## Contributed talks session B (Monday)

#### Event Title

Pseudospectra of elements of reduced Banach algebras

#### Location

Cupples I Room 218

#### Start Date

7-18-2016 2:30 PM

#### End Date

18-7-2016 2:50 PM

#### Description

Let $A$ be a Banach algebra with identity $1$. For $\epsilon>0$, the $\epsilon$-pseudospectrum of an element $a\in A$ is defined as $\Lambda_{\epsilon}(A,a)=\sigma(A,a)\cup \{\lambda\in \mathbb{C}:\|(\lambda-a)^{-1}\|\geq \frac{1}{\epsilon}\}$, where $\sigma(A,a)$ denotes the spectrum of $a$ in $A$. Let $p\in A$ be a non-trivial idempotent. Let $q=1-p$. Then $pAp$ and $qAq$ are Banach algebras with identities $p$ and $q$ respectively. These algebras are called reduced Banach algebras. Suppose $a\in A$ such that $ap=pa$. We consider the relationship between the pseudospectrum of $a\in A$ and the pseudospectra of $pap\in pAp$ and $qAq\in qAq$. We show the following: \begin{enumerate} \item $\sigma(A,a)=\sigma(pAp,pap)\cup \sigma(qAq,qaq)$. \item For $\epsilon>0, \Lambda_{\frac{\epsilon}{2}}(A,a)\subseteq\Lambda_{\epsilon}(pAp,pap)\cup\Lambda_{\epsilon}(qAq,qaq)\subseteq \Lambda_{\text{max }\{\|p\|,\|q\|\}\epsilon}(A,a)$. In particular, if $\|p\|=\|q\|=1, \Lambda_{\frac{\epsilon}{2}}(A,a)\subseteq \Lambda_{\epsilon}(pAp,pap)\cup\Lambda_{\epsilon}(qAq,qaq)\subseteq \Lambda_{\epsilon}(A,a)$. \item If $A=B(H)$ for a Hilbert space $H$, and $p\in A$ is an orthogonal projection ($p=p^{*}=p^{2}$), then $\Lambda_{\epsilon}(A,a)=\Lambda_{\epsilon}(pAp,pap)\cup\Lambda_{\epsilon}(qAq,qaq)$. \item If $A=B(X)$ for a Banach space $X$, and $p\in A$ is a projection onto a closed subspace of $X$ such that $\|p\|=\|q\|=1$, then $\Lambda_{\epsilon}(A,a)=\Lambda_{\epsilon}(pAp,pap)\cup\Lambda_{\epsilon}(qAq,qaq)$. \item If $a\in A$ is of $G_{1}$ class, i.e., $\|(\lambda-a)^{-1}\|=\frac{1}{d(\lambda,\sigma(A,a))}\, \forall \lambda\notin \sigma(A,a)$, and $\|p\|=\|q\|=1$, then $\Lambda_{\epsilon}(A,a)=\Lambda_{\epsilon}(pAp,pap)\cup\Lambda_{\epsilon}(qAq,qaq)$. \item Suppose $p_{1},\cdots, p_{n}$ are idempotents in $A$ such that $ap_{i}=p_{i}a \, \forall i$ and ${\sum_{i=1}^{n} p_{i}=1}$. Then $\sigma(A,a)=\cup_{i=1}^{n} \sigma(p_{i}Ap_{i},p_{i}ap_{i})$ and $\Lambda_{\frac{\epsilon}{n}}(A,a)\subseteq\cup_{i=1}^{n} \Lambda_{\epsilon}(p_{i}Ap_{i},p_{i}ap_{i})\subseteq \Lambda_{\underset{i}{\text{max}}\{\|p_{i}\|\}\epsilon}(A,a)$. \end{enumerate} The talk is based on joint work with S. H. Kulkarni.

#### Share

COinS

Jul 18th, 2:30 PM Jul 18th, 2:50 PM

Pseudospectra of elements of reduced Banach algebras

Cupples I Room 218

Let $A$ be a Banach algebra with identity $1$. For $\epsilon>0$, the $\epsilon$-pseudospectrum of an element $a\in A$ is defined as $\Lambda_{\epsilon}(A,a)=\sigma(A,a)\cup \{\lambda\in \mathbb{C}:\|(\lambda-a)^{-1}\|\geq \frac{1}{\epsilon}\}$, where $\sigma(A,a)$ denotes the spectrum of $a$ in $A$. Let $p\in A$ be a non-trivial idempotent. Let $q=1-p$. Then $pAp$ and $qAq$ are Banach algebras with identities $p$ and $q$ respectively. These algebras are called reduced Banach algebras. Suppose $a\in A$ such that $ap=pa$. We consider the relationship between the pseudospectrum of $a\in A$ and the pseudospectra of $pap\in pAp$ and $qAq\in qAq$. We show the following: \begin{enumerate} \item $\sigma(A,a)=\sigma(pAp,pap)\cup \sigma(qAq,qaq)$. \item For $\epsilon>0, \Lambda_{\frac{\epsilon}{2}}(A,a)\subseteq\Lambda_{\epsilon}(pAp,pap)\cup\Lambda_{\epsilon}(qAq,qaq)\subseteq \Lambda_{\text{max }\{\|p\|,\|q\|\}\epsilon}(A,a)$. In particular, if $\|p\|=\|q\|=1, \Lambda_{\frac{\epsilon}{2}}(A,a)\subseteq \Lambda_{\epsilon}(pAp,pap)\cup\Lambda_{\epsilon}(qAq,qaq)\subseteq \Lambda_{\epsilon}(A,a)$. \item If $A=B(H)$ for a Hilbert space $H$, and $p\in A$ is an orthogonal projection ($p=p^{*}=p^{2}$), then $\Lambda_{\epsilon}(A,a)=\Lambda_{\epsilon}(pAp,pap)\cup\Lambda_{\epsilon}(qAq,qaq)$. \item If $A=B(X)$ for a Banach space $X$, and $p\in A$ is a projection onto a closed subspace of $X$ such that $\|p\|=\|q\|=1$, then $\Lambda_{\epsilon}(A,a)=\Lambda_{\epsilon}(pAp,pap)\cup\Lambda_{\epsilon}(qAq,qaq)$. \item If $a\in A$ is of $G_{1}$ class, i.e., $\|(\lambda-a)^{-1}\|=\frac{1}{d(\lambda,\sigma(A,a))}\, \forall \lambda\notin \sigma(A,a)$, and $\|p\|=\|q\|=1$, then $\Lambda_{\epsilon}(A,a)=\Lambda_{\epsilon}(pAp,pap)\cup\Lambda_{\epsilon}(qAq,qaq)$. \item Suppose $p_{1},\cdots, p_{n}$ are idempotents in $A$ such that $ap_{i}=p_{i}a \, \forall i$ and ${\sum_{i=1}^{n} p_{i}=1}$. Then $\sigma(A,a)=\cup_{i=1}^{n} \sigma(p_{i}Ap_{i},p_{i}ap_{i})$ and $\Lambda_{\frac{\epsilon}{n}}(A,a)\subseteq\cup_{i=1}^{n} \Lambda_{\epsilon}(p_{i}Ap_{i},p_{i}ap_{i})\subseteq \Lambda_{\underset{i}{\text{max}}\{\|p_{i}\|\}\epsilon}(A,a)$. \end{enumerate} The talk is based on joint work with S. H. Kulkarni.