Closed unit ball of positive bounded operator space and its extreme point

Question

Let $H$ be infinite dimensional Hilbert space. Then the closed unit ball of positive bounded operator space $B(H)^+$ is not the convex hull of the projections of $B(H)$. Please help me. Thanks.

Answer 1

Edit: Many thanks to Martin for bringing to my attention a flaw in my earlier argument.

Here is an entirely new argument. In what follows, $ n \in \Bbb{N} $ and $ [n] \stackrel{\text{df}}{=} \Bbb{N}_{\leq n} $.

• Let $ P_{1},\ldots,P_{n} $ be projections in $ B(\mathcal{H}) $. Let $ \lambda_{1},\ldots,\lambda_{n} \in \Bbb{R}_{> 0} $ satisfy $ \displaystyle \sum_{k = 1}^{n} \lambda_{i} \leq 1 $.
• We claim that if $ \displaystyle T \stackrel{\text{df}}{=} \sum_{k = 1}^{n} \lambda_{k} P_{k} $ is compact, then it has finite rank.
• Observe that $ \lambda_{k} P_{k} \leq T $ for each $ k \in [n] $.
• This implies that $ \text{Range}(P_{k}) \subseteq \text{Range}(\sqrt{T}) $ for each $ k \in [n] $.
• If $ T $ is compact, then $ \sqrt{T} $ is too, and so $ P_{k} $ has finite rank for each $ k \in [n] $.
• Hence, $ T $ also has finite rank.
• Therefore, any positive compact operator with infinite rank and norm $ \leq 1 $ cannot lie in the convex hull of the projections of $ B(\mathcal{H}) $.