Multiplier Algebra: Double Centralizer?

Question

Given a C*-algebra $A$.

Consider operators: $$(L,R)\in\mathcal{B}(A):\quad L(a)b=aR(b)$$

Does that imply: $$L(ab)=L(a)b\quad R(ab)=aR(b)$$

How to check this?

Reference: Lawrence G. Brown, Stable isomorphism of hereditary subalgebras of a $C^*$-algebra.

Answer 1

For any $c\in A$, $$ L(ab)c=abR(c)=aL(b)c.$$ As this holds for any $c\in A$, it follows that $L(ab)=aL(b)$. Similarly, $$ cR(ab)=L(c)ab=cR(a)b, $$ so $R(ab)=R(a)b$.