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 supnorm), such that T has dense range, but is not surjective?

13$\begingroup$ Despite the two quick votes to close, I don't find this a trivial question. Am I missing something? $\endgroup$ – Bill Johnson Jul 3 '12 at 21:41

1$\begingroup$ I guess you mean also that $A$ should be closed and map its domain back into itself. I would have to review semigroup theory (or think more than I care to right now) to see if that is correct. Anyway, how do you get such an $A$? $\endgroup$ – Bill Johnson Jul 3 '12 at 22:48

1$\begingroup$ Surely, Yemon; for the weak$^*$ topology the problem is trivial. $\endgroup$ – Bill Johnson Jul 3 '12 at 23:25

3$\begingroup$ There is a theorem of Lotz stating that there are no strongly continuous semigroups on $l^{\infty}$, meaning that if semigroup is strongly continuous, then the generator is bounded. $\endgroup$ – András Bátkai Jul 4 '12 at 17:27

2$\begingroup$ Here is a much easier question: Does there exist a non surjective bounded linear operator from some Banach space into $\ell_\infty$ that has dense range? I see how to do this but the argument uses something that is not elementary. Is there a simple reason such an operator exists? $\endgroup$ – Bill Johnson Jul 5 '12 at 16:21
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 quasicomplementary copies of $\ell_\infty$ in $\ell_\infty$ that are not complementary?
(Recall that two closed subspaces of a Banach space are said to be quasicomplementary if their sum is dense and their intersection is $\{0\}$.)

$\begingroup$ @Bill: The existence of "quasicomplementary but not complementary subspaces" in $l_{\infty}$ is equivalent to $l_{\infty}$ has infinite dimension separable quotient. \\ But the existence of "quasicomplementary but not complementary subspaces" in $l_{\infty}$ is not equivalent to the existence of "one to one bounded linear operator form $l_{\infty}$ into itself that has dense range but is not surjective". $\endgroup$ – Qingping Zeng Aug 27 '12 at 13:58

$\begingroup$ The former condition is in fact equivalent to the existence of "one to one bounded linear operator form some Banach space $X$ into $l_{\infty}$ that has dense range but is not surjective". \\ And, it is well known that $l_{\infty}$ has infinite dimension separable quotient. So, the existence of "one to one bounded linear operator form some Banach space $X$ into $l_{\infty}$ that has dense range but is not surjective" can also be deduced from the above facts. $\endgroup$ – Qingping Zeng Aug 27 '12 at 14:03

$\begingroup$ @Qingping: What I said is that the existence of two quasicomplementary subspaces of
$\ell_\infty$
that are isomorphic to$\ell_\infty$
and are not complementary is equivalent to the existence of a one to one bounded linear operator from$\ell_\infty$
into itself that has dense range but is not surjective. $\endgroup$ – Bill Johnson Dec 7 '12 at 18:34
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!

1$\begingroup$ For the sake of completeness, let me add the reference to the published version: <cite authors="Johnson, William; Nasseri, Amir Bahman; Schechtman, Gideon; Tkocz, Tomasz">_Johnson, William; Nasseri, Amir Bahman; Schechtman, Gideon; Tkocz, Tomasz_, Injective Tauberian operators on $L_1$ and operators with dense range on $\ell_\infty$, Can. Math. Bull. 58, No. 2, 276280 (2015). ZBL1330.47003.</cite> $\endgroup$ – Olaf Teschke Oct 6 '17 at 12:55
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.

2$\begingroup$ If you merge your existing accounts then you will be able to add this information to the existing question. Please try not to use the "answer" boxes for comments, as MO is not set up like a forum or blog thread. $\endgroup$ – Yemon Choi Jul 8 '12 at 0:36

1$\begingroup$ I merged the two users, but this will not be very useful since Amir does not have a registered user. @Amir: Please consider registering a user; it will help make sure that you can edit all of your posts in the future. $\endgroup$ – Anton Geraschenko Jul 8 '12 at 18:23