most recent 30 from http://mathoverflow.net 2013-05-18T16:39:18Z http://mathoverflow.net/feeds/question/58753 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/58753/center-of-the-algebra-of-bounded-operators ivo 2011-03-17T13:53:04Z 2011-10-14T12:27:40Z

<p>Suppose that X is Banach space. How to prove that the center of algebra B(X)( bounded operators on X) consists only of operators aI, where a is scalar and I is identity operator?</p>

http://mathoverflow.net/questions/58753/center-of-the-algebra-of-bounded-operators/58758#58758 Sven Raum 2011-03-17T14:34:24Z 2011-03-17T14:34:24Z

<p>If I see it correctly, then take a Hamel basis $(v_i)$ with $i \in I$ of the Banach space $X$. Then the coordinate switch $T_{i,j}:v_k \mapsto \delta_{i,k}v_j$ is bounded, because finite dimensional subspaces have a complement in a Banach space. Writing down any $S$ in the centre of $\mathscr{B}(X)$ as a matrix with rows and columns indexed by $I$, commutativity of $S$ and $T_{i,i}$ shows that is must be diagonal and commutativity of $S$ and $T_{i,j}$ shows that all diagonal entries must be the same. Hence $S$ is a scalar multiple of the identity.</p> <p>The essence of the argument is that finite dimensional subspaces are complemented in Banach spaces.</p>

http://mathoverflow.net/questions/58753/center-of-the-algebra-of-bounded-operators/58759#58759 BS 2011-03-17T14:41:04Z 2011-03-17T14:41:04Z

<p>Maybe this is hw ? Anyway, you always have rank one operators $x\mapsto L(x)v$, $v\in X$, $L\in X^*$, and any operator which commutes with all of them is easily seen to be scalar.</p>

http://mathoverflow.net/questions/58753/center-of-the-algebra-of-bounded-operators/58760#58760 Stefan Waldmann 2011-03-17T14:51:52Z 2011-03-17T14:51:52Z

<p>The argument of BS works also in the case where $X$ is a Hausdorff locally convex space since the topological dual still separates points (by Hahn-Banach). This is enough to show that the (continuous) finite rank operators act transitively on non-zero vectors from which it follows that their center is already trivial. But then all bounded operators have at most the center of the finite rank ones and the multiples of the identity.</p> <p>So if you want to go beyond trivial center you have to consider topological vector spaces which are not lc, pretty weird ones ;) I don't know the answer for $L^p$ spaces with $p &lt; 1$...?</p> <p>I guess the more interesting problem is whether there are more bounded operators than the continuous finite rank operators and the multiples of the identity...</p>

http://mathoverflow.net/questions/58753/center-of-the-algebra-of-bounded-operators/78126#78126 tetrapharmakon 2011-10-14T12:27:40Z 2011-10-14T12:27:40Z

<p>What about the center of the algebra of possibly unbounded operator? (sorry for posting another question but I think it's not worth a new post)</p>