda,\delta}\|^2=\int_{\lambda-\delta}^{\lambda+\delta}(x-\lambda)^2f_{\lambda,\delta}^2dx \le 2\delta^2\|f_{\lambda,\delta}\|^2. $$ So $\|(M-\lambda I)f_{\lambda,\delta}\|\rightarrow 0$ as $\delta\rightarrow 0$ while $\|f_{\lambda,\delta}\|=1$.
The same type of spectrum occurs for the differentiation operator $\frac{1}{i}\frac{d}{dx}$ on $L^2(\mathbb{R})$ where you would like to identify $e^{isx}$ as an eigenvector with eigenvalue $s$, but it's not in the space. However, $s$ is in the continuous spectrum, a.k.a. approximate point spectrum. This is one of the earliest cases that motivated these definitions, where you could think of $$ F(x)=\int_{-\infty}^{\infty}c(s)e^{isx}ds $$ as a "continuous" (i.e. integral) sum of approximate eigenfunctions of the differentiation operator, with a coefficient function $c(s)$ defined on the continuum. And, when considered in $L^2$, the operator $\frac{1}{i}\frac{d}{dx}$ has continuous spectrum $\mathbb{R}$.
Question 2: For your second question, you don't care about a dense domain for $R_{\lambda}(T)$ if it is bounded because you can then extended it to the whole space. And, if $T$ is closable, then the extension of $R_{\lambda}(T)$ and the closure of $T$ remain inverses of each other. However, if $\lambda$ is an approximate eigenvalue and $T$ is closed, then $R_{\lambda}(T)$ must be unbounded, and you're basically stuck with a dense domain for $R_{\lambda}(T)$ because of the closed graph theorem.
Best Answer
When dealing with differential operators, the most common condition is where the resolvent operator is compact. Resolvents of differential operators are often compact on bounded domains because they map uniformly bounded sets of continuous functions to sets of functions with uniformly bounded derivatives, which means the image is an equicontinuous set of functions. Some argument of that type is usually given to prove a resolvent is compact. Showing compactness is trickier on infinite domains, and the resolvent need not be compact, which is the case for the simplest differential operator, the momentum operator.
Not all operators with point spectrum have some compact resolvent, though. When considered on $L^{2}$, the Fourier transform has 4 eigenvalues and a complete orthonormal basis of eigenfunctions (the Hermite functions.) If the dimensions of any of the eigenspaces is infinite, then an operator cannot be compact, unless the eigenvalue is $0$. Or, if the eigenvalues cluster somewhere other than at $0$, the operator cannot be compact.
If the resolvent $R(\lambda) = (A-\lambda I)^{-1}$ of an operator $A : \mathcal{D}(A)\subset \mathcal{H}\rightarrow\mathcal{H}$ on a separable infinite-dimensional Hilbert space $\mathcal{H}$ is compact for some $\lambda$, then the eigenvalues cluster only at $0$, which forces the eigenvalues of $A$ to cluster only at $\infty$.