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$1$:I know that if ‎$‎‎F$ is a ‎locally convex ‎compact ‎space ‎then ‎‎$‎‎‎\overline{co}(‎Ext (F))=F$‎

($Ext$: means extreme point)

$2$:I ‎know ‎that ‎if ‎‎$‎‎M$ ‎is a ‎Von ‎Neumann ‎algebra ‎then ‎‎$‎\overline{co}(Proj(M))=‎‎‎Ball_1(M_+)$‎‎

$3$:I know that ‎$‎‎Ext(Ball_1 (B(H)^+))=Proj(B(H))$‎

$4$:$B(H)$ is Von Neumann algebra then by $2$ I can say‎‎$‎\overline{co}(Proj(B(H)))=‎‎‎Ball_1(B(H)^+)$‎‎

by ‎these ‎information I‎ ‎want ‎to ‎know ‎that ‎‎

Q:if ‎$‎‎dim (H)=‎\infty‎$ ‎then ‎‎‎$Ball_1 (B(H)^+)‎‎‎‎\neq ‎‎{‎co}(Proj(M))$‎‎

alex v
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1 Answers1

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Let $\{e_n\}$ be an orthonormal sequence in $H$ and denote $p_n$ by the rank one projection onto $\mathbb{C}e_n$. Then the positive element $x=\sum \frac{1}{n}p_n$ works. To see it, assume $x$ is a convex combination of projections $q_1,\cdots,q_k$, namely $x=t_1q_1+\cdots t_kq_k$. Then there is $j$ such that $q_jH$ contains an infinite subset $E$ of the sequence $\{e_n\}$ which is impossible. Because $xE$ is infinite (in $\mathbb{R}^+$) and $(t_1q_1+\cdots t_kq_k)(E)$ is finite.

ABB
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  • Your $x $ cannot be written as a convex combination of projections (not even as a positive combination). But this is a nontrivial fact. Every invertible positive operator can be written as $t_1q_1+\cdots +t_nq_n $ with the $t_j $ positive. And every operator can be written as a linear combination of projections. – Martin Argerami Jan 21 '16 at 13:32