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 


$\|\cdot\|\leq\|\cdot\|\_{1}$. 


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 $\|f\|\_{1}=\|f\|+\|\partial f\|$.



Ideals in smooth subalgebras of C*-algebras
$\newcommand{\norm}[1]{\Vert#1\Vert}$


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 ${\mathcal I}$.




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[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](http://www.ams.org/journals/tran/1964-111-02/S0002-9947-1964-0161177-1/)
-- 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.)


