Let $A$ be a $C^{\ast}$-algebra. We say $A$ is weakly commutative if $ab^\*c=cb^\*a$ for all $a,b,c \in A$ and define weak center of $A$ as $$Z\_w(A)= \{ v \in A : av^\*c=cv^\*a \;\forall a,c \in A \}.$$



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> Are these notions of weak commutativity and weak center the same as the usual notions of commutativity and center in $C^{\ast}$-algebras?
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By using approximate identity, I have managed to prove that both notions weakly commutative and commutative are same. It is also clear that center of $A$, i.e., $Z(A)$, contains $Z\_w(A)$ but reverse inclusion is not clear. Any ideas?



Weak center is same as center for $C^{\ast}$-Algebra?
Not the same. The center of $M\_2$ is $\mathbb{C}\cdot I\_2$, but its weak center is $\{0\}$. E.g. $I\_2$ is not in the weak center because not all $a,c \in M\_2$ commute.


