# Example of a densly defined positive self adjoint unbounded operator.

I know example of a densely defined positive self-adjont unbounded operator with discrete spectrum. What is the example of a self-adjoint positive unbound operator with continuous spectrum?

The operator $L=\frac{1}{i}\frac{d}{dt}$ defined on the domain $\mathcal{D}(L)$ consisting of periodic absolutely continuous functions $f$ on $[-\pi,\pi]$ with $f' \in L^2[-\pi,\pi]$ is a densely defined linear operator with discrete spectrum. The eigenvalues are the integers, and $\{ \frac{1}{\sqrt{2\pi}}e^{inx}\}$ is an orthonormal basis of eigenvectors.

The same operator on the $L^{2}(\mathbb{R})$ has continuous spectrum. That is, let $\mathcal{D}(L)$ consist of all absolutely continuous functions $f$ on $\mathbb{R}$ for which $f,f' \in L^2(\mathbb{R})$. Then $L$ is a densely-defined selfadjoint linear operator with $\sigma(L)=\mathbb{R}$ consisting only of continuous spectrum. And $L^2$ is positive, with only absolutely continuous spectrum.

• It is not positive operator @TrialAndError
– Jana
Commented Nov 30, 2015 at 16:57
• @AjoyJana : I added a comment for you: $L^2$ is positive and has only absolutely continuous spectrum, where $\mathcal{D}(L^2)=\{ f \in \mathcal{D}(L) : Lf \in \mathcal{D}(L) \}$. Commented Nov 30, 2015 at 16:59
• @AjoyJana : Now you know how I chose my name. :) Commented Nov 30, 2015 at 17:02
• Yes. @TrialAndError
– Jana
Commented Nov 30, 2015 at 17:04

One classical example is $-\Delta$ with the Hilbert space $L^2(\mathbf R)$. Take $T:(u,v)\mapsto \displaystyle\int_{\mathbf R}u' v' \mathrm d \lambda$ where $u,v\in H^1(\mathbf R)$, you can define $-\Delta$ as the operator associated to the bilinear form $T$. Then $-\Delta$ is densely defined, positive and self adjoint.

Now for the spectrum, take $\lambda\in \mathbf C^*$ and $\chi_n$ some nice cut-off functions such that $\chi_n=1$ on $[-n;n]$ and $0$ on $]-\infty;-n-1]\cup[n+1;+\infty[$. Now the functions $\psi_n:x\mapsto \sin(\lambda x)\chi_n(x)$ are "almost" eigenfunctions of $-\Delta$. That is, $\|\psi_n\|\rightarrow \infty$ when $n$ goes to infinity but $\|(-\Delta-\lambda^2\mathrm {Id})\psi_n\|$ stay bounded, so $\lambda^2$ is in the spectrum of $-\Delta$.

This worked because $\mathbf R$ is not bounded, but the laplacian over a riemannian compact manifold (with boundary or not) has a discrete spectrum.