About commutative semisimple Banach algebra

Question

Is the norm on a unital semi-simple commutative Banach algebra with $\|I\|=1$, unique? ($I$ denotes the identity element)

Answer 1

Let $A := C^1([0,1])$. Then you can consider the two norms on $A$ given by $$ \lVert f \rVert_1 = \lVert f \rVert_\infty + \lVert f' \rVert _\infty $$ and $$ \lVert f \rVert_2 = \lvert f(0) \rvert + \lVert f'\rVert_\infty. $$ They both make $A$ into a Banach algebra and $C^1([0,1])$ is semisimple.

Answer 2

in general we have this theorem from [General Theory of Banach Algebras Book by Charles Earl Rickart]$\;\;[corollary(2.8.17)]$

$\mathbf{theorem}:$ if $A$ is a semisimple annihilator banach algebra then $A$ has an unique norm topology.

in other words every two norm $\parallel .\parallel_1\;$,$\;\parallel .\parallel_2$ that makes $A$ a semisimple annihilator banach algebra, are equivalent.

$\mathbf{remark}:$ note that it is not necessary that $A$ be commutative or even have an identity.

we note that a banach algebra $\frak{A}$ is called an annihilator banach algebra if, for arbitrary closed left ideal $\frak{L}$ and closed right adeal $\frak{R}$ in $\frak{A}$, both of the following conditions are satisfied:

$(1):\mathcal{A}_l(\frak{L})=\{0\}$ if and only if ${\frak{L}}=\frak{A}$.

$(2):\mathcal{A}_r(\frak{R})=\{0\}$ if and only if ${\frak{R}}=\frak{A}$.