surjectivity of operators on l^infty - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T09:54:53Z http://mathoverflow.net/feeds/question/101253 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/101253/surjectivity-of-operators-on-linfty surjectivity of operators on l^infty Amir 2012-07-03T20:26:26Z 2012-07-07T23:57:17Z <p>Can anyone give me an example of an bounded and linear operator <code>$T:\ell^\infty\to \ell^\infty$</code> (the space of bounded sequences with the usual sup-norm), such that T has dense range, but is not surjective?</p> http://mathoverflow.net/questions/101253/surjectivity-of-operators-on-linfty/101509#101509 Answer by Bill Johnson for surjectivity of operators on l^infty Bill Johnson 2012-07-06T15:16:26Z 2012-07-06T15:16:26Z <p>Here is an answer to an easier but related question.</p> <p>Proposition. There is a one to one operator $T$ from $\ell_1(2^{\aleph_0})$ into $\ell_\infty$ that has dense range.</p> <p>Of course, such an operator cannot be surjective because $\ell_1(2^{\aleph_0})$ is not isomorphic to $\ell_\infty$. </p> <p>My proof of the Proposition uses an old result of Bill Davis and mine (Remark 4 in</p> <p>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):</p> <p>$\ell_\infty$ has a biorthogonal system $(x_\alpha,x_\alpha^*)_{\alpha&lt;2^{\aleph_0}}$ with $\|x_\alpha\|=1$ and $\sup_\alpha \|x_\alpha^*\|&lt;\infty$ such that the linear span of $(x_\alpha)$ is dense in $\ell_\infty$.</p> <p>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.</p> <p>Here is a variation on the OP's question:</p> <p>Is there a one to one bounded linear operator from $\ell_\infty$ into itself that has dense range but is not surjective? </p> <p>The interest in the variation is that this question is easily seen to be equivalent to:</p> <p>Are there quasi-complementary copies of $\ell_\infty$ in $\ell_\infty$ that are not complementary?</p> <p>(Recall that two closed subspaces of a Banach space are said to be quasi-complementary if their sum is dense and their intersection is ${0}$.)</p> http://mathoverflow.net/questions/101253/surjectivity-of-operators-on-linfty/101612#101612 Answer by Amir for surjectivity of operators on l^infty Amir 2012-07-07T23:57:17Z 2012-07-07T23:57:17Z <p>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.</p>