I have absolutely no idea about Spectral theory and want to ask some fundamental questions.
1.) What does it mean that the resolvent of an operator is Hilbert-Schmidt? (Cause I saw a theorem that was like: The resolvent of our self-adjoint operator is Hilbert-Schmidt, hence all eigenvalues are purely discrete and the eigenfunctions are simple and form an orthonormal basis). I don't see how this Hilbert-Schmidt is related to any of the other things.
2.) Can we always represent the resolvent of a self-adjoint operator with a Hilbert-Schmidt integral operator and is the Hilbert Schmidt integral kernel the same as the Green's function?
If anything is unclear, please let me know.
A bounded linear operator $A$ on a separable Hilbert space $X$ with orthonormal basis $\{ e_{j} \}_{j=1}^{\infty}$ is a Hilbert-Schmidt operator if $$ \sum_{j=1}^{\infty}\|Ae_{j}\|^{2} < \infty. $$ This condition is true for one orthonormal basis iff it is true for every other orthonormal basis.
If you're studying $X=L^{2}[a,b]$, then an integral operator $$ (A f)(x) = \int_{a}^{b} K(x,y)f(y)\,dy $$ is Hilbert-Schmidt iff $$ \int_{a}^{b}\int_{a}^{b}|K(x,y)|^{2}\,dxdy < \infty. $$ To see that this is the case, let $\{ e_{j}(x) \}_{j=1}^{\infty}$ be an orthonormal basis of $L^{2}[a,b]$, and assume that $\{ e_{j}(x)\overline{e_{k}(y)}\}_{j,k}$ is an orthonormal basis of $L^{2}([a,b]\times[a,b])$. Then Parseval's identity applied to both $L^{2}[a,b]$ and $L^{2}([a,b]\times[a,b])$ gives $$ \begin{align} \sum_{j}\|Ae_{j}\|^{2} & = \sum_{j}\sum_{k}|(Ae_{j},e_{k})|^{2} \\ & = \sum_{j,k}|(K,e_{j}(x)\overline{e_{k}(y)})_{L^{2}([a,b]\times[a,b])}|^{2} \\ & = \|K\|^{2}_{L^{2}([a,b]\times[a,b])} \\ & = \int_{a}^{b}\int_{a}^{b}|K(x,y)|^{2}\,dx\,dy. \end{align} $$ This generalizes to infinite intervals and various measures.
Hilbert-Schmidt operators are compact, which gives you a lot of information about their spectra. Not all resolvents of selfadjoint operators are Hilbert-Schmidt. The resolvent definitely can't be Hilbert-Schmidt if the operator has an interval of continuous spectrum. The classical example of continuous spectrum for a differential operator is on $L^{2}(\mathbb{R})$, with $$ Lf = \frac{1}{i}\frac{d}{dx}f = -if'(x), $$ where $\mathcal{D}(L)$ consists of all absolutely continuous functions on $\mathbb{R}$ with $f$, $f' \in L^{2}(\mathbb{R})$. The resolvent is represented as a Green function $G(x,y)$ which is not square-integrable on $\mathbb{R}\times\mathbb{R}$. The resolvent $(\lambda I-L)^{-1}$ of $L$ exists as a bounded linear operator on $L^{2}(\mathbb{R})$ iff $\lambda \notin\mathbb{R}$. To find the resolvent, let $f\in L^{2}(\mathbb{R})$ be given, and notice that $g=(\lambda I -L)^{-1}f$ is the unique solution of $$ \lambda g(x) -\frac{1}{i}g'(x) =f(x),\;\;\; g \in L^{2}, g' \in L^{2}\\ g'-i\lambda g = -if \\ (e^{-i\lambda t}g(t))'=-ie^{-i\lambda t}f(t). $$ The unique solution of this equation is $$ (\lambda I-L)^{-1}f = g(x) = \left\{\begin{array}{ll} -i\int_{-\infty}^{x}e^{i\lambda(x-t)}f(t)\,dt, & \Im\lambda > 0 \\ i\int_{x}^{\infty}e^{i\lambda(x-t)}f(t)\,dt, & \Im\lambda < 0. \end{array}\right. $$ To see that the above defines a bounded $(\lambda I-L)^{-1}$, consider $\Im\lambda > 0$ first. Then $$ \begin{align} \int_{-\infty}^{\infty}|(\lambda I-L)^{-1}f|^{2}\,dx & \le \int_{-\infty}^{\infty}\left(\int_{-\infty}^{x}e^{-\Im\lambda(x-t)}|f(t)|\,dt\right)^{2}\,dx \\ & \le \int_{-\infty}^{\infty}\int_{-\infty}^{x}e^{-\Im\lambda(x-t)}\,dt \int_{-\infty}^{x}e^{-\Im\lambda(x-t)}|f(t)|^{2}\,dt\,dx \\ & \le \frac{1}{\Im\lambda}\int_{-\infty}^{\infty}\int_{-\infty}^{x}e^{-\Im\lambda(x-t)}|f(t)|^{2}\,dt\,dx \\ & = \frac{1}{\Im\lambda}\int_{-\infty}^{\infty}\int_{t}^{\infty} e^{-\Im\lambda(x-t)}\,dx\,|f(t)|^{2}dt \\ & = \frac{1}{(\Im\lambda)^{2}}\int_{-\infty}^{\infty}|f(t)|^{2}\,dx =\frac{\|f\|^{2}}{(\Im\lambda)^{2}}. \end{align} $$ Eventually you find that $\|(\lambda I-A)^{-1}\| \le 1/|\Im\lambda|$ for $\lambda\notin\mathbb{R}$.
This operator $(\lambda I-A)^{-1}$ is bounded, it comes from a Green function, but the Green function is not Hilbert-Schmidt. For $\Im\lambda > 0$, $$ G(x,t) = ie^{i\lambda(x-t)}H(x-t), $$ where $H$ is the Heaviside step function. This is not square integrable on $\mathbb{R}^{2}$ because $|G(x,t)|^{2} = e^{-2\Im\lambda(x-t)}H(x-t)$.