# Tagged Questions

**0**

votes

**2**answers

270 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

**1**answer

90 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 ...

**1**

vote

**1**answer

140 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

**2**answers

345 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

**1**answer

372 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

**8**answers

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

**3**answers

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 ...