#### Event Title

#### Location

Cupples I Room 115

#### Start Date

7-21-2016 5:00 PM

#### End Date

21-7-2016 5:20 PM

#### Description

We say that a domain $D$ in the complex plane is a spectral set for $T$, a bounded operator on a Hilbert space $H$, if a form of von Neumann's inequality holds; that is, if for any rational function $f$ with poles off of $D$, $\|f(T)\| \leq \|f\|$, where the left hand norm is the operator norm of $f(T)$ and the right hand norm is the supremum norm of $f$ over $D$. It is called a complete spectral set if the inequality also holds for $f$ matrix valued for any size square matrix. The classical von Neumann inequality says that the unit disk is a (complete) spectral set for any operator with its spectrum in this set. If one tries to extend this result to more complicated domains, the attempt often fails. One potential way around this is to relax the inequality by requiring instead that $\|f(T)\| \leq K \|f\|$ for a constant $K$. Following in the footsteps of Mascioni, Stessin, Stampfli, Badea-Beckerman-Crouzeix and others, we consider planar domains that are comprised of the intersection of level sets of moduli of complex functions under certain regularity conditions. We also consider applications to a variation of the rational dilation problem. The key tool is our work on the relationship between the generation of uniform algebras of analytic functions in planar domains and the extension to the polydisk of a bounded analytic function defined on an analytic variety inside the polydisk. This is joint work with Dmitry Yakubovich and Daniel Est\'evez.

COinS

Jul 21st, 5:00 PM
Jul 21st, 5:20 PM

Tests for complete $K$-spectral sets

Cupples I Room 115

We say that a domain $D$ in the complex plane is a spectral set for $T$, a bounded operator on a Hilbert space $H$, if a form of von Neumann's inequality holds; that is, if for any rational function $f$ with poles off of $D$, $\|f(T)\| \leq \|f\|$, where the left hand norm is the operator norm of $f(T)$ and the right hand norm is the supremum norm of $f$ over $D$. It is called a complete spectral set if the inequality also holds for $f$ matrix valued for any size square matrix. The classical von Neumann inequality says that the unit disk is a (complete) spectral set for any operator with its spectrum in this set. If one tries to extend this result to more complicated domains, the attempt often fails. One potential way around this is to relax the inequality by requiring instead that $\|f(T)\| \leq K \|f\|$ for a constant $K$. Following in the footsteps of Mascioni, Stessin, Stampfli, Badea-Beckerman-Crouzeix and others, we consider planar domains that are comprised of the intersection of level sets of moduli of complex functions under certain regularity conditions. We also consider applications to a variation of the rational dilation problem. The key tool is our work on the relationship between the generation of uniform algebras of analytic functions in planar domains and the extension to the polydisk of a bounded analytic function defined on an analytic variety inside the polydisk. This is joint work with Dmitry Yakubovich and Daniel Est\'evez.