On operator factorization of linear relations (Q395603)

Let \(\mathcal X\)\(,\mathcal Y\) both be real or complex vector spaces. A linear relation \(\mathcal A\) from \(\mathcal X\) to \(\mathcal Y\) is a subspace of the Cartesian product \(\mathcal X\times \mathcal Y\) (whose elements are denoted by \((x,y)\), \(x \in X\), \(y \in Y\)). Let \(\mathcal L(\mathcal X,\mathcal Y)\) denote the set of all linear relations from \(\mathcal X\) to \(\mathcal Y\) and \(\mathcal O(\mathcal X,\mathcal Y)\) denote the set of all linear transformations (operators) from \(\mathcal X\) into \(\mathcal Y\). For a linear relation \(\mathcal A\) from \(\mathcal X\) to \(\mathcal Y\), define the domain, range, kernel and the multi-valued part of \(\mathcal A\), respectively, by: \(\operatorname{dom}A=\{x: (x,y) \in \mathcal A\}\), \(\operatorname{ran}A=\{y: (x,y) \in \mathcal A\}\), \(\operatorname{ker}A=\{x: (x,0) \in \mathcal A\}\) and \(\operatorname{mul}A=\{y: (0,y) \in \mathcal A\}\), all of which are subspaces of \(\mathcal X\times \mathcal Y\). The main result of the article under review (generalizing a well known result of \textit{R. G. Douglas} [Proc. Am. Math. Soc. 17, 413--415 (1966; Zbl 0146.12503)]) is the following Theorem. Let \(\mathcal X\), \(\mathcal Y\), \(\mathcal Z\) be vector spaces, \(A\in \mathcal L(\mathcal X,\mathcal Z)\) and \(B \in \mathcal L(\mathcal X,\mathcal Y)\). Then there exists \(T \in \mathcal O(\mathcal X,\mathcal Y)\) satisfying \(A=TB\) if and only if \(A,B\) and \(T\) satisfy: \(\operatorname{dom}A \subseteq\operatorname{dom}B\), \(\operatorname{ker}B \subseteq\operatorname{ker}A\), and there exists a subspace \(\mathcal D\subseteq\operatorname{mul}B\) and a surjective linear transformation \(T_{\operatorname{mul}}: \mathcal D\rightarrow\operatorname{mul}A\).