Errata to: “Subalgebras of 𝐵(𝑋)”
Albert Wilansky · Proceedings of the American Mathematical Society · 1972
In Theorem 4 of this paper it is stated that the set of all operators with ascent and descent 0 or 1 is uniformly closed in B(H).This is not true as can be seen by the following example [P.R. Halmos, A Hilbert space problem book, Problem 85].If for each k=\, 2, • • • , Ak is the weighted shift on the Hilbert space of two-way square-summable sequences, with sequence of weights (• • • , 1, 1, Ijk, 1, 1, ■ • ■), then \\Ak-^"J-^O where/I ĥas its sequence of weights (• • • , 1, 1, 0, 1, 1, • ■ •)• Each Ak, being invertible, is of ascent and descent 0 or 1, but A oe is not of ascent 0 or 1, since A^e_l=Aoa(le0)=0 whereas A ODe_1=e09i0.In the proof of Theorem 4 we argue that since R(T*n) = R(Tt2), Iim"_M(x, Tïy)=\[mn^{x, T*2z) = 0.This argument breaks down since the vector z is dependent on n.