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Suppose I have a self-adjoint pseudo-differential operator $A$ on $\mathbb{R}^n$ and a continuous function $f$ (possibly bounded, or Schwartz, or compactly supported) on its spectrum. Then I can consi …

Let $A$ be a real $n\times n$ matrix and let $\mu_1, \dots, \mu_n$ the (generalized, complex) eigenvalues of $A$. Assume that $$ 0 < \alpha < \mathrm{Re}(\mu_1) < \dots < \mathrm{Re}(\mu_n).$$ I am i …

I have been trying to prove this for a while but failed so far. Let $A$ and $B$ are two positive, self-adjoint operators with compact resolvent on a Hilbert space $H$ defined on the same dense domai …

I learned that there is a holomorphic functional calculus for closed operators: If $T$ is a closed operator on a Hilbert space, and $f$ is a function that is holomorphic on some open subset $\Omega$ o …

The Wikipedia article (http://en.wikipedia.org/wiki/Mathieu_function) is quite extensive, I believe, and in general there is NO way to explicitly calculate either eigenvalues or eigenfunctions of the …

Roughly, the trick is not to view $L$ as an operator on $L^2$, but on $C^0$ I will use the following version of Krein-Rutmann which is proven in "Du, Yihong: Order Structure and Topological Methods i …

I would like to know if there are theorems that state under which circumstances spectra of operator families depend smoothly on the parameter. To clarify, suppose I have a 1-parameter family $T_h$ of …

You just need to check that $$ \int_M H_k(x, y) \mathrm{d}y = 1 + O(t^{k+1})$$ for all $x \in M$ and for all $k$. For example, this follows directly from the method of stationary phase (just take a ge …

The resolvent set is the set of all $\zeta \in \mathbb{C}$ for which $T-\zeta$ is invertible (which means especially that the Range is all of $H$). The spectrum $\Sigma$ is the complement of the resol …

I find some statements your post quite curious. If you look at the Laplace operator on $L^2(-\infty,\infty)$, it has a complete set of generalized eigenvectors, namely $\cos(\sqrt{\lambda}x)$, $\sin …