Ultraweak closure inside a closed ball - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T21:55:57Z http://mathoverflow.net/feeds/question/102328 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/102328/ultraweak-closure-inside-a-closed-ball Ultraweak closure inside a closed ball Michael 2012-07-16T03:34:27Z 2012-07-17T01:14:10Z <p>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 $\mathcal{B}(H)$. </p> <p>If $S$ is a subalgebra of $\mathcal{B}(H)$, Do we have $\overline{S\cap B_r} = \bar S \cap B_r$ ?</p> http://mathoverflow.net/questions/102328/ultraweak-closure-inside-a-closed-ball/102411#102411 Answer by Narutaka OZAWA for Ultraweak closure inside a closed ball Narutaka OZAWA 2012-07-17T01:14:10Z 2012-07-17T01:14:10Z <p>The answer is NO for general non-selfadjoint subalgebras. With a subspace $X\subset \mathcal{B}(H)$, one can associate a subalgebra $$S_X= \langle \begin{pmatrix} \alpha &amp; x \cr 0 &amp; \alpha \end{pmatrix} : \alpha \in \mathbb{C}I_H,\ x \in X \rangle \subset{\cal B}(H\oplus H).$$ 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 $\overline{S_X\cap B_r} \neq \overline{S_X}\cap B_r$.</p> <p>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. </p>