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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?

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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.

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Yes, it's a homomorphism: take $x,a,b\in A$. Then $$T_{a\cdot b}(x)=(ab)x=a(bx)=T_a(bx)=T_a(T_b(x)),$$ hence $T_{a\cdot b}=T_a\circ T_b$ and $\varphi(ab)=\varphi(a)\circ \varphi(b)$. We also have $$\varphi(a+b)(x)=(a+b)\cdot x=ax+bx=\varphi(a)(x)+\varphi(b)(x),$$ and $\varphi(e)=Id$.

It's continuous, as $$\lVert T_a-T_b\rVert_{B(A)}=\sup_{\lVert x\rVert=1}\lVert T_ax-T_bx\rVert=\sup_{\lVert x\rVert=1}\lVert ax-bx\rVert\leq \lVert a-b\rVert.$$

All has been done without the assumption that $A$ is commutative.

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