# Non strictly-singular operators and complemented subspaces

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 $Y$?

• If you send me an email I'll write up an answer to your question. – Bill Johnson Jul 26 '15 at 11:50
• I don't think it's exactly what you want, but the example on page 13 of this paper may be of interest. – user19038 Jul 26 '15 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 $(I-P)X$ 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.(GR-ATHN); Freeman, D.(1-TX); Haydon, R.(4-OXBR); Odell, E.(1-TX); Raikoftsalis, Th.(GR-ATHN); Schlumprecht, Th.(1-TXAM); Zisimopoulou, D.(GR-ATHN) 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].

• 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 $(L(X\oplus Y)$. – Bill Johnson Jul 27 '15 at 13:08
• 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 $C[0,1]$. – Bill Johnson Jul 27 '15 at 13:13
• Welcome to MO Spiros! You should be aware that there is an ask-johnson tag that folks have found very useful :) – Kevin Beanland Jul 28 '15 at 17:58
• Kevin, in my email aliases I have "askSpiros". – Bill Johnson Jul 29 '15 at 0:46