Left multipliers on commutative $C^*$-algebras are right multipliers and vice versa

Question

In showing the multiplier algebra $M(A)$ of a $C^*$-algebra $A$ is commutative, we are (in the exercise) first to show that any left multiplier $L : A → A$ (i.e., $\forall a,b ∈ A: L(ab) = L(a)b$) is also a right multiplier (i.e., $\forall a,b ∈ A: L(ab) = aL(b)$). But is this really generally true? If so, why make the distinction between left and right in the first place?

Answer 1

Since $A$ is commutative, $L(a)b = L(ab) = L(ba) = L(b)a = aL(b)$. You can see from this the importance of commutativity. It was used in the second equality and the last. In a non-commutative $C^*$ algebra, these two steps will fail, so in general left and right multipliers will not always agree.

Answer 2

The other answer shows that if $A$ is commutative, then every left multiplier is also a right multiplier.

The converse is also true, i.e. if every left multiplier is also a right multiplier, then $A$ is commutative. Indeed, if $x\in A$, write $L_x$ for left multiplication with $x$. Then $L_x$ is a left multiplier, and thus also a right multiplier. Hence, if $y,z \in A$, $$xyz = L_x(yz) = yL_x(z)= yxz.$$ Since $z$ is arbitrary, we can conclude $xy=yx$. Hence, $A$ is commutative.

In conclusion, $A$ is commutative if and only if every left multiplier is also a right multiplier.