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Let $A$ be a $C^{*}$-algebra, represented on a Hilbert space $H$, and $D$ a selfadjoint unbounded operator on $H$ (note that we do not impose that $D$ have compact resolvent). Let

$\mathcal{A}:=${$a\in A : [D,a]\in B(H) $}

and topologize by the spectral invariant norm $\|a\|_{1}:=\|a\|+\|[D,a] \|$. Let $\mathcal{I}$ be a two sided ideal in $\mathcal{A}$, and denote by $I$ its closure in $A$. Let $0\leq h\leq k\in\mathcal{A}$ be such that $h\mathcal{I}$ is dense in $\mathcal{I}$ (in its Banach algebra topology).

Question: Is $k\mathcal{I}$ dense in $\mathcal{I}$?

Note that the hypotheses imply that $h$ and hence $k$ are strictly positive $\mod I$ in $A$. Therefore both $hI$ and $kI$ are dense in $I$. Also, in case $h$ and $k$ commute, or when $h,k\in\mathcal{I}$, the answer to this question is yes.

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This question keeps getting bumped. Do you still have a counter-example, as claimed in your comment below –  Yemon Choi Aug 3 '12 at 20:39
    
Why does this question keep getting bumped? –  Jon Bannon Sep 18 '12 at 12:31
    
I think it's because no answer has received an upvote. I am going to try a downvote since the OP appears to have swanned off –  Yemon Choi Sep 18 '12 at 16:34

1 Answer 1

${\cal I}$ is dense in $I$, since $I$ is the closure of ${\cal I}$.

Therefore $k{\cal I}$ is dense in $kI$, using continuity of multiplication.

You note $kI$ is dense in $I$.

Since density is transitive, $k{\cal I}$ is dense in $I$. Since ${\cal I} \subseteq I$, $k{\cal I}$ is also dense in ${\cal I}$. In the topology of $A$. Do you require density in the stronger topology of ${\cal A}$?

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The question is about density in the stronger topology of $\mathcal{A}$. Actually, the statement is false. I have a counterexample but it takes some time to write down properly. I will do that later. –  alterationx10 Dec 9 '11 at 12:40

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