Strictly singular operator

Remarks

Example: every compact operator is strictly singular. Indeed, suppose that $T|_M$ had a bounded inverse on $T(M)$. Then the closed unit ball of $M$ would be the image of a relatively compact set under a bounded map, so it would be compact, and $M$ would be finite-dimensional. For $1 \le p < q < \infty$, the inclusion $\ell^p \to \ell^q$ is strictly singular but not compact. Non-example: the identity map of an infinite-dimensional Banach space $X$ is not strictly singular, since it is an isomorphism on $M = X$. On Hilbert spaces, the strictly singular operators are exactly the compact operators. For a Banach space $X$, the strictly singular operators from $X$ to $X$ form a closed two-sided ideal of the algebra of bounded operators. If $F$ is a Fredholm operator and $S$ is strictly singular, then $F + S$ is Fredholm with the same index.