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The purpose of this question is to collect sufficient conditions on a unital $\ast$-subalgebra $\mathcal{A}$ of the algebra of bounded linear operators $B(\mathcal{H})$ on a separable Hilbert space $\mathcal{H}$ that guarantee that $\mathcal{A}$ is actually a $C^{*}$ algebra (is closed in the operator norm). Please provide links and references. At least, I'd like a reference or proof for the following:

"Thm:" If $\mathcal{A}$ is a unital $\ast$-subalgebra of $B(\mathcal{H})$ and whenever $A\in\mathcal{A}$ is self-adjoint it follows that $A_{+}$ and $A_{-}$ both lie in $\mathcal{A}$, then $\mathcal{A}$ is norm-closed.

(Here, $A_{+}$ and $A_{-}$ live naturally in the $C^{*}$-algebra generated by $A$ and $I$, isomorphic to $C(\sigma(A)))$, where $A$ corresponds to the function $f(x)=x$, $A_{+}$ corresponds to $max[f,0\]$ and $A_{-}$ to $min[f,0]$.)

(Edit: Nik has pointed out that the "Thm" is false. The broader question stands: Is there any other interesting abstract characterization of a C*-algebra that doesn't obviously say the algebra is norm-closed?)

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Jon, I think your "Theorem" is false. For example, $A$ could be the algebra of complex-valued Lipschitz functions on $[0,1]$, acting as multiplication operators on $L^2[0,1]$. That's a unital *-subalgebra of $B(H)$ which is stable under lattice operations, but not closed in operator norm.

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    $\begingroup$ Here's another example: let $A$ be the set of all eventually constant sequences of complex numbers. This is a unital $*$-subalgebra of $l^\infty$ that is not norm closed. Not only is it stable under lattice operations, it's stable under the continuous functional calculus. $\endgroup$ – Nik Weaver Aug 17 '12 at 1:34
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    $\begingroup$ ... the Borel functional calculus, even. $\endgroup$ – Nik Weaver Aug 17 '12 at 4:25
  • $\begingroup$ In light of your comment, it seems like an answer to my broader question seems unlikely...my hope was that closure under the functional calculi plus some other condition should force the algebra to be C*...Thanks again for the examples! $\endgroup$ – Jon Bannon Aug 17 '12 at 10:34

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