# When is a $\ast$-algebra a $C^{\ast}$-algebra?

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.
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. – Nik Weaver Aug 17 '12 at 1:34