Showing that $\|1\| = 1.$

Question

Let $A$ be a Banach algebra. A map $a \mapsto a^*$ is called an involution if for all $a,b \in A$ and $\lambda \in \mathbb C,$

$(1)$ $(a^*)^* = a,$

$(2)$ $(a + b)^* = a^* + b^*,$

$(3)$ $(\lambda a)^* = \overline {\lambda}\ a^*,$

$(4)$ $(ab)^* = b^* a^*.$

A Banach algebra $A$ with an involution is called a $C^*$-algebra if $\|a^* a\| = \|a\|^2.$

With the above definition in mind I am trying to prove the following exercise $:$

$\textbf {Exercise} :$ Let $A$ be a unital $C^*$-algebra. Show that $\|1\| = 1.$

If we can somehow show that $1^* = 1$ then by the definition of $C^*$-algebra we are through. But I am struggling to prove this. Could anyone please give me some small hint in this regard?

Thanks a bunch.

Answer 1

From Operator Theoretic Aspects of Ergodic Theory, page 509:

An involution on a complex Banach algebra $A$ is a map $x \mapsto x*$ satisfying $$ (x^*)^* = x, \quad (x+y)^* = x^* + y^*, \quad (\lambda x)^* = \bar{\lambda}x^*, \quad (xy)^* = y^*x^* $$ for all $x, y \in A$, $\lambda \in \mathbb{C}$. In an algebra with involution one has $e^* = e$ since $$ e^* = e^*e = (e^*e)^{**} = (e^*e^{**})^* = (e^*e)^* = (e^*)^* = e. $$

Answer 2

To see that $1^*=1$, it suffices to show that $1^*a=a=a1^*$. This is easy:

$$1^*a=(a^*1)^*= a^{**}=a$$ and similarly $a1^*=a$.

Then note that

$\|1\|=\|1^*1\|=\|1\|^2$ whence $\|1\|=0$ or $\|1\|=1$. If $\|1\|=0$, then $1=0$ and the algebra is trivial, i.e. $A=\{0\}$. In all other cases, the norm of the unit is $1$.