Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $H$ be a Hilbert space, and $S\subseteq \mathcal{B}(H)$. We denote $\bar S$ the ultraweak closure of $S$, and $B_r$ the closed ball of center 0 and radius $r>0$ of the normed space $\mathcal{B}(H)$.

If $S$ is a subalgebra of $\mathcal{B}(H)$, Do we have $\overline{S\cap B_r} = \bar S \cap B_r$ ?

share|improve this question
2  
The equality does not hold in general. Take $S$ to be unit sphere in $H$. Then $\bar S=B_1$, whereas $S \cap B_r = \varnothing$ for $r<1$. Maybe you want $S$ convex? –  Andreas Thom Jul 16 '12 at 5:14
    
Indeed, the case I'm interested in is when $S$ is a subalgebra of $\mathcal{B}(H)$. I modified the question. –  Michael Jul 16 '12 at 13:48
2  
If $S$ is self-adjoint, then yes: this is the Kaplansky density theorem. –  Yulia Kuznetsova Jul 16 '12 at 14:16
    
Thanks, do we need to suppose $S$ unital ? What if $S$ is just supposed to be convex ? –  Michael Jul 17 '12 at 0:08
add comment

1 Answer 1

up vote 2 down vote accepted

The answer is NO for general non-selfadjoint subalgebras. With a subspace $X\subset \mathcal{B}(H)$, one can associate a subalgebra $$S_X= \langle \begin{pmatrix} \alpha & x \cr 0 & \alpha \end{pmatrix} : \alpha \in \mathbb{C}I_H,\ x \in X \rangle \subset{\cal B}(H\oplus H).$$ Now choose a state $\omega$ which eliminates ${\cal K}(H)$ and let $X = \langle x \in B(H) : 2 x_{11} = \omega(x) \rangle$. Then, $\overline{S_X}=S_{\mathcal{B}(H)}$. However, since $x\in X\cap B_r$ implies $|x_{11}|\le r/2$, one has $\overline{S_X\cap B_r} \neq \overline{S_X}\cap B_r$.

In passing, I'll explain why Kaplansky's density theorem holds for $\mathrm{C}^\ast$-algebras. If $S \subset \mathcal{B}(H)$ is a subspace, then the inclusion extends to a weak$^\ast$-ultraweak continuous contraction $\pi\colon S^{\ast\ast}\to\mathcal{B}(H)$. By continuity and the Goldstine's theorem, one has $\overline{S\cap B_r}=\pi(S^{\ast\ast}\cap B_r)$. But if one knows $S^{\ast\ast}$ is actually a $\mathrm{C}^\ast$-algebra and $\pi$ is a $\ast$-homomorphism, then one gets $\pi(S^{\ast\ast}) \cong S^{\ast\ast}/\ker\pi$ isometrically, which implies $\pi(S^{\ast\ast}\cap B_r)=\pi(S^{\ast\ast})\cap B_r = \overline{S}\cap B_r$. This proof is probably circular, because Kaplansky's density theorem would be needed to justify some results quoted above.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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