TITLE: Surjectivity of operators on $\ell^\infty$
QUESTION [28 upvotes]: Can anyone give me an example of an bounded and linear operator $T:\ell^\infty\to \ell^\infty$ (the space of bounded sequences with the usual sup-norm), such that T has dense range, but is not surjective?

REPLY [10 votes]: In this paper by Amir Bahman Nasseri, Gideon Schechtman, Tomasz Tkocz, and me it is shown that there is an injective, dense range, non surjective operator on $\ell_\infty$.  The proof is not technically difficult but has some interest.  From the theory of Tauberian operators one deduces that such an operator exists if and only if there is a dense range, injective, non surjective operator Tauberian operator (``Tauberian" in this context just means that the second adjoint of the operator is injective) on $L_1(0,1)$. Although  $L_1(0,1)^* = L_\infty(0,1)$ is isomorphic to $\ell_\infty$, the a priori equivalence is not obvious because of the lack of reflexivity, but we need only the obvious implication.  The main tool for constructing the  $L_1$ operator is a finite dimensional lemma proved by computer scientists. So, in some sense, the original question about operators on a non separable Banach space is connected to computer science!