An overview of some recent developments on the invariant subspace problem (Q2851039)

According to Paul Halmos, ``one of the most important, most difficult and most exasperating unsolved problems of operator theory is the problem of invariant subspaces'' ([\textit{P. R. Halmos}, A Hilbert space problem book. 2nd ed. New York--Heidelberg--Berlin: Springer-Verlag (1982; Zbl 0496.47001)], Problem 191).NEWLINENEWLINEThe paper under review presents an overview of some recent developments on the invariant subspace problem with which the authors have been concerned.NEWLINENEWLINEThe authors provide in Section~1 a brief historical account of results related to the invariant subspace problem.NEWLINENEWLINEIn Section~2, they consider the issue of universal operators. A~bounded linear operator \(U\) on a Hilbert space \({\mathcal H}\) is said to be \textit{universal} if, for each non-zero bounded linear operator \(T\) on \({\mathcal H}\), there is a scalar \(\lambda \neq 0\) and an invariant subspace \({\mathcal M}\) for \(U\) such that the restriction \(U|_{{\mathcal M}}\) is similar to \((\lambda T)\).NEWLINENEWLINEThere is a strong connection between universal operators and the invariant subspace problem: every bounded linear operator on \({\mathcal H}\) has a non-trivial invariant subspace if and only if the minimal non-trivial invariant subspaces of a given universal operator are all one-dimensional.NEWLINENEWLINE\textit{R. S. Caradus} [Proc. Am. Math. Soc. 23, 526--527 (1969; Zbl 0186.19204)] showed that, if a bounded linear operator \(U\) on \({\mathcal H}\) satisfies that (1) \(\ker U\) is infinite-dimensional and (2) \(U\) is surjective, then \(U\) is universal. The authors apply this criterion to test universality for the adjoint of a composition operator on the Hardy space. Let \(\varphi : \mathbb D \to \mathbb D\) be a holomorphic map and consider the composition operator \(C_\varphi\) defined on the Hardy space \(H^2(\mathbb D)\) by the expression \(C_\varphi f = f \circ \varphi\) for all \(f \in H^2(\mathbb D)\). Define a measure \(\mu_\varphi\) on the Borel subsets of \(\mathbb T\) by the expression \(\mu_\varphi (E)=m(\varphi^{-1}(E)),\) where \(m\) is the Lebesgue measure. Then \(\mu_\varphi\) is absolutely continuous with respect to \(m\) and we may consider the Radon-Nikodým derivative NEWLINE\[NEWLINE g_\varphi = \frac{d\mu_\varphi}{dm}. NEWLINE\]NEWLINE The authors prove the apparently new result that, if \(g_\varphi\) is essentially bounded away from zero and \(\varphi\) is not univalent, then \(C_\varphi^\ast\) is universal. The result applies to any finite Blaschke product \(\varphi\) of degree \((\geq 2)\). The rest of Section 2 surveys several results regarding the universality of operators of the form \(T-\lambda I\), where \(T\) is a composition operator on the Hardy space or \(T\) is a bilateral weighted shift.NEWLINENEWLINEIn Section~3, the authors consider Bishop operators. Let \(0<\alpha<1\). The \textit{Bishop operator} \(T_\alpha\) is defined on the Hilbert space \(L^2[0,1]\) by the expression NEWLINE\[NEWLINE (T_\alpha f)(t) = t f(\{t+\alpha\}), NEWLINE\]NEWLINE where the curly brackets denote the fractional part of a number. \textit{A. M. Davie} [Bull. Lond. Math. Soc. 6, 343--348 (1974; Zbl 0287.47003)] showed that, for every \(\alpha\) that is not a Liouville number, the Bishop operator \(T_\alpha\) has a non-trivial closed hyperinvariant subspace. Recall that \(\alpha\) is a Liouville number if for each \(n\) there exist \(p\) and \(q\) such that \(q>1\) and \(|\alpha -p/q| < q^{-n}\). It is an open question the existence of invariant subspaces for \(T_\alpha\) when \(\alpha\) is a Liouville number.NEWLINENEWLINEIn Section~4, the authors consider finitely strictly singular operators on Banach spaces. Let \((X,Y)\) be two Banach spaces. A bounded linear operator \(T : X \to Y\) is said to be \textit{strictly singular} if, for each \(\varepsilon >0\) and for each infinite-dimensional subspace \(Z \subseteq X\), there is a vector \(z \in Z\) with \(\|z\|=1\) and \(\|Tx\| < \varepsilon\). A bounded linear operator \(T : X \to Y\) is said to be \textit{finitely strictly singular} if, for each \(\varepsilon >0\), there is an \(N \in \mathbb N\) such that, for every subspace \(Z \subseteq X\) with \(\dim X \geq N\), there is a \(z \in Z\) with \(\|z\|=1\) and \(\|Tx\| < \varepsilon\). The following implications hold: NEWLINE\[NEWLINE T \text{ compact} \Rightarrow T \text{ finitely strictly singular} \Rightarrow T \text{ strictly singular.} NEWLINE\]NEWLINE A.\,Pełczyński posed the question of the existence of invariant subspaces for strictly singular operators, inspiring \textit{C. J. Read} to investigate and answer this question negatively by constructing a stricly singular operator without non-trivial invariant subspaces [Stud. Math. 132, No. 3, 203--226 (1999; Zbl 0929.47004)]. The existence of invariant subspaces for finitely strictly singular operators remained an open problem. The authors of the present paper provide a negative solution to this problem by showing that Read's strictly singular operator is, in fact, finitely strictly singular.