Orthogonality of compact operators (Q2407567)

Let \(X,Y\) be real Banach spaces. We recall that, for \(x,y \in X\), \(x\) is Birkhoff-James orthogonal to \(y\) (write \(x \perp_B y\)) if \(\|x+\lambda y\| \geq \|x\|\) for all scalars \(\lambda\). Let \(T, S: X \rightarrow Y\) be bounded linear operators such that \(\|T(x)\|=\|T\|\) for a unit vector \(x\). Suppose \(T(x) \perp_B S(x)\). Then, for any scalar \(\lambda\), \(\|T+ \lambda S\| \geq \|T(x)+\lambda S(x)\|\geq \|T(x)\| = \|T\|\) and thus \(T \perp_B S\). This paper answers the converse question when \(X\) is reflexive and \(Y\) a smooth and strictly convex reflexive space, under the additional hypothesis that the set of points where \(T\) attains its norm is either connected or has only two points. This improves a result of \textit{R. Bhatia} and \textit{P. Šemrl} [Linear Algebra Appl. 287, No. 1--3, 77--85 (1999; Zbl 0937.15023)], proved in the case of finite-dimensional Hilbert spaces. There are misprints on page 89. There is some overlap with the work of \textit{P. Ghosh} et al. [Linear Algebra Appl. 500, 43--51 (2016; Zbl 1350.46014)].