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 $x\in E$ $$ \langle x\zeta,\zeta\rangle\leq \lambda< \langle a\zeta,\zeta\rangle$$


$2 \times 2$ 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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.