# $R$ is a right multiplier and $R(a)b=a\overset{?}{\implies} A$ is unital

Let $A$ be a $C^*$-algebra, and $R:A\to A$ its right multilplier. Is it true that $$\exists b\in A\quad \forall a\in A \quad R(a)b=a\qquad$$ implies $A$ is unital. I know this is true if A is a weak$^*$ dense ideal of $W^*$-algebra. But what about the general case?

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• [deleted premature comment] – Yemon Choi Aug 11 '14 at 12:44
• Are you assuming that $R$ is one half of a double centralizer $(L,R)$, or merely that $R:A\to A$ is linear and satisfies $R(xa)=xR(a)$ for all $x$ and $a$ in $A$? – Yemon Choi Aug 11 '14 at 12:50
• @YemonChoi, if $R$ is a right half of a multiplier $(L,R)$, then $L(b)$ is a right unit. Since $A$ have two sided BAI, then $A$ is unital. Therefore I just assume $R$ to be a left $A$-module map. It is automatically continuous [C*-algebras and their authomorphism groups. G. K. Pedersen, lemma 3.12.2] – Norbert Aug 11 '14 at 19:25

$A$ is unital. Something a little more general is true: if a C*-algebra $A$ has an element $b$ that is a right divisor of all its elements, then $A$ is unital. Proof: We have $(b^*b)^{1/4}=xb$ for some $x\in A$. So $(b^*b)^{1/2}=b^*x^*xb\leq \|x\|^2(b^*b).$ The inequality $(b^*b)^{1/2}\leq \|x\|^2(b^*b)$ holds in the C*-algebra generated by $b^*b$, and this implies that 0 can only be an isolated point of the spectrum of $b^*b$. So the C*-algebra $C^*(b^*b)$ has a unit $p$. Since $b(b^*b)^{1/n}\to b$ and $p$ is a unit for $(b^*b)^{1/n}$, we also have $bp=b$. Any other element of $A$ is a left multiple of $b$, so $p$ is a (right) unit for $A$.