# Positive extension of functionals on a subset of the state space of a $C^*$ algebra

Let $A$ be a finite dimensional $C^*$ algebra and $S(A)$ the state space. Let $K\subset A$ be an intersection of $S(A)$ with a vector subspace $J\subset A$ and let $f$ be a positive affine functional on $K$.

I am dealing with the following questions: Can $f$ be extended to a positive linear functional on $A$? And if $F$ is a completely positive map on $J$, under which conditions can it be extended to a completely positive map on $A$? Is there anything else than Arveson's extension theorem?

Is there a good reference where these questions are treated in the finite dimensional case?

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can you write more details, with what exactly do you intersect a subspace $J\subset A$? – Kate Juschenko Feb 20 '11 at 15:53
I guess Kate is asking: if $J\subset A$ then how can we intersect $S(A)$ with $J$, as $S(A)$ is a subset of the dual of $A$, not of $A$... – Matthew Daws Feb 20 '11 at 19:46
Since $A$ is finite dimensional, the states can be identified with density operator with respect to a fixed trace, that is, positive operators with trace 1. Then $S(A)\subset A$. In the general case, the question has to be formulated in the dual space, but in finite dimensions, this is all the same. – Anna Jenčová Feb 20 '11 at 20:36

The intersection of a subspace $J$ with the set of positive matrices of trace one is convex, say equal to $V$. Obviously, we can extend $f$ to a cone generated by $V$, denote $c(V)$ (in fact V is a convex combination of finite set of positive operators with trace equal to $1$). Then $f$ has natural positive extension to $c(V)-c(V)$ and also to $c(V)-c(V)+i(c(V)-c(V))$. Since now $f$ is defined on a self-adjoint subspace of a C*-algebra by Krein's theorem it has a positive extension to the whole C*-algebra. for the last 2 questions: positive functional on a C*-algebra is automatically completely positive.

EDIT: the proof of the extension theorem for the unital case is simple: if $\phi:S\rightarrow \mathbb{C}$ is positive then $\phi(1)\geq ||\phi||>0$. Let $\phi'=\phi \cdot \phi(1)^{-1}$ is positive and $||\phi'||=\phi'(1)=1$. By Hahn-Banach $\phi'$ has contractive extension $\overline{\phi}$ to the C*-algebra, since $\overline{\phi}$ is contraction and unital, it is automatically positive.

To get the question for the non-unital case one can argue as follows. If $1$ is not in the subspace generated by $V$, then the following extension is well defined:

$f(t1+(1-t)a)=t\lambda +(1-t)f(a)$ for $a\in V$.

In order to have a positive extension to $c(V)-c(V)$ one needs to choose $\lambda$ properly.

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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. – Anna Jenčová Feb 21 '11 at 9:54
Does it mean that one can do this for any convex subset of $S(A)$? – Anna Jenčová Feb 21 '11 at 13:25
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. – Kate Juschenko Feb 21 '11 at 13:48
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 $Mf_2-f_1<0$. – Anna Jenčová Feb 21 '11 at 14:43
there is very good collection of positive/completely positive maps in the book of Vern Paulsen, amazon.com/Completely-Bounded-Maps-Operator-Algebras/dp/… – Kate Juschenko Feb 21 '11 at 15:26

Kate's argument can be finished as follows:

Let us denote $L=c(V)-c(V)$. We suppose that $L$ is such that $L\cap A^+=c(V)$. Let $\tilde L$ be the smallest real vector subspace, containing $c(V)-c(V)$ and $1$.

Let us denote$f_m=\sup\{ f(x), x\in L, x\le 1\}$ and $f_M=\inf\{f(x), x\in L, x\ge 1\}$. Then $f_m\ge \sup \{f(a), a\in V\}$. If $x\in L$ is such that $x\ge 1$, then for any $y\in L$, $y\le 1\le x$ we have $f(y)\le f(x)$ and hence $f_m\le f_M$. If there is no elemment in $L$ greater than $1$, then $f_M=\infty$, so that we always have $0\le f_m\le f_M$.

Choose some $\lambda$ such that $f_m\le \lambda\le f_M$. We extend $f$ to a linear functional $\tilde f$ on $\tilde L$, by $\tilde f(x+t)=f(x)+t\lambda$, for $x\in L$, $t\in \mathbb R$. Then it is easy to see that if $-1\le x+t \le 1$, then $-\lambda \le \tilde f(x+t)\le \lambda$, so that $\|\tilde f\|\le \lambda=\tilde f(1)$. Extending $\tilde f$ to $\tilde L+i\tilde L$, $\tilde f$ must be positive and we are back at the unital case.

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now I see... from the fact that you intersect the subspace $J$ with $S(A)$ follows that $(c(V)-c(V))^+\subseteq c(V)$, this I've missed this fact for the non-unital case. – Kate Juschenko Mar 5 '11 at 17:08