Second conjugate operators to operators on $c_0$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T05:33:29Z http://mathoverflow.net/feeds/question/83977 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/83977/second-conjugate-operators-to-operators-on-c-0 Second conjugate operators to operators on $c_0$ BSalkas 2011-12-20T23:50:39Z 2011-12-21T19:02:15Z <p>I posted my question at MS but unfortunately it is still without a response, so let me ask it here.</p> <p>We can think about a bounded operator $T\colon c_0\to c_0$ as a double-infinite matrix $[T_{mn}]_{m,n\geq 1}$ which acts on a sequence $a=[a_1, a_2, a_3, \ldots ]\in c_0$ in the same way as usual (finite) matrices act on vectors ($n$-tuples of scalars), i.e.</p> <p>$$Ta= [T_{mn}][a_n] = \left[ \sum_{n=1}^\infty T_{mn}a_n\right]$$</p> <p>Suppose $a=[a_1, a_2, a_3, \ldots ]\in \ell^\infty = (c_0)^{**}$. Does the following formula still hold:</p> <p>$$T^{**}a= [T_{mn}][a_n] = \left[ \sum_{n=1}^\infty T_{mn}a_n\right]$$</p> <p>EDIT: Second question deleted, since I wanted to ask, in fact, about something else.</p> http://mathoverflow.net/questions/83977/second-conjugate-operators-to-operators-on-c-0/83979#83979 Answer by Bill Johnson for Second conjugate operators to operators on $c_0$ Bill Johnson 2011-12-21T00:08:29Z 2011-12-21T19:02:15Z <p>Yes. The reason is that the unit vector basis for $c_0$ is a shrinking basis, which means that the biorthogonal functionals to the basis are a Schauder basis for $c_0^* = \ell_1$. This implies that the unit vector basis for $c_0$ is a weak$^*$ Schauder basis for $c_0^{**} = \ell_\infty$, which means that the expansion of a vector in $\ell_\infty$ in terms of the unit vector basis must converge in the weak$^*$ topology. This is enough, since the adjoint of an operator on $\ell_1$ is weak$^*$ continuous on $\ell_\infty$. </p> <p>I'll take this opportunity to say something more about a space with shrinking bases. R. C. James, who introduced the concept, realized that if $(e_n)$ is a shrinking basis for a space $X$, then <code>$X^{**}$</code> can be represented as the space of sequences $(a_n)$ s.t. the partial sums of the series $\sum a_n e_n$ are uniformly bounded. Once he realized this, he very quickly constructed the space we now call J, which is isomorphic to its second dual but is non reflexive; in fact, $\dim J^{**}/J =1$.</p> <p>You can read about the basics of basis theory in many books; Albiac-Kalton is a good choice.</p>