If $H$ is a separable Hilbert space, then how can I show that the Banach algebra of bounded operators on $H$ also has a countable dense subset? I tried to construct a countable set of bounded operators but cannot find an appropriate way. Could anyone help me?
2026-03-25 09:26:21.1774430781
The Banach algebra of bounded operators on a separable Hilbert space has a countable dense subset
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It's not true. Hint: Let $H=L^2(\Bbb R)$. For $m\in L^\infty$ let $T_m$ be the operator defined by multiplication by $m$: $$T_mf=mf.$$Then $||T_m||=||m||_\infty$ (hence $||T_{m_1}-T_{m_2}||=||m_1-m_2||_\infty$.)