1
vote
1answer
48 views

Integral representation of joint projection valued measures

Given two positive $\sigma$-finite measures $\mu_{1/2}$ on the spaces $X_{1/2}$ one can define the product measure $\mu_1\otimes\mu_2$ on the product space $X_1\times X_2$. It can be proved that the ...
0
votes
2answers
279 views

Spectral decomposition of compact operators

Let T be a compact operator on an infinite dimensional hilbert space H. I am proving the theorem which says that $Tx=\sum_{n=1}^{\infty}{\lambda}_{n}\langle x,x_{n}\rangle y_{n}$ where ($x_{n}$) is an ...
0
votes
1answer
94 views

Does spectral theory assume separability

On an infinite dimensional space, the spectral theorem for compact normal operators says that the eigenvectors form an orthonormal basis which, from wikipedia, is equivalent to the space being ...
6
votes
1answer
239 views

A doubt about the parts of the spectrum of tensor products

Let $\mathcal{H}$ be any complex Hilbert space of infinite dimensional. By an operator $T$ I mean a linear bounded transformation from $\mathcal{H}$ into $\mathcal{H}$, i.e, ...
1
vote
1answer
141 views

Weyl asymptotics vs. form perturbations

Consider Hilbert spaces $V$,$H$; a closed quadratic form $a$ with domain $V$; and its associated operator $A$ on $H$. (If necessary, the form can be assumed to be coercive.) For the sake of ...
1
vote
2answers
353 views

Lebesgue integral with respect to vector measures?

Good evening, I'm reading some papers of Jim Agler and Nicholas Young, in which they prove a formula of integral representation with respect to a vector measure, but the integration is in the sense ...
2
votes
1answer
379 views

orthonormal basis of eigenvectors for laplacian on a concave polygon

I am interested in the Laplace operator $\Delta$ on a concave polygon. When the polygon is convex, it is known that $\Delta: H^2(\Omega) \rightarrow L^2(\Omega)$ is boundedly invertible. In addition, ...
9
votes
8answers
4k views

Can a self-adjoint operator have a continuous set of eigenvalues?

This should be a trivial question for mathematicians but not for typical physicists. I know that the spectrum of a linear operator on a Banach space splits into the so-called "point," "continuous" ...
2
votes
3answers
1k views

Infinite hermitian matrix

Suppose we have a finite square n x n matrix of complex numbers H that is Hermitian and skew-symmetric: $H^\dagger = H$ and $H^T = -H$. (T denotes transpose, $\dagger$ denote conjugate transpose. I ...