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 Johnson Jul 26 '15 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. $\endgroup$ – user19038 Jul 26 '15 at 22:51

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].

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.