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.

  • $\begingroup$ 1) I seem to recall that these matters are discussed in one of the volumes of Reed ans Simon. 2) Many self-adjoint operators do not have an associated Hilbert-Schmidt integral kernen, in particular those with purely continuous spectrum. Hilbert-Schmidt kernels are indeed an example of a Green's function but there are other operators with an associated Green's function. $\endgroup$ – Urgje Jul 22 '14 at 14:29
  • $\begingroup$ In my opinion the point is that the resolvent of an operator is Hilbert-Schmidt if the Green's function is square integrable over the product space. $\endgroup$ – user38355 Jul 22 '14 at 14:32
  • $\begingroup$ I see, so thanks to both of you. you almost completely answered the question, if anybody of you could explain now how the existence of this Hilbert Schmidt operator representation is related to the spectrum, you probably could post this as an answer. Unfortunately, I am not at all an expert in Functional Analysis,so it is actually to tough for me to read about this in Reed/Simon. Sorry, this has nothing to do with laziness, but it would take me about a week or two, to get there. $\endgroup$ – user159356 Jul 22 '14 at 14:37

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 $|G(x,t)|^{2} = e^{-2\Im\lambda(x-t)}H(x-t)$.

  • $\begingroup$ an amazing answer. by the way, I currently try to write down a few questions that I encounter in our chatroom, so that I don't forget about them, so don't be confused about this. $\endgroup$ – user159356 Jul 23 '14 at 10:46
  • 1
    $\begingroup$ I corrected a wrong sign in the formula for $(\lambda I -A)^{-1}f$. The minus had to be in front of the part with $\Im\lambda > 0$. I found out because I wanted to make this point: if you integrate $\frac{1}{2\pi i}\int (\lambda I-A)^{-1}f\,d\lambda$ counterclockwise around a closed rectangle enclosing the real interval $(a,b)$ (omit the vertical sides), and if you let $\Im\lambda \downarrow 0$, you get the spectral measure associated with that interval. This is general. Here you get $(\chi_{[a,b]}f^{\wedge})^{\vee}$. Titchmarsh proved $f=(f^{\wedge})^{\vee}$ for general transforms this way. $\endgroup$ – DisintegratingByParts Jul 23 '14 at 11:35
  • $\begingroup$ just another point about the orthogonality: The eigenfunctions to a self-adjoint operator are always orthonormal, right? So, how can I find the appropriate inner-product in general? Because as I indicated in my new question in our chatroom, it appears to be strange sometimes? - But maybe, we should continue this discussion in our chatroom. Thanks again! :-) $\endgroup$ – user159356 Jul 23 '14 at 11:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy