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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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# Ultraweak closure inside a closed ball

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 $\mathcal{B}(H)$.

Do we have $\overline{S\cap B_r} = \bar S \cap B_r$ ?