# Norm of a multiplier of a right-ideal in C*-algebras

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 $$a \in A$$?

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 $$J$$.
• I'm trying to understand your argument. Why do we have $v=pv$? – AlexE Jun 3 at 7:36
• 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 $pA^{**}q$. – Narutaka OZAWA Jun 3 at 10:36
• 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? – AlexE Jun 7 at 12:59
• 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 $\|va\|^2=\|vaa^*v^*\|\geq(\|a\|-\varepsilon)^2\|vv^*\|$. – Narutaka OZAWA Jun 8 at 2:11
• How to show that the $\sigma(A^{**},A^*)$-closure of $I$ is of the form $pA^{**}$ for some projection $p$? – mathbeginner Jul 8 at 18:34