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Let $H$ be a Hilbert space. We denote $K(H)$ by the space of compact operators on $H$ which is a two sided ideal in $B(H)$.

Let $E$ be a norm closed convex subset of positive operators in $K(H)$ and let $a$ be a non-zero positive compact operator where $a\notin E$.

Q: Is there any vector $\zeta\in H$ which separates $a$ and $E$, I mean there is a a positive number $\lambda$ such that for all $x\in E$ $$ \langle x\zeta,\zeta\rangle\leq \lambda< \langle a\zeta,\zeta\rangle$$

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$2 \times 2$ counterexample, $E = \left\{\left[\matrix{\lambda& 0\cr 0&\lambda}\right]: \lambda \geq 0\right\}$ and $A = \left[\matrix{1&1\cr 1&1}\right]$. Then for any nonzero $\zeta$ we have $\{\langle B\zeta,\zeta\rangle: B \in E\} = [0,\infty)$, so no $\zeta$ can separate.

The general idea is that you can separate $E$ and $a$ with a bounded linear functional on $K(H)$, i.e., tracing against some trace class operator, but you can't expect to do it with a single vector.

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  • $\begingroup$ As a coda to the above answer: what you need for the argument to work is that the set be closed not just for the norm but in the so-called weak operator topology (even in the strong operator topology). $\endgroup$ – oeiras Apr 3 '16 at 16:27
  • $\begingroup$ Dear Nik, You mean $E$ and $a$ can be separated by a positive trace class operator? $\endgroup$ – Ali Bagheri Apr 3 '16 at 16:43
  • $\begingroup$ @oeiras: no, we're talking about subsets of $K(H)$, not $B(H)$. The weak* topology isn't relevant. $\endgroup$ – Nik Weaver Apr 3 '16 at 18:25
  • $\begingroup$ @AliBagheri: no way. If $E$ is a cone, then tracing against any positive operator will yield either $\{0\}$ or $[0,\infty)$. Getting $\{0\}$ is too much to ask for. $\endgroup$ – Nik Weaver Apr 3 '16 at 18:25
  • $\begingroup$ @NikWeaver The spaces $K(H)$ and the finite dimensional operators form a dual pair in the natural way and so we can apply the Hahn-Banach theorem to this situation. $\endgroup$ – oeiras Apr 3 '16 at 19:31

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