Let $X$ be a Banach space. An operator $T:X\to X$ is called strictly singular iff for any infinite dimensional subspace $Y\subseteq X,$ $T|_{Y}:Y\to T(Y)$ is not an isomorphism.

It is known that for $X=\ell_p,$ $1\leq p<\infty,$ an operator is strictly singular iff it is compact. Also $T:\ell_\infty\to\ell_\infty$ is strictly singular iff $T$ is weakly compact. Can someone provide me proofs for these facts? I could not really locate proofs of the above mentioned facts in literature.


1 Answer 1


[J. Lindenstrauss and L. Tzafriri. Classical Banach spaces I. Sequence spaces. Springer 1977]. In page 76, after Prop. 2.c.3, it says that the proof of 2.c.3 shows that an operator $T:\ell_p\to\ell_p$ is strictly singular if and only if it is compact.

[F. Albiac and N. Kalton. Topics in Banach space theory. Springer 2006] Theorem 5.5.1 says that a weakly compact operator $T:C(K)\to X$ is strictly singular, and Theorem 5.5.3 says that a non-weakly compact operator $T:C(K)\to X$ is not strictly singular.

Note that $\ell_\infty$ is a $C(K)$ space with $K$ the Stone-Cech compactification of the set of positive integers.


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