# surjectivity of operators on l^infty

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?

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Despite the two quick votes to close, I don't find this a trivial question. Am I missing something? –  Bill Johnson Jul 3 '12 at 21:41
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 $A$? –  Bill Johnson Jul 3 '12 at 22:48
Surely, Yemon; for the weak$^*$ topology the problem is trivial. –  Bill Johnson Jul 3 '12 at 23:25
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. –  András Bátkai Jul 4 '12 at 17:27
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? –  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 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 $\{0\}$.)

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@Bill: The existence of "quasi-complementary but not complementary subspaces" in $l_{\infty}$ is equivalent to $l_{\infty}$ has infinite dimension separable quotient. \\ But the existence of "quasi-complementary 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". –  Qingping Zeng Aug 27 '12 at 13:58
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. –  Qingping Zeng Aug 27 '12 at 14:03
@Qingping: What I said is that the existence of two quasi-complementary 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. –  Bill Johnson Dec 7 '12 at 18:34
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.