Lemma 1

Let A be a normed algebra. The left regular representation of A and the right regular representation of A are continuous of norm $\leqslant 1$.

If the algebra A is unital, the mappings $x\mapsto \|\boldsymbol{\gamma }_x\|$ and $x\mapsto \|\boldsymbol{\delta }_x\|$ are norms on A, defining the topology of A. They satisfy the inequality (1).

If the unital normed algebra A is non-zero, that is, if $e\not= 0$, we also have $\|\boldsymbol{\gamma }_e\|=\|\boldsymbol{\delta }_e\|= 1$.

It is immediate that $\|\boldsymbol{\gamma }_x\|\leqslant \|x\|$ and that $\|\boldsymbol{\delta }_x\|\leqslant \|x\|$.

If A has a unit element $e$, then we have $x=\boldsymbol{\gamma }_xe=\boldsymbol{\delta }_xe$, hence: (2) $\|x\|\leqslant \|\boldsymbol{\gamma }_x\| \cdot \|e\|\|x\|\leqslant \|\boldsymbol{\delta }_x\| \cdot \|e\|$.

In this case, $x\mapsto \|\boldsymbol{\gamma }_x\|$ and $x\mapsto \|\boldsymbol{\delta }_x\|$ are therefore norms equivalent to the norm of A. They satisfy (1) since $\boldsymbol{\gamma }$ and $\boldsymbol{\delta }$ are representations.

The last assertion follows from the fact that $\boldsymbol{\gamma }_e=\boldsymbol{\delta }_e=$ Id$_A$.