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The following question was first posted on Math.Stackexchange.com but unfortunately I didn’t get any answer. This might be obvious for many researchers but I can’t see how this is so, thus I am asking it here:

Let $V$ be a closed subspace of $B(H,K)$ such that $xy^*z \in V$ for all $x,y,z \in V$. Let $I$ be an ideal $(IV^*V+VV^*I \subset I)$ of $V$. Let $C(I)$ denotes the $C^{\ast}$-algebra generated by $II^{\ast}$. In a paper the following inclusion is stated without proof:

$VI^{\ast} \subset C(I)$

I don’t see how this can be true, since $C(I)$ is generated by $II^{\ast}$ therefore how could $VI^*$ be included inside $C(I)$?

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    $\begingroup$ What paper is this from? I don't immediately see the answer. $\endgroup$
    – Nik Weaver
    Commented Jan 11, 2020 at 16:09
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    $\begingroup$ Why don't you just email her for an explanation? Assuming it isn't explained in the thesis. $\endgroup$
    – Nik Weaver
    Commented Jan 11, 2020 at 20:09
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    $\begingroup$ My guess is that there are extra assumptions in this part of Kaur's thesis, which are used to justify this inclusion and the subsequent result that $A(I)$ is an ideal in $A(V)$. It would help if you quote the whole statement of Prop 2.2.4 $\endgroup$
    – Yemon Choi
    Commented Jan 12, 2020 at 6:39
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    $\begingroup$ By the way, it it is good practice to link to the MSE question, so that people here can see what has been said over there: math.stackexchange.com/questions/3504653/… Likewise, at MSE you should link to this MO post, so that the people responding at MSE can see what has been said in comments here $\endgroup$
    – Yemon Choi
    Commented Jan 12, 2020 at 6:43
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    $\begingroup$ @MathLover think about how your last comment would be received by someone asking for more details and context. Does the thesis not specify what $V$ is? Yemon wonders if there are extra assumptions in the thesis and your comment does not answer this. $\endgroup$
    – Nik Weaver
    Commented Jan 12, 2020 at 14:04

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