If $T$ is a bounded operator which is not strictly singular, acting on a separable Banach space $X$, can one always find an infinite dimensional, closed and complemented, subspace $Y$ such that $T$ restricted to $Y$ is an isomorphism on $Y$?

$\begingroup$ If you send me an email I'll write up an answer to your question. $\endgroup$– Bill JohnsonCommented Jul 26, 2015 at 11:50

2$\begingroup$ I don't think it's exactly what you want, but the example on page 13 of this paper may be of interest. $\endgroup$– user19038Commented Jul 26, 2015 at 22:51
2 Answers
Thanks for the email, Markus.
Let’s agree that “space” means “infinite dimensional Banach space” so that subspaces are always infinite dimensional.
A Banach space $X$ is decomposable if it is the direct sum of two subspaces; in other words, if there is a (bounded, linear) projection $P$ on $X$ s.t. $PX$ and $(IP)X$ are both infinite dimensional. The first indecomposable Banach space was constructed by Gowers and Maurey; in fact, their space is hereditarily indecomposable. Now we know that indecomposable spaces are very common; see [AFHORSZ] and references therein. In particular, $\ell_p$, $1<p<\infty$, is a subspace of a separable indecomposable space.
For an example that gives a negative answer to Markus’ problem, take an indecomposable space $X$ that contains a decomposable subspace $Y$ ($Y$ can be a Hilbert space). Take a projection $P$ on $Y$ that has infinite dimensional range and infinite dimensional kernel. Extend $P$ to an operator $T$ from $X$ into some injective space that contains $X$. $T$ is obviously not strictly singular since $T$ is the identity on $PY$. Also, the kernel of $T$, being infinite dimensional, has infinite dimensional intersection with every finite codimensional subspace. But since $X$ is indecomposable, all complemented subspaces of $X$ are finite codimensional.
I could not have answered this natural and basic (though I never thought of it until reading this post) question a few years ago. $$ $$ [AFHORSZ] Argyros, S. A.(GRATHN); Freeman, D.(1TX); Haydon, R.(4OXBR); Odell, E.(1TX); Raikoftsalis, Th.(GRATHN); Schlumprecht, Th.(1TXAM); Zisimopoulou, D.(GRATHN) Embedding uniformly convex spaces into spaces with very few operators. (English summary) J. Funct. Anal. 262 (2012), no. 3, 825–849. $$ $$
EDIT: After I posted this, user19038 gave a reference in a comment above that shows that the OP's question was raised by Vitali Milman in a 1970 paper and solved in the linked paper. The example involves only classical spaces; it is the inclusion mapping from $L \log^\lambda L$ into $L_1$ with $\lambda < 1/2$.
I think that the correct Markus' question concerns operators in L(X).In this case the answer is positive for separable C(K). I do not know what happens in the case of spaces with an unconditional basis. Moreover the following seems interesting. Let X be a separable reflexive space and T in L(X). Does there exist indecomposable Y containing isomorphically X and S in L(Y) that extends T? I also do not know what is the answer if in the previous question if we replace the indecomposable Y by the space C[o,1].

1$\begingroup$ For general spaces there is no difference between the OP's question for operators on a space and for operators between two spaces. because a counterexample for $L(X,Y)$ gives a counterexample for $(L(X\oplus Y)$. $\endgroup$ Commented Jul 27, 2015 at 13:08

1$\begingroup$ The answer to your last question is yes. Consider $X$ to be a subspace of $L_\infty$. Use a back and forth argument to extend the operator on $X$ to an operator on some separable sublattice $Y$ of $L_\infty$ that contains $X$ and the constant functions. $Y$ is isometric to a separable $C(K)$ space and thus is norm one complemented in $C[0,1]$. $\endgroup$ Commented Jul 27, 2015 at 13:13

$\begingroup$ Welcome to MO Spiros! You should be aware that there is an askjohnson tag that folks have found very useful :) $\endgroup$ Commented Jul 28, 2015 at 17:58

1$\begingroup$ Kevin, in my email aliases I have "askSpiros". $\endgroup$ Commented Jul 29, 2015 at 0:46