Timeline for Positive extension of functionals on a subset of the state space of a $C^*$ algebra
Current License: CC BY-SA 2.5
20 events
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| Feb 24, 2011 at 1:23 | history | edited | Kate Juschenko | CC BY-SA 2.5 |
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| Feb 23, 2011 at 23:46 | comment | added | Kate Juschenko | $\lambda$ should be changed for the norm of $f$, after extension $f$ to the subspace $S$, I don't know how to use algebraically interior points for such type of extension. | |
| Feb 23, 2011 at 23:41 | history | edited | Kate Juschenko | CC BY-SA 2.5 |
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| Feb 23, 2011 at 18:36 | comment | added | Anna Jenčová | OK, but I don't see that if $x,y\in c(V)$ with $x-y\le 1$, then $f(x-y)\le \lambda$ | |
| Feb 23, 2011 at 16:10 | comment | added | Kate Juschenko | I've added the proof to the answer, please check it. | |
| Feb 22, 2011 at 13:16 | history | edited | Kate Juschenko | CC BY-SA 2.5 |
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| Feb 21, 2011 at 17:52 | comment | added | Kate Juschenko | if one needs to have the property for the convex subset $V$ of $S(A)$, then one needs to put extra condition on $V$, namely, if $t\in c(V)-c(V)$ is positive, then $\lambda t$ is in $V$ for some $\lambda\geq 0$. | |
| Feb 21, 2011 at 17:48 | history | edited | Kate Juschenko | CC BY-SA 2.5 |
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| Feb 21, 2011 at 17:42 | comment | added | Kate Juschenko | I've added non-unital case, I hope it helps. | |
| Feb 21, 2011 at 17:41 | history | edited | Kate Juschenko | CC BY-SA 2.5 |
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| Feb 21, 2011 at 16:24 | comment | added | Anna Jenčová | Yes, I have this book, but I only found extension theorem for an operator system, which is a self adjoint subspace, containing the unit. But I wonder if there is an extension theorem for other situations, e.g. when $J$ is self-adjoint, but does not contain the unit (but maybe a positive invertible element). | |
| Feb 21, 2011 at 15:26 | comment | added | Kate Juschenko | there is very good collection of positive/completely positive maps in the book of Vern Paulsen, amazon.com/Completely-Bounded-Maps-Operator-Algebras/dp/… | |
| Feb 21, 2011 at 14:43 | comment | added | Anna Jenčová | I don't think one can find a positive extension for any convex subset. Let $V$ be a convex subset generated by two positive trace 1 elements $\rho_1$ and $\rho_2$. Then for any $f_1,f_2\ge 0$ there is a positive affine functionals on $V$ with $f_1=f(\rho_1)$ and $f_2=f(\rho_2)$. Suppose that $\rho_1\le M\rho_2$ for some $M>1$. Then $M\rho_2-\rho_1$ is positive, but one can always find $f_1,f_2\ge 0$ such that $Mf_2-f_1<0$. | |
| Feb 21, 2011 at 13:48 | comment | added | Kate Juschenko | If you define positivity of $f$ on $M_n(J)$ as operator that sends positive matrix with coefficients in J to a positive operator, then this should not be enough for its extension, since $M_n(C(V)-c(V)+i(c(V)-C(V)))$ might have more positive elements, than those that come from matrix with coefficients in J. there should be an example, where there is no extension to completely positive map. | |
| Feb 21, 2011 at 13:25 | comment | added | Anna Jenčová | Does it mean that one can do this for any convex subset of $S(A)$? | |
| Feb 21, 2011 at 9:54 | comment | added | Anna Jenčová | Yes, thanks very much. For the last 2 questions: I was asking about a cp map $J\to B(H)$, for a Hilbert space $H$, not a functional. | |
| Feb 21, 2011 at 2:04 | history | edited | Kate Juschenko | CC BY-SA 2.5 |
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| Feb 21, 2011 at 0:31 | history | edited | Kate Juschenko | CC BY-SA 2.5 |
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| Feb 20, 2011 at 22:27 | history | edited | Kate Juschenko | CC BY-SA 2.5 |
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| Feb 20, 2011 at 22:09 | history | answered | Kate Juschenko | CC BY-SA 2.5 |