Question

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?

Answer 1

Here is an answer to an easier but related question.

Proposition. There is a one to one operator $T$ from $\ell_1(2^{\aleph_0})$ into $\ell_\infty$ that has dense range.

Of course, such an operator cannot be surjective because $\ell_1(2^{\aleph_0})$ is not isomorphic to $\ell_\infty$.

My proof of the Proposition uses an old result of Bill Davis and mine (Remark 4 in

Davis, W. J.; Johnson, W. B. On the existence of fundamental and total bounded biorthogonal systems in Banach spaces. Studia Math. 45 (1973), 173–179):

$\ell_\infty$ has a biorthogonal system $(x_\alpha,x_\alpha^*)_{\alpha<2^{\aleph_0}}$ with $\|x_\alpha\|=1$ and $\sup_\alpha \|x_\alpha^*\|<\infty$ such that the linear span of $(x_\alpha)$ is dense in $\ell_\infty$.

To prove the Proposition, define $T$ to be the norm one linear extension of the map $e_\alpha \mapsto x_\alpha$, where $(e_\alpha)$ is the unit vector basis for $\ell_1(2^{\aleph_0})$. This mapping obviously has dense range and is one to one because every biorthogonal system is countably linearly independent.

Here is a variation on the OP's question:

Is there a one to one bounded linear operator from $\ell_\infty$ into itself that has dense range but is not surjective?

The interest in the variation is that this question is easily seen to be equivalent to:

Are there quasi-complementary copies of $\ell_\infty$ in $\ell_\infty$ that are not complementary?

(Recall that two closed subspaces of a Banach space are said to be quasi-complementary if their sum is dense and their intersection is ${0}$.)

Answer 2

I could prove that if $T$ has dense then $T$ is surjective, in the cases where $T=S^{*}+W$, $W$ is weakly compact and $S:l^1\rightarrow l^1$ or when $T$ has totally disconnected spectrum.