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. $$ $$

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 $\lambda < 1/2$.