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? 


This question on [MSE](https://math.stackexchange.com/questions/895050/r-mbox-is-a-right-multiplier-and-rab-a-overset-implies-a-mbox-is-unit).



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


