Proposition 6.1.4.label If $A$ is a Banach algebra with an identity $1$ then there exists a norm $\norm{\cd}{0}$ on $A$ such that
- (1)
$\norm{\cd}{0}$ is equivalent to $\norm{\cd}{}$
- (2)
$(A,\norm{\cd}{0})$ is a Banach algebra
- (3)
$\norm{1}{0}=1$
Proof. For each $x\in A$, let $L_{x}$ denote the operator $y\in A\mto xy\in A$. The map $x\mto L_{x}$ is then injective since $L_{x}1=x$. Put $\norm{x}{0}=\norm{L_x}{}$, $x\in A$. By the inequality $\norm{xy}{}\leq \norm{x}{}\norm{y}{}$, we have $\norm{x}{0}\leq \norm{x}{}$. OTOH, we have
\begin{align*}\norm{x}{0}=\norm{L_x}{}=\sup\curl{\norm{xy}{}:\norm{y}{}\leq 1}\geq\frac{\norm{x}{}}{\norm{1}{}}\end{align*}
Hence the norm $\norm{\cd}{0}$ is equivalent to the original norm. The last 2 assertions are automatic.$\square$