Analytically invariant spectral resolvents of closed operators (Q1084634)

For a closed operator T on a Banach space X with domain \(D_ T\), the spectral resolvent is defined to be a map E:G\(\to Inv.T\), such that \(E(G)\subset D_ T\), if G is relatively compact; for every open subset G of \({\mathbb{C}},\sigma [T| E(G)]\subset \bar G\) and for any \((G_ i)^ n_{i=0}\) cover of \(\sigma\) (T) with \(G_ 0\) a neighborhood of \(\infty\), \(x=\sum^{n}_{i=0}E(G_ i)\), where Inv.T is the lattice of the invariant subspaces of X under T. Also the author defines T-absorbent invariant subspaces, \(\mu\)-spaces of T and analytically invariant subspaces under T, then he proved the following two main results. Theorem 1. For T with a spectral resolvent E, the following are equivalent. a) E is analytically invariant under T. b) for every \(G\subset {\mathbb{C}}\), E(G) is a \(\mu\)-space of T, c) for every \(G\subset {\mathbb{C}}\), E(G) is T-absorbent in G. Theorem 2. For T with a spectral resolvent E, G open in \({\mathbb{R}}\) and any open cover \((G_ i)^ n_{i=0}\) in \({\mathbb{R}}\), of \(\bar G,\) with \(G_ 0\) a neighborhood of \(\infty\), \(G_ i\) is relatively compact (1\(\leq i\leq n)\), one has \(E(G)=\sum^{n}_{i=0}E(G)\cap E(G_ i)\).