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Let $A$ be a $C^*$-algebra.

If $I$ is an essential two-sided ideal in $A$, then it is fact that for every $a \in A$ we have $\|a\| = \sup_{x \in I, \|x\|=1} \|xa\|$. The argument is that we have an injective (since the ideal is essential) $C^*$-map of $A$ into the multiplier algebra of $I$, which due to injectivity must be isometric.

I need now the corresponding result for right-ideals, i.e., assume now that $I$ is an essential right-ideal in $A$. Do we still have $\|a\| = \sup_{x \in I, \|x\|=1} \|xa\|$ for every $a \in A$?

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Yes. The $\sigma(A^{**},A^*)$-closure of $I$ in the second dual von Neumann algebra $A^{**}$ is an ultraweakly closed right ideal, which is of the form $pA^{**}$ for some projection $p$. (In fact $p$ is the ultrastrong limit of any left approximate unit of $I$.) Thus,
$$\sup_{x\in I,\ \|x\|=1}\| xa \| = \sup_{x\in A^{**},\ \|x\|=1}\| pxa \|$$ for every $a\in A^{**}$. Consider the central support projection $z$ of $p$ in $A^{**}$. Since $Ia$ is nonzero for every nonzero $a\in A$, the $*$-homomorphism $A\ni a\mapsto za \in zA^{**}$ is faithful. Now, let $a\in A$ be given. For any $\epsilon>0$, the projection $q:=z1_{[\|a\|-\epsilon,\|a\|]}(|aa^*|^{1/2})$ in $zA^{**}$ is nonzero and so there is a nonzero partial isometry $v$ in $zA^{**}$ such that $v=pv=vq$. It follows that $$\sup_{x\in A^{**},\ \|x\|=1}\| pxa \| \geq \| pva \| \geq \|a\|-\epsilon.$$ Since $\epsilon>0$ was arbitrary, we are done.

Another proof is to use Kadison's transitivity theorem in combination with the following fact: if $J$ is an essential ideal in $A$, then $\|a \|=\sup_\pi\|\pi(a)\|$, where $\pi$ runs over irreducible $*$-representations of $A$ which do not kill $J$.

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  • $\begingroup$ I'm trying to understand your argument. Why do we have $v=pv$? $\endgroup$
    – AlexE
    Commented Jun 3, 2020 at 7:36
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    $\begingroup$ The existence of $v$ follows from comparison theory of projections, but here's a proof. Since $q$ is dominated by the central cover $z$ of $p$, one has $pA^{**}q\neq\{0\}$. Pick any nonzero $x\in pA^{**}q$ and consider the polar decomposition $x=v|x|$. The nonzero partial isometry $v$ belongs to $pA^{**}q$. $\endgroup$
    – Narutaka OZAWA
    Commented Jun 3, 2020 at 10:36
  • $\begingroup$ Sorry to bother you again, but I'm confused by the last step. We have $\|pva\| = \|vqa\|$ and I see that $\|qa\| \ge \|a\|-\varepsilon$. But how do you let the $v$ disappear for this estimate? $\endgroup$
    – AlexE
    Commented Jun 7, 2020 at 12:59
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    $\begingroup$ By construction, $qa$ is near to a partial isomery and so any nonzero "segment" of $qa$ has norm at least $\|a\|-\varepsilon$. To be precise, since $aa^*\geq(\|a\|-\varepsilon)^2q$, one gets $\|va\|^2=\|vaa^*v^*\|\geq(\|a\|-\varepsilon)^2\|vv^*\|$. $\endgroup$
    – Narutaka OZAWA
    Commented Jun 8, 2020 at 2:11
  • $\begingroup$ How to show that the $\sigma(A^{**},A^*)$-closure of $I$ is of the form $pA^{**}$ for some projection $p$? $\endgroup$
    – mathbeginner
    Commented Jul 8, 2020 at 18:34

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