English (United Kingdom)

https://curated-unify.zendy.io/wp-json/zendy-region/v1/featured_content/oa?rat=en

https://curated-unify.zendy.io/wp-json/zendy-region/v1/highlighted_journal/

Global: Zendy Plus annual

Presents the access of premium content as premium feature

Premium Content

Presents the keyphrase highlighting as premium feature

Keyphrase Highlighting

Presents the summarisation as premium feature

Summarisation

Insights

Presents the pdf analysis as premium feature

PDF Analysis

Presents the zaia usage as premium feature

ZAIA

Global: Zendy Plus monthly

Global: Zendy Tools annual

Global: Zendy Tools monthly

Global: Zendy Free Plan

We give a characterization of commutative semispectral measures by means ofFeller and Strong Feller Markov kernels. In particular:  {itemize} we show that a semispectral measure $F$ is commutative if and onlyif there exist a self-adjoint operator $A$ and a Markov kernel$\mu_{(\cdot)}(\cdot):\Gamma\times\mathcal{B}(\mathbb{R})\to[0,1]$,$\Gamma\subset\sigma(A)$, $E(\Gamma)=\mathbf{1}$, such that$$F(\Delta)=\int_{\Gamma}\mu_{\Delta}(\lambda)\,dE_{\lambda},$$ \noindent and$\mu_{(\Delta)}$ is continuous for each $\Delta\in R$ where,$R\subset\mathcal{B}(\mathbb{R})$ is a ring which generates the Borel$\sigma$-algebra of the reals $\mathcal{B}(\mathbb{R})$. Moreover,$\mu_{(\cdot)}(\cdot)$ is a Feller Markov kernel and separates the points of$\Gamma$. we prove that $F$ admits a strong Feller Markov kernel$\mu_{(\cdot)}(\cdot)$, if and only if $F$ is uniformly continuous. Finally, weprove that if $F$ is absolutely continuous with respect to a regular finitemeasure $\nu$ then, it admits a strong Feller Markov kernel. {itemize} Themathematical and physical relevance of the results is discussed giving aparticular emphasis to the connections between $\mu$ and the imprecision of themeasurement apparatus.

Semispectral Measures and Feller markov Kernels