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 $a \in A$?



Norm of a multiplier of a right-ideal in C*-algebras
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 $J$.


