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Apr 25, 2021 at 17:56 comment added Bill Johnson To get an example, take two HI spaces $U$ and $V$ s.t. every bounded linear operator from $U$ to $V$ is strictly singular and apply the Edelstien-Wojtasczcyk theorem (see vol.1 of Lindenstrauss-Tzafriri Theorem 2.c.13) to $X := U \oplus V$,
Apr 25, 2021 at 17:54 comment added Bill Johnson Tomek, the OP wants $U$ to contain a complemented copy of $V$ or $V$ to contain a complemented copy of $U$ and $X=U\oplus V$.
Apr 25, 2021 at 16:37 comment added Tomasz Kania Yes, of course. A subspace is complemented whenever there is a continuous linear projection thereonto. By the Hahn--Banach theorem, finite-dimensional subspaces are always complemented, so such decompositions are never unique as you may interchange finite-dimensional pieces.
Apr 25, 2021 at 16:32 comment added Jack L. @Tomasz Kania. I believe I should have added some conditions. I expect the map $T$ to be surjective if the kernel is complemented and to be injective if the closed image is complemented. Equivalently, $T$ should either have a continuous right inverse or a left inverse (I have re-edited for clarity). Would (a modification of) your suggestion still work in this case?
Apr 25, 2021 at 16:23 history edited Jack L. CC BY-SA 4.0
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Apr 25, 2021 at 15:31 comment added Tomasz Kania Finite-dimensional subspaces are always complemented, so write $U = U_1 \oplus F_1$ with $F_1$ finite-dimensional and take $U^\prime = U_1$ and $V^\prime = V + F_1$.
Apr 25, 2021 at 15:13 comment added Jack L. @Tomasz Kania. Could I kindly ask that you expand on your suggestion. (Seems I should easily see it but I’m not getting it right away ).
Apr 25, 2021 at 15:04 comment added Tomasz Kania Sure, you can always adjoin some finite-dimensional subspace of $V$ to $U$.
Apr 25, 2021 at 14:42 history asked Jack L. CC BY-SA 4.0