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 $\overline{S\cap B_r} = \bar S \cap B_r$ ?

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    $\begingroup$ 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? $\endgroup$ – Andreas Thom Jul 16 '12 at 5:14
  • $\begingroup$ Indeed, the case I'm interested in is when $S$ is a subalgebra of $\mathcal{B}(H)$. I modified the question. $\endgroup$ – Michael Jul 16 '12 at 13:48
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    $\begingroup$ If $S$ is self-adjoint, then yes: this is the Kaplansky density theorem. $\endgroup$ – Yulia Kuznetsova Jul 16 '12 at 14:16
  • $\begingroup$ Thanks, do we need to suppose $S$ unital ? What if $S$ is just supposed to be convex ? $\endgroup$ – Michael Jul 17 '12 at 0:08

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.

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