Let $B$ be a $C^{*}$-algebra and $\mathcal{B}$ a dense *-subalgebra stable under holomorphic functional calculus and $C^{1}$-functional calculus for selfadjoint elements. Also, $\mathcal{B}$ is a Banach algebra in a norm $\|\cdot\|_{1},$ satisfying


Also, there is a countable bounded approximate unit $u_{n}$ for $\mathcal{B}$ which is a contractive, increasing approximate unit for $B$. Let $\mathcal{I}$ be a closed two sided ideal in $\mathcal{B}$, and denote by $I$ its closure in $B$.

Is it true that $\mathcal{I}=I \cap \mathcal{B}$ ?

The pertinent examples are Lipschitz functions on the circle and on the real line, both with norm $\|f\|_{1}=\|f\|+\|\partial f\|$.

  • $\begingroup$ This is an interesting question. Can you prove this for commutative C* algebras? $\endgroup$ – Paul Siegel Mar 17 '11 at 14:58
  • $\begingroup$ No, not yet, I am trying to do so, but I fail to see a general method for establishing this. It wouldn't surprise me if the statement is false, but in that case I would like to see where the problems arise. $\endgroup$ – alterationx10 Mar 17 '11 at 15:45


In general, I think the answer to your question is no. Take ${\mathcal B}=C^1[-1,1]$ with the norm $\norm{f}= \norm{f}_\infty+\norm{f'}_\infty$ and let ${\mathcal I}$ be the closed ideal consisting of those $C^1$-functions which vanish at $x=0$ and whose 1st derivative vanishes at $x=0$. Then $I\cap {\mathcal B}$ contains the function $f(x)=x$ which is evidently not in ${\mathcal I}$.

[Some general remarks follow, in a rambling style owing to lack of sleep. I may try to edit these later.]

In the commutative unital setting, taking $B=C(X)$, we know what the closed ideals of $B$ are (they are precisely the "kernels" of closed subsets of $X$, in the language of hulls and kernels).

If your subalgebra ${\mathcal B}$ also has maximal ideal space (homeo to) $X$, then your question is related to -- perhaps is equivalent to, I have not thought in detail -- the following one:

Can I find a closed two sided ideal in ${\mathcal B}$ which is not the kernel of its hull?

Without your restrictions on stability-under-func-calc, this kind of question has been much studied for commutative examples, and I think also for certain noncommutative examples related to group algebras.

For little Lipschitz algebras (on the circle) the answer is no -- this ought to be in a paper of Sherbert from the 1970s -- so I expect the answer to your original question is "yes". (For the "big" Lipschitz algebras my suspicion is that the counter-example I gave for $C^1[-1,1]$ would also work.)

  • $\begingroup$ Thanks a lot. This simple counterexample is very clarifying. Thanks as well for the reference. $\endgroup$ – alterationx10 Mar 18 '11 at 16:38

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