Multiplication operators Consider a commutative Banach algebra $A$ and the Banach algebra of bounded operators $B(A)$ on $A$. Associate to each $a\in A$ the multiplication operator $T_ax =ax$ ($x\in A$). Is always the mapping $\varphi(a)=T_a$, $\varphi\colon A\to B(A)$ a continuous algebra homormorphism? What if $A$ is not commutative?

Since $\|T_a x\| \le \|a \| \|x\|$, we have $\|T_a\| \le \|a\|$, so it is continuous.
That it is a homomorphism is easy.  Even if $A$ is not commutative, since $T_{ab} x = abx = T_a T_b x$ it is a homomorphism.
