Relation between completely continuous operator and weak-to-norm continuous operator

Question

In Banach spaces $X$ and $Y$ what is the relation between completely continuous operator and weak-to-norm continuous operator? The followings are what I know:

1. Compact operator is always completely continuous
2. If a linear operator is weak-to-norm continuous, then is is bounded with finite rank
3. Not all compact operator has finite rank

Based on the above three points, completely continuous is not equivalent to weak-to-norm continuous, otherwise we conclude that all the compact operators have finite rank.

However, based on the definition of complete continuity and weak-to-norm continuity (both of them maps weak convergent sequence to norm convergen sequence), I feel like they are the same concept. What's wrong here?

Answer 1

A weak-to-norm continuous operator is always a completely continuous operator, but not the other way around. The problem is, not every Banach space is a sequential space when equipped with its weak topology. Examples include infinite-dimensional spaces with Shur's property, i.e. where strong sequential convergence and weak sequential convergence are the same. The classic example of such a space is $\ell^1$, see here for example.

As an example of a completely continuous, but not weak-to-norm continuous operator, consider the identity map on $\ell^1$.