Tensor products, multiplications and Weyl's theorem
Let \(T\in B(\mathcal X)\) be a bounded linear operator on a Banach space. Let \(\sigma(T)\), \(\sigma_w(T)\), \(\pi_0(T)\) and \(\pi_{00}(T)\) denote the spectrum, the Weyl spectrum, the set of all Riesz points, and the set of all isolated eigenvalues of \(T\), respectively. \(T\) satisfies Weyl's theorem (briefly, \(T\in(W)\)) if \(\sigma(T)\setminus\sigma_w(T)=\pi_{00}(T)\) and it satisfies Browder's theorem if \(\sigma(T)\setminus\sigma_w(T)=\pi_{0}(T)\). Now, let \(T_i\in B(\mathcal X_i)\), \(i=1,2\), \(T_1\in B(\mathcal X_1)\), \(T_2\in B(\mathcal X_2)\) and \(Z\in B(B(\mathcal X_1,\mathcal X_2))\) be either \(T_1\otimes T_2\) or the multiplication operator \(Z(X)=T_2 X T_1\). In the present paper, it is proved that if (i)~\(T_i\) has the single valued extension property at points \(\lambda \notin\sigma_w(T_i)\) and (ii)~\(T_i\) is of Kato type at isolated points of \(\sigma(T_i)\), then both \(Z\in (W)\) and its conjugate operator \(Z^*\in(W)\). This result is interesting since it is known that (i) and (ii) imply \(T_i\in (W)\), but \(T_1\in(W)\) and \(T_2\in(W)\) do not imply \(Z\in(W)\).
scientific article; zbMATH DE number 5121528
scientific article